MA3403 Lecture Notes

Algebraic Topology
Lecture Notes
Fall 2018
Gereon Quick
Please send suggestions for corrections to gereon.quick@ntnu.no.
Contents
Lecture 1. Introduction 5
Lecture 2. Cell complexes and homotopy 17
Lecture 3. Singular chains and homology 39
Lecture 4. Singular homology, functoriality and H0 49
Lecture 5. Relative homology and long exact sequences 59
Lecture 6. The Eilenberg-Steenrod Axioms 71
Lecture 7. Generators for Hn (S n ) and first applications 79
Lecture 8. Calculating degrees 89
Lecture 9. Local vs global degrees 95
Lecture 10. Homotopies of chain complexes 99
Lecture 11. Homotopy invariance of singular homology 105
Lecture 12. Locality and the Mayer-Vietoris sequence 115
Lecture 13. Cell complexes 123
Lecture 14. Homology of cell complexes 133
Lecture 15. Computations of cell homologies and Euler characteristic 141
Lecture 16. Designing homology groups and homology with coefficients 151
Lecture 17. Tensor products, Tor and the Universal Coefficient Theorem 161
Lecture 18. Singular cohomology 173
Lecture 19. Ext and the Universal Coefficient Theorem for cohomology 185
Lecture 20. Cup products in cohomology 199
Lecture 21. Applications of cup products in cohomology 211
3
4 CHAPTER 0. CONTENTS
Lecture 22. Poincaré duality and intersection form 221
Lecture 23. Classification of surfaces 229
Lecture 24. More on Poincaré duality 239
Bibliography 249
Appendix A. Exercises 251
1. Exercises after Lecture 2 251
5. Exercises on the Hurewicz map 259
LECTURE 1
Introduction
Organization:
Lectures: Tuesdays and Thursdays, both days at 12.15-14.00 in S21 in Sen-
tralbygg 2.
Exercises: We will have some suggested exercises, hopefully on a weekly
basis. You should try to solve as many exercises as possible, not just the ones
I suggest, but also all that you find in other textbooks. We will not have an
exercise class though. But you can discuss exercises with me at any time!
Course webpage: wiki.math.ntnu.no/ma3403/2018h/start
where all news about the class will be announced. You will also find lecture notes
a few hours after class on the webpage.
Office hours: Upon request.
Just send me an email: gereon.quick@ntnu.no
Textbooks: We will not follow just one book... but there are many good
texts out there. For example, you can look at
[H] A. Hatcher, Algebraic Topology. It’s available online for free. It contains
much more than we have time for during one semester.
[Mu] J.R. Munkres, Elements of Algebraic Topology.
[V] J.W. Vick, Homology Theory - An Introduction to Algebraic Topology.
Two books that you can use as an outlook to future topics:
[Ma] J.P. May, A Concise Course in Algebraic Topology. It’s also online
somewhere.
[MS] J.W. Milnor, J. Stasheff, Characteristic Classes.
5
6 CHAPTER 1. INTRODUCTION
There are many other good books and lecture notes out there. Ask me if you
need more.
What is required?
I will assume that you are starting your third year at NTNU (or more). You
should have taken the equivalent of Calculus 1-3 or MA1101-1103, MA 1201-
1202. So you should be familiar with Euclidean space Rn , multivariable calculus
and linear algebra. Ideally, you have taken TMA4190 Introduction to Topology
and/or General Topology.
You should also know a bit about algebra, like what is a group, an abelian
group, a field, ideally also what is a ring and module over a ring.
Finally, it would be good if you knew what a topoogical space is and you
would know what the words open, closed, compact, etc mean. But, in fact,
you could also just have some few examples of topologial spaces in mind, like
n-spheres, torus etc. without knowing too many abstract stuff. For, the class is
much more about the ideas and methods we develop than anything else. And
these methods are useful almost everywhere.
Nevertheless, if you want to refresh your knowledge on Topology, you may
want to have a look at the book
[J] K. Jänich, Topology.
What this class is about:

Note: If some of the following words do not yet make sense to you, no worries!
For the moment we are just waving our hands and use fancy words. We will make
sense of all this during the semester...
Very roughly speaking, topology studies spaces up to continuous trans-
formation of one space into the other.
The correct place to do this is the category of topological spaces whose
objects are topological spaces and whose morphisms are continuous maps. The
isomorphisms in this category are called homeomorphisms, i.e., a continuous
map with a continuous (left- and right-) inverse, or a continuous bijective map
with a continuous inverse.
7
We can then describe topology as the science which studies properties of

spaces which do not change under homeomorphisms. You have seen many such
properties already, e.g. compactness (to be recalled in a bit).
This gives rise to a typical question in topology:
Typical question in topology

Given two topological spaces X and Y . Are X and Y homeomorphic, i.e.,
≈
is there a homeomorphism ϕ : X −→Y?
Let us look at a familiar example. Fix two natural numbers n < m.

∼
=
• Is there a linear isomorphism (of vector spaces) Rm − → Rn ?
Answer: No, since linear algebra tells us that isomorphic vector spaces
have equal dimension.
• Is there a diffeomorphism (bijective differential map with differentiable
∼
=
inverse) Rm − → Rn ?
Answer: No, since otherwise the derivative at 0 would be a linear iso-
∼
=
morphism Rm − → Rn between tangent spaces.
∼
=
• Is there a bijective map (of sets) Rm −→ Rn ?
Answer: Yes. Surprisingly enough one can construct such maps, and it
is actually not that difficult.
≈
• Is there a homeomorphism Rm − → Rn ?
The answer to the last question is: No. But it is not so simple to show.
In fact, one of the goals of algebraic topology is to develop tools that help us
decide similar questions. For example, is there a homeomorphism between the
2-dimensional sphere S 2 and the torus? The answer is no. But how can we
prove that? Both spaces are compact and (in some sense) two-dimensional and
oriented...
Algebraic Topology in a nutshell

Translate problems in topology into problems in algebra which are (hope-
fully) easy to answer.
Key idea: develop algebraic invariants (numbers, groups, rings etc and
homomorphisms between them) which decode the topological problem.
This should be done such that homeomorphic spaces should have the
same invariants (that is where the name comes from).
In particular, this implies: if we find values of an invariant that differ for X

and Y , then they cannot be homeomorphic.
Remark: We will later see that all the invariants we construct are preserved
under homotopy equivalences, a weaker notion than homeomorphisms. This
will finally lead to the idea of the stable homotopy category being the motive of
topological spaces. We will not discuss this in class, but feel free to ask me about
it. :)
For example, the first important tool that we are going to define soon is
singular homology. It will allow us to use a simple algebraic argument to show
≈
that there cannot be a homeomorphism Rm − → Rn .
Just to make you taste a little more of what algebraic topology can do:
Multiplicative Structures on Rn
Let Rn × Rn → Rn be a bilinear map with two-sided identity element e ̸= 0
and no zero-divisors. Then n = 1, 2, 4, or 8.
What we are looking for is a ”multiplication map”. You know the cases
n = 1 and n = 2 very well. It’s just R and C ∼ = R2 . These are actually
fields.
For n = 4, there are the Hamiltonians, or Quaternions, H ∼ = R4 with a
multiplication which as almost as good as the one in C and R, but it is
not commutative. (You add elements i, j, k to R with certain multiplication
rules.)
For n = 8, there are the Octonions O ∼ = R8 . The multiplication is not
associative and not commutative.
And that’s it!
This is a really deep result!

The crucial and, at first glance maybe surprising, point to prove this fundamental
result is that the statement has something to do with the behavior of tangent
spaces on spheres. That’s a topological problem. Frank Adams was the first to
solve it.
In this class we will start to walk on the path towards a proof of this problem.
Unfortunately, we won’t make it to the finish line within one semester. So, if you
like, learn more about it in Advanced Aglebraic Topology...
9
• Before we move on, let us play a game and see an invariant in action.
The rules: Take a piece of paper and draw two crosses, i.e. spots with four
free ends.
Each move involves joining two free ends with a curve which does not cross
any existing line, and then putting a short stroke across the line to create two
new free ends. The players play alternating moves.
If there are no legal moves left, the player who made the last legal move
wins.
Let us assume that we know that the game ends after a finite number of
moves, say m moves. At the end we will have created a connected, closed planar
graph, in particular, a figure which has vertices, edges and faces.
We claim that no matter how you play, what strategy you use etc, there are
always 8 moves before the game stops, it is always the second player who wins
and there is a fixed number of vertices, edges and faces!
Why? Well, the number of moves and everything about the figure we create
is determined by Euler’s formula v − e + f = 2. The number 2 is an example
of an algebraic invariant.
To understand how this works, we need to determine how the number v of
vertices, the number e of edges, and the number f of faces depends on the number
m of moves.
For the vertices, when we start the game we have two vertices. In each move,
we create one new vertex. Thus we get
v = 2 + m.
For the edges, when we start the game we have no edges. In each move, we
create one line, but we split it into two edges by adding a vertex in the middle.
Hence in each move, we create 2 edges. Thus we get
e = 2m.
11
For the faces, we have to think backwards. At the end, there is exactly one
free (or loose) end pointing into each face that we created. For, if there was a
face with two free ends pointing into one face, then we could connect these two
ends within that face and the game would not have stopped. Note that there is
also exactly one loose end pointing out of the figure. (Again, of there were two
we could connect them by going around the figure.)
Now we need to check how many free ends we produce. We start with 4 free
ends per cross, that is 8 free ends. In each move, we connect two free ends, but we
also create two new ones. Thus the number of free ends does not change during
the whole game. Hence we get
f = 8.
In total we get
2=v−e+f
2 = 2 + m − 2m + 8
0 = −m + 8
m = 8.
Hence no matter how we play, the game ends after 8 moves. Since this number
is even, the second player always wins. Moreover, we always get v = 10, e = 16,
and f = 8.
Alternative: Changing the starting setup changes the outcome of the game.
For if we start with n crosses (or nodes), then we get with the same reasoning as
above
v =n+m
e = 2m
f = 4n.
Euler’s formula then yields
2=v−e+f
2 = n + m − 2m + 4n
m = 5n − 2.
Thus the game ends after m = 5n − 2 moves. For example, if n = 3, this is an
odd number and the first player always wins.
Here is the idea why Euler’s formula holds: We can draw any connected
planar graph (a graph we can draw in the plane such that its edges only intersect
in the vertices and we can walk along the edges between any two vertices) as
follows:
1) We start with a graph consisting of just one vertex and no edges, so
v = 1 and e = 0. And we have one face, the outer face or the plane around
the vertex, so f = 1. So in total the formula holds v−e+f = 1−0+1 = 2.
2) Now we can extend the graph by either
a) adding one vertex and connect it via an edge to the first one; that
is we change v → v + 1 and e → e + 1 or
b) draw an edge from the existing vertex to itself; this way we create
a new face as well, hence we change e → e + 1 and f → f + 1.
Thus after both operations the formula v − e + f = 2 still holds. Now
we continue this process until we have created the planar graph we had
in mind.
Here is another example of the use of an algebraic invariant:

• Football pattern:
Question: How many pentagons and hexagons are there on a classical foot-
ball?
We set P := # of pentagons and H := # of hexagons. The collection of
all vertices, edges and faces of all the pentagons and hexagons on the football
forms a graph on the surface of the football. This graph and therefore the
pattern on the football is governed by Euler’s formula v − e + f = 2. Hence we
need to calculate the number of vertices v, the number of edges e and the number
of faces f .
The number of faces is obviously given by
f = P + H.
To calculate the number of edges e we observe that every pentagon has 5

edges and every hexagon has 6 edges. That yields 5P + 6H edges. But we have
counted too many edges. For at each edge, there are two faces which meet.
Thus we need to divide our number by 2 and get
5P + 6H
e= .
2
13
To calculate the number of vertices v we observe again that every pentagon

has 5 vertices and every hexagon has 6 vertices. That yields 5P + 6H edges. But
again we have counted too many vertices. For at each vertex, there are three
faces which meet. Thus we need to divide our number by 3 and get
5P + 6H
v= .
3
Now we apply Euler’s formula:
v−e+f =2
5P + 6H 5P + 6H
− +P +H =2 (multiply by 6)
3 2
10P + 12H − 15P − 18H + 6P + 6H = 2 (simplify)
P = 12.
To get H, we count how many hexagons there are per pentagon: Each penta-
gon is surrounded by 5 hexagons which would yield H = 5P . But each hexagon
is attached to 3 pentagons at the same time. Hence we have counted three times
as many hexagons as there really are. This yields
5P 5 · 12 60
H= = = = 20.
3 3 3
• •
3 • 2 P = # pentagons = ?
•
• H = # hexagons = ?
1 • • •
4 2
• • 1 • • f-e+v=2
• 5 • • •
f= P+H
• • 3
• • e = ( 5P + 6H )/2
•
v = ( 5P + 6H )/3
P must be 12
H = ( 5P )/3
H must be 20
The Euler characteristic:
The number v − e + f = 2 is an example of the Euler characteristic of a
surface, i.e., a two-dimensional manifold. In fact, 2 is the Euler characteristic
of the sphere. The Euler characteristic can be defined for any topological space
X. It is denoted by χ(X) and it is always an integer number. More generally,
χ(X) is a topological invariant which means it does not change if we transform
X continuously.
Here are some examples for surfaces:
• For a sphere X = S 2 , it is 2: χ(S 2 ) = 2.
• For a torus X = T 2 , it is 0: χ(T 2 ) = 0.
• For a surface with two holes, it is −2.
• In general, for a surface with g holes, it is 2 − 2g.
Now assume we have two spaces X and Y , defined in some complicated way
which makes it difficult to understand how they look. But let us assume we can
calculate their Euler characteristics by some method. Then, if χ(X) ̸= χ(Y ), we
know that we cannot transform X continuously into Y .
And there are also more positive examples. It often happens that an invariant
defined one way turns out to encode a lot of other information as well.
You will learn more about these things soon...
v-e+f=2 v-e+f=0
sphere torus
“Euler characteristic of the surface”

15
The rough initiating idea for our algebraic invariants

Now back to the general situation. Let us try to get a first idea of how
algebraic topologists think and come up with their fancy invariants. Let us say
we have a space X and we want to characterise it, or at least be able to distinguish
it from other spaces.
The initiating idea is to study X by taking test spaces we understand
well and looking at the space of all maps from these test spaces into X. This
may sound like an awkward detour, but it turns out to be pretty smart.
So what are those test spaces? The most simple space is a point. So let T = •
the one-point space and let C(•,X) be the set/space of all (continuous) maps
from • into X. Since any map in C(•,X) is determined by the one-point image,
we just get that C(•,X) is the set of points of X.
So what happens if we take a one-dimensional test space like the unit
interval [0,1]? A continuous map γ : [0,1] → X is a path in X from γ(0) to γ(1).
Each γ(0) and γ(1) also gives us also an element in C(•,X). Hence if we look at
C(•,X) modulo those which can be connected by a path in C([0,1],X), then we
can read off how many “pieces” X has. Making this more precise gives us the set
π0 (X) of connected components of X. The set π0 (X) is the first example of an
algebraic invariant.
Let us keep going with this. A two-dimensional test space might be the
square [0,1] × [0,1]. Given a continuous map α : [0,1] × [0,1] → X, the two
restrictions α(0,t) and α(1,t) define two paths in X. If we assume that they have
the same start and end points, then we get a relation on the set C([0,1],X) of
paths in X.
This leads, by looking only at paths which are loops, to the fundamental
group π1 (X) of X, the next algebraic invariant.
Continuing this way and to look at maps from an n-dimensional test space
modulo relations that come from maps from a corresponding n + 1-dimensional
test space we can produce a sequence of algebraic invariants π2 (X), π3 (X), . . .
which are called the higher homotopy groups of X. (Actually, for the nth
homotopy group one uses the n-dimensional sphere, the n-dimensional space with
the maximal symmetry.)
The collection of all homotopy groups encodes a lot of information about X.
In fact, in many cases it contains all the information about X up to homotopy,
i.e., continuous deformation of X.
However, homotopy groups are notoriously difficult to compute. That
is why one also uses a different type of test spaces.
Starting again with the unit interval [0,1] in dimension one, we could also
proceed as follows. In dimension two we take an equilateral triangle, called a
two-simplex. In dimension three we take a regular tetrahedron, called a three-
simplex. In dimension four, we continue with a regular four-dimensional simplex
and so forth.
This leads to the singular homology groups Hn (X) of X. These groups
will be the main object of our studies for a while. in general, they carry less
information than homotopy groups. However, their big advantage is that they
are computable! During the next couple of weeks we will develop the machinery
to compute homology groups. Along the way we will witness many fundamental
ideas that turned out to be extremely useful in many areas of mathematics...
LECTURE 2
Cell complexes and homotopy
Our main goal today is to introduce cell complexes as an important type of

topological spaces and the conecpt of homotopy which is a fundamental idea to
simplify problems.
We start with a super brief recollection of some basic notions in topology.
A crash course in topology
Roughly speaking, a topology on a set of points is a way to express that points
are near to each other as a generalization of a space with a metric, i.e., a concrete
distance function.
You know the fundamental example of a metric space. For, recall from Cal-
culus 2 that the norm of a vector x = (x1 , . . . ,xn ) ∈ Rn is defined by
q
|x| = x21 + x22 + · · · + x2n ∈ R.
For any n, the space Rn with this norm is called n-dimensional Euclidean
space. The norm induces a maetric, i.e., a distance function by
d(x,y) := |x − y| for x,y ∈ Rn .
This turns Rn into a metric space and therefore an example of a topological
space in the following way:
Open sets in Rn
• Let x be a point in Rn and r > 0 a real number. The ball
Br (x) = {y ∈ Rn : |x − y| < r}
with radius ϵ around x is an open set in Rn .
• The open balls Br (x) are the prototypes of open sets in Rn .
• A subset U ⊆ Rn is called open if for every point x ∈ U there
exists a real number ϵ > 0 such that Bϵ (x) is contained in U .
• A subset Z ⊆ Rn is called closed if its complement Rn \ Z is open
in Rn .
17
18 CHAPTER 2. CELL COMPLEXES AND HOMOTOPY
• Familiar examples of open sets in R are open intervals, e.g. (0,1) etc.
• The cartesian product of n open intervals (an open rectangle) is open in
Rn .
• Similarly, closed intervals are examples of closed sets in R.
• The cartesian product of n closed intervals (a closed rectangle) is closed
in Rn .
• The empty set ∅ and Rn itself are by both open and closed sets.
• Not every subset of Rn is open or closed. There are a lot of subsets
which are neither open nor closed. For example, the interval (0,1] in R;
the product of an open and a closed interval in R2 .
The set of open sets in Rn
TRn = {U ⊆ Rn open}
is a subset of all subsets of Rn and has the following properties:

• ∅, Rn ∈ TRn
• Uj ∈ TRn for all j ∈ J ⇒ ∪j∈J Uj ∈ TRn
• U1 , U2 ∈ TRn ⇒ U1 ∩ U2 ∈ TRn .
We take these three properties as the model for a topology:
19
Definition: Topological spaces

Let X be a set together with a collection TX of subsets which satisfy
(i) ∅, X ∈ TX
(ii) Uj ∈ TX for all j ∈ J ⇒ ∪j∈J Uj ∈ TX
(iii) U1 , U2 ∈ TX ⇒ U1 ∩ U2 ∈ TX .
(Note that in (ii), J can be an arbitrary indexing set.)
Then we say that the pair (X,TX ) is a topological space and the sets in
TX are called open. We also say that TX defines a topology on X. We
often drop mentioning TX and just say X is a topological space (when the
topology TX is given otherwise). The complement of an open set is called a
closed set.
Here are some examples of topological spaces which also demonstrate that
some topologies are more interesting than others:
• Rn with TRn as described above.
• An arbitrary set X with the discrete topology TX = P(X), where
P h(X) is the power set of X, i.e., the set of all subsets of X. In the
discrete topology, all subsets are open and hence all subsets are also
closed.
• On an arbitrary set X, there is always the coarse topology TX =
{∅, X}.
• Let (X, d) be a metric space. Then we can imitate the construction of
the standard topology on Rn and define the induced topology as the
set of all U ⊆ X such that for each x ∈ U there exists an r > 0 so that
B(x, r) ⊆ U . Here B(x, r) = {y ∈ X : d(x, y) < r} is the metric ball of
radius r centred at x.
• Let (X,TX ) be a topological space, let Y ⊂ X be an arbitrary subset.
The induced topology or subspace topology of Y is defined by
TY := {V ⊂ Y : there is a U ∈ TX such that V = U ∩ X}.

Warning
It is important to note that that the property of being an open subset
really depends on the bigger space we are looking at. Hence open always
refers to being open in some given space.
For example, a set can be open in a space X ⊂ R2 , but not be open in R2 ,
see the picture.
Open sets are nice for a lot of reasons. First of all, they provide us with a
way to talk about things that happen close to a point.
Definition: Open neighborhoods

We say that a subset V ⊆ X containing a point x ∈ X is a neighborhood
of x if there is an open subset U ⊆ V with x ∈ U . If V itself is open, we
call V an open neighborhood.
The type of maps that preserve open sets are the continuous maps:
21
Definition: Continuous maps

Let (X, TX ) and (Y, TY ) be topological spaces. A map f : X → Y is con-
tinuous if and only if, for every V ∈ TY , f −1 (V ) ∈ TX , i.e., the preimages
of open sets are open.
We denote the set of continuous maps X → Y by C(X,Y ).
Topological spaces form a category with morphisms given by continuous
maps.
Examples of continuous maps include:

• Continuous maps Rn → Rm that you are familiar with from Calculus 2.
• If X carries the discrete topology then every map f : X → Y is
continuous.
• If Y carries the coarse topology then every map f : X → Y is contin-
uous.
Definition: Homeomorphisms
A continuous map f : X → Y is a homeomorphism if it is one-to-one and
onto, and its inverse f −1 is continuous as well. Homeomorphisms preserve
the topology in the sense that U ⊂ X is open in X if and only if f (U ) ⊂ Y
is open in Y .
Homeomorphisms are the isomorphisms in the category of topological
spaces.
Some examples are:

• tan : (−π/2,π/2) → R is a homeomorphism.
• f : R → R, x 7→ x3 is a homeomorphism.
But not every continuous bijective map is a homeomorphism. Here is an
example:
Example: A bijection which is not a homeomorphism

Let
S 1 = {(x,y) ∈ R2 : x2 + y 2 = 1} ⊂ R2
be the unit circle considered as a subspace of R2 . Define a map
f : [0,1) → S 1 , t 7→ (cos(2πt), sin(2πt)).
We know that f is bijective and continuous from Calculus and Trigonometry

class. But the function f −1 is not continuous. For example, the image
under f of the open subset U = [0, 41 ) (open in [0,1)!) is not open in S 1 .
For the point y = f (0) does not lie in any open subset V of R2 such that
V ∩ S 1 = f (U ).
Here is an extremely important property a subset in a topological space can

have. We are going to use it quite often.
Definition: Compactness
Let X be a topological space. A subset Z ⊂ X S is called compact if for
any collection {Ui }i∈I , Ui ⊂ X open, with Z ⊂ i∈I Ui there exist finitely
many i1 , . . . , in ∈ I such that Z ⊂ Ui1 ∪ · · · ∪ Uin .
In other words, a subset Z in a topological space is compact iff every open
cover {Ui }i of Z has a finite subcover.
• By the Theorem of Heine-Borel, a subset Z ⊂ Rn is compact if and

only if it is closed and bounded. Being bounded means, that there is
some (possibly huge) r >> 0 such that Z ⊂ Br (0).
• In patricular, neither R nor any Rn is compact.
• The n-dimensional disk Dn = {x ∈ Rn : |x| ≤ 1} and the n-sphere
S n = {x ∈ Rn+1 : |x| = 1} are compact.
• Finite sets, i.e., a subset which contains only finitely many elements, are
always compact.
• If X carries the discrete topology, then a subset Z ⊂ X is compact if
and only if it is finite.
• If X carries the coarse topology, then every Z ⊂ X is compact.
23
Definition: Connectedness
A topological space X is called connected if it is not possible to split it
into the union of two non-empty, disjoint subsets which are both open and
closed at the same time.
In other words, a space is connected if and only if the empty set and the
whole space are the only subsets which are both open and closed.
Note that the image f (X) of a connected space X under a continuous map
f : X → Y is again connected.
Simple examples of connected spaces are given by intervals in R.
Definition: Hausdorff spaces

A topological space X is called Hausdorff if, for any two distinct points
x,y ∈ X, there are two disjoint open subsets U,V ⊂ X such that x ∈ U
and y ∈ V .
In other words, in a Hausdorff space we can separate points by open subsets.
Every subspace of RN (with the relative topology) is a Hausdorff space. More-

over, basically all the spaces we look at will be Hausdorff. However, there are
spaces which are not Hausdorff.
For a typical counter-example, consider two copies of the real line Y1 :=
R × {1} and Y2 := R × {2} as subspaces of R2 . On Y1 ∪ Y2 , we define the
equivalence realtion (x,1) ∼ (x,2) for all x ̸= 0.
Let X be the set of equivalence classes. The topology on X is the quotient
topology defined as follows (see also below): a subset W ⊂ X is open in X if and
only if both its preimages in R × {1} and R × {2} are open.
Then X looks like the real line except that the origin is replaced with two
different copies of the origin. Away from the double origin, X looks perfectly nice
and we can separate points by open subsets. But every neighborhood of one of
the origins contains the other. Hence we cannot separate the two origins by open
subsets, and X is not Hausdorff.
Here are some useful facts about compact spaces:
Lemma: Closed in compact implies compact

1) Let X be a compact topological space. Let Z ⊂ X be a closed subset.
Then Z is compact.
2) Let Y be a Hausdorff space. Then any compact subset of Y is closed.
Let us prove the first assertion. The other one is left as a little exercise.
Proof: Let {Ui }i∈I be an open cover of Z. We set U := X \ Z. Then
{U, Ui }i∈I is an open cover of X. Since X is compact, there exist i1 , . . . ,in such
that X ⊂ U ∪ Ui1 ∪ . . . ∪ Uin and hence, by the definition of U , we have Z ⊂
Ui1 ∪ . . . ∪ Uin . QED
Another useful fact:
Lemma: Continuous images of compact sets are com-

pact
Let f : X → Y be continuous. Let K ⊂ X be compact. Then f (K) ⊂ Y is
compact.
But, in general, if Z ⊂ Y is compact, then f −1 (Z) ⊂ X does not have to
be compact.
As a consequence we can deduce a useful criterion for when continuous bijec-

tions are homeomorphisms:
Lemma: Continuous bijection from compact to Haus-

dorff is a homeomorphism
Let X be a compact space and Y be Hausdorff. If f : X → Y is a continuous
bijection, then f is a homeomorphism.
25
Proof: Since f is a bijection, there is a set-theoretic inverse map which we

denote by g := f −1 : Y → X. We need to show that g is continuous. So let
K ⊂ X be a closed subset. We are going to show that g −1 (K) = f (K) ⊂ is
closed in X. Since X is compact, K is also compact as a closed subset. Hence
its image f (K) ⊂ Y is compact. Since Y is Hausdorff, this implies that f (K) is
closed in Y . QED
Compactness, being Hausdorff, and being connected are important examples
of topological properties:
Homeomorphisms preserve topological properties

Slogan: Topology is the study of properties which are preserved under
homeomorphisms. From this point of view, a topological property is by
definition a property that is preserved under homeomorphisms.
Hence, roughly speaking, from the point of view of a topologist, two spaces
which are homeomorphic are basically the same.
For example, if f : X → Y is a homeomorphism, then X is compact if
and only if Y is compact. For, both f and its inverse f −1 are continuous
and surjective maps. Hence if X is compact, so is f (X) = Y ; and if Y is
compact, so is f −1 (Y ) = X.
We will remind ourselves of many other important topological properties along

the way.
Constructing new spaces out of old
There are sveral ways to construct topological spaces. Here are two important
constructions that we are going to use:
Definition: Product topology

Let X and Y be two topological spaces. The product topology on X × Y
is the coarsest topology, i.e., the topology with fewest open sets, such that
the projection maps X × Y → X and X × Y → Y are both continuous.
More concretely, a subset W ⊂ X × Y is open in the product topology if
for every point w = (x,y) ∈ W there are open subsets x ∈ U ⊂ X and
y ∈ V ⊂ Y with U × V ⊂ W .
Definition: Disjoint unions or sums of spaces

Let X and Y be two topological spaces. We denote by X ⊔ Y the disjoint
union (or sum) of X and Y . Recall that as a set we can define X ⊔ Y as
X ⊔ Y = X × {0} ∪ Y × {1}.
(In other words, we take one copy of X and one copy of Y and by the
indexing we make sure that we keep them apart.)
The disjoint union inherits a topology by defining
TX⊔Y = {U ⊔ V : U ∈ TX , V ∈ TY }.
Another important construction for producing new topological spaces is to

take quotients.
• Quotient Spaces
Let X be a topological space. Let ∼ be an equivalence relation on X. For
any x ∈ X let [x] be the equivalence class of x. We denote as usually the set of
equivalence classes by
X/ ∼:= {set of equivalence classes under ∼} = {[x] : x ∈ X}.
Let π : X → X/ ∼, x 7→ [x] be the natural projection. The quotient topology
is defined by
U ⊂ X/ ∼ open ⇐⇒ π −1 (U ) ⊂ X open.
Note that the map π : X → X/ ∼ is continuous by definition.
The quotient topology is the coarsest topology, in the sense that it has fewest
open sets, such that the quotient map π is continuous.
The quotient topology has the following universal property: For any topo-
logical space Y and for any maps f : X → Y which descends to a map f¯: X/ ∼→
Y , i.e., f is constant on equivalence classes, such that the diagram
f
X / Y
<
π
f¯
X/ ∼
commutes, the map f is continuous iff f¯ is continuous.

Many important examples of spaces that we will study arise as follows:
27
• Take a subset X ⊂ Rn and consider it with the induced topology as a

subset.
• Consider an interesting equivalence relation ∼ on X and take the quo-
tient topological space X/ ∼.
Let us look at some examples of this procedure:
Torus
We start with the square
S := {(x,y) ∈ R2 : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} ⊂ R2
with the subspace topology induced from the topology of R2 . Now we would
like to glue opposite sides to each. This corresponds to taking the quotient
T := S/((x,0) ∼ (x,1) and (0,y) ∼ (1,y)).
Real projective space

Real projective space RPn is the space of lines in Rn+1 through the origin.
As a topological space it can be constructed as follows:
We define the equivalence relation ∼ on the n-sphere S n by identifying
antipodal points, i.e., x ∼ y ⇐⇒ y = −x. Then we have
RPn = S n / ∼
and equip it with the quotient topology. Since S n is compact and RPn is
the continuous image of S n (under the quotient map), we see that RPn is
compact.
There is also a complex version:

Complex projective space

Again, complex projective space CPn is the space of one-dimensional C-
vector subspaces in Cn+1 . It can be topologized as follows:
We define the equivalence relation ∼ on the sphere S 2n+1 by x ∼ y if and
only if there is a λ ∈ C with |λ| = 1 such that y = λx where we think S 2n+1
as the subspace of points x in Cn with |x| = 1. Then we have
CPn = S 2n+1 / ∼
and equip it with the quotient topology. Since S 2n+1 is compact, CPn is
compact.
Aside: Note that the “topological dimension” of CPn is 2n (we have not
said what that means though). The n rather refers to the dimension as a
complex manifold.
Projective spaces play an important role in geometry and topology. We will

meet them quite frequently during this course (and future courses).
It happens also that it might be necessary to present a well-known space in a
different form. For example, we can write spheres as quotients. We will see that
this is just one example of a whole class of interesting spaces.
Sphere as a quotient
For every n ≥ 1, there is a homeomorphism
≈
ρ̄ : Dn /∂Dn −
→ S n.
There are in fact many different ways to construct such a homeomorphism.
Let us write down one in concrete terms for the special case n = 2. The
general case follows by throwing in more coordinates.
We define a continuous map ρ : D2 → S 2 such that
(
ρ(0,0) = (0,0, − 1) and
ρ(x,y) = (0,0, + 1) for all (x,y) ∈ ∂D2 = S 1 .
Since ρ will be constant on ∂Dn , it will induce a map ρ̄ on the quotient
D2 /∂D2 .
We define ρ as a rotation invariant map which sends the inner part of D2 of
points with radius less than 1/2 mapping onto the lower hemisphere of S 2
and the outer part of D2 of points with radius greater than 1/2 mapping
29
onto the upper hemisphere ρ : D2 → S 2 by

 p
 2x,2y, − 1 − 4(x2 + y 2 ) if x2 + y 2 ≤ 1/4
ρ(x,y) = p
 f (x,y)x,f (x,y)y, 1 − f (x,y)2 (x2 + y 2 ) if x2 + y 2 ≥ 1/4
p
where we denote f (x,y) = 4 − 4 x2 + y 2 (to make the formula fit in a
frame). This map is well-defined also for points with x2 + y 2 = 1/4. More-
over, ρ is contiunuous, as a composite of continuous functions, and constant
on ∂D2 .
An inverse map can be defined by

1 1
( 2 x, 2 y)
 if − 1 ≤ z ≤ 0
2 2 2
S → D /∂D , (x,y,z) 7→ (g(x,y)x,g(x,y)y) if 0 ≤ z < 1
class of ∂D2

for (0,0,1)
q √
1− 1− x2 +y 2
where we denote g(x,y) = √ . Note that this map is well-defined
22 2
x +y
2 2
also for z = 0, since then x + y = 1 and g(x,y) = 1/2.
• Compactifications
The concrete maps we wrote down in the previous example are kind of ugly.
But there is another way to show that there is such a homeomorphism Dn /∂Dn ≈
S n.
For we can also consider S n as the one-point compactification of Rn . Let
us first say what that means:
Definition: One-point compactification

1) If Y is a compact Hausdorff space and X ⊂ Y is a proper subspace whose
closure equals Y , then Y is called a compactification of X.
If Y \ X consists of a single point, then Y is called the one-point com-
pactification of X.
2) Let X be a topological space with topology TX . Let ∞ denote an abstract
point which is not in X and let X̂ := X ∪ {∞}. We define a topology TX̂
on X̂ as follows:
• each open set in X is an open set in X̂, i.e., TX ⊂ TX̂ and
• for each compact subset K ⊆ X, define an open subset UK ∈ TX̂
by UK := (X \ K) ∪ {∞}.
Then X̂ is a one-point compactification of X.
To see that X̂ actually is compact, take any open cover of X̂. Then at least
one of the open sets contains ∞. Hence that set covers (X \ K) ∪ {∞} for
some compact set K. Since K is compact, finitely many of the remaining
open sets suffice to cover K and therefore all of X̂.
Examples of one-point compactifications are spheres. For S n is the one-

point compactification of Rn . For n = 1, one can think of S 1 as taking the real
number line and connect the two ends at infinity in one point ∞ to close the
circle. More generally, one can construct a homeomorphism via stereographic
projection.
As an application, we give a new proof Dn /∂Dn ≈ S n :
Sphere as a quotient revisited

For every n ≥ 1, there is a homeomorphism
≈
ρ̄ : Dn /∂Dn −
→ S n.
Since S n ≈ Rn ∪ {∞}, it suffices to construct homeomorphism
(
x
≈ if |x| < 1
ρ : Dn −
→ Rn ∪ {∞}, x 7→ 1−|x|
∞ if |x| = 1.
We claim that the map ρ is continuous. To show this, we use the sequential
criterion of continuity. Let (an ) be a sequence in Dn with limn→∞ an =
c. If c ∈ Dn \ ∂Dn is an interior point, then ρ(c) ∈ Rn and we know
limn→∞ ρ(an ) = ρ(c), since the restriction of ρ to Dn \ ∂Dn is a composite
of continuous maps and the an will all be in Dn \ ∂Dn for n sufficiently
31
large. If c ∈ ∂Dn is a boundary point, then ρ(c) = ∞. Since an → c, the

sequence (ρ(an )) is unbounded, since the denominator of ρ(an ) tends to 0
while the norm of the nominator tends to 1.
Hence for any compact subset K in Rn , i.e., for any closed and bounded
K ⊂ Rn , there is a natural number N (K) such that ρ(an ) ∈ / K for all
n ≥ N (K). That means that the sequence (ρ(an )) converges in the
topology of Rn ∪ {∞} to ρ(c) = ∞. This shows that ρ is continuous.
We also know that ρ̄ is bijective, since the restriction ρ : Dn \ ∂Dn → Rn
is bijective and ρ sends ∂Dn to ∞. Hence ρ̄ is a continuous bijection from
a compact space to a Hausdorff space. As we have seen above, this implies
that ρ̄ is a homeomorphism.
• Cell complexes
Another way to think of the above procedure is the following. The sphere
consists of two parts that we glue together:
• an open n-disk, i.e., the open interior Dn \ ∂Dn ,
• and a single point, which corresponds to the class of the boundary ∂Dn ;
on S 2 we can picture this point as the northpole (the light blue dot in
the above picture).
Topologists think of such building blocks as the cells of a space. However,
not all spaces can be built this way. So let us make precise what is needed:
Definition: Cell complexes

A cell complex or CW -complex is a space X which results from the
following inductive procedure:
(1) Start with a discrete set X 0 . The points of X 0 will be the 0-cells
of X.
(2) If X n−1 is defined, we construct the n-skeleton X n by attaching
n-cells enα to X n−1 via continuous maps φα : S n−1 → X n−1 . This
means that X n is the quotient space of the disjoint union X n−1 ⊔α
Dαn of X n−1 with a collection of n-disks Dαn under the identifications
x ∼ φα (x) for x ∈ ∂Dαn and φα : ∂Dαn−1 = Sαn−1 → X n−1 . Thus, as
a set, X n consists of X n−1 together with a union of n-cells enα each
of which is an open n-disk Dαn \ ∂Dαn .
(3) If this process stops after finitely many steps, say N , then X =
X N . But it is also allowed to continue with S the inductive process
indefinitely. In this case, one defines X = n X n and equips X
with the weak topology, i.e., a set A ⊂ X is open (or closed) if

and only if A ∩ X n is open (or closed) in X n for each n.
We have already seen some examples of cell complexes:

• The sphere S n is a cell complex with just two cells: one 0-cell e0 (that
is a point) and one n-cell en which is attached to e0 via the constant
map S n−1 → e0 . Geometrically, this corresponds to expressing S n as
Dn /∂Dn : we take the open n-disk en = Dn \ ∂Dn and collapse the
boundary ∂Dn to a single point which is e0 .
• Real projective space RPn is a cell complex with one cell in each
dimension up to n. To show this we proceed inductively. We know
that RP0 consists of a single point, since it is S 0 whose two antipodal
points are identified. Now we would like to understand how RPn can be
constructed from RPn−1 : We embed Dn as the upper hemisphere into
S n , i.e., we consider Dn as {(x0 , . . . , xn ) ∈ S n : x0 ≥ 0}. Then
RPn = S n /x ∼ −x = Dn /(x ∼ −x for boundary points x ∈ ∂Dn ).
But ∂Dn is just S n−1 . Thus the quotient map
S n−1 → S n−1 / ∼ = RPn−1
attaches an n-cell en , the open interior of Dn , at RPn−1 . Thus we obtain
RPn from RPn−1 by attaching one n-cell via the quotient map S n−1 →
RPn−1 . Summarizing, we have shown that RPn is a cell complex with
one cell in each dimension from 0 to n:
RPn = e0 ∪ e1 ∪ · · · ∪ en .
• We can continue this process and build the infinite projective space
∞ n
S
RP := n RP . It is a cell complex with one cell S in each dimension.
We can think of RP∞ as the space of lines in R∞ = n Rn .
The torus is a cell complex with one 0-cell, two 1-cells and one
2-cells. This should be apparent from the construction of the torus as
a quotient of a square that we have seen above. Starting with X 0 being
a point p, the red dot in the picture below. Then we attach two open
1-cells e1a , e1b ⊂ D1 via the two constant maps
φ1a , φ1b : S 0 → X 0
where we think of e1 = (0,1) ⊂ [0,1] = D1 as the open unit interval. (In
the picture they look like two straight lines, but we should think of the
end points being attached to p.)
33
Finally, we attach an open 2-cell e2 ⊂ D2 via the attaching map


1
x ∈ e a
 if y > 0
φ2 : S 1 → X 1 , φ2 (x, y) = x ∈ e1b if y < 0

p if y = 0.
Note that this is a well-defined map, since we have identified the end-
points in X 1 with p and hence (1,0) and (−1,0) are sent to the same
point p.
• Actually, every compact smooth manifold can be turned into a fi-

nite cell complex. This illustrates the vast scope and importance of cell
complexes in algebraic topology.
What makes topology unique

Note that the ability to build spaces by gluing together cells (or other
specific spaces) makes life as a topologist particularly comfortable. For
example, we will see that this procedure will often allow us to create spaces
with given algebraic invariant. This flexibility together with the concept of
homotopy, which we will explore next, puts algebraic topologists in a
unique position and led to the solution of a lot of problems, not just
in topology. Geometry, in its various forms, is usually much more rigid
and does not allow us to perform such manoeuvres.
Note that there is a direct way to define the Euler characteristic of cell
complexes. We will later see the reason why this is the correct definition using
homology. Right now we can already check at the example of a tetrahedron that
this definition agrees with Euler’s formula we saw in the first lecture.
Definition: Euler characteristic for cell complexes

The Euler number of a cell complex X (with cells in dimension at most n)
is defined to be the integer
Xn
χ(X) = (−1)k #{k − dimensional cells that are attached to X k−1 }.
k=0
For example, the Euler characteristic of S n is

(
2 n even
χ(S n ) = 1 + (−1)n =
0 n odd
For real projective n-space we get

(
0 n even
χ(RPn ) = 1 − 1 + 1 − · · · + (−1)n =
1 n odd
For the torus, we get
χ(T ) = 1 − 2 + 1 = 0.
To compare this definition with Euler’s formula we used in the first lecture,
let us look at the tetrahedron which is also a cell complex:
• Homotopy
Homotopy is a fundamental notion in topology. Let us start with a definition
and then try to make sense of this.
35
Definition: Homotopies
Let f0 ,f1 : X → Y be two continuous maps. Then f0 and f1 are called
homotopic, denoted f0 ≃ f1 , if there is a continuous map h : X ×[0,1] → Y
such that, for all x ∈ X,
h(x,0) = f0 (x), and h(x,1) = f1 (x).
Homotopy defines an equivalence relation (exercise!) on the set of continu-
ous maps from X to Y . The set of equivalence classes of continuous maps
from X to Y modulo homotopy is denoted by [X,Y ].
Definition: Homotopy equivalences and contractible

spaces
• A continuous map f : X → Y is called a homotopy equivalence
if there is a continuous map g : Y → X such that g ◦ f ≃ idX and
f ◦ g ≃ idY .
• Two spaces X and Y are called homotopy equivalent if there
exists a homotopy equivalence f between X and Y . This is often
denoted by X ≃ Y .
• A space which is homotopy equivalent to a one-point space is called
contractible.
For example, Rn is contractible, since
h : Rn × [0,1] → Rn , (t,x) 7→ (1 − t)x
defines a homotopy between the identity map on Rn and the constant map Rn →
{0} ⊂ Rn to the one-point space consisting of the origin. For the same reason,
the n-disk Dn is contractible.
However, it is not always obvious which spaces are homotopy equivalent to
each other. So it will be useful to develop some intuition for homotopy equiva-
lences. There is a particular type that is easier to spot:
Definition: Deformation retracts

Let X be a topological space and A ⊂ X a subspace.
• Then A is called a retract of X if there is a retraction ρ : X → A,
i.e., there is a continuous map ρ : X → A with ρ|A = idA .
• Note that we can consider ρ also as a map X → X via the
ρ
inclusion X → − A ⊂ X. If ρ is then in addition homotopic to the
identity of X, then A is called a deformation retract of X. In
this case, ρ is called a deformation retraction. Note that in this
case, ρ and the inclusion A ⊂ X are mutual homotopy inverses.
• If this homotopy between ρ and idX can be chosen such that all
points of A remain fixed, i.e., the homotopy h(t,a) = a for all
a ∈ A and all t ∈ [0,1], then ρ is called a strong deformation
retraction and A is called a strong deformation retract of X.
For a deformation retraction, one can think of the homotopy h as a map which
during the time from 0 to 1 pulls back all the points of X into the subspace A,
and leaves the whole time the points in A fixed. Here are some examples:
• The origin {0} is a strong deformation retract fo Rn and of the n-disk
Dn .
• For any topological space Y , the product Y ×{0} is a strong deformation
retract of Y × Rn and Y × Dn . For example, the circle S 1 × {0} is a
strong deformation retract of the solid torus S 1 × D2 .
• The n-sphere S n is a strong deformation retract of the punctured disk
Dn+1 \ {0} and also of Rn+1 \ {0}.
37
Why homotopy?
The simplest reason why we consider the homotopy relation is that it
works. It is fine enough such that all the tools that we are going to
define are invariant under homotopy, i.e., they are constant on equivalence
classes. But it is also coarse enough that it identifies enough things such
that many problems become simpler and in fact solvable.
With respect to first point, one can consider the homotopy category
hoTop of spaces, i.e., the category whose objects are topological spaces
and whose sets of morphisms from X to Y are the sets of homotopy classes
of maps [X,Y ], satisfies a universal property for invariants.
With respect to the second point, we just indicate that life in hoTop is
much easier because there are much fewer morphisms. For example,
there are many and complicated continuous maps S 1 → C \ {0}. But there
are very few homotopy classes of such maps, since [S 1 ,C \ {0}] = Z, up to
homotopy a map S 1 → C \ {0} is determined by the winding number, i.e.,
the number of times it goes around the origin.
To convince ourselves that homotopy actually works, we remark that homo-
topy is even fine enough to detect diffeomorphism classes between smooth
manifolds and helped for example to classify manifolds up to bordism. But
this is a story we save for a future lecture/class.
If you are still not convinced, then let us remark that to study things up-
to-homotopy is so useful that mathematicians work hard to find analogs
of the homotopy relation and the homotopy category in many different
areas. If you want to learn more about this, have a look at Quillen’s highly
influential book on Homotopical Algebra. You will also see an example
in homological algebra where one talks about homotopies between chain
complexes.
LECTURE 3
Singular chains and homology
We would like to make the idea to study a topological space X by considering

all continuous maps from test spaces into X precise. We start with defining an
important class of test spaces:
Definiton: The standard n-simplex

For n ≥ 0, the standard n-simplex ∆n is the set ∆n ⊂ Rn+1 defined by
n
X
n n+1
∆ = {(t0 , . . . ,tn ) ∈ R : ti = 1, ti ≥ 0 for all i}.
i=0
n
Another way to describe ∆ is to say that it is the convex hull of the
standard basis {e0 , . . . ,en } in Rn+1 :
( )
X X
∆n = ti ei : ti = 1, ti ≥ 0 .
i i
The ti are called barycentric coordinates.
It will be convenient to keep both these descriptions in mind.
The standard simplices are related by face maps for 0 ≤ i ≤ n which can be
described as
ϕni (t0 , . . . ,tn−1 ) = (t0 , . . . ,ti−1 ,0,ti , . . . ,tn−1 )
with the 0 inserted at the ith coordinate (t0 is the 0th coordinate).
39
40 CHAPTER 3. SINGULAR CHAINS AND HOMOLOGY
Using the standard basis, ϕni can be described as the affine linear map (a
translation plus a linear map)
(
ej j<i
ϕni : ∆n−1 ,→ ∆n determined by ϕni (ej ) =
ej+1 j ≥ i.
A short way of expressing the above formula for ϕni is that it embedds ∆n−1
into ∆n by omitting the ith vertex (that is what the hat in the following formula
means):
ϕni = [e0 , . . . ,ei−1 ,b
ei ,ei+1 , . . . ,en ] : ∆n−1 → ∆n .
Definiton: Faces
Note that ϕni maps ∆n−1 onto the subsimplex opposite to the ith corner, or
in the standard basis, opposite to ei . We call the image of ϕni the ith face
of ∆n (which is opposite to ei ).
Note that the union of the images of all the face inclusions is the boundary
of ∆n .
The face maps satisfy a useful identity, sometimes called simplicial identity:
Lemma: A useful identity

For all 0 ≤ j < i ≤ n + 1 we have
(1) ϕni ◦ ϕn−1
j = ϕnj ◦ ϕn−1
i−1 .
41
The first composition, ϕni ◦ ϕn−1

j , results in a 0 at the jth and i + 1st place.
n n−1
The second composition, ϕj ◦ϕi−1 , has the effect to insert a 0 at the (i−1)st
place and then one at the jth place. But since j < i, this means that, in
both cases, we have an extra 0 at the jth and at the (i + 1)st spot.
Thus both compositions yield
ϕni ◦ ϕn−1
j (t0 , . . . ,tn−2 )
=(t0 , . . . ,tj−1 ,0,tj , . . . ,ti−2 ,0,ti−1 , . . . ,tn−2 )
=ϕnj ◦ ϕn−1
i−1 (t0 , . . . ,tn−2 ).
We are going to study a topological space X by looking at all the continuous

maps from simplices into X. We give those sets of maps a name:
Definition: Singular n-simplices

Let X be any topological space. A singular n-simplex in X is a continu-
ous map σ : ∆n → X. We denote by Singn (X) the set of all n-simplices in
X. For example, Sing0 (X) is just the set of points of X. But, in general,
Singn (X) carries more interesting information for n ≥ 1.
For 0 ≤ i ≤ n, we can use the face maps ϕni to define maps
dni : Singn (X) → Singn−1 (X), σ 7→ σ ◦ ϕni
by sending an n-simplex σ to the n − 1-simplex defined by precomposition with

the ith face inclusion. The image dni (σ) = σ ◦ ϕni is called the ith face of σ. We
will sometimes use the notation σ (i) := σ ◦ ϕni for the ith face.
Since the collection of all face inclusions ϕni forms the boundary of ∆n , we can
use the maps dni to talk about the boundary of an n-simplex. The boundaries
of simplices will actually play a crucial role in the story.
We need to make this precise. First let us look at a simple example. Let X
be some space and σ : ∆1 → X be a 1-simplex in X. Assume σ(e0 ) = x0 ̸= x1 =
σ(e1 ). Then we would like to say that the boundary of σ is given by x0 and x1 .
Now let us assume that σ : ∆1 → X is another 1-simplex in X which forms a
closed loop, i.e., σ(e0 ) = σ(e1 ) = x ∈ X. Now we would like to say that σ has
no boundary (since it is a loop). Our face maps express σ(e0 ) = σ(e1 ) as
d10 (σ) = d11 (σ).
It would be nice if we had a short way to formulate that the boundary of σ
vanishes. For example, it would be nice if we were allowed to rewrite this equation
as
∂(σ) = d10 (σ) − d11 (σ) = 0.
But, so far, Sing0 (X) is just a set and we are not allowed to add or subtract
elements. We are now going to remedy this defect, since algebraic operations
make life much easier. Therefore, we formally extend Singn (X) into an abelian
group.
The general way to turn a set B into an abelian group, is to form the associated
free abelian group. The idea is to add the minimal amount of structure
and relations to turn B into an abelian group. Since this is an important
construction, we recall how this works:
Good to know about free abelian groups

• Any abelian group A can be seen as a Z-module with n · a :=
a+· · ·+a (n summands), for n ∈ N and a ∈ A, and (−n)·a := −n·a.
Thus, abelian groups are in bijection with Z-modules. An abelian
group A is called free over a subset B ⊂ A if B is a Z-basis, i.e.,
if any element a ∈ Z can be written uniquely as a Z-linear
combination of elements in B. The cardinality of a basis is the
same for any choice of basis and is called the rank of A.
• The group Zr is free abelian with basis {e1 , . . . ,er } with ei =
(0, . . . ,i,0, . . . ,0) (the 1 in the ith position).
• Note that, for example, the group Z/2Z is not free, since it does
not admit a basis: the vector 1 ∈ Z/2Z cannot be in a basis since
2 · 1 = 0.
43
• Given a set B, there is an associated free abelian group ZB

with basis B which is characterized by the following universal
property: any map f : B → A of sets into an arbitrary abelian
group A can be extended uniquely to a group homomorphism
ϕ : ZB → A with ϕ(b) = f (b) for all b ∈ B.
In terms of category theory, this means that the functor
AbGroups → Sets which forgets the group structure, is right
adjoint to the functor
Sets → AbGroups, B 7→ ZB.
In other words,
HomSets (B,A) = HomAbGroups (ZB,A).
• Any subgroup of a free abelian group F is a free abelian group.
We apply this construction to the set B = Singn (X):
Definition: Singular n-chains

The group Sn (X) of singular n-chains in X is the free abelian group
generated by n-simplices
Sn (X) := ZSingn (X).
Thus an n-chain is a finite Z-linear combination of simplices
k
X
ai σi , ai ∈ Z, σi ∈ Singn (X).
i=1
Note: If n < 0, Singn (X) is defined to be empty and Sn (X) is the trivial
abelian group {0}. So whenever we talk about n-chains, n will be assumed to be
nonnegative.
Definition: Boundary operators

We define the boundary operator by
n
X n
X
∂n : Singn (X) → Sn−1 (X), ∂(σ) = (−1)i dni σ = (−1)i σ (i) .
i=0 i=0
We can then extend this to a homomorphism, which we also call boundary

operator, by additivity, i.e.,
m
! m
X X
∂n : Sn (X) → Sn−1 (X), ∂ aj σj := aj ∂(σj ).
j=1 j=1
Note that we will often just write ∂ instead of ∂n .
In particular, for the loop σ we considered above we are allowed to write in

S0 (X)
∂1 (σ) = d10 (σ) + (−1)d11 (σ) = d10 (σ) − d11 (σ) = 0.
A loop is an example of a particularly important class of chains. For, the

equation ∂(σ) = 0 expresses algebraically that σ has no boundary. We give
such chains a special name:
Definition: Cycles
An n-cycle in X is an n-chain c ∈ Sn (X) with ∂n c = 0. We denote the
group of n-cycles by
Zn (X) := Ker (∂n : Sn (X) → Sn−1 (X))
= {c ∈ Sn (X) : ∂n (c) = 0} ⊆ Sn (X).
Note that the group of 0-cycles is all of S0 (X), since every 0-chain is mapped
to 0:
Z0 (X) = S0 (X).
To find another example of a 1-cycle we could consider a 1-chain c = α+β+γ

where α, β, γ : ∆1 → X are singular 1-simplices such that
α(e1 ) = β(e0 ), β(e1 ) = γ(e0 ), γ(e1 ) = α(e0 ).
For then we get
∂(c) = d0 (α) − d1 (α) + d0 (β) − d1 (β) + d0 (γ) − d1 (γ)
= α(e1 ) − α(e0 ) + β(e1 ) − β(e0 ) + γ(e1 ) − γ(e0 )
= 0.
As the notation suggests, we are going to think of a chain of the form ∂(c) as
the boundary of c:
45
Definition: Boundaries
An n-dimensional boundary in X is an n-chain c ∈ Sn (X) such that
there exists an (n + 1)-chain b with ∂n+1 b = c. We denote the group of
n-boundaries by
Bn (X) := Im (∂n+1 : Sn+1 (X) → Sn (X))
= {c ∈ Sn (X) : there is a b ∈ Sn+1 (X) with ∂n+1 (b) = c}.
As an aside, here is another way of thinking of the algebraic process.
Signs are like orientations... just not exactly

We want to express the fact that a loop has no boundary by saying that
the signs of the boundary points cancel out. The following picture
illustrates that the something similar happens when several vertices are
involved:
In general, we can think of the signs as giving the faces of the simplices
an orientation. And if an n-simplex is a face of an (n + 1)-simplex, then it
inherits an induced orientation which is determined by how it fits into the
bigger simplex. Going down two steps of inherited signs means things
cancel out.
However, thinking of signs as orientations is formally not correct as we will
notice in an example below. But, as we will see soon, we can algebraically
remedy this defect.
As the above picture suggests, every boundary is a cycle:
Theorem: Boundaries of boundaries vanish

For every topological space X, the boundary operator satisfies ∂ ◦ ∂ = 0, or
more precisely
∂n ◦ ∂n+1 = 0 : Sn+1 (X) → Sn−1 (X).
Proof: It suffices to check this for an (n + 1)-simplex σ. The general case
follows, since each ∂ is a homomorphism. For σ, we just calculate:
n+1
! n+1
X X
i n+1
∂n ◦ ∂n+1 (σ) = ∂n (−1) di σ = ∂n (σ ◦ ϕn+1
i )
i=0 i=0
n+1
X n
X
= (−1)i (−1)j σ ◦ ϕn+1
i ◦ ϕnj
i=0 j=0
X X
= (−1)i+j σ ◦ ϕn+1
i ◦ ϕnj + (−1)i+j σ ◦ ϕn+1
i ◦ ϕnj
0≤j<i≤n+1 0≤i≤j≤n
(∗) X X ′ ′
= (−1)i+j σ ◦ ϕn+1
j ◦ ϕni−1 + (−1)j +i −1 σ ◦ ϕn+1
j′ ◦ ϕni′ −1
0≤j<i≤n+1 0≤j ′ <i′ ≤n+1
= 0.
Note that at (∗) we applied identity (1) to the left hand sum and just changed
the labels of the indices as i → j ′ and j → i′ − 1. Since both sums run over the
same indices (it does not matter how we label them) and the right hand sum is
the left hand sum multiplied by (−1), both sums cancel out. QED
As an immediate consequence we get:
Corollary: Every boundary is a cycle

For every n ≥ 0, we have
Bn (X) ⊆ Zn (X).
This basic resut shows that the sequence {Sn (X), ∂n }n has an important prop-
erty:
Definition: Chain complexes

A graded abelian group is a sequence of abelian groups, indexed by the
integers. A chain complex is a graded abelian group {An }n together with
homomorphisms ∂n : An → An−1 with the property that ∂n−1 ◦ ∂n = 0.
Hence we have shown that we obtain for any topological space X a complex
of (free) abelian groups
∂ n ∂ ∂n−1
2 1∂ 0 ∂ ∂
··· →
− Sn (X) −→ Sn−1 (X) −−−→ · · · −
→ S1 (X) −
→ S0 (X) −
→ 0.
47
It is called the singular chain complex of X. We will see next lecture what
such chain complexes are good for.
LECTURE 4
Singular homology, functoriality and H0
Recall that we constructed, for any topological space X, the singular chain
complex of X
∂ n∂ ∂n−1 2∂ 1 ∂ 0 ∂
··· →
− Sn (X) −→ Sn−1 (X) −−−→ · · · −
→ S1 (X) −
→ S0 (X) −
→ 0.
The homomorphisms ∂n satisfy the fundamental rule: ∂ ◦ ∂ = 0.
The following definition of homology groups applies to any chain complex.
However we formulate it only for the singular chain complex:
Definition: Singular homology

The nth singular homology group of X is defined to be the quotient
group of n-cycles modulo n-boundaries:
Zn (X) Ker (∂ : Sn (X) → Sn−1 (X))
Hn (X) = = .
Bn (X) Im (∂ : Sn+1 (X) → Sn (X))
We are going to say that two cycles whose difference is a boundary are
homologous.
Let us make a first attempt to understand what is going on here:
Singular what?
In algebra, homology is a way to measure the difference between cycles
and boundaries. Singular homology is an application of homology in order
to understand the structure of a space.
Given a space X, the group of n-cycles measures how often we can map
an n-dimensional simplex into X without collapsing it to any of its n − 1-
dimensional faces.
Let σ(∆n ) be the image in X of such a cycle. If we can even map an (n + 1)-
dimensional simplex σ ′ (∆n+1 ) into X whose boundary is σ(∆n ), then we can
continuously collapse all of σ(∆n ) to a point. In this case, we would like to
49
50 CHAPTER 4. SINGULAR HOMOLOGY, FUNCTORIALITY AND H0
forget about this σ. For, from an n-dimensional point of view, this σ(∆n ) is
not interesting. That is what it means geometrically/topologically to take
the quotient by Bn (X).
But if we cannot find an n + 1-dimensional simplex such that σ(∆n ) is
its boundary, then σ(∆n ) potentially carries interesting n-dimensional
information about X.
The slogan is: Hn (X) measures n-dimensional wholes in X.
Before we see some examples of homology groups we go back to the idea of

“orientations of simplices” and see why taking the quotient by boundaries is
a good thing. We said that we think of the signs as orientations, but this is not
completely correct. But modulo boundaries we are good:
Orientations revisited
Let X be some space, and suppose we have a one-simplex σ : ∆1 → X.
Define
ϕ : ∆1 → ∆1 , (t,1 − t) 7→ (1 − t,t).
Precomposing with ϕ gives another singular simplex σ̄ = σ◦ϕ which reverses
the orientation of σ. It is not true that σ̄ = −σ in S1 (X).
However, we claim that
σ̄ ≡ −σ mod B1 (X).
This means that there is a 2-chain in X whose boundary is σ̄ + σ. If d0 (σ) =
d1 (σ) such that σ ∈ Z1 (X) is a 1-cycle, then σ̄ and σ are homologous and
[σ̄] = [σ] in H1 (X).
To prove the claim we need to construct an appropriate 2-chain. Let
π : ∆2 → ∆1 be the affine map determined by sending e0 and e2 to e0
51
and e1 to e1 . For x ∈ X and n ≥ 0, we write κnx : ∆n → X for the constant

map with value x.
Now we calculate
∂(σ ◦ π) = σ ◦ π ◦ ϕ20 − σ ◦ π ◦ ϕ21 + σ ◦ π ◦ ϕ22 = σ̄ − κ1σ(0) + σ.
Hence up to the term −κ1σ(0) we get what we want. So we would like to elim-
inate this term. To do that we define the constant 2-simplex κ2σ(0) : ∆2 → X
at σ(0). Its boundary is
∂(κ2σ(0) ) = κ1σ(0) − κ1σ(0) + κ1σ(0) = κ1σ(0) .
Thus
σ̄ + σ = ∂(π ◦ σ + κ2σ(0) )
which proves the claim.
Actually, we will have to get back to orientations almost regularly, in par-

ticular when we talk about simplicial complexes, and step by step improve our
understanding and control.
Aside: The sequence of homology groups {Hn (X)}n also forms a graded
abelian group. Note that even though Zn (X) and Bn (X) are free abelian groups
because they are subgroups of the free abelian group Sn (X), the quotient Hn (X)
is not necessarily free. Moreover, while Zn (X) and Bn (X) may be uncount-
ably generated, Hn (X) turns out to be finitely generated for the spaces we are
interested in.
Let us look at two simple examples:
(1) Let X = ∅. Then Sing∗ (∅) = ∅ and S∗ (∅) = 0 is just the trivial abelian
group by convention. Hence · · · → S2 → S1 → S0 is the zero chain
complex and Z∗ (∅) = B∗ (∅) = 0. The homology in all dimensions is
therefore 0.
(2) Let X = pt be a one-point space. Then, for each n, there is only
one singular n-simplex, namely the constant map σn : ∆n → pt. In
other words, Sn (X) = Z · σn is generated by a single element. Hence
(i)
σn = σn ◦ ϕni = σn−1 and

X n X n 0
 n odd
i (i)
∂σn = (−1) σn = (−1)σn−1 = σn−1 n even

i=0 i=0 0 n = 0.
For cycles and boundaries this means

(
Z · σn n odd or n = 0
Zn (X) =
0 n even and n ̸= 0,
and
(
Z · σn n odd or
Bn (X) =
0 n even.
For the homology groups we get

(
Z n=0
Hn (pt) ∼
=
̸ 0.
0 n=
To complete the picture, the singular chain complex looks like

∂=id ∂=0 ∂=id ∂=0
· · · −−→ Z −−→ Z −−→ Z −−→ Z → 0.
• Functoriality
Now that we have defined homology we can ask how it behaves under contin-
uous maps. So let X and Y be topological spaces and f : X → Y be a continuous
map. Since singular simplices are just maps, we can define an induced map
f∗ : Singn (X) → Singn (Y ), σ 7→ f ◦ σ
just by composition with f .
The same construction yields an induced map on chains:
m
X m
X
f∗ = Sn (f ) : Sn (X) → Sn (Y ), aj σj 7→ aj (f ◦ σj ).
j=1 j=1
The induced map is compatible with the boundary operator in the following
way:
53
Lemma: The singular chain complex is natural

For every n ≥ 0, we have a commutative diagram
Sn (f )
Sn (X) / Sn (Y )
∂X ∂Y

Sn−1 (X) / Sn−1 (Y ).
Sn−1 (f )
Proof: We just calculate and check that both ways have the same outcome
for any singular n-simplex σ on X:
n
X
∂Y (Sn (f ))(σ) = (−1)i (f ◦ σ) ◦ ϕni
i=0
Xn
= (−1)i f ◦ (σ ◦ ϕni )
i=0
n
!
X
i
= Sn−1 (f ) (−1) σ ◦ ϕni
i=0
= Sn−1 (f )(∂X σ).
□
The lemma has the important consequence that
Sn (f )(Zn (X)) ⊂ Zn (Y ) and Sn (f )(Bn (X)) ⊂ Bn (Y ).
For, if c ∈ Zn (X), then

∂Y (Sn (f )(c)) = Sn−1 (f )(∂X (c)) = Sn−1 (f )(0) = 0;
and, if c ∈ Bn (X), then there is a b ∈ Bn+1 (X) with ∂X (b) = c and
∂Y (Sn+1 (f )(b)) = Sn (f )(∂X (b)) = Sn (f )(c),
i.e., there is an element, b′ = Sn+1 (f )(b), with ∂Y (b′ ) = Sn (f )(c).
Proposition: Homology is functorial

Thus we get a well defined induced homomorphism on homology groups
Hn (f ) : Hn (X) → Hn (Y ), [c] 7→ [Sn (f )(c)].
The homomorphisms Sn (f ) and Hn (f ) have the following properties:
• Sn (idX ) = idSn (X) and Hn (idX ) = idHn (X)

• Sn (f ◦ g) = Sn (f ) ◦ Sn (g) and Hn (f ◦ g) = Hn (f ) ◦ Hn (g).
To summarize our observations: Sn (−) and Hn (−) are functors from the
category of topological spaces to the category of abelian groups. For the sequence
of all Sn (−) even more is true: S∗ (−) is a functor from the category of topological
spaces to the category of chain complexes of abelian groups (with chain maps as
morphisms).
Invariance
As a consequence, if f : X → Y is a homeomorphism, then Hn (f ) is an
isomorphism of abelian groups. In other words, homology groups only
depend on the topology of a space.
In fact, we will soon see that homology is a coarser invariant in the sense that
homotopic maps induce the same map in homology.
• The homology group H0
Let us try to understand the simplest of the homology groups.
Lemma: Augmentation
For any topological space X, there is a homomorphism
ϵ : H0 (X) → Z
which is nontrivial whenever X ̸= ∅.
Proof: If X = ∅, then H∗ (∅) = 0 by definition. In this case, we define ϵ to

be the zero homomorphism.
Now let X ̸= ∅. Then there is a unique map X → pt from X to the one-point
space. By functoriality, it induces a homomorphism
ϵ : H0 (X) → H0 (pt) = Z.
□
Let us try to understand this ϵ a bit better. The map X → pt induces a
homomorphism of chain complexes S∗ (X) → S∗ (pt) which sends any 0-simplex
55
σ : ∆0 → X to the constant map

σ
κ0 : ∆0 →
− X → pt
which is the generator of S0 (pt) = Z. Hence we get a map σ 7→ 1 which extends

to a homomorphism ϵ̃ : S0 (X) → Z by additivity, i.e.
X X
ϵ̃( aj σ j ) = aj ∈ Z.
j j
To double check that this map descends to a homomorphism ϵ on H0 (X)

we need to show that it maps boundaries to 0. (We know this already, but let us
do it anyway.)
So let
P b be a 0-chain which is the boundary of a 1-chain c, i.e., b = ∂c, and
let c = j aj γj with finitely many 1-simplices γj : ∆ → X. Then each γj ◦ ϕ10
1
and γj ◦ ϕ11 are 0-simplices and are sent to 1 by ϵ̃. Thus we get
X X X
ϵ(b) = ϵ(∂c) = ϵ( aj (γj ◦ ϕ10 − γj ◦ ϕ11 )) = aj − aj = 0.
j j j
We learn from this discussion that, since a 0-simplex ∆0 → X can be identified

with its image point, ϵ counts the points on X, with multiplicities. And if
two points can be connected by a 1-simplex, i.e., by a path in X, then they add
up to 0. This leads us to:
Theorem: H0 for path-connected spaces

If X is path-connected and non-empty, then ϵ is an isomorphism
∼
=
ϵ : H0 (X) −
→ Z.
Proof: Since X is non-empty, there is a point x ∈ X. The 0-simplex σ = κ0x

with value x is an element in S0 (X) which is sent to 1 ∈ Z. Additivity implies that
ϵ is surjective. To show that ϵ is also injective, we need to show that the classes
of the 0-simplices given by constant maps at any two points are homologous.
So let y ∈ X be another point. Since X is path-connected, there is a path
ω : [0,1] → X with ω(0) = x and ω(1) = y. We define a 1-simplex σω by
σω (t0 ,t1 ) := ω(1 − t0 ) = ω(t1 ) for t0 + t1 = 1, 0 ≤ t0 ,t1 ≤ 1.

The boundary of σω is
∂(σω ) = d0 (σω ) − d1 (σω ) = σω (e1 ) − σω (e0 )

= σω (0,1) − σω (1,0)(= ω(0) − ω(1))
= κ0x − κ0y
(where we identify 0-simplices and their image points). Hence the 0-simplices κ0x
and κ0y are homologous. Since 0-simplices generate H0 (X) and ϵ is a homomor-
phism, this implies that ϵ is injective. QED
Corollary: H0 is generated by path components

F
If X is a disjoint union X = i∈I Xi where each Xi is path-connected and
non-empty, then, for all n ≥ 0,
Hn (X) ∼
M
= Hn (Xi ).
i∈I
In particular, for n = 0 we get
H0 (X) ∼
M
= Z.
i∈I
In other words, H0 (X) is the free abelian group generated by the set of
path-components of X.
Proof: If σ : ∆n → X is an n-simplex, then its image lies in exactly one

connected component Xi . Otherwise, we could write ∆n as the disjoint union of
two open and closed subsets contradicting the fact that ∆n is connected. Hence
σ factors into ∆n → Xi ,→ X.
57
Since singular n-chains are freely generated by n-simplices, this shows that
the singular chain complex of X splits into a direct sum
M
S∗ (X) = S∗ (Xi ).
i∈I
For the same reason the boundary operators
∂ : Sn (X) ∼ Sn−1 (Xi ) ∼
M M
= Sn (Xi ) → = Sn−1 (X)
i∈I i∈I
split into components ∂Xi : Sn (Xi ) → Sn−1 (Xi ). Hence we get an isomorphism.
The statement for n = 0 then follows from the previous result on path-connected
spaces. □
LECTURE 5
Relative homology and long exact sequences
If we want to show that singular homology groups are useful, we need to be

able to compute them. For H0 that was not so difficult. But for n ≥ 1, we need
to develop some techniques.
In general, if you would like to compute something for spaces, it is always a
good idea to think about the relation to subspaces. Maybe the information
on smaller subspaces provides insides on the whole space. That is the idea we
are going to exploit now for homology groups.
Let X be a topological space and let A ⊂ X be a subset. We can consider
(X,A) as a pair of spaces. If (Y,B) is another such pair, then we denote by
C((X,A),(Y,B)) := {f ∈ C(X,Y ) : f (A) ⊂ B}
the set of continuous maps which respect the subspaces. In fact, we get a category
Top2 of pairs of topological spaces.
Given a pair of spaces A ⊂ X, any n-simplex of A defines an n-simplex on X:

σ σ
(∆n →
− A) 7→ (∆n →
− A ⊂ X).
The induced map Sn (A) → Sn (X) is injective. Hence we are going to identify
Sn (A) with its image in Sn (X) and consider Sn (A) as a subgroup of Sn (X).
59
60 CHAPTER 5. RELATIVE HOMOLOGY AND LONG EXACT SEQUENCES
Definiton: Relative chains

We define the group of relative n-chains by
Sn (X,A) := Sn (X)/Sn (A).
The group Sn (X,A) is free, since the quotient map sends basis elements to
basis elements, and is generated by the classes of n-simplices of X whose
image is not entirely contained in A.
Since the boundary operator is defined via composition with the face maps,
it satisfies
∂(Sn (A)) ⊂ Sn−1 (A) ⊂ Sn−1 (X).
For, if the image of σ : ∆n → X lies in A, then so does the image of the composite
∆n−1 ,→ ∆n → X.
Thus ∂ induces a homomorphism ∂¯ on Sn (X,A) and we have a commuta-
tive diagram
Sn (X) / Sn (X,A)
∂ ∂¯

Sn−1 (X) / Sn−1 (X,A).
Since ∂ ◦ ∂ = 0 and since Sn (X) → Sn (X,A) is surjective, we also have
∂¯ ◦ ∂¯ = 0.
We define relative n-cycles and relative n-boundaries by

Zn (X,A) := Ker (∂¯ : Sn (X,A) → Sn−1 (X,A)) and
Bn (X,A) := Im (∂¯ : Sn+1 (X,A) → Sn (X,A)).
61
Definition: Relative homology

The nth relative homology group of the pair (X,A) is defined as
Hn (X,A) := Zn (X,A)/Bn (X,A).
Roughly speaking, relative homology groups measure the difference between

the homology of X and the homology of A. Let us try to make this more precise.
That an n-chain c in Sn (X) represents a relative n-cycle means that ∂(c) ¯ =0
in Sn−1 (X)/Sn−1 (A), i.e., ∂(c) ∈ Sn−1 (A). Hence it just means that the image of
the boundary of c lies in A.
So let us consider the preimage of Zn (X,A) under the quotient map Sn (X) →
Sn (X,A) and define
Zn′ (X,A) := {c ∈ Sn (X) : ∂(c) ∈ Sn−1 (A)}.
Similarly, that an n-chain c in Sn (X) represents a relative n-boundary means

that there is an n + 1-chain b such that
c ≡ ∂(b) mod Sn (A), i.e., c − ∂(b) ∈ Sn (A).
Hence the preimage of Bn (X,A) under the quotient map is

Bn′ (X,A) := {c ∈ Sn (X) : ∃ b ∈ Sn+1 (X) such that c − ∂(b) ∈ Sn (A)}.
Now we observe that Zn (X,A) = Zn′ (X,A)/Sn (A) (since Sn (X,A) is Sn (X)/Sn (A))
and Bn (X,A) = Bn′ (X,A)/Sn (A). Hence we get
Zn (X,A) Z ′ (X,A)/Sn (A) Z ′ (X,A)
Hn (X,A) = = n′ = n′ .
Bn (X,A) Bn (X,A)/Sn (A) Bn (X,A)
In other words, we could also have used the latter quotient to define Hn (X,A).
Empty subspaces
As a special case with A = ∅ we get
Zn′ (X,∅) = Zn (X), Bn′ (X,∅) = Bn (X), and Hn (X,∅) = Hn (X).
Now let us have a look at two examples to see how the images of simplices
in Hn (X) and Hn (X,A) can differ.
Example: Relative cycles on the cylinder

Let X = S 1 × [0,1] be a cyclinder over the circle, and let the subspace
A = S 1 × 0 ⊂ X be the bottom circle.
We construct an element in the relative homology H1 (X,A) by taking a
1-simplex
σ : ∆1 → X,
(te1 , (1 − t)e0 ) 7→ (cos(2πt), sin(2πt),1).
Since σ is a closed curve in X, we have σ(e0 ) = σ(e1 ). Hence its boundary

vanishes:
∂(σ) = σ(e1 ) − σ(e0 ) = 0.
Therefore, σ ∈ Z1 (X) ⊂ Z1′ (X,A). We will see very soon, that σ, in fact,
represents a nontrivial class in H1 (X).
However, the image of σ in the relative homology group H1 (X,A) vanishes.
For, consider the 2-chains τ1 and τ2 as indicated in the picture. Then we
have
∂(τ1 + τ2 ) = d0 (τ1 ) − d1 (τ1 ) + d2 (τ1 ) + d0 (τ2 ) − d1 (τ2 ) + d2 (τ2 )
=σ−β+γ+β−γ+α
= σ + α with α ∈ S1 (A).
Hence, modulo S1 (A), we have σ ∈ B1 (X,A) and
[σ] = 0 in H1 (X,A).
63
And the second example:
Example: Relative cycles on ∆n

Let us look the standard n-simplex X = ∆n as a space on its own. We
would like to study it relative to its boundary
[
∂∆n := Im ϕni ≈ S n−1
i
which is homeomorphic to the n − 1-dimensional sphere.
There is a special n-simplex in Singn (∆n ) ⊂ Sn (∆n ), called the universal
n-simplex, given by the identity map ιn : ∆n → ∆n . It is not a cycle,
since its boundary ∂(ιn ) ∈ Sn−1 (∆n ) is the alternating sum of the faces of
the n-simplex each of which is a generator in Sn−1 (∆n ):
X
∂(ιn ) = (−1)i ϕni (∆n−1 ) ̸= 0.
i
However, each of these singular simplices lies in ∂∆n , and hence ∂(ιn ) ∈
Sn−1 (∂∆n ).
Thus the class ῑn ∈ Sn (∆n ,∂∆n ) is a relative cycle. We will see later
that the relative homology group Hn (∆n ,∂∆n ) is an infinite cyclic group
generated by [ῑn ].
• Long exact sequences
Back to the general case. So let (X,A) be a pair of spaces. We know that
the inclusion map i : A ,→ X induces a homomorphism Hn (i) : Hn (A) → Hn (X).
Moreover, the map of pairs j : (X,∅) → (X,A) induces a homomorphism
Hn (j) : Hn (X) ∼
= Hn (X,∅) → Hn (X,A).
We claim that there is yet another interesting map.
Connectng homomorphism
For all n, there is a connecting homomorphism, which is often also called
boundary operator and therefore usually also denoted by ∂,
∂ : Hn (X,A) → Hn−1 (A), [c] 7→ [∂(c)]
with c ∈ Zn′ (X,A).
Let us try to make sense of this definition: We just learned that we can
represent an element in Hn (X,A) by an element c ∈ Zn′ (X,A). Then ∂(c) is an
element in Sn−1 (A). In fact, ∂(c) is a cycle, since it is a boundary and therefore
∂(∂(c)) = 0.
In particular, ∂(c) represents a class in the homology Hn−1 (A). Hence we can
send [c] under the connecting homomorphism to be the class [∂c] ∈ Hn−1 (A).
It remains to check that this is well-defined, i.e., if we choose another
representative for the class [c] we need to show that we obtain the same class
[∂(c)].
Another representative of [c] in Zn′ (X,A) has the form c + ∂(b) + a with
b ∈ Sn+1 (X) and a ∈ Sn (A). Then we get
∂(c + ∂(b) + a) = ∂(c) + ∂(a).
But, since ∂(a) ∈ Bn−1 (A), we get
[∂(c)] = [∂(c) + ∂(a)] in Hn−1 (A).
Thus, the connecting map is well-defined. And it is a homomorphism, since ∂
is a homomorphism.
Hence we get a sequence of homomorphisms
Hn (i) Hn (j) ∂
Hn (A) −−−→ Hn (X) −−−→ Hn (X,A) →
− Hn−1 (A).
65
It is an exercise to check that the connecting homomorphism is natural:
The connecting homomorphism is functorial

For any f ∈ C((X,A),(Y,B)), the following diagram commutes
Hn (f )
Hn (X,A) / Hn (Y,B)
∂ ∂

Hn−1 (A) / Hn−1 (B).
Hn−1 (f|A )
In fact, the existence of the connecting map, the above sequence and its
properties can be deduced by a purely algebraic process, that we will recall below.
For, the relative chain complex fits into the short exact sequence of chain
complexes
0 → S∗ (A) → S∗ (X) → S∗ (X,A) → 0.
Such a sequence induces a long exact sequence in homology of the form
Hn+1 (i)
··· / Hn+1 (X,A)
∂
t Hn (j)
Hn (A) / Hn (X) / Hn (X,A)
Hn (i)
∂
t
Hn−1 (A) / ···
Hn−1 (i)
A digression to homological algebra
Maps of chain complexes

Let A∗ and B∗ be two chain complexes. A morphism of chain complexes
f∗ : A∗ → B∗ is a sequence of homomorphisms {fn }n∈Z
fn : An → Bn such that fn−1 ◦ ∂nA = ∂nB ◦ fn for all n ∈ Z.
A homomorphism of chain complexes induces a homomorphism on homol-
ogy
Hn (f ) : Hn (A∗ ) → Hn (B∗ ), [a] 7→ [fn (a)].
Check, as an exercise, that this is well-defined.
Consider a short exact sequence of chain complexes
f∗ g∗
(2) 0 → A∗ −
→ B∗ −
→ C∗ → 0,
i.e., for every n, the corresponding sequence of abelian groups is exact.
Since f∗ and g∗ are homomorphisms of chain complexes, they induce maps
on homology groups
Hn (f ) Hn (g)
(3) Hn (A∗ ) −−−→ Hn (B∗ ) −−−→ Hn (C∗ ).
Since gn ◦ fn = 0, we know Hn (g) ◦ Hn (f ) = 0.
But is the sequence exact at Hn (B∗ ), i.e., is Ker (Hn (g)) = Im (Hn (f ))?
Let us look at an extended picture of the short exact sequence:
fn+1 gn+1
(4) 0 / An+1 / Bn+1 / Cn+1 / 0
∂A ∂B ∂C
fn gn
0 / An / Bn / Cn / 0
∂A ∂B ∂C
fn−1 gn−1
0 / An−1 / Bn−1 / Cn−1 / 0
Let [b] ∈ Hn (B∗ ) such that Hn (g)([b]) = 0. In fact, [b] is represented by a

cycle, i.e., some b ∈ Bn with ∂B (b) = 0. Since Hn (g)([b]) = 0, there is some
c̄ ∈ Cn+1 such that ∂C (c̄) = gn (b). By exactness of (2), gn+1 is surjective and
there is some b̄ ∈ Bn+1 with gn+1 (b̄) = c̄.
Now we can consider ∂B (b̄) ∈ Bn , and have gn (∂B (b̄)) = ∂C (c̄) in Cn . What is
the difference b − ∂B (b̄)?
Well, it maps to 0 in Cn . By exactness of (2), there is some a ∈ An such
that fn (a) = b − ∂B (b̄). Is a a cycle, and hence does it represent a homology
class?
We know
fn−1 (∂A (a)) = ∂B (fn (a)) = ∂B (b − ∂B (b̄)) = ∂B (b) − ∂B (∂B (b̄)) = ∂B (b).
But we assumed ∂B (b) = 0. Thus fn−1 (∂A (a)) = 0. Since fn−1 is injective,
this implies ∂A (a) = 0. Hence a is indeed a cycle, and therefore represents a
homology class [a] ∈ Hn (A∗ ). It also follows
Hn (f )([a]) = [b − ∂B (b̄)] = [b].
67
Thus sequence (3) is exact.

Hn (f∗ )
However, the map Hn (A∗ ) −−−−→ Hn (B∗ ) may fail to be injective and the
Hn (g∗ )
map Hn (B∗ ) −−−−→ Hn (C∗ ) may fail to be surjective. That means sequence
(3) does not fit into a short exact sequence, in general.
But we can connect all these sequences together for varying n and obtain a
long exact sequence:
The homology long exact sequence

f∗ g∗
Let 0 → A∗ −→ B∗ − → C∗ → 0 be a short exact sequence of chain complexes.
Then, for each n, there is a functorial homomorphism
∂ : Hn (C∗ ) → Hn−1 (A∗ )
such that the sequence
Hn+1 (f∗ )
··· / Hn+1 (C∗ )
∂
t Hn (g∗ )
Hn (A∗ ) / Hn (B∗ ) / Hn (C∗ )
Hn (f∗ )
∂
t
Hn−1 (A∗ ) / ···
Hn−1 (f∗ )
is exact.
Proof: This is a typical example of a diagram chase. We will illustrate it by

constructing the connecting homomorphism ∂ and leave the rest as an exercise.
It is more fun and, in fact, easier to do it yourself than to read it. All we need is
to look again at the extended picture (4) of the short exact sequence above.
To construct ∂ : Hn (C∗ ) → Hn−1 (A∗ ), let c ∈ Cn be a cycle. Since gn is
surjective, there is a b ∈ Bn with gn (b) = c. Since ∂C (c) = 0 and the diagram
commutes, we get gn−1 (∂B (b)) = ∂C (gn (b)) = ∂C (c) = 0. Since the horizontal
sequences are exact, this implies there is an a ∈ An−1 with fn−1 (a) = ∂B (b). In
fact, there is a unique such a because fn−1 is injective.
Moreover, we claim that this a is a cycle. For, since the diagram commutes,
we have fn−2 (∂A (a)) = ∂B (fn−1 (a)) = ∂B (∂B (b)) = 0. Since fn−2 is injectve,
this implies ∂A (a) = 0.
This means a represents a homology class and we define ∂ by sending the
class of c to the class of a.
But we need to check that this does not depend on the choices we have made.
So let b′ ∈ Bn be another element with gn (b′ ) = c, and let a′ ∈ An−1 be the
element that we find as above. Then we need to show [a′ ] = [a] in Hn−1 (A∗ ), i.e.,
that a′ − a is a boundary. So we need an ā ∈ An such that ∂A (ā) = a′ − a. We
know gn (b′ −b) = c−c = 0. By exactness, there is an ā ∈ An with fn (ā) = b′ −b.
Since the diagram commutes, we have fn−1 (∂A (ā)) = ∂B (b′ ) − ∂B (b). But we also
have fn−1 (a′ − a) = ∂B (b′ ) − ∂B (b) by definition of a′ and a. Hence, since fn−1 is
injective, we must have ∂A (ā) = a′ − a.
Finally, it is also clear from the construction that if c is a boundary, then a is
zero.
This proves the existence of ∂. Moreover, we know already that the induced
homology sequence is exact at Hn (B∗ ). It remains to check exactness at Hn (A∗ )
and Hn (C∗ ). This is left as an exercise. QED
Why do we care about long exact sequences?

Well, they are extemely useful. For example, for a pair of space (X,A), if
we can show Hn+1 (X,A) = 0 and Hn (X,A) = 0, then Hn (A) ∼ = Hn (X).
Long exact sequences will be one of the main computational tools for
studying interesting homology groups.
Furthermore, there is the famous Five Lemma (here in one of its variations):
Five Lemma
Suppose we have a commutative diagram
A1 / A2 / A3 / A4 / A5
f1 f2 f3 f4 f5

B1 / B2 / B3 / B4 / B5
with exact rows. Then
• If f2 and f4 are surjective and f5 injective, then f3 is surjective.
• If f2 and f4 are injective and f1 surjective, then f3 is injective.
In particular, if f1 , f2 , f4 , and f5 are isomorphisms, then f3 is an isomor-
phism.
69
The proof is another diagram chase and left as an exercise. You should defi-
nitely do it, it’s fun!
Here we rather state two consequences:
• Given a map of short exact sequences
0 / A′ / A / A′′ / 0
f′ f f ′′

0 / B′ / B / B ′′ / 0
in which f ′ and f ′′ are isomorphisms. Then f is an isomorphism.
• Back in topology, let f : (X,A) → (Y,B) be a map of pairs of spaces. If
any two of A → B, X → Y and (X,A) → (Y,B) induce isomorphisms,
then so does the third. This observation will simplify our life a lot.
LECTURE 6
The Eilenberg-Steenrod Axioms
Singular homology can in fact be unquely characterized by a quite short list

of properties some of which we have already checked. This list of properties is
called the Eilenberg-Steenrod Axioms. We are now going to formulate them
in general and will then discuss the relation to singular homology as we defined
it.
First some preparations:
We denote by Top2 the category of pairs of topological spaces. Two continuous
maps f0 ,f1 : (X,A) → (Y,B) between pairs are called homotopic, denoted f0 ≃
f1 , if there is a continuous map
h : X × [0,1] → Y
such that, for all x ∈ X,
h(x,0) = f0 (x), h(x,1) = f1 (x), and h(A × [0,1]) ⊂ B.
For A ⊂ X, we call
[
A◦ = U with U open in X
U ⊂A
the interior of A and

\
Ā = Z with Z closed in X
A⊂Z
the closure of A.
Eilenberg-Steenrod Axioms
A homology theory (for topological spaces) h consists of:
• a sequence of functors hn : Top2 → Ab for all n ∈ Z and
• a sequence of functorial connecting homomorphisms
∂ : hn (X,A) → hn−1 (A,∅) =: hn−1 (A)
which satisfy the following properties:
71
72 CHAPTER 6. THE EILENBERG-STEENROD AXIOMS
• Dimension Axiom: hq (pt) is nonzero only if q = 0.

• Long exact sequences: For any pair (X,A), the sequence
∂ hn (i) hn (j) ∂ ∂
··· →
− hn (A) −−−→ hn (X) −−−→ hn (X,A) →
− hn−1 (A) →
− ···
is exact, where we write hn (X) := hn (X,∅).
• Homotopy Axiom: If f0 ,f1 : (X,A) → (Y,B) are homotopic, then
the induced maps on homology
hn (f0 ) = hn (f1 ) : hn (X,A) → hn (Y,B)
for all n ∈ Z.
• Excision Axiom: For every pair of spaces (X,A) and every U ⊂ A
with Ū ⊂ A◦ the homomorphism
hn (k) : hn (X \ U, A \ U ) → hn (X,A)
induced by the inclusion map k : (X \ U, A \ U ) ,→ (X,A) is an
isomorphism.
• Additivity Axiom: If X = ⊔α Xα is a disjoint union, then the
inclusion maps iα : Xα ,→ X induce an isomorphism for every n
M ∼
=
⊕α hn (iα ) : hn (Xα ) −
→ hn (⊔α Xα ).
α
We have already shown that singular homology satisfies the dimension axiom
and the connectiong homomorphism fits into long exact sequences. It remains to
check homotopy invariance and excision. But before we do that we will assume
these properties for a moment and use them to compute some homology groups.
First an important consequence of the homotopy axiom:
Proposition: Homotopy invariance of homology

Let f : (X,A) → (Y,B) be a map of pairs which is a homotopy equiva-
lence, i.e., there is a map g : (Y,B) → (X,A) such that g ◦ f ≃ id(X,A) and
f ◦ g ≃ id(Y,B) . Then f induces an isomorphism
∼
=
Hn (f ) : Hn (X,A) −
→ Hn (Y,B)
in homology for all n.
In other words, homology is invariant under homotopy equivalences,
not just homeomorphisms.
73
Recall that, for n ≥ 1, we write Dn for the n-dimensional unit disk

X
Dn = {x = (x1 , . . . ,xn ) ∈ Rn : |x|2 = x2i ≤ 1}.
i
Recall that D is homotopy equivalent to a point. For, the constant map Dn →

n
{0} is a strong deformation retraction with homotopy

h : Dn × [0,1] → Dn , (x,t) 7→ (1 − t)x
between the identity map of Dn and the constant map.
As a consequence of the homotopy axiom and our computation of Hn (pt)
we get:
Corollary for contractible spaces

Recall that a space which is homotopy equivalent to a one-point space is
called contractible. For every contractible space X, we have
(
Z q=0
Hq (X) =
0 q ̸= 0.
Before we look at an application of the axioms, a remark on homology theories

with a brief outlook to the future (of your studies in algebraic topology):
A remark on homology theories

In fact, if we require h0 (pt) = Z the above properties or axioms characterize
singular homology uniquely. In other words, if h satisfies the Eilenberg-
Steenrod axioms, then h must be singular homology.
We will see later that we can define variations of singular homology with
coefficients different from Z. If R is a commutative ring with unit and M an
R-module, then there are singular homology groups Hn (X; M ) which fit into
a homology theory which satisfies the dimension axiom with h0 (pt) = M .
We can even go a step further (in a different class) where we drop the
dimension assumption allow hq (pt) ̸= 0 for infinitely many n. This leads to
generalized homology theories, for example K-theory or cobordism,
which are extremely useful for the solution of many fundamental problems,
not just in topology. For example, complex K-theory can be used to
show the theorem me mentioned in the notes of the first lecture: only in
dimensions 1, 2, 4, and 8 there is a nice multiplication on Rn .
• Homology of the sphere
As a fundamental example we are going to compute the homology of the k-
dimensional sphere S n . Actually, we already know one case. For, S 0 is just the
disjoint union of two points. Hence Hq (S 0 ) ∼
= Z ⊕ Z for q = 0 is 0 for all other n.
Theorem: Homology of the sphere

For n ≥ 1, we have
(
Z if q = 0 or q = n
Hq (S n ) =
0 otherwise
and
(
Z q=n
Hq (Dn ,S n−1 ) =
0 otherwise.
Proof of the Theorem: For n ≥ 1, the n-sphere S n is path-connected.

By our previous result, that implies H0 (S n ) ∼
= Z.
The proof will proceed by induction using the long exact sequence in homol-
ogy for pairs of spaces. This explains why we compute Hq (S n ) and Hq (Dn ,S n−1 )
at the same time.
For n = 1 and q = 0, the pair (D1 ,S 0 ), with i : S 0 ,→ D1 , is equipped with
the exact sequence
H0 (i)
H0 (S 0 ) / H0 (D1 ) / H0 (D1 ,S 0 ) / 0
Z⊕Z / Z / ? / 0.
The map H0 (i) is induced by the map S0 (S 0 ) → S0 (D1 ) which sends a 0-

simplex ∆0 → S 0 to the composite ∆0 → S 0 ,→ D1 .
The image of S 0 in D1 consists of the two endpoints of D1 and both points
are homologous as 0-simplices of D1 . Hence they both represent the class of the
generator of H0 (D1 ). Hence the map H0 (i) sends each generator of H0 (S 0 )
to the generator of H0 (D1 ). Writing (1,0) and 0,1) for the generators of Z ⊕ Z,
any (a,b) ∈ Z ⊕ Z is of the form a · (1,0) + b · (0,1). Hence (a,b) is sent to a + b
under Hn (i).
75
This implies that H0 (i) is surjective. Since the above sequence is exact, this
implies
H0 (D1 ,S 0 ) = 0.
For n ≥ 2, the exact sequence becomes

H0 (i)
H0 (S n−1 ) / H0 (Dn ) / H0 (Dn ,S n−1 ) / 0
Z / Z / ? / 0.
Since both S n−1 and Dn are path-connected, their 0th homology is isomorphic
to Z and the generator of H0 (S n−1 ), the class of any constant map κ0x : ∆0 → S n−1 ,
is sent to the generator of H0 (Dn ), the class of κ0x : ∆0 → Dn corresponding to
the image point x ∈ S n−1 ⊂ Dn . Hence H0 (i) is surjective and we have again
H0 (Dn , S n−1 ) = 0.
This finishes the argument for H0 .
For q = 1, we start with the exact sequence
H0 (i)
(5) H1 (D1 ) / H1 (D1 , S 0 ) / H0 (S 0 ) / H0 (D1 )
0 / ? / Z⊕Z / Z.
Since the sequence is exact, this shows that H1 (D1 ,S 0 ) is isomorphic to the kernel
of
Hn (i) : Z ⊕ Z → Z, (a,b) 7→ a + b.
Thus
(6) H1 (D1 , S 0 ) ∼
= Z.
For n ≥ 2, we get the sequence

∼
H1 (Dn ) / H1 (Dn ,S n−1 ) / H0 (S n−1 )
= / H0 (Dn )
∼
0 / ? / Z
= / Z.
Since the right most map is an isomorphism, we get
(7) H1 (Dn ,S n−1 ) = 0 for all n ≥ 2.
In order to study further groups, we consider the subspaces
n
D+ := {(x0 , . . . ,xn ) ∈ S n : x0 ≥ 0} and D−
n
:= {(x0 , . . . ,xn ) ∈ S n : x0 ≤ 0}
which correspond to the upper and lower hemisphere (including the equator),
respectively, of S n .
For n ≥ 1, we have the exact sequence
∼ ∼
n
H1 (D− ) / H1 (S n )
= / H1 (S n ,D−
n
)
∂=0 / n
H0 (D− )
= / H0 (S n )
∼
0 Z
= / Z.
n n
Since D− is contractible, we know H1 (D− ) = 0. Hence the map H1 (S n ) →
n n
H1 (S ,D− ) is injective. Since the map
Z∼ = H0 (Dn ) → H0 (S n ) ∼
=Z
−
is an isomorphism, the connecting hmomorphism ∂ is 0. Since the sequence

is exact, this implies that the map H1 (S n ) → H1 (S n ,D−
n
) is also surjective.
Thus, in total, we have an isomorphism
∼
=
H1 (S n ) −
→ H1 (S n ,D−
n
).
To finish the analysis for q = 1, we consider the open subspace

U−n := {(x0 , . . . ,xn ) ∈ S n : x0 < −1/2} ⊂ D−
n
.
n
Its closure is still contained in the open interior of D− , i.e.,
n ◦
Ū−n ⊂ (D− )
77
Hence we can apply the excision axiom to the inclusion of pairs

n
k : (S n \ U−n , D− \ U−n ) ,→ (S n , D−
n
)
and obtain an isomorphism
∼
=
Hq (k) : Hq (S n \ U−n , D−
n
\ U−n ) −
→ Hn (S n , D−
n
).
But we also know

≈
(S n \ U−n , D−
n
\ U−n ) ≃ (D+
n
, S n−1 ) −
→ (Dn , S n−1 )
where the last homeomorphism is given by vertical projection, and the
homotopy equivalence is the natural retraction.
In particular, we get an isomorphism
(8) Hq (S n , Dn ) ∼
= Hq (Dn , S n−1 ).
−
For H1 , this implies

(
Z if n = 1 by (6).
H1 (S n ) ∼ n ∼
= H1 (S n , D− ) = H1 (Dn , S n−1 ) =
0 else by (7).
This finishes the case q = 1. In particular, we now know H1 (S 1 ) ∼
= Z.
Finally, for q ≥ 2, we proceed by induction. The pair (S n ,D−
n
) yields the
exact sequence
∼
(9) n
Hq (D− ) / Hq (S n )
= / Hq (S n , D−
n
) / Hq−1 (Dn )
0 0.
Whereas the pair (Dn ,S n−1 ) yields the exact sequence

∼
Hq (Dn ) / Hq (Dn , S n−1 )
= / Hq−1 (S n−1 ) / Hq−1 (Dn )
0 0.
Together with isomorphism (8), we conclude

Hq (S n ) ∼
= Hq (S n , Dn ) ∼
= Hq (Dn , S n−1 ) ∼
− = Hq−1 (S n−1 ).
Hence knowing H1 (S 1 ) = Z implies H2 (S 2 ) = Z and H2 (D2 ,S 1 ) = Z. Con-

tinuing by induction on q yields the assertion of the theorem. □
LECTURE 7
Generators for Hn (S n ) and first applications
Generators for Hn (S n )
Last time we calculated the homology groups of S n and the pair (Dn ,S n−1 ).
To make this calculation a bit more concrete, let us try to figure out the generators
of the infinite cyclic groups Hn (Dn , S n−1 ) and Hn (S n ):
• On the standard n-simplex, there is a special n-chain Sn (∆n ), called the fun-
damental n-simplex, given by the identity map ιn : ∆n → ∆n . We observed
in a previous lecture that ιn is not a cycle, since its boundary ∂(ιn ) ∈ Sn−1 (∆n )
is the alternating sum of the faces of the n-simplex each of which is a generator
in Sn−1 (∆n ).
X
∂(ιn ) = (−1)i ϕni (∆n−1 ) ̸= 0.
i
However, each of these singular simplices lies in ∂∆n , and hence

∂(ιn ) ∈ Sn−1 (∂∆n ).
Thus the image of ιn in Sn (∆n ,∂∆n ) is a relative cycle. Let us denote its image
also by ιn and its class in H n (∆n , ∂∆n ) by [ιn ].
If Hn (∆n , ∂∆n ) is nontrivial, then [ιn ] must be a nontrivial generator. For,
if σ : ∆n → ∆n is any n-simplex of ∆n which defines a nontrivial class [σ] in
Hn (∆n , ∂∆n ), then
[σ] = Hn (σ)([ιn ]).
This is because ιn is the identity map and Hn (σ)([ιn ]) is defined by composing σ
and ιn . Hence if [ιn ] was trivial, then any class in Hn (∆n , ∂∆n ) would be trivial.
• Now we use this observation to find a generator of Hn (Dn ,S n−1 ). The
standard n-simplex ∆n and the unit n-disk Dn are homeomorphic. In order to
find a homeomorphism we just need to smoothen out the corners of ∆n . (Note
that we cannot ask for a diffeomorphism, since ∆n is not a smooth manifold.)
79
80 CHAPTER 7. GENERATORS FOR HN (S N ) AND FIRST APPLICATIONS
In fact, we can choose a homeomorphism of pairs
≈
φn : (∆n , ∂∆n ) −
→ (Dn , S n−1 )
which maps ∂∆n homeomorphically to S n−1 . We will construct a concrete home-

omorphism below. For the moment, let us accept that we have such a homeo-
morphism φn for every n.
Then φn induces an isomorphism

∼
=
Hn (φn ) : Hn (∆n , ∂∆n ) −
→ Hn (Dn , S n−1 ) with [ιn ] 7→ [φn ].
Since we now know Hn (Dn , S n−1 ) ∼

= Z, we also have Hn (∆n , ∂∆n ) ∼
= Z and [ιn ]
n n−1 ∼
as a generator. Thus [φn ] is a generator of Hn (D , S ) = Z.
• Recall that we showed last time that the connecting homomorphism

∼
=
∂ : Hn (Dn , S n−1 ) −
→ Hn−1 (S n−1 )
an isomorphism. The image of [φn ] under ∂ is a generator. In other words, [∂(φn )]

is a generator of Hn−1 (S n−1 ) for all n ≥ 2.
Constructing φn
1
For each ∆n the point c = (t0 , . . . ,tn ) with ti = n+1 for all i is the barycen-
n
ter of ∆ .
For every point x ∈ ∆n which is not c, there is a unique ray from c to x.
We denote the unique point where this ray hits ∂∆n by f (x). In particular,
if x ∈ ∂∆n , then f (x) = x.
81
Now we define the map

(
x−c
if x ̸= 0
φn : ∆n → Dn , x 7→ |f (x)−c|
0 if x = c.
It is clear that φn is continuous except possibly at x = c. But since there is a
strictly positive lower bound for |f (x) − c| > 0, we know |φn (x)| ≤ M |x − c|
for some real number M . This implies that φn is also continuous at x = c.
Moreover, φn is a bijection, since it is one restricted to each ray. Since ∆n
is compact and φn is a continuous bijection, it is a homeomorphism.
Finally, we write down a generator for the unit circle.
A concrete generator of H1 (S 1 )
We just learned that the class [∂(φ2 )] is a generator of H1 (S 1 ). We can
describe this class as follows:
By definition, ∂(φ2 ) is the 1-cycle
∂(φ2 ) = d0 φ2 − d1 φ2 + d2 φ2
= φ2 ◦ ϕ20 − φ2 ◦ ϕ21 + φ2 ◦ ϕ22 .
Recall that φ2 maps ∂∆2 homeomorphically to S 1 . With this in mind, the
summands look like
1 2
φ2 ◦ ϕ20 (1 − t, t) = eiπ(− 6 +t 3 )
7 2
φ2 ◦ ϕ21 (1 − t, t) = eiπ( 6 −t 3 )
7 2
φ2 ◦ ϕ22 (1 − t, t) = eiπ( 6 +t 3 ) .
We proved in Lecture 03 that the 1-simplex

∆1 → ∆1 , t 7→ φ2 ◦ ϕ21 (1 − t,t)
is homologous to the 1-simplex
∆1 → ∆1 , t 7→ φ2 ◦ ϕ21 (t,1 − t)
which reverses the direction of the walk from one vertex to the other.
In an exercise, we will show that after splitting a path into different steps,
the 1-chain associated to the initial path is homologous to the sum of the
1-chains associated to the parts. This result implies that the 1-cycles ∂φ2
is homologous to the 1-cycle corresponding to the familiar path
γ : ∆1 → S 1 , (1 − t, t) 7→ e2πit
which walks once around the circle.
In summary, we showed that [γ] = [∂φ2 ] is the desired generator of
H1 (S 1 ).
First applications
The calculation of the homology of spheres has many interesting consequences.

We will discuss some of them today and will see many more soon.
We start with a result we advertised in the first lecture:
Theorem: Invariance of dimension

For n ̸= m, the space Rn is not homeomorphic to Rm .
Proof: Assume there was a homeomorphism f : Rn → Rm . Then the

restricted map
f : Rn \ {0} → Rm \ {f (0)}
is also a homeomorphism, since these are open subsets and f|Rn \{0} and (f −1 )|Rm \{f (0)}
are still continuous mutual inverses.
We showed as an exercise that, for any k ≥ 1, S k−1 is a strong deformation
retract of Rk \{0}. In particular, we showed S k−1 ≃ Rn \{0}. Since the translation
Rn → Rn , y 7→ y + x is a homeomorphism for any x ∈ Rn , this implies that
S k−1 ≃ Rk \ {x} for every x ∈ Rk .

83
Hence, if the homeomorphism f existed, we would get an induced isomor-

phism
∼
Hq (S n−1 ) ∼ → Hq (Rm \ {f (0)}) ∼
=
= Hq (Rn \ {0}) − = Hq (S m−1 ).
But by our calculation of the Hq (S n−1 ), such an isomorphism can only exist
if n − 1 = q = m − 1. This contradicts the assumption n ̸= m. QED
We can also give a short proof of Brouwer’s famous Fixed-Point Theorem:
Brouwer Fixed-Point Theorem

Let f : Dn → Dn be a continuous map of the closed unit disk into itself.
Then f must have a fixed point, i.e., there is an x ∈ Dn with f (x) = x.
Before we prove the theorem, let us have a look at dimension one, where the
result is very familiar:
Brouwer FPT is familiar in dimension one

Note that you have seen this theorem for n = 1 in Calculus 1. Let f : [0, 1] →
[0,1] be a continuous map. Then it must have a fixed point. For, if not, then
g(x) = f (x) − x is a continuous map defined on [0, 1]. We have g(0) ≥ 0
and g(1) ≤ 0, since f (0) ≥ 0 and f (1) ≤ 1.
If g(0) = 0 or g(1) = 0, we are done. But if g(0) > 0 and g(1) < 0, then the
Intermediate Value Theorem implies that there is an x0 ∈ (0, 1) with
g(x0 ) = 0, i.e., f (x0 ) = x0 .
Proof of Brouwer’s FPT: Since we know the theorem for n = 1, we assume

n ≥ 2. Suppose that there exists an f without fixed points, i.e., f (x) ̸= x for
all x ∈ Dn . Then, for every x ∈ Dn , the two distinct points x and f (x)
determine a line. Let g(x) be the point where the line segment starting at f (x)
and passing through x hits the boundary ∂Dn . This defines a continuous map
g : Dn → ∂Dn .
Let i : S n−1 = ∂Dn ,→ Dn denote the inclusion map. Note that if x ∈ ∂Dn ,
then g(x) = x. In other words,
g ◦ i = idS n−1 .
Applying the homology functor yields a commutative diagram
idH n−1 )
n−1 (S
Z∼
= Hn−1 (S n−1 ) / Hn−1 (S n−1 ) ∼
=Z
5
Hn (i) ) Hn (g)
Hn−1 (Dn ) = 0.
But the identity homomorphism on Z cannot factor through 0. This

contradicts the assumption that f has no fixed point. □
85
Typical application of homology theory
The previous arguments are in fact a typical examples of proves in Algebraic

Topology:
≈
• The topological assumption that a homeomorphism Rn − → Rm exists, is
translated by applying homology to a statement about an isomorphism
of groups. For groups, the existence of such an isomorphism is easily checked
to be false.
• The geometric assumption that there is no fixed point be expressed in
terms of maps and their compositions. Applying the homology functor
translates this statement into an analogous statement about groups and
homomorphisms and their compositions. Since the resulting statement
about groups is obviously false, the original statement about spaces must be
false as well.
• The degree of a map S n → S n

The calculation of the homology of the sphere leads to another important
algebraic invariant.
Definition: The degree

For n ≥ 1, let f : S n → S n be a continuous map. Then the induced homo-
morphism
n
Hn (S )
Z∼
= Hn (S n ) −−−−→ Hn (S n ) ∼
=Z
is given by multiplication with an integer, the image of 1.
We denote this integer by deg(f ) and call it the degree of f .
Let us calculate a first example:

Theorem: The degree of a reflection

Let r : S n → S n be the reflection map defined by reversal of the first
coordinate
r : (x0 ,x1 , . . . , xn ) 7→ (−x0 , x1 , . . . , xn ).
Then deg(r) = −1.
Before we start the proof, let us have a look at what happens for the reflection
map
r : D1 = [−1, 1] → [−1, 1] = D1 , t 7→ −t
and its restriction to S 0 . Recall that S 0 consists of just two points, x = 1 and
y = −1 (on the real line R). The effect of r on S 0 is to interchange x and y.
The inclusion maps ix and iy induce an isomorphism
∼
=
→ H0 (S 0 ).
H0 ({x}) ⊕ H0 ({y}) −
Thus H0 (r) can be viewed as
H0 (r) : H0 (S 0 ) → H0 (S 0 ), (a, b) 7→ (b, a).
During the calculation of Hn (S n ), we remarked that the map

ϵ := H0 (i) : H0 (S 0 ) → H0 (D1 )
induced by the inclusion i : S 0 ,→ D1 can be identified with the homomorphism
ϵ : Z ⊕ Z → Z, (a, b) 7→ a + b.
Let Ker (ϵ) = {(a, −a) ∈ H0 (S 0 ) : a ∈ Z} be the kernel of ϵ. Then we get

that the effect of H0 (r) on Ker (ϵ) is given by multiplication by −1:
H0 (r) : Ker (ϵ) → Ker (ϵ), (a, −a) 7→ (−a, a) = −(a, −a).
87
Now we can address the actual proof.

Proof of the Theorem: For n ≥ 1, let
n
D+ := {(x0 , . . . ,xn ) ∈ S n : xn ≥ 0} and D−
n
:= {(x0 , . . . ,xn ) ∈ S n : xn ≤ 0}
be the upper and lower hemispheres on S n , respectively. We also denote by
n n n n
r the reflection map on D+ and D− . (Note that we defined D+ and D− using a
n n n n
different coordinate than for defining r so that r(D+ ) ⊂ D+ and r(D− ) ⊂ D− .)
Then we have a commutative diagram
∼ ∼ ∼
H1 (S 1 )
= / 1 o
H1 (S 1 , D+ )
= 1
H1 (D− , S 0)
= / Ker (ϵ)
H1 (r) H1 (r) H1 (r) H0 (r)
∼ ∼ ∼
H1 (S 1 )
= / 1 o
H1 (S 1 , D+ )
= 1
H1 (D− , S 0)
= / Ker (ϵ).
The right hand square commutes, since the isomorphism

1 ∼
=
H1 (D− , S 0) −
→ Ker (ϵ)
1
is part of the exact sequence induced by the pair (D− , S 0 ):
1
H1 (D− , S 0) / H0 (S 0 )
ϵ / 1
H0 (D− ) / 1
H0 (D− ,S 0 )
Z / Z⊕Z
ϵ / Z / 0.
We know that the left hand and central squares commute, since the inclusion
and the reflection commute. We know that the horizontal maps are isomorphisms
from the calculation of these groups.
Thus, knowing H0 (r) = −1 on Ker (ϵ), we see that H1 (r) is also multipli-
cation by −1 on H1 (S 1 ).
Now we can proceed by induction: For n ≥ 2, we have again a commutative
diagram from the calculation of Hn (S n ):
∼ ∼ ∼
Hn (S n )
= / n o
Hn (S n , D+ )
= n
Hn (D− , S n−1 )
= / Hn−1 (S n−1 )
Hn (r) Hn (r) Hn (r) Hn−1 (r)
∼ ∼ ∼
Hn (S n )
= / n o
Hn (S n , D+ )
= n
Hn (D− , S n−1 )
= / Hn−1 (S n−1 ).
The right most square commutes by an exercise from last week. The left hand
and central squares commute, since the inclusion and the reflection commute.
Assuming the assertion for n − 1, i.e., Hn−1 (r) is multiplication by −1, we see
that Hn (r) is also multiplication by −1. □
LECTURE 8
Calculating degrees
In last week’s exercises we showed many useful properties of the degree and
calculated the degree of some interesting maps. Today, we are going to continue
our study of the degree.
But before we move on, another reason why the degree is so important:
Brouwer degree
Let p be an arbitrary point in S n . We consider p as the base point of
S n . Let C(S n ,S n )∗ be the set of pointed continuous maps, i.e., maps
f : S n → S n with f (p) = p. Pointed homotopy defines an equivalence
relation on this set. Hence we can define the quotient set
[S n , S n ]∗ := C(S n , S n )∗ / ≃
where we identify f and g if they are homotopic to each other f ≃ g.
Now the degree defines a function from C(S n ,S n )∗ to the integers Z. Since
the degree is invariant under homotopy, i.e., f0 ≃ f1 implies deg(f0 ) =
deg(f1 ), it induces a function
deg : [S n , S n ]∗ → Z, f 7→ deg(f ).
This function is actually an isomorphism of abelian groups. In fancier
language, we write πn (S n ) = [S n , S n ]∗ , call it the nth homotopy group
of S n and say that the degree completely determines πn (S n ).
Now let us see what kind of maps between spheres there are. Actually, such
maps arise quite naturally. For, every invertible real n × n-matrix A defines a
≈
homeomorphism between Rn − → Rn . It extends to a homeomorphism on the
one-point compactification S n of Rn and therefore defines a map
A : S n → S n.
A more direct way to produce a map is to assume we have an orthogonal

matrix:
89
90 CHAPTER 8. CALCULATING DEGREES
Proposition: Orthogonal matrices

Let O(n) denote the group of orthogonal (real) n × n-matrices, i.e.,
O(n) = {A ∈ M (n × n, R) : AT A = I}
where I is the identity matrix. The restriction to S n−1 of any A in O(n)
defines a map
A : S n → S n , x 7→ Ax.
The degree of this map is det A, i.e., deg(A) = det(A) = ±1.
Proof: This follows from the fact that every othogonal matrix is the product
of reflections (at appropriate hyperplanes in Rn ). A reflection has determinant
−1, but it also has degree −1 as we have shown before. Since both deg and det
are multiplicative, the result follows. QED
Now let A ∈ GLn (R) be an invertible n × n-matrix. It defines a map
f : Rn \ {0} → Rn \ {0}, f (x) := Ax.
It induces a map
Hn−1 (f ) : Hn−1 (Rn \ {0}) → Hn−1 (Rn \ {0}).
Since S n−1 ,→ Rn \ {0} is a deformation retract, we know

Z∼
= Hn−1 (S n−1 ) ∼
= Hn−1 (Rn \ {0}).
Hence the effect of Hn−1 (A) is given by multiplication by an integer.
Proposition: It’s the sign

Hn−1 (A) = sign(det(A)) where sign(det(A)) denotes the sign, i.e., 1 or −1,
of det(A).
Proof: Recall from linear algebra that any invertible matrix A has a po-
lar decomposition A = BC with B a symmetric matrix with only positive
eigenvalues and C ∈ O(n). Since we already know that the assertion is true
if A ∈ O(n), it suffices to show that B is homotopic to the identity as maps
Rn → Rn .
Since all eigenvalues of B are positive, we know det(B) > 0. Hence B and I
lie both in the component GLn (R)+ of the matrices with det > 0. The continuous
91
map
Γ : [0,1] → GLn (R)+ , t 7→ tI + (1 − t)B
defines a homotopy between I and B.
To check that Γ(t) is in GLn (R)+ for all t, we observe that the eigenvalues of
Γ(t) are all strictly positive. For, let λ be an eigenvalue of B. Then t + (1 − t)λ
is an eigenvalue of Γ(t), since all nonzero vectors are eigenvectors of tI. This
implies det(Γ(t)) > 0. QED
For n > 1, we know Hn (Rn , Rn \ {0}) ∼
= Z, since (Dn , S n−1 ) ,→ (Rn , Rn \ {0})
is a deformation retract. For A as above, we obtain a commutative diagram from
the long exact sequences of pairs
Hn (A)
Hn (Rn , Rn \ {0}) / Hn (Rn , Rn \ {0})

Hn−1 (Rn \ {0}) / Hn−1 (Rn \ {0}).
Hn−1 (A)
The vertical connecting homomorphisms are isomorphisms, since they are iso-
morphisms for the pair (Dn , S n−1 ). Since the diagram commutes, we deduce the
following consequence from the previous result:
Corollary
The effect of the map
Hn (A) : Hn (Rn , Rn \ {0}) → Hn (Rn , Rn \ {0})
is given by multiplication with sign(det(A)).
Now we would like to apply this to a situation familiar from Calculus. First
a brief observation:
Lemma
Let U ⊂ Rn be an open subset and x ∈ U . Then
Hn (U, U \ {x}) ∼
= Z.
Proof: Let Z be the complement of U in Rn . Since U is open, Z is closed.

Hence Z̄ = Z ⊂ Rn \ {x} = (Rn \ {x})◦ . Hence we can apply excision to the
inclusion of pairs (U, U \ {x}) ,→ (Rn , Rn \ {x}) and get
∼
→ Hn (Rn , Rn \ {x}) ∼
=
Hn (U, U \ {x}) − = Z.
Proposition: Degree of smooth maps

Let U ⊂ Rn open with 0 ∈ U . Let
f : U → Rn
be a smooth map (or say twice differentiable with continuous second deriva-
tives) with f −1 (0) = 0 and Df (0) ∈ GLn (R).
For such a map f , the effect of the homomorphism
Hn (f ) : Hn (U, U \ {0}) → Hn (Rn , Rn \ {0})
is given by multiplication with sign(det(Df (0))).
Proof: • By the Taylor expansion of a differentiable map, we can write f

as
f (x) = Ax + g(x) with A = Df (0) and g(x)/|x| → 0 for x → 0.
• In particular, we can assume |g(x)| < |x|/2 for x small enough. By excision,
we can shrink U to become small enough such that still 0 ∈ U and |g(x)| < |x|/2
for all x ∈ U .
• We can further assume that A = I is the identity. For if not, we can replace
f with A−1 f and use the functoriality of Hn .
• Now we have |f (x) − x| < x/2 for all x ∈ U . Hence the map
h : U × [0,1] → Rn , (x, t) 7→ tf (x) + (1 − t)x
satisfies h(x, t) ̸= 0 for all (x,t). This implies that Dh(0, t) is in GLn (R)+ for all
t. Thus h defines a homotopy between f and the identity map and the effect of
Hn (f ) is the same as the one of the identity map. □
Local degree
Often the effect of a map can be studied by focussing on the neighborhood
of certain interesting points. We would like to exploit this idea for studying the
degree.
93
For n ≥ 1, let f : S n → S n be a map with the property that there is a point

y ∈ S n such that f −1 (y) consists of finitely many points. (Note that almost all
maps have this property.)
We label these points by x1 , . . . , xm . Now we choose small disjoint open
neighborhoods U1 , . . . , Um of each xi such that each Ui is mapped into an open
neighborhood V of y in S n . (We could choose V first, and then intersect f −1 (V )
with small open disks around xi ...).
Since xi ∈ Ui and the different Uj s are disjoint, we have
f (Ui \ {xi }) ⊂ V \ {y} for each i.
For any given i, the obvious inclusions of pairs induce the following diagram:
Hn (f|Ui )
(10) Hn (Ui , Ui \ {xi }) / Hn (V, V \ {y})
deg(f |xi )
∼
=
ki ∼
=
t pi
Hn (f )
Hn (S n , S n \ {xi }) o Hn (S n , S n \ f −1 (y)) / Hn (S n , S n \ {y})
j O O
j ∼
=
∼
=
Hn (f )
Hn (S n ) / Hn (S n ).
By the excision axiom applied as in the proof of the lemma below and by
an exercise, we know that the diagonal maps on the left and the vertical maps
on the right are isomorphisms, as indicated in (10).
Definition: Local degree

The source and target of the dotted top horizontal arrow in (10) are iden-
tified with Z. Hence the effect of this homomorphism is given by multipli-
cation by an integer. We denote this integer by deg(f |xi ) and call it the
local degree of f at xi .
Let us calculate some examples:

• If f is a homeomorphism, then any y has a unique preimage x. In this
case, all maps in daigram (10) are isomorphisms and we have
deg(f ) = deg(f |x) = ±1.
• If f maps each Ui homeomorphically to V , then we have deg(f |xi ) = ±1
for each i.
The latter observation can be used to calculate the degree of f in many
interesting situations. For we have the following result which connects global and
local degrees:
Proposition: Global is the sum of local

With the above assumptions we have
m
X
deg(f ) = deg(f |xi ).
i=1
We are going to prove this result in the next lecture. In the diagram above
we claimed that some maps are isomorphisms. Here is an explanation why:
Lemma
(a) Let U ⊂ S n be an open subset and x ∈ U . Then there is an isomorphism
∼
→ Hn (S n , S n \ {x}) ∼
=
Hn (U, U \ {x}) − = Z.
(b) Let x1 , . . . , xm be m distinct points in S n and U1 , . . . , Um disjoint open
neighborhoods with xi ∈ Ui . Then there is an isomorphism
∼
→ Hn (S n , S n \ {x1 , . . . , xm }) ∼
=
⊕m Hn (Ui , Ui \ {xi }) −
i=1 = ⊕m Z.
i=1
Proof: (a) Let Z be the complement of U in Rn . Since U is open, Z is

closed. Hence Z̄ = Z ⊂ S n \ {x} = (S n \ {x})◦ = S n \ {x}. Hence we can apply
excision to the inclusion of pairs (U, U \ {x}) ,→ (S n , S n \ {x}) and get the above
isomorphism.
(b) Let U := ∪i Ui . Then Z := S n \ U is closed. As above, we can apply
excision to the inclusion of pairs (U, U \ {x1 , . . . , xm }) ,→ (S n , S n \ {x1 , . . . , xm })
and get an isomorphism
∼
=
→ Hn (S n , S n \ {x1 , . . . , xm }).
Hn (U, U \ {x1 , . . . , xm }) −
Since U is actually a disjoint union and each xi ∈ Ui , we know the inclusions
induce an isomorphism
∼
=
⊕m
i=1 Hn (Ui , Ui \ {xi }) −
→ Hn (U, U \ {x1 , . . . , xm }).
Together with (a) this proves the assertion. □
LECTURE 9
Local vs global degrees
Last time we defined the local degree of a map. The situation was as follows:
For n ≥ 1, let f : S n → S n be a map with the property that there is a point
y ∈ S n such that f −1 (y) = {x1 , . . . ,xm } consists of finitely many points.
We choose small disjoint open neighborhoods U1 , . . . ,Um of each xi such that
each Ui is mapped into an open neighborhood V of y in S n . (We could choose V
first, and then intersect f −1 (V ) with small open disks around xi ...).
Since xi ∈ Ui and the different Uj s are disjoint, we have
f (Ui \ {xi }) ⊂ V \ {y} for each i.
For any given i, the obvious inclusions of pairs induce the following diagram:
Hn (f|Ui )
(11) Hn (Ui , Ui \ {xi }) / Hn (V, V \ {y})
deg(f |xi )
∼
=
ki ∼
=
t pi
Hn (f )
Hn (S n , S n \ {xi }) o Hn (S n , S n \ f −1 (y)) / Hn (S n , S n \ {y})
j O O
j ∼
=
∼
=
Hn (f )
Hn (S n ) / Hn (S n ).
By the excision axiom applied as in the proof of the lemma below and by
an exercise, we know that the diagonal maps on the left and the vertical maps on
the right are isomorphisms, as indicated in (11). Then we made the following
definition:
95
96 CHAPTER 9. LOCAL VS GLOBAL DEGREES
Definition: Local degree

The source and target of the dotted top horizontal arrow in (10) are iden-
tified with Z. Hence the effect of this homomorphism is given by multipli-
cation by an integer. We denote this integer by deg(f |xi ) and call it the
local degree of f at xi .
We have the following result which connects global and local degrees:
Proposition: Global is the sum of local

With the above assumptions we have
m
X
deg(f ) = deg(f |xi ).
i=1
Let us finally prove this result.

Proof: • As explained in the lemma below, excision implies that
∼
=
→ Hn (S n , S n \ f −1 (y)) = ⊕i Z
⊕i ki : ⊕i Z = Hn (Ui , Ui \ {xi }) −
is an isomorphism. HenceforthP we are going to identify the groups in diagram
(11) with Z or the direct sum i Z, respectively.
• We know that pi ◦ ki is the diagonal isomorphism. Hence ki corresponds to
the inclusion of the ith summand, and pi corresponds to the projection to the
ith summand.
• Since the lower triangle commutes, the composite pi ◦ j satisfies
pi ◦ j(1) = 1.
Since we also know
X
pi ◦ ki (1) = 1 and ( ki )(1) = (1, . . . , 1)
i
we must have
j(1) = (1, . . . ,1)
as well.
• Since the upper square in diagram (11) commutes, we know
Hn (f )(ki (1)) = deg(f |xi ).
97
• Together with the above, this shows

X
Hn (f )(j(1)) = deg(f |xi ).
i
• Since the lower square in diagram (11) commutes and since the lower
horizontal map is given by the degree of f , the asserted formula follows. QED
We have used the following lemma in diagram (11) the above proof:
Lemma
(a) Let U ⊂ S n be an open subset and x ∈ U . Then there is an isomorphism
∼
→ Hn (S n , S n \ {x}) ∼
=
Hn (U, U \ {x}) − = Z.
(b) Let x1 , . . . , xm be m distinct points in S n and U1 , . . . , Um disjoint open
neighborhoods with xi ∈ Ui . Then there is an isomorphism
∼
→ Hn (S n , S n \ {x1 , . . . , xm }) ∼
=
⊕m Hn (Ui , Ui \ {xi }) −
i=1 = ⊕m Z.
i=1
We proved this lemma in a previous lecture. Now let us apply what we learned
in an example:
Example: Degree on the unit circle

Let S 1 ⊂ C be the unit circle, k ∈ Z, and let
fk : S 1 → S 1 , z 7→ z k .
We claim deg(fk ) = k.
98 CHAPTER 9. LOCAL VS GLOBAL DEGREES
• We know this is true for for k = 0 when f0 is the constant map and for
k = 1 when f1 is the identity.
• We know it also for k = −1, since z 7→ z −1 is a reflection at the real
axis.
• It suffices to check the remaining cases for k > 0, since the cases for k < 0
follow from composition with z 7→ z −1 and the multiplicativity of the
degree.
• So let k > 0. For any y ∈ S n , fk−1 (y) consists of k distinct points
x1 , . . . ,xk . Each point xi has an open neighborhood Ui which is mapped
homeomorphically by f to an open neighboorhood V of y. This local
homeomorphism is given by stretching (by the factor k) and a rotation
in positive direction.
• Stretching by a factor is homotopic to the identity near xi . Hence the
local degree of the stretching is +1.
A rotation is a homeomorphism and its global and local degree at any point
agree.
Since the rotation is in the positive direction, it is homotopic to the identity
and has therefore degree +1.
• Hence deg(f |xi ) = 1.
• Thus we can conclude by the proposition that
k
X
deg(f ) = deg(f |xi ) = k.
i=1
LECTURE 10
Homotopies of chain complexes
We still need to prove the Homotopy Axiom and Excision Axiom for singular
homology. The proof will follow from constructing a homotopy between chain
complexes, a concept we are now going to explore.
Recall that a chain complex K∗ = (K∗ ,∂ K ) constists of a sequence of abelian
groups
K K K
∂n+2 ∂n+1 n ∂K ∂n−1
· · · −−−→ Kn+1 −−−→ Kn −→ Kn−1 −−−→ · · ·
together with homomorphisms ∂nK : Kn → Kn−1 with the property that ∂n−1 ◦∂n =
0. Our main example is the singular chain complex.
Just to make sure that we understand the definition, let us look at an example
of a sequence of groups that is not a chain complex. Consider the sequence of
maps
2 2 2 2 2
··· →
− Z→
− Z→
− Z→
− Z→
− ···
where each map consists of multiplication by 2. This is not a chain complex,

since 2 · 2 = 4, i.e., ∂n−1 ◦ ∂n = 4 ̸= 0.
Recall the definition of a map of chain complexes from Lecture 5:
Maps of chain complexes

Let K∗ = (K∗ , ∂ K ) and L∗ = (L∗ , ∂ L ) be two chain complexes. A mor-
phism of chain complexes f∗ : K∗ → L∗ , also called chain map, is a
sequence of homomorphisms {fn }n∈Z
(12) fn : Kn → Ln such that fn−1 ◦ ∂nK = ∂nL ◦ fn for all n ∈ Z.
A homomorphism of chain complexes induces a homomorphism on homol-
ogy
Hn (f ) : Hn (K∗ ) → Hn (L∗ ), [x] 7→ [fn (x)].
99
100 CHAPTER 10. HOMOTOPIES OF CHAIN COMPLEXES
We need to check that this is well-defined. Since we hopped over this point
in Lecture 5, let us do it now.
There are two things we should check. We need to know that f sends cycles
to cycles and boundaries to boundaries.
First, let x ∈ Kn be a cycle in K, i.e., ∂nK (x) = 0. Then (12) implies
∂nL (fn (x)) = fn−1 (∂nK ((x)) = fn−1 (0) = 0.
Thus fn (x) is a cycle in L and we get fn (Zn (K∗ )) ⊂ Zn (L∗ ).
K
Second, let a x ∈ Kn be a boundary, say ∂n+1 (y) = x. Then (12) implies
K L
fn (x) = fn (∂n+1 (y)) = ∂n+1 (fn+1 (y)).
Thus fn (x) is a boundary in L and we get fn (Bn (K∗ )) ⊂ Bn (L∗ ). This shows
that f induces a well-defined homomrphism between the homologies of K∗ and
L∗ .
We would like to transfer the notion of homotopies between maps of spaces
to the homotopies between maps of chain complexes. This follows the general
slogan: Homotopy is a smart thing to do.
Why? The notion of an isomorphism in a category, e.g. the category of
topological spaces or the category of chain complexes, is often too rigid. There
are too few isomorphism such that classifying objects up to isomorphism is too
difficult. Therefore, one would like to relax the conditions. For many situations,
homotopy turns out to provide the right amount of flexibility and rigidity
at the same time. Moreover, many invariants, in fact all invariants in Algebraic
Topology, do not change if we alter a map by a homotopy.
In other words, our invariants only see the homotopy type.
Actually, this is exactly what we are going to show for singular homology
today. It is also true in Homological Algebra. The homology of a chain complex
only depends on the homotopy type of the complex.
So let us define homotopies between chain maps:
Definition: Homotopies of chain maps

Let f, g : K∗ → L∗ be two morphisms of chain complexes. A chain homo-
topy between f and g is a sequence of homomorphisms
hn : Kn → Ln+1
101
such that
L
(13) fn − gn = ∂n+1 ◦ hn + hn−1 ◦ ∂nK for all n ∈ Z.
Kn+1 / L
; n+1
K
hn L
∂n+1 ∂n+1
fn −gn
Kn / Ln
;
K
∂n L
∂n
hn−1
Kn−1 / Ln−1
If such a homotopy exists, we are going to say that f and g are homotopic
and write f ≃ g.
We say that f is null-homotopic if f ≃ 0.
As for topological spaces, this yields an equivalence relation:
Lemma: Homotopy is an equivalence relation

(1) Chain homotopy is an equivalence relation on the set of all morphisms
of chain complexes.
(2) If f ≃ f ′ : K∗ → L∗ and g ≃ g ′ : L∗ → M∗ , then g ◦ f ≃ g ′ ◦ f ′ .
Proof: (1) We need to show that homotopy is reflexive, symmetric and

transitive:
• We obtain f ≃ f with h = 0 being the zero map.
• If h is a homotopy which gives f ≃ g, then −h is a homotopy which
shows g ≃ f .
• If h is a homotopy which gives f ≃ g : K∗ → L∗ and h′ is a homotopy
which shows g ≃ k : K∗ → L∗ , then h + h′ is a homotopy which shows
f ≃ k. For
fn − kn = fn − gn + gn − kn
L
= ∂n+1 ◦ hn + hn−1 ◦ ∂nK + ∂n+1
L
◦ h′n + h′n−1 ◦ ∂nK
L
= ∂n+1 ◦ (hn + h′n ) + (hn−1 + h′n−1 ) ◦ ∂nK .
(2) Let h be a homotopy which shows f ≃ f ′ and k be a homotopy which
shows g ≃ g ′ . Composition with g on the left and using that g is a chain map
yields
gn ◦ (fn − fn′ ) = gn ◦ (∂n+1
L
◦ hn + hn−1 ◦ ∂nK )
M
= ∂n+1 ◦ (gn+1 ◦ hn ) + (gn ◦ hn−1 ) ◦ ∂nK .
This shows that the sequence of maps gn+1 ◦ hn defines a homotopy g ◦ f ≃ g ◦ f ′ .
Composition with f ′ on the right and using that f ′ is a chain map yields
(gn − gn′ ) ◦ fn′ = (∂n+1
M
◦ kn + kn−1 ◦ ∂nL ) ◦ fn′
M
= ∂n+1 ′
◦ (kn ◦ fn′ ) + (kn−1 ◦ fn−1 ) ◦ ∂nK .
This shows that the sequence of maps kn ◦ fn′ defines a homotopy g ◦ f ′ ≃ g ′ ◦ f ′ .
Summarizing we have shown
g ◦ f ≃ g ◦ f ′ ≃ g′ ◦ f ′.
By transitivity, this shows the desired result. □
Now we are ready to show an important fact in homological algebra:
Homology identifies chain homotopies

If f ≃ g : K∗ → L∗ are homotopic morphisms of chain complexes, then
Hn (f ) = Hn (g) for all n ∈ Z.
Proof: This follows immediately from the fact that fn − gn is just given by
boundaries which, by definition, vanish in homology.
More concretely, let x ∈ Kn be an arbitrary cycle in Kn and let h be a
homotopy which gives f ≃ g. Then we get by using the definition of homotopies
L
Hn (f )([x]) = [fn (x)] = [gn (x) + ∂n+1 (hn (x)) + hn−1 (∂nK (x))] = [gn (x)] = Hn (g)([x])
L
where we use that ∂n+1 (hn (x)) is obviously a boundary in Ln and that hn−1 (∂nK (x)) =
0, since x is a cycle in Kn by assumption. □
Now we can also mimic the notion of homotopy equivalences.
Chain homotopy equivalences

A morphism of chain complexes f : K∗ → L∗ is called a homotopy equiv-
alence if there exists a morphism of chain complexes g : L∗ → K∗ such that
g ◦ f ≃ idK∗ and f ◦ g ≃ idL∗ .
103
If such a homotopy equivalence exists, we write K∗ ≃ L∗ and say that K∗

and L∗ are homotopy equivalent.
In particular, by adopting language from algebraic topology, if the identity

map on a chain complex K∗ is homotopy equivalent to the zero map, then we say
that K∗ is contractible. For example, if X is a contractible space, then its sin-
gular chain complex S∗ (X) is a contractible chain complex. Note that a chain
complex K∗ with at least one nonzero homology group cannot be contractible.
As a consequence of what we proved we get:
Chain homotopy equivalences

• If K∗ ≃ L∗ , then Hn (K∗ ) ∼ = Hn (L∗ ).
• Given two chain complexes K∗ and L∗ we denote the set of morphisms
of chain complexes by Mor(K∗ ,L∗ ). Let [K∗ ,L∗ ] := Mor(K∗ ,L∗ )/ ≃ denote
the set of equivalence classes under the relation given by chain homotopies.
Then we can define a new category whose objects are chain complexes and
whose sets of morphisms from K∗ → L∗ are homotopy classes of chain maps,
i.e., the sets [K∗ ,L∗ ]. Let us call this category K.
Since the homotopy relation respects composition, we obtain that homology
defines a functor
K → Ab, K∗ 7→ Hn (K∗ )
where Ab denotes the category of abelian groups.
Let us look at some examples:

• Let K∗ be the chain complex
0
··· → 0 → Z →
− Z → 0 → ··· .
Since all maps are trivial, we have Hn (K∗ ) = Kn for all n. Hence K∗ has
exactly two nonzero homology groups, both being Z. In particular, it is
not contractible.
1
··· → 0 → Z →
− Z → 0 → ··· .
This complex is actually an exact sequence. Thus Hn (K∗ ) = 0 for all n.
Moreover it is contractible. We can write down a homotopy by
0
id / 0
?
0

id /
Z ?Z
1
1 1

id /
Z ?Z
0

0
id / 0.
2
··· → 0 → Z →
− Z → 0 → ··· .
This complex has one nonzero homology group H1 (K∗ ) = Z/2. It is
therefore not contractible.
2 2 2 2 2
··· →
− Z/4 →
− Z/4 →
− Z/4 →
− Z/4 →
− ··· .
The homology of K∗ vanishes, since, at each stage, the image and the
kernel of the differential is 2Z/4. Nevertheless, K∗ is not contractible.
For if there was a homotopy between idK∗ and the zero map, it would
like this
Z/4
id / Z/4
=
hn
2 2

Z/4
id / Z/4
=
2 2
hn−1
Z/4 / Z/4
id
and satisfy id = 2hn + hn−1 2. But 2hn + hn−1 2 can only produce even
numbers modulo 4. Hence it cannot be the identity map on Z/4.
After all this abstract stuff we should better demonstrate that the notion of
chain homotopies is useful for our purposes. We are going to do this by showing
that homotopies between maps of spaces induces a chain homotopy. By what we
have just seen, this will prove the Homotopy Axiom for singular homology.
LECTURE 11
Homotopy invariance of singular homology
We are going to prove the Homotopy Axiom for singular homology. The proof
will follow from constructing a homotopy between chain complexes.
Let f, g : X → Y be two homotopic maps and let h : X × [0,1] → Y be a
homotopy between them. Let σ : ∆n → X be an n-simplex on X. Then h induces
a map
σ×id h
∆n × [0,1] −−−→ X × [0,1] →
− Y
which defines a homotopy between f ◦ σ and g ◦ σ.
Our goal is to turn this into a geometrically induced chain homotopy
between Sn (f ) and Sn (g). By our result from the previous lecture, this will
imply the Homotopy Axiom.
So let us have a closer look at the space ∆n × [0,1]. For n = 1, it looks just
like a square. Via the diagonal we can divide it into two triangles which look like
∆2 . For n = 2, ∆2 × [0,1] looks like a prism which we can divide into three copies
of ∆3 .
In general, ∆n × [0,1] looks like a higher dimensional prism which we can
divide into n + 1 copies of ∆n+1 . We should make this idea more precise:
Simplices on a prism
For every n ≥ 0 and 0 ≤ i ≤ n, we define an injective map
pni : ∆n+1 → ∆n × [0,1],
(t0 , . . . ,tn+1 ) 7→ ((t0 , . . . ,ti−1 ,ti + ti+1 ,ti+2 , . . . ,tn+1 ), ti+1 + · · · + tn+1 ).
We can consider each pni as an n + 1-simplex on the space ∆n × [0,1]. When
n is clear we will often drop it from the notation.
For n = 0, we have only one map

p00 : ∆1 → ∆0 × I = {e0 } × [0,1], (t0 ,t1 ) 7→ (0,t1 ).
105
106 CHAPTER 11. HOMOTOPY INVARIANCE OF SINGULAR HOMOLOGY
Let us have a look at what happens for n = 1: Then the effect of p10 and
p11 is given by
p0 : ∆2 → ∆1 × [0,1],(t0 ,t1 ,t2 ) 7→ (t0 + t1 ,t1 + t2 )


e0 = (1,0,0) 7→ (1,0,0) = (e0 ,0)

e1 = (0,1,0) 7→ (1,0,1) = (e0 ,1)

e = (0,0,1) 7→ (0,1,1) = (e ,1)
2 1
and
p1 : ∆2 → ∆1 × [0,1],(t0 ,t1 ,t2 ) 7→ (t0 ,t1 + t2 ,t2 )


e0 = (1,0,0) 7→ (1,0,0) = (e0 ,0)

e1 = (0,1,0) 7→ (0,1,0) = (e1 ,0)

e = (0,0,1) 7→ (0,1,1) = (e ,1).
2 1
107
In general, the effect of pni on the vertex ek of ∆n+1 for 0 ≤ k ≤ n + 1 is given

by
(
(ek ,0) if 0 ≤ k ≤ i
(14) pni (ek ) =
(ek−1 ,1) if k > i.
In fact, we could define pni as the unique affine map which satisfies (14).
Let j0 and j1 be the two inclusions
j0 : ∆n ,→ ∆n × [0,1], x 7→ (x,0)
j1 : ∆n ,→ ∆n × [0,1], x 7→ (x,1)
determined by the endpoints of [0,1].
For the next result, recall our formulae for the face maps on standard sim-
plices:
For 0 ≤ i ≤ n + 1 which can be described as
ϕn+1
i (t0 , . . . ,tn ) = (t0 , . . . ,ti−1 ,0,ti , . . . ,tn )
with the 0 inserted at the ith coordinate.
Using the standard basis, ϕin+1 can be described as the affine map (a trans-
lation plus a linear map)
(
ek k<i
(15) ϕn+1
i : ∆n ,→ ∆n+1 determined by ϕn+1
i (ek ) =
ek+1 k ≥ i.
Lemma: Prism and face maps

We have the following identifications of maps:
(16) pn0 ◦ ϕn+1

0 = j1 ,
(17) pnn ◦ ϕn+1

n+1 = j0 ,
(18) pni ◦ ϕn+1

i = pni−1 ◦ ϕn+1
i for 1 ≤ i ≤ n,
(19) pnj+1 ◦ ϕn+1

i = (ϕni × id) ◦ pjn−1 for j ≥ i.
(20) pnj ◦ ϕn+1 n n−1

i+1 = (ϕi × id) ◦ pj for j < i,
Proof: (16) We check the effect of pn0 ◦ ϕn+1

0
pn0 (ϕn+1
0 (t0 , . . . ,tn )) = p0 (0,t0 , . . . ,tn )
n
!
X
= (t0 , . . . ,tn ), ti
i=0
= (t0 , . . . ,tn ,1) = j1 (t0 , . . . ,tn ).
(17) Similarly, we calculate
pnn (ϕn+1 n
n+1 (t0 , . . . ,tn )) = pn (t0 , . . . ,tn ,0)
= (t0 , . . . ,tn ,0) = j0 (t0 , . . . ,tn ).
(18) We calculate and compare:
pni (ϕn+1
i (t0 , . . . ,tn ))
= pni (t0 , . . . ,ti−1 ,0,ti , . . . ,tn )
= ((t0 , . . . ,ti−1 ,0 + ti ,ti+1 , . . . ,tn ), 0 + ti + · · · + tn )
n
!
X
= (t0 , . . . ,tn ), tj ,
j=i
109
and
pni−1 (ϕn+1
i (t0 , . . . ,tn ))
= pni−1 (t0 , . . . ,ti−1 ,0,ti , . . . ,tn )
= ((t0 , . . . ,ti−2 ,ti−1 + 0,ti ,ti+1 , . . . ,tn ), ti + · · · + tn )
n
!
X
= (t0 , . . . ,tn ), tj .
j=i
Hence both maps agree.

(19) For j ≥ i, the assertion amounts to showing that the following diagram
commutes:
pn−1
∆n+1 j
/ ∆n × [0,1]
8
ϕn+1
i
∆n
pn
j+1 &
∆n−1 × [0,1] / ∆n × [0,1].
ϕn
i ×id
To check this, we are going to use formulae (14) and (15). Since the affine maps
involved are determined by their effect on the ek s, this will suffice to prove the
formulae.
For k < i, we get
pnj+1 ◦ ϕn+1
i (ek ) = pnj+1 (ek ) = (ek ,0)
= (ϕni × id)(ek ,0) = (ϕni × id) ◦ pn−1
j (ek ).
For i ≤ k ≤ j, we get
pnj+1 ◦ ϕn+1
i (ek+1 ) = pnj+1 (ek ) = (ek+1 ,0)
= (ϕni × id)(ek ,0) = (ϕni × id) ◦ pn−1
j (ek ).
And for k > j, we get

pnj+1 ◦ ϕn+1
i (ek ) = pnj+1 (ek+1 ) = (ek ,1)
= (ϕni × id)(ek−1 ,1) = (ϕni × id) ◦ pn−1
j (ek ).
(20) follows by a similar argument. □

Definition: Induced prism operator

• For every n ≥ 0 and 0 ≤ i ≤ n, the map pni induces a group
homomorphism
Pin : Sn (X) → Sn+1 (X × [0, 1])
which is defined on generators by composition with pni
pn σ×id
Pin (σ) = (σ × id) ◦ pni : ∆n+1 −→
i
∆n × [0, 1] −−−→ X × [0, 1]
and extended Z-linearly.
• This construction descends to a map Pin on relative chains for
any pair (X, A).
• We define a group homomorphism, often called prism operator,
n
X
P n : Sn (X,A) → Sn+1 (X × I,A × I), P n = (−1)i Pin .
i=0
Let jtX denote the inclusion X ,→ X × [0, 1], x 7→ (x, t). The prism operator
is the desired chain homotopy:
Chain homotopy lemma

The homomorphisms P n provide a chain homotopy between the two mor-
phisms of chain complexes
S∗ (j0X ) ≃ S∗ (j1X ) : S∗ (X,A) → S∗ (X × I,A × I).
111
Proof: We need to show Sn (j0X ) − Sn (j1X ) = ∂n+1 ◦ P n + P n−1 ◦ ∂n . Let σ be

an n-simplex. Then we calculate
n
!
X
n−1 n−1
P ◦ ∂n (σ) = P (−1)i σ ◦ ϕni
i=0
X
= (−1)i+j (σ × id) ◦ (ϕni × id) ◦ pn−1
j
0≤j<i≤n
X
+ (−1)i+j (σ × id) ◦ (ϕni × id) ◦ pn−1
j
0≤i≤j≤n−1
X
=− (−1)i+j+1 (σ × id) ◦ pnj+1 ◦ ϕn+1
i by (20)
0≤j<i≤n
X
− (−1)i+j+1 (σ × id) ◦ pnj ◦ ϕn+1
i+1 by (19).
0≤i≤j≤n−1
On the other hand, we have
n
!
X
∂n+1 ◦ P n (σ) = ∂n+1 (−1)j (σ × id) ◦ pnj
j=0
n X
X n+1
= (−1)i+j (σ × id) ◦ pnj ◦ ϕn+1
i
j=0 i=0
X
= (−1)i+j (σ × id) ◦ pnj ◦ ϕn+1
i (i < j)
0≤i<j≤n
Xn
+ (σ × id) ◦ pni ◦ ϕn+1
i (i = j)
i=0
n+1
X
− (σ × id) ◦ pni−1 ◦ ϕn+1
i (i = j + 1)
i=1
X
+ (−1)i+j (σ × id) ◦ pnj ◦ ϕn+1
i (i > j + 1)
1≤j+1≤i≤n+1
′
X
= (−1)i+j +1 (σ × id) ◦ pnj′ +1 ◦ ϕn+1
i
0≤i≤j ′ ≤n−1
+ (σ × id) ◦ j1 − (σ × id) ◦ j0
′
X
+ (−1)i +j+1 (σ × id) ◦ pnj ◦ ϕn+1
i′ +1 .
0≤j≤i′ ≤n
For the final step we used again the trick to relabel the indices and wrote j ′ = j −1
and i′ = i − 1. By comparing the two calculations, we see that all summands
cancel out except for (σ × id) ◦ j1 − (σ × id) ◦ j0 .
Thus we can conclude:
∂n+1 ◦ P n (σ) + P n−1 ◦ ∂n (σ) = (σ × id) ◦ j1X − (σ × id) ◦ j0X

= j1X ◦ σ − j0X ◦ σ
= Sn (j1X )(σ) − Sn (j0X )(σ). □
As a consequence we get the Homotopy Axiom:
Theorem: Homotopy Invariance

If f ≃ g : (X,A) → (Y,B), then
Sn (f ) ≃ Sn (g) : Sn (X,A) → Sn (Y,B)
for all n. Hence f ≃ g implies Hn (f ) = Hn (g).
Proof: Let h be a homotopy between f and g. We can write this as
f = h ◦ j1X and g = h ◦ j0X .
Then the previous lemma yields
Sn (f ) − Sn (g)
= Sn (h) ◦ Sn (j1X ) − Sn (h) ◦ Sn (j0X )
= Sn (h) ◦ (∂n+1 ◦ P n ) + Sn (h) ◦ (P n−1 ◦ ∂n )
= ∂n+1 ◦ (Sn+1 (h) ◦ P n ) + (Sn (h) ◦ P n−1 ) ◦ ∂n since S∗ (h) is a chain map.
Thus the sequence of homomorphisms Sn+1 (h)◦P n is a chain homotopy between

Sn (f ) and Sn (g).
Applying the theorem about chain homotopic maps and their induced maps
on homology implies the last statement. □
We have already used homotopy invariance of singular homology at numerous
occasions. Here is yet another one:
113
Proposition: Homology of weak retracts

Let i : A ,→ X be a weak retract, i.e., assume there is continuous map
ρ : X → A such that ρ ◦ i ≃ idA . Then
Hn (X) ∼ = Hn (A) ⊕ Hn (X,A) for all n.
Proof: Functoriality and homotopy invariance tell us that ρ ◦ i ≃ idA implies
Hn (ρ) ◦ Hn (i) = Hn (ρ ◦ i) = Hn (idA ) = idHn (A) for all n.
Hence Hn (i) is injective for all n. That means that the sequence
Hn (i)
0 → Hn (A) −−−→ Hn (X)
is exact. Since Hn−1 (i) is injective, the exactness of the sequence
∂ Hn−1 (i)
· · · → Hn (X,A) →
− Hn−1 (A) −−−−→ Hn−1 (X) → · · ·
implies that the connecting homomorphism ∂ : Hn (X,A) → Hn−1 (A) is the zero
map. Thus we get a short exact sequence
Hn (i) Hn (j)
(21) 0 → Hn (A) −−−→ Hn (X) −−−→ Hn (X,A) → 0.
Since Hn (ρ) is a left-inverse of Hn (i), this sequence splits, i.e.,

∼
=
Hn (X) −
→ Hn (A) ⊕ Hn (X,A), [c] 7→ (Hn (ρ)([c]), Hn (j)([c]).
is an isomorphism.
We can describe an inverse of this map as
Hn (A) ⊕ Hn (X,A) → Hn (X),

([a], [b]) 7→ Hn (i)([a]) + [c′ ] − Hn (i ◦ ρ)([c′ ])
where [c′ ] is any class with Hn (j)([c′ ]) = [b].

We need to check that the choice of [c′ ] does not matter. So let [c′′ ] be
another class with Hn (j)([c′′ ]) = [b]. Then we have
Hn (j)([c′ ] − [c′′ ]) = 0, i.e., [c′ ] − [c′′ ] ∈ Ker (Hn (j)).

By the exactness of (21), this implies that there exists a class [ã] ∈ Hn (A) with
Hn (i)([ã]) = [c′ ] − [c′′ ]. Thus
[c′ ] − Hn (i ◦ ρ)([c′ ]) − ([c′′ ] − Hn (i ◦ ρ)([c′′ ]))
= [c′ ] − [c′′ ] − Hn (i ◦ ρ)([c′ ] − [c′′ ])
= Hn (i)([ã]) − Hn (i ◦ ρ)(Hn (i)([ã]))
= Hn (i)([ã]) − (Hn (i) ◦ Hn (ρ) ◦ Hn (i))([ã])
= Hn (i)([ã]) − Hn (i)([ã]) since Hn (ρ) ◦ Hn (i) = idHn (A)
= 0. □
Homotoy invariance revisited

Let hoTop denote the homotopy category of Top, i.e., the category
whose objects are topological spaces and whose morphisms are homotopy
classes of continuous maps:
MorhoTop (X, Y ) = MapTop (X, Y )/ ≃=: [X, Y ].
where MapTop (X, Y ) denotes the set of continuous maps from X to Y . The
result we have just shown implies that singular homology descends to a
functor on hoTop:
Hn
Top /
: Ab
Hn

hoTop.
This observation applies to almost all algebraic invariants in Topology. In

other words, invariants in Algebraic Topology distinguish neither between
homotopic maps nor between homotopy equivalent spaces.
However, many topological properties are not invariant under homotopy.
For example, compactness is not invariant under homotopy. In other
words, if X is compact, then it may well be the case that a space which
is homotopy equivalent to X is not compact. To convince ourselves of this
fact, it suffices to take X = {0} and Y = Rn . A bit more interesting is
X = S n and Y = Rn+1 \ {0}. While X is compact, Y is not, but the
inclusion S n ,→ Rn+1 \ {0} is a homotopy equivalence.
This observation should make us aware of the scope of our abilities. The
tools we develop in this class are great. But they are not the end of
the story...
LECTURE 12
Locality and the Mayer-Vietoris sequence
We are going to discuss the Excision Axiom for singular homology and some
consequences. Let us first recall what it says:
Excision Axiom of singular homology

Let (X,A) be a pair of spaces and let Z ⊂ A be a subspace the closure
of which is contained in the interior of A, in formulae Z̄ ⊆ A◦ . Then the
inclusion map k : (X − Z, A − Z) ,→ (X,A) induces an isomorphism
Hn (k) : Hn (X − U, A − U ) → Hn (X,A) for all n.
We are going to deduce the excision property of homology from the following
locality principle.
Let X be a topological space and let A = {Aj }j∈J be a cover of X, i.e., a
collection of subsets Aj ⊆ X such that X is the union of the interiors of the
Aj s.
A-small chains
• An n-simplex σ : ∆n → X is called A-small if the image of σ is contained
in one of the Aj s. P
• An n-chain c = i ni σi if X is called A-small if, for every i, there is a
Aj such that σi (∆n ) ⊂ Aj .
• We are going to denote the subgroup of A-small n-chains by
SnA (X) := {c ∈ Sn (X) : σ is A − small}.
• For a subspace A ⊂ X, we write
SnA (X)
SnA (X,A) := .
SnA (A)
115
116 CHAPTER 12. LOCALITY AND THE MAYER-VIETORIS SEQUENCE
If, for each j, ιj : Aj ,→ X denotes the inclusion map, then we can describe
SnA (X) also as
!
M ⊕j Sn (ιj )
SnA (X) = Im Sn (Aj ) −−−−−→ Sn (X) .
j∈J
The point of A-small chains is that we can use their chain complex to compute
singular homology:
Locality Principle/Small Chain Theorem

For any cover A of X, the inclusion of chain complexes
S∗A (X,A) ⊂ S∗ (X,A)
induces an isomorphism in homology.
The proof of this theorem takes quite an effort and we will postpone it for a
moment. Instead we will now explain how the excision property follows from the
theorem.
• Proof of the Excision Axiom using small chains:
Since Z̄ ⊆ A◦ , we have (X − Z)◦ ∪ A◦ = X. Thus, if we set B := X − Z,
A = {A, B} is a cover of X.
Moreover, we can rewrite
(X − Z, A − Z) = (B, A ∩ B).
Hence our goal is to show that

S∗ (B,A ∩ B) → S∗ (X,A)
induces an isomorphism in homology.
The inclusion of chain complexes S∗A (X) ⊂ S∗ (X) induces a morphism of
short exact sequences of chain complexes
0 / S∗ (A) / S∗A (X) / S∗A (X)/S∗ (A) / 0

0 / S∗ (A) / S∗ (X) / S∗ (X)/S∗ (A) / 0.
The middle vertical map induces an isomorphism in homology by the

Small Chain Theorem. The induced long exact sequences in homology and
117
the Five-Lemma imply that the right-hand vertical map induces an isomor-
phism in homology as well. Thus we are reduced to compare S∗ (B,A ∩ B) and
S∗A (X)/S∗ (A).
Now we observe
S∗A (X) = S∗ (A) + S∗ (B) ⊂ S∗ (X)
and hence
S∗ (B) S∗ (B) = S∗ (A) + S∗ (B)
∼ S A (X)
= −
→ = ∗
S∗ (A ∩ B) S∗ (A) ∩ S∗ (B) S∗ (A) S∗ (A)
where the middle isomorphism follows from the general comparison of quo-
tients of sums and intersections of abelian groups.
Thus the chain map
S∗ (B,A ∩ B) → S∗A (X)/S∗ (A)
induces an isomorphism in homology and the excison axiom holds. QED
• The Mayer-Vietoris sequence

The above proof inspires us to look at the following situation which will lead
to an important computational tool.
Assume that A = {A,B} is a cover of X. Consider the diagram
jA
A∩B / A
jB iA

B / X.
iB
For every n, these maps induce homomorphisms in homology

Hn (jA )
αn : Hn (A ∩ B) → Hn (A) ⊕ Hn (B), αn =
−Hn (jB )
x 7→ (Hn (jA )(x), −Hn (jB )(x))
and

βn : Hn (A) ⊕ Hn (B) → Hn (X), βn = Hn (iA ) Hn (iB )
(a,b) 7→ Hn (iA )(a) + Hn (iB )(b).
Theorem: Mayer-Vietoris sequence

For any cover A = {A,B} of X, there are natural homomorphisms
∂nM V : Hn (X) → Hn−1 (A ∩ B) for all n
which fit into an exact sequence
βn+1
··· / Hn+1 (X)
MV
∂n+1
r βn
Hn (A ∩ B) / Hn (A) ⊕ Hn (B) / Hn (X)
αn
MV
∂n
r
Hn−1 (A ∩ B) / ···
αn−1
Proof: From the proof of the Excision Axiom we remember that there is a
short exact sequence of chain complexes
 

S∗ (jA )  h i
−S∗ (jB ) S∗ (iA ) S∗ (iB )
0 → S∗ (A ∩ B) −−−−−−−−→ S∗ (A) ⊕ S∗ (B) −−−−−−−−−−−−→ S∗A (X) → 0.
Note that the exactness at the right-hand term was part of the proof of the
Excision Axiom and the exacntess at the middle term can be easily checked by
looking long enough at the commutative diagram
∆n
$ jA "
A∩B / A
jB iA
% /

B iB
X.
119
It induces a long exact sequence in homology

A
βn+1
··· / A
Hn+1 (X)
A
∂n+1
s A
βn
Hn (A ∩ B) / Hn (A) ⊕ Hn (B) / HnA (X)
αn
A
∂n
r
Hn−1 (A ∩ B) / ···
αn−1
By definition of small chains, the homomorphism βn factors through small chains,

in other words, it is induced by the composition
h i
Sn (iA ) Sn (iB )
Sn (A) ⊕ Sn (B) −−−−−−−−−−−−→ SnA (X) ,→ Sn (X).
Thus we can apply the inverse of the isomorphism of the Small Chain The-
orem and define ∂nM V to be
∼
= ∂A
→ HnA (X) −→
∂nM V : Hn (X) − n
Hn−1 (A ∩ B).
Then the following sequence

MV
∂n
βn
Hn (A) ⊕ Hn (B) / Hn (X) / Hn−1 (A ∩ B)
O 7
∼
=
A
βn ' A
∂n
HnA (X)
is exact at Hn (X), since the triangles commute.
This yields the sequence of homomorphisms and the desired long exact se-
quence. QED
The MVS is an extremely useful tool

The Mayer-Vietoris sequence (MVS) is an important computational tool.
Its power relies on the simple idea: If you want to understand a big space,
split it up into smaller spaces you understand and then put the information
back together.
The MVS tells us how the homology of X is built out of homologies of the
cover by A and B.
Let us apply this new insight to some concrete examples:

• Let us calculate the homology of X = S 1 yet another time. Let x = (0,1)
and y = (0, − 1) on S 1 . We set A = S 1 − {x} and B = S 1 − {y}. Then A and
B are two open subsets which cover S 1 . We observe that both A and B are
contractible.
The intersection A ∩ B contains the points p = (−1,0) and q = (1,0). In fact,
the inclusion
{p, q} ,→ A ∩ B
is a deformation retract.
Since A = {A,B} is a cover of S 1 , we can write down the corresponding MVS.
For n ≥ 2, all the homology groups hitting and being hit by Hn (S 1 ) are zero, since
Hn (A) ⊕ Hn (B) = Hn ({x}) ⊕ Hn ({y}) = 0 and Hn−1 (A ∩ B) = Hn−1 ({p,q}) = 0.
Thus
Hn (S 1 ) = 0 for all n ≥ 2.
Since S 1 is path-connected, we know H0 (S 1 ) = Z. It remains to check n = 1.

The MVS for n = 1 looks like
0 / H1 (S 1 ) / H0 (A ∩ B) / H0 (A) ⊕ H0 (B)
0 / H1 (S 1 ) / Z⊕Z   / Z⊕Z

1 1
−1 −1
where we obtain the lower right-hand map by observing that all summands are
of the form H0 (pt) and hence each generator in H0 (A ∩ B) is sent to (1, − 1) by
[H0 (jA ), −H0 (jB )]. Thus H1 (S 1 ) is the kernel of this map:

1 ∼ 1 1
H1 (S ) = Ker = {(x, −x) ∈ Z ⊕ Z} ∼
= Z.
−1 −1
• For n ≥ 2, let A = S n − {S} and B = S n − {N } where N and S are the

north- and south-pole of S n , respectively,. We observe that both A and B are
contractible. Moreover, the inclusion of j : S n−1 ,→ A ∩ B as the equator is a
strong deformation retract. In particular, j is a homotopy equivalence.
Together with the inverse of the isomorphism Hq−1 (j), we get
∼
=
Hq−1 (j)−1 ◦ ∂qM V : Hq (S n ) −
→ Hq−1 (S n−1 )
121
is an isomorphism for all q ≥ 2. Since we know Hq (S 1 ) for all q, this yields Hq (S n )

by induction.
• Let K be the Klein bottle which can be constructed from a square by
gluing the edges as indicated in the following picture:
The outcome of this procedure is the twisted surface whose 3-dimensional

shadow we see in the next picture which is taken from wikipedia.org:
(Note that we should really think of K as an object in R4 where it does not

self-intersect.)
We observe that K can be constructed by taking two Möbius bands A
and B and gluing them together by a homeomorphism between their boundary
circles. Hence K = A ∪ B and A ∩ B ≈ S 1 . In the exercises we are going to
caculate the homology of the Möbius strip. It is given by H0 (M ) = H1 (M ) = Z
and H2 (M ) = 0.
We would like to use this information to calculate the homology of K.
Since K is path-connected as a quotient of a path-connected space, we know
H0 (K) = Z.
Now we apply the MVS: We observe that Hn (A), Hn (B) and Hn (A ∩ B)
vanish for n ≥ 2. Hence Hn (K) = 0 for all n ≥ 3.
The remaining MVS looks like this:
∂M V φ1
0 → H2 (K) −−−→ H1 (A ∩ B) −→ H1 (A) ⊕ H1 (B) → H1 (K) → 0.
The 0 on the right-hand side is justified by the fact that
φ0
H0 (A ∩ B) ∼= Z −→ Z ⊕ Z ∼ = H0 (A) ⊕ H0 (B)
is injective.
The map φ1 is given by
φ1
Z −→ Z ⊕ Z, 1 7→ (2, − 2),
since
H1 (A ∩ B) = H1 (S 1 ) → H1 (M ) = H1 (A)
wraps the circle around the boundary of M twice, and
H1 (A ∩ B) = H1 (S 1 ) → H1 (M ) = H1 (B)
does that too, but with reversed orientation. (We will understand this fact
better after we have done the exercises.) Hence on the second factor we use the
map z 7→ z −2 to produce a Möbius band.
In particular, φ1 is injective and hence
H2 (K) = 0.
Moreover, H1 (K) is the cokernel of φ1 . If we choose the basis

{b1 := (1,0), b2 := (1, − 1)} for Z ⊕ Z,
then we see that φ1 maps 1 ∈ Z to 2b2 in Zb1 ⊕ Zb2 . Hence the cokernel of φ1 is
isomorphic to Zb1 ⊕ Zb2 /2b2 . Thus
H1 (K) ∼
= Z ⊕ Z/2.
LECTURE 13
Cell complexes
We return to an important type of topological spaces, called CW- or cell

complexes, that is particularly convenient for our purposes in many respects.
It will turn out that this type of spaces both appears very frequently and is
quite accessible for calculations. In particular, we will learn next week that the
homology of a cell complex is quite easy to compute.
The idea of creating a cell complex is to successively glue cells to what has
already been built. The general procedure for doing this is the following:
Gluing a space along a map

Suppose we have a space X and a pair (B,A) of spaces. We define a space
X ∪f B, often also denoted X ∪A B if the map f is either understood or
just the inclusion, which fits into the diagram
f
A _ / X

B / X ∪f B
φ
by
X ∪f B := (X ⊔ B)/(a ∼ f (a) for all a ∈ A).
We say that X ∪f B arises from attaching B to X along f , or along A,
and f is called an attaching map.
By its construction, there are two types of equivalence classes in X ∪f B:

• classes which consist of single points of B − A,
• classes which consist of sets {x} ⊔ f −1 (X) for any point x ∈ X.
123
124 CHAPTER 13. CELL COMPLEXES
Note that the lower horizontal map φ : B → B ∪f X arises as part of the
construction. It is given by
(
b if b ∈ B − A
φ : B → B ∪f X, b 7→
[b] if b ∈ A.
In particular, this shows that φ|B−A is a homeomorphism.

The topology of X ∪f B is the quotient topology and is characterized by
the universal property: whenever there is a diagram of solid arrows of the form
f
A _ / X

B / X ∪f B
#,
Y
then there is a unique dotted arrow which makes all triangles commute. We
can reformulate this fact by saying that X ∪f B is the pushout of the solid
diagram.
For example:
• if X = ∗ consists of just a point, then
X ∪f B = ∗ ∪f B = B/A;
• if A = ∅, then X ∪f B = X ⊔ B is just a disjoint union.
A more important example is the following:
Attaching a cell
We consider the pair (Dn , S n−1 ) of an n-disk and its boundary. We are
going to think of Dn as an n-cell.
Suppose we are given a map f : S n−1 → X. Then we can attach an n-cell
to X via f as
f
S n−1
_ / X

Dn / X ∪f Dn .
125
We could speed up this process by attaching several cells at once:

f
n−1 /X
`
α∈J Sα
_
` `
n / X ∪f Dαn .
α∈J Dα α∈J
Let us look at some examples:

• Let us start with n = 0 and write (D0 ,S −1 ) for (∗,∅)). Attaching 0-cells
to a space X just means adding a set of discrete points to X:
a
X ∪f Dα0 = X ⊔ J
α∈J
where J is a set with the discrete topology.

• Now let us attach two 1-cells to a point X = ∗:
f
S 0 ⊔ _ S 0 / ∗

D1 ⊔ D1 / ∗ ∪f (D1 ⊔ D1 ).
Since there is only one choice for f , we get a figure eight: we start with
two 1-disks D1 and then we identify all four boundary points with the
0-cell. We denote this space by S 1 ∨ S 1 .
• We continue with this space and attach one 2-cell: We can think of S 1 ∨S 1
as an empty square where we glue together the horizontal edges and the
vertical edges. Then we glue in a 2-cell into the square by attaching its
boundary to the edges a, b, a−1 , and b−1 , i.e., by walking clockwise:
f =aba−1 b−1
S 1 _ / S1 ∨ S1

D2 / (S 1 ∨ S 1 ) ∪f D2 = T 2 .
The result of this procedure is a two-dimensional torus.
This example motivates the following key concept:
Cell complex
A cell complex, or CW-complex, is a space X equipped with a sequence
of subspaces
∅ = Sk−1 X ⊆ Sk0 X ⊆ Sk1 X ⊆ Sk2 X ⊆ · · · X
such that
• X is the union of the Skn Xs,
• for all n, Skn X arises from Skn−1 by attaching n-cells, i.e., there
is a pushout diagram
n−1 fn /
`
α∈Jn Sα Skn−1 X
_

n /
`
α∈Jn Dα φn Skn X.
The space Skn X is called the n-skeleton of X.
In our example of the torus T 2 the skeleta are

Sk0 T 2 = ∗, Sk1 T 2 = S 1 ∨ S 1 , Sk2 T 2 = T 2 .
127
Before we study more examples, we fix more terminology and list some facts
which should help clarify the picture:
• The topology of a cell complex is determined by its skeleta, i.e., a subset
U ⊂ X is open (closed) if and only if U ∩ Skn X is open (closed) for all
n.
• In fact, the topology on X is determined by its cells, i.e., U is open

(closed) in X if and only its intersection with each cell is open (closed),
or equivalently, if φ−1 n
α (U ) is open (closed) in each Dα . This topology is
called the weak topology and explains the W in CW-complex.
• That implies that a map g : X → Y is continuous if and only if its

restriction to each skeleton is continuous, or equivalently, if and only if
g ◦ φα : Dαn → Y is continuous for all Dαn .
• For any n-cell Dαn , the induced map φα : Dαn → X is called the charac-
teristic map of the cell. As we explained before, the restriction to the
open interior (Dαn )◦ = Dαn − Sαn−1
(φα )|(Dαn )◦ → X
is a homeomorphism onto its image.
• We will call the image of Dαn under φα in X a closed n-cell of X. We

will refer to n as the dimension of the cell. Since Dn is compact, it is
a compact subset.
• The image of the interior (Dαn )◦ of Dαn in X is often called an n-cell or

open n-cell of X and will be denoted by enα . Note that this subset is
not necessarily an open subset of X.
• The C in CW-complex stands for closure finite which means that, for
every cell, φα (Sαn−1 ) is contained in finitely many cells (of dimension at
most n − 1).
• A cell complex X is called finite-dimensional if there is an n such that

X = Skn X. The smallest such n is called the dimension of X, i.e., the
unique n such that Skn X = X and Skn−1 X ⊊ X.
• A cell complex is called of finite type if each indexing set Jn is finite,

i.e., if only finitely many cells are attached in each step.
• A cell complex is called finite if it is finite-dimensional and of finite type,
i.e., if it has only finitely many cells.
• The dimension of a cell complex is a topological invariant, i.e., it is invari-

ant under homeomorphisms. Moreover, every cell complex is Hausdorff.
• However, a cell complex is compact if and only if it is finite.
• Note that every nonempty cell complex must have at least one 0-cell.
• The cell structure of a cell complex is in general not unique. Often there
are many different cell structures. We will observe this for example for
the n-sphere.
Here is an important theorem which demonstrates the wide range and impor-
tance of cell complexes:
Compact smooth manifolds are cell complexes

Every compact smooth manifold can be given the structure of a cell complex.
Here some important examples:
• A simple example is given by surfaces of a three-dimensional cube: it

has eight 0-cells, twelve 1-cells, six 2-cells.
129
• Similarly, every n-simplex is a cell complex. For example, ∆3 has four

0-cells, six 1-cells, four 2-cells, and one 3-cell.
• The sphere S n is a cell complex with just two cells: one 0-cell e0 (that
is a point) and one n-cell which is attached to e0 via the constant map
S n−1 → e0 . Geometrically, this corresponds to expressing S n as Dn /∂Dn :
we take the open n-disk Dn \ ∂Dn and collapse the boundary ∂Dn to a
single point which is, say, the north pole N = e0 .
• The n-sphere X = S n can also be equipped with a different cell struc-

ture:
We start with two 0-cells which give us the 0-skeleton
≈
→ S 0.
Sk0 X −
≈
Now we attach two 1-cells via the homeomorphism f : S 0 −→ Sk0 X. This
gives us one 1-cell as the upper half-circle and one 1-cell as the lower
half-circle and
≈
S1 −
→ Sk1 X.
Then we attach two 2-cells as the upper and lower hemisphere along
≈
the map S 1 −
→ Sk1 , i.e., this gives us
≈
→ S2
Sk2 X −
with Sk1 X ≈ S 1 as the equator of S 2 . Now we continue this procedure
until we reach S n .
Hence, in this cell structure on S n , there are exactly two k-cells in
each dimension k = 0, . . . , n.
• Real projective space RPn is a cell complex with one cell in each
dimension up to n. To show this we proceed inductively. We know that
RP0 consists of a single point, since it is S 0 whose two antipodal points
are identified.
Now we would like to understand how RPn can be constructed from
n−1
RP :
We embed Dn as the upper hemisphere into S n , i.e., we consider
D as {(x0 , . . . ,xn ) ∈ S n : x0 ≥ 0}. Then
n
RPn = S n /(x ∼ −x) = Dn /(x ∼ −x for boundary points x ∈ ∂Dn ).
But ∂Dn is just S n−1 . Hence the quotient map

S n−1 → S n−1 / ∼ = RPn−1
attaches an n-cell en , the open interior of Dn , at RPn−1 .

Thus we obtain RPn from RPn−1 by attaching one n-cell via the
quotient map S n−1 → RPn−1 .
Summarizing, we have shown that RPn is a cell complex with one
cell in each dimension from 0 to n.
• We can continue this process and build the infinite projective space
RP∞ := n RPn . It is a cell complex with one cell S
S
in each dimension.
∞ ∞
We can think of RP as the space of lines in R = n Rn .
• Complex projective space CPn is a cell complex.

Let (z0 : . . . : zn ) denote the homogeneous coordinates of a point in
CP . Let φ : D2n → CPn be given by
n
n−1
X
(z0 , . . . ,zn−1 ) 7→ (z0 : z1 : . . . : zn−1 : 1 − ( |zi2 |)1/2 ).
i=0
Then φ sends ∂D2n to the points with zn = 0, i.e., into CPn−1 .

Let f denote the restriction of φ to S 2n−1 = ∂D2n . Then φ factors
through D2n ∪f CPn−1 , i.e., we get a commutative diagram with an
induced dotted arrow
f
S 2n−1
_ / CPn−1

D2n / CPn−1 ∪f D2n
g
φ '
-
CPn .
The induced map
g : D2n ∪f CPn−1 → CPn
131
Since we can rescale the nth coordinate, this map is bijective. Hence
it is a continuous bijection defined on a compact space. We learned
earlier that this implies that g is a homeomorphism.
We conclude that CPn is a cell complex with exactly one i-cell in
each even dimension up to 2n.
• Again we could continue this process and build infinite complex pro-
jective space CP∞ which is a cell complex with one i-cell in each even
dimension.
Finally, we would like to have a good notion of subspace in a cell complex
which respects the cell structure. It turns out that it is not sufficient to just
require to have a subspace. Though not much more is actually required. For,
a subspace A ⊂ X is subcomplex, or sub-CW-complex, if it is closed and a
union of cells of X.
These conditions imply that A is a cell complex on its own. For, since A
is closed the characteristic maps of each cell of A has image in A and so does
each attaching map. Hence the cells with their characteristic maps which lie in
A provide A with a cell structure.
A more technical definition sounds like this:
Subcomplexes
Let X be a cell complex with attaching maps {fα : Sαn−1 → Skn−1 X : α ∈
Jn , n ≥ 0}.
A subcomplex A of X is a closed subspace A ⊆ X such that for all n ≥ 0,
there is a subset Jn′ ⊂ Jn so that Skn A := A ∩ Skn X turns A into a cell
complex with attaching maps {fβ : β ∈ Jn′ , n ≥ 0}.
A pair (X,A) which consists of a CW-complex X and a subcomplex A is
called a CW-pair.
Examples of CW-pairs are given by

• each skeleton Skn X of a cell complex X;
• RPk ⊂ RPn for every k ≤ n;
• CPk ⊂ CPn for every k ≤ n;

• the spheres S ⊂ S n for every k ≤ n but only for the second cell
k
structure with two i-cells in each dimension.

With the first cell structure on S n with one 0-cell and one n-cell, S k is
not a subcomplex of S n .
The next step is to study the homology of cell complexes...
LECTURE 14
Homology of cell complexes
We are going to show that there is a relatively simple procedure to determine

the homology of a cell complex.
Before we start this endeavour we need an auxiliary result which is a conse-
quence of the excision property of singular homology:
Lemma: Homology after collapsing a subspace

Let A ⊂ X be a subspace. Suppose there is another subspace B of X such
that
(a) Ā ⊆ B ◦ and
(b) A ,→ B is a deformation retract.
Then
∼
=
Hn (X,A) −
→ Hn (X/A,∗)
is an isomorphism for all n.
Proof: We have a commutative diagram
j
(X,A)
i / (X,B) o (X − A, B − A)
k
j̄

(X/A,∗)
ī / (X/A,B/A) o (X/A − ∗, B/A − ∗).
Our goal is to show that the left-hand vertical map induces an isomorphism
in homology. We will achieve this by showing that all the other maps induce
isomorphisms in homology:
• The map k is a homeomorphism of pairs and hence induces an iso-
morphism in homology.
• The map j induces an isomorphism in homology by the assumption (a)
and excision.
133
134 CHAPTER 14. HOMOLOGY OF CELL COMPLEXES
• The map i induces a homomorphism of long exact sequences
··· / Hn (A) / Hn (X) / Hn (X,A) / ···

∼
=

··· / Hn (B) / Hn (X) / Hn (X,B) / ···
By assumption (b), the left-hand vertical arrow is an isomorphism for

all n. By the Five-Lemma this implies that i induces an isomorphism
in homology.
• For the map ī, we observe that the retraction ρ : B → A ,→ B induces
a map ρ̄ : B/A → A/A = ∗ ,→ B/A.
Moreover, the homotopy B × I → B between ρ and the identity of B
is constant on A. Thus it induces a homotopy B/A × I → B/A between
ρ̄ and the identity of B/A.
In other words, ∗ → B/A is a deformation retract. Hence the long
exact sequence and the Five-Lemma imply that ī induces an isomorphism
in homlogy.
• Finally, we have ∗¯ ⊂ (B/A)◦ by definition of the quotient topology.
Hence map j̄ induces an isomorphism in homology by excision.
QED
Corollary: Homology of a bouquet of spheres

For any indexing set J, let us write α∈J Sαk for the quotient
W
a a _
Sαk−1 ,→ Dαk → Sαk .
α∈J α∈J α∈J
The homology of this space, often called bouquet of k-spheres, is given by
(
Z[J] if q = k
Sαk ,∗) ∼
_
Hq ( =
α∈J
0 if q ̸= k
where Z[J] denotes the free abelian group on the set J.
(Note that the relative homology group in the statement is an example of a

reduced homology that we introduced in last week’s exercises.)
Proof: Each summand Sαk−1 is a subspace of Dαk for which there is an open
neighborhood Uα such that Sαk−1 ,→ Uα is a deformation retract (we could even
135
take Uα = Dαn − {0}). Hence we can apply the previous result to conclude
a a ∼
=
_
H∗ ( Dαk , Sαk−1 ) −
→ H∗ ( Sαk ,∗).
α α α
Hence we are reduced to calculate the relative homology on the left-hand side.
To do this, we can apply the long exact sequence of a pair to deduce that
a a ∼
=
a
∂ : Hq ( Dαk , Sαk−1 ) −
→ Hq−1 ( Sαk−1 ,∗)
α α α
is an
Lisomorphism for all q. Finally, we know that the latter group is isomorphic
to α∈J Z = Z[J] when q = k and 0 otherwise. □
Now we would like to apply this observation to a cell complex X. If we write
Xk = Skk X for the k-skeleton of X, then we get the following commutative
diagram
` k−1
/ k /
` W k
(22) α Sα α Dα α Sα
f φ φ̄

Xk−1 / Xk = Xk−1 ∪f ( α Dαk ) /
`
Xk /Xk−1 .
where the right-hand vertical map is induced by φ and taking quotients. Since
the restriction of φ to the open interior of the n-disks is a homeomorphism
onto its image, this implies that the dotted arrow φ̄ is a homeomorphism.
Hence we deduce from the previous result on bouquets of spheres:
(
Z[Jn ] if q = k
Hq (Xk , Xk−1 ) ∼
= Hq (Xk /Xk−1 ,∗) ∼
=
0 if q ̸= k
where Jn denotes the indexing set of the attached k-cells.
In other words, the relative homology group Hk (Xk , Xk−1 ) keeps track of
the k-cells of X.
This group will play a crucial role for us today. Let us analyse some conse-
quences of what we have found out about this group.
Let us look at a piece of the long exact sequence of the pair (Xk ,Xk−1 ):
Hq+1 (Xk , Xk−1 ) → Hq (Xk−1 ) → Hq (Xk ) → Hq (Xk , X k−1 ).
For q ̸= k, the last term Hq (Xk , Xk−1 ) = 0 vanishes and hence the map
Hq (Xk−1 ) → Hq (Xk ) is surjective.
For q ̸= k − 1, the first term Hq+1 (Xk , Xk−1 ) = 0 vanishes and hence the map
Hq (Xk−1 ) → Hq (Xk ) is injective.
Hence we have shown that the inclusion Xk−1 ,→ Xk induces an isomorphism

∼
=
(23) Hq (Xk−1 ) −
→ Hq (Xk ) for q ̸= k, k − 1.
Hence, for a fixed q > 0, we can observe how Hq (Xk ) varies when we let Xk
go through all skeleta of X:
• Hq (X0 ) = 0 since X0 is a discrete set and the higher homology groups
of points vanish.
• For k = 0, . . . , q − 1, Hq (Xk ) = 0 remains trivial by (23).
• As a consequence, we observe that Hn (Xk ) = 0 whenever n > k.
• For k = q, Hq (Xq ) is a subgroup of the free abelian group Hq (Xq , Xq−1 ),

and therefore it is free abelian as well.
• For k = q + 1, Hq (Xq+1 ) may not be free anymore, i.e., there might be

relations induced by the exact sequence
Hq+1 (Xq+1 , Xq ) → Hq (Xq ) → Hq (Xq+1 ) → 0.
• For k ≥ q + 1, Hq (Xk ) remains stable, i.e., the inclusions of skeleta

induce a sequence of isomorphisms
∼
= ∼
=
Hq (Xq+1 ) −
→ Hq (Xq+2 ) −
→ ··· .
• If X is finite-dimensional, there is a d such that X = Xd . The above

sequence of isomorphisms then implies the inclusion Xq+1 ,→ X induces
an isomorphism
Hq (Xk ) ∼
= Hq (X) for q < k.
∼
=
• Still, for X finite-dimensional, since Hq (Xq+1 ) −
→ Hq (X) and since
Hq (Xq ) → Hq (Xq+1 ) → Hq (Xq+1 , Xq ) = 0
is exact, we see that
Hq (Xq ) → Hq (X) is surjective.
137
• If X is infinite-dimensional, the group Hq (Xk ) still maps isomorphi-

cally into Hq (X) for q < k. For, the image of a standard simplex is
compact and therefore lands in a finite subcomplex. Hence the union
of the images of a finite collection of standard simplices is still compact
and therefore also lands in a finite subcomplex. Hence it lands in a
finite skeleton. Thus any q-chain in X is the image of a chain in a
finite skeleton. For the same reason, if c ∈ Sq (X) is a boundary, then
it is a boundary in Sq (Xm ) for some m ≥ q.
• In summary, all the q-dimensional homology of X is created in the q-

skeleton Xq , and all the relations in Hq (X) occur in the q + 1-skeleton
Xq+1 .
The key points of this discussion are:
Proposition: The homology is governed by the skeleta

For any k, q ≥ 0 and cell complex X, we have
• Hq (Xk ) = 0 for k < q and
∼
=
• Hq (Xk ) −→ Hq (X) for k > q.
In particular, Hq (X) = 0 if q is bigger than the dimension of the cell complex
X.
Now we would like to find an efficient way to calculate the homology of our
cell complex X. Apparently, the group Hn (Xn , Xn−1 ) carries crucial information
about X. Therefore, we are going to give it a new name:
Cellular n-chains
The group of cellular n-chains in a cell complex X is defined to be
Cn (X) := Hn (Xn , Xn−1 ).
We claim that these groups sit inside a sequence of homomorphisms who form
a chain complex. The differential
dn : Cn (X) → Cn−1 (X)

is defined as the composite
dn
Cn (X) = Hn (Xn , Xn−1 ) / Hn−1 (Xn−1 , Xn−2 ) = Cn−1 (X)
4
∂n ) jn−1
Hn−1 (Xn−1 )
where ∂n is the connecting homomorphism in the long exact sequence of pairs and
jn−1 is the homomorphism induced by the inclusion (Xn−1 , ∅) ,→ (Xn−1 , Xn−2 ).
To show that dn ◦ dn+1 = 0 we consider the commutative diagram:
Cn+1 (X) = Hn+1 (Xn+1 , Xn ) Hn−1 (Xn−2 ) = 0

dn+1
∂n+1
+
Hn (Xn )
jn ∂n
/ Cn (X) = Hn (Xn , Xn−1 ) /1 Hn−1 (X
n−1 ) _
=0 dn
jn−1
+
Hn (Xn+1 ) Cn−1 (X) = Hn−1 (Xn−1 , Xn−2 )

0 = Hn (Xn+1 , Xn )
Since j and ∂ are part of long exact sequences, we know j ◦ ∂ = 0 and get
dn ◦ dn+1 = (jn−1 ◦ ∂n ) ◦ (jn ◦ ∂n+1 ) = 0.
Cellular chain complex

Thus (C∗ (X), d) is a chain complex. It is called the cellular chain com-
plex.
Now we would like to determine the homology of this chain complex.

• To do this we need to understand the kernel of d:
Ker (dn ) = Ker (jn−1 ◦ ∂n ).
Since jn−1 is injective, we get

Ker (dn ) = Ker (∂n ) = Im (jn ) = Hn (Xn )
where the middle identity is implied by the exactness of the long exact se-
quence these maps are part of, and the last identity is implied by the fact that
jn : Hn (Xn ) → Hn (Xn , Xn−1 ) is injective.
139
• For the image of d, we use again that jn is injective and get

Im (dn+1 ) = jn (Im (∂n+1 )) ∼
= Im (∂n+1 ) ⊆ Hn (Xn ).
Since the left-hand column in the above big diagram is exact, we know
Hn (Xn )/Im (∂n+1 ) ∼
= Hn (Xn+1 ).
In other words, we just proved:

Hn (C∗ (X)) = Hn (Xn )/Im (∂n+1 ) ∼
= Hn (Xn+1 ).
But we had already showed Hn (Xn+1 ) ∼

= Hn (X). Hence we proved the fol-
lowing important result:
Theorem: Cellular Homology

For a cell complex X, there is an isomorphism
H∗ (C∗ (X)) ∼= H∗ (X)
which is functorial with respect to filtration-preserving maps between
cell complexes.
In this theorem we are referring to maps which preserve the skeleton structure
of cell complexes. We should better make this concept precise:
Maps between cell complexes

• A filtration on a space X is a sequence of subspaces
X0 ⊆ X1 ⊆ . . . ⊆ Xn ⊆ Xn+1 ⊆ . . . ⊆ X.
such that X can be written as the union of these subspaces. If X
is a space together with a filtration, we call X a filtered space.
• For example, every cell complex has a filtration by its skeleta.
• Let X and Y be filtered spaces. A continuous map f : X → Y is
called filtration-preserving if f (Xp ) ⊂ Yp for all p.
• A map between cell complexes is called cellular if it preserves the
filtration by skeleta.
In other words, if we are given two cell complexes and care about their cell
structure, we should only consider filtration-preserving maps between them.
An immediate and very useful consequence of the above theorem is:
Corollary: Homology of even cell complexes

Let X be a cell complex with only even cells, i.e., the inclusion X2k ,→ X2k+1
is an isomorphism for all k. Then
H∗ (X) ∼= C∗ (X).
In particular, Hn (X) is free abelian for all n, Hn (X) = 0 for odd n, and the
rank of Hn (X) for even n is the number of n-cells.
For example, recall that complex projective n-space CPn has exactly one cell
in each even dimension up to 2n. Hence as an application we can read off the
homology of complex projective space:
(
Z for 0 ≤ k ≤ 2n and k even
Hk (CPn ) =
0 for k odd.
Note to the theorem and corollary

We should keep in mind that the homology of X is independent of any cell
structures. We defined it long before we knew that cell complexes exist.
The theorem shows that knowing a cell structure on X can nevertheless be
very helpful for computing H∗ (X).
Moreover, we learned that the cell structure on any given cell complex may
not be unique. We saw for example two different cell structures on S n .
However, the theorem tells us that any cell structure one can construct on X
has to obey certain constraints what are induced by the fact the homology
of the cellular chain complex is H∗ (X).
LECTURE 15
Computations of cell homologies and Euler characteristic
• Homology of real projective n-space

As another application of the theorem on the cellular chain complex and
the homology of cell complexes we are going to compute the homology of real
projective space.
First of all, recall that attaching and characteristic maps assemble to a com-
mutative diagram
( Dn , S n−1 ) / (Xn , Xn−1 )
` `
(W
n
( S , ∗).
We have shown that all these maps induce isomorphisms in homology. In partic-
ular,
a a ∼
=
(24) Hn ( Dn , S n−1 ) −
→ Cn (X) = Hn (Xn , Xn−1 ).
We are now going to exploit this fact for the computation of H∗ (RPn ).
Recall that the cell structure of RPn is such that
• Skk (RPn ) = RPk and
• there is exactly one k-cell in each dimension k = 0, . . . ,n. We denote
this k-cell by ek .
Hence the cellular chain complex looks like this:
dn d2 d1
0 / Cn (RPn ) / ··· / C1 (RPn ) / C0 (RPn ) / 0
0 / Z[en ] / ··· / Z[e1 ] / Z[e0 ] / 0

dn d2 d1
141
142
CHAPTER 15. COMPUTATIONS OF CELL HOMOLOGIES AND EULER CHARACTERISTIC
In order to compute the homology of this chain complex, we need to determine
the differentials dn :
• We know that H0 (RPn ) is Z = Z[e0 ]. That implies that the differential
d1 must be trivial.
• For k > 1, the differential dk is defined as the top row in the following
commutative diagram
∂k jn−1
Ck (RPn ) = Hk (RPk , RPk−1 ) / Hk−1 (RPk−1 ) / Hk−1 (RPk−1 , RPk−2 ) = Ck−1 (RPn )
O O O
∼
= Hk−1 (π) ∼
=
∂k
Hk (Dk , S k−1 ) / Hk−1 (S k−1 ) / Hk−1 (S k−1 ,∗)
The map π : S k−1 → RPk−1 is the attaching map of the k-cell in

n
RP . The outer vertical maps are isomorphisms by our discussion of
diagram (24). We also know that the lower differential ∂k is an isomor-
phism by our original calculation of the homology of the sphere.
Hence, in order to understand the effect of the differential dk , we
need to understand the effect of the maps in the following commutative
diagram:
q
(25) RPOk−1 / RPk−1 /RPk−2
π ≈

S k−1 / S k−1
g
where q is the quotient map. In other words, we need to calculate the

degree of the lower horizontal map g.
The composite
π q
S k−1 →
− RPk−1 →
− RPk−1 /RPk−2
pinches the subspace S k−2 ⊂ S k−1 to a point.
Hence the lower horizontal map g in (25) is given by
q
S k−1
π / RPk−1 / S k−1
5
µ
pinch equator )
S k−1 /S k−2 = S−k−1 ∨ S+k−1 .
143
To determine the effect of µ, we observe that the subspace S k−1 −S k−2

consists of two components. Let us denote these two components by
(S k−1 − S k−2 )+ and (S k−1 − S k−2 )− , respectively. The restriction of
q ◦ π to each component of S k−1 − S k−2 is a homeomorphism onto
RPk−1 − RPk−2 .
Let us write (q ◦ π)+ and (q ◦ π)− for the restrictions of q ◦ π to the
subspaces (S k−1 − S k−2 )+ and (S k−1 − S k−2 )− , respectively:
(S k−1 − S k−2 )+
(q◦π)+
)
α RPk−1
5 − RP
k−2
(q◦π)−
(S k−1 − S k−2 )−
By definition of RPk−1 , both (q ◦ π)+ and (q ◦ π)− are homeomor-

phisms and they differ by precomposing with the antipodal map.
Hence the map µ is the identity on one copy of S k−1 and the an-
tipodal map α on the other copy of S k−1 .
Thus the effect of g on homology is given by
Hk−1 (g) : Hk−1 (S k−1 ) → Hk−1 (S k−1 ), σ 7→ σ + Hk−1 (α)(σ).
But we know what the effect of Hk−1 (α) is. Namely, it is given by
Hk−1 (α) = (−1)k−1 . Hence
(
2 if k is even
Hk−1 (g) = 1 + (−1)k =
0 if k is odd.
144
Summarizing, we have shown that the cellular chain complex of RPn looks
like:
2 0 0 2 0
0→Z→
− Z→
− ··· →
− Z→
− Z→
− Z → 0 if n is even
0 2 0 2 0
0→Z→
− Z→
− ··· →
− Z→
− Z→
− Z → 0 if n is odd
where the left-hand copy of Z is in dimension n and the right-hand one is in

dimension 0.
And in words: in real projective space, odd cells create new generators,
whereas even cells create torsion (except for the zero-cell) in the previous
dimension.
Homomology of RPn
The homology of real projective n-space is given by


 Z k=0

Z k = n is odd
Hk (RPn ) =

 Z/2 0 < k < n and k is odd

0 otherwise.

• What homology sees and does not see

The example of RPn indicates what kind of structure of a cell complex singular
homology can detect and what it cannot detect and also how we can calculate
the differential in the cellular chain complex.
Let X be a cell complex. Its cell structure is determined by attaching maps
a
Sαn−1 → Xn−1 .
Knowing these maps, up to homotopy, determines the homotopy type of the cell
complex X.
However, homology does not record all of the information of the attaching
maps. For, homology only sees the effect of the composite obtained by pinching
145
out Xn−1 :
Sαn−1 / /
`
α∈Jn Xn−1 Xn−1 /Xn−2
≈
W' w
β∈Jn−1 Sβn−1 .
In other words, homology only records what is going on modulo sub-

skeleta. However, we will see now that homology does a pretty good job at this
recording.
Let us try to understand this picture a bit better. For each α, the left-hand
diagonal map can be described as the composite
fα qαβ
Sαn−1 −→ Xn−1 −−→ Xn−1 /(Xn−1 − enβ ) =: Sβn−1
where qαβ is the quotient map. Moreover, we identify the quotient Xn−1 /(Xn−1 −
enβ ) with the boundary Sβn−1 of the cell enβ . Note that this map might be trivial
from some (or all) β.
The sum of the effect of these maps in homology is actually the differential
dn in the cellular chain complex. For we have a commutative diagram
P
∂n Hn−1 (qαβ ◦fα )
Hn ( α Sαn−1 ,∗) / Hn−1 ( α Sαn−1 )
α,β
/ Hn ( β Sβn−1 ,∗)
W ` W
Hn (φ̄) ∼
= ∼
=

Hn (Xn , Xn−1 ) / Hn−1 (Xn−1 ) / Hn−1 (Xn−1 , Xn−2 ).
∂n jn−1 4
dn
We conclude from this discussion:
Cellular differentials are sums of degrees

With the above notations, the effect of the cellular differential on the
generator in Cn (X) which corresponds to the cell enα in X is given by the
sum of degrees
X X
dn ([enα ]) = Hn−1 (qαβ ◦ fα )([enα ]) = deg(qαβ ◦ fα ) · [enα ].
β β
In other words, in order to compute the cellular differential we need to calcu-

late the degrees of various maps.
146
• Euler characteristic of cell complexes
Euler characteristic of finite CW -complexes

Let X be a finite cell complex. Let ck denote the number of k-cells in X.
Then the Euler characteristic of X is defined to be the integer given by
the finite sum
X
χ(X) = (−1)k ck .
k
The main result on χ(X) we are going to prove today is that it only de-
pends on the homotopy type of X and is, in particular, independent of the
given cell structure of X. We are going to prove this by showing that χ(X)
can be computed using the singular homology of X.
Recall that we have seen an Euler number for polyhedra in the first lecture. It
was defined as the number of vertices minus the number of edges plus the number
of faces. This fits well with the above definition for a finite cell complex.
For, if we assume the invariance of χ for a moment, then we get χ(S 2 ) = 2
using the standard cell structure on S 2 , i.e., one 0-cell and one 2-cell. This
implies that Euler’s polyhedra formula holds.
Corollary: Euler’s polyhedra formula

For any cell structure on the 2-sphere S 2 with F 2-cells, E 1-cells and V
0-cells, we have the formula
F − E + V = 2.
As a preparation, we recall some facts about abelian groups.

Let A be an abelian group. Recall that the set of torsion elements is defined
as
Torsion(A) = {a ∈ A : na = 0 for some n ̸= 0}.
This set is in fact a subgroup of A. A group is called torsion-free if Torsion(A) =

0. The quotient A/Torsion(A) is always torsion-free.
Now we assume that A is finitely generated. Then Torsion(A) is a finite
abelian group and A/Torsion(A) is a finitely generated free abelian group and
147
therefore isomorphic to Zr for some integer r. The number r is called the rank
of A denoted by rank(A).
In fact, by choosing generators of A/Torsion(A), we can construct a homomo-
prhism A/Torsion(A) → A which splits the projection map A → A/Torsion(A).
Thus if A is finitely generated abelian, then
A∼
= Torsion(A) ⊕ Zr .
We are going to use the following lemma from elementary algebra without
proving it:
Lemma: Ranks in exact sequences

• Let 0 → A → B → C → 0 be a short exact sequence of finitely generated
abelian groups. Then the ranks of these groups satisfy
rank(B) = rank(A) + rank(C).
• More generally, for a long exact sequence of finitely generated abelian
groups
0 → An → An−1 → . . . → A1 → A0 → 0
the ranks satisfy
n
X
0= (−1)i rank(Ai ).
i=0
Now we are euqipped for the proof of the above mentioned result:
Theorem: Euler characteristic via homology

Let X be a finite cell complex. Then the Euler characteristic of X satisfies
X
χ(X) = (−1)k rank(Hk (X)).
k
Proof: Let ck be again the number of k-cells in the given finite cell structure
of X. Let C∗ := C∗ (X) denote the cellular chain complex of X. To simplify
the notation let us denote by Z∗ , B∗ , and H∗ the cycles, boundaries and homology,
respectively, in this complex.
148
By their definition, they fit into two short exact sequences:
0 → Zk → Ck → Bk−1 → 0
and
0 → Bk → Zk → Hk → 0,
By our previous study of the cellular chain complex, we know

ck = rank(Ck ).
Hence, using the above discussion, we can rewrite χ(X) as follows:
X
χ(X) = (−1)k rank(Ck )
k
X
= (−1)k (rank(Zk ) + rank(Bk−1 ))
k
X
= (−1)k (rank(Bk ) + rank(Hk ) + rank(Bk−1 )).
k
When we take the sum over all k, the summands rank(Bk ) and rank(Bk−1 )
will cancel out. Thus we get
X
χ(X) = (−1)k rank(Hk ).
k
But by the theorem on the homology of the cellular chain complex, Hk is exactly
the singular homology group Hk (X) of X. QED
Note that the numbers rank(Hk (X) are called the Betti numbers of X.
They had already played an important role in mathematics, before homology
groups had been systematically developed. As the theorem shows, these numbers
are an interesting invariant of a space.
The description of the Euler number in the theorem now generalizes easily:
Definiton: Euler characteristic revisited

Let X be a topological space such that each Hn (X) has finite rank and
that there is an d such that Hn (X) = 0 for all n > d. Then the Euler
characteristic of X is defined to be the integer given by the finite sum
X
χ(X) = (−1)k rank(Hk (X)).
k
149
• Designing cell complexes

For example, for m ∈ Z, we can easily construct a space X with Hn (X) = Z/m
and H̃i (X) = 0 for i ̸= n. We start with S n and attach an n + 1-cell to it via a
map f : S n → S n of degree m. The cellular chain complex of this space is
m
0→Z−
→Z→0
with the copies of Z in dimensions n + 1 and n, respectively. The homology of
this space is exactly what we wanted.
This procedure can easily be generalized.
Theorem: Moore spaces

Let A∗ be any graded abelian group with An = 0 for n < 0. Then there
exists a cell complex X with H̃∗ (X) = A∗ .
We are going to prove this result in the next lecture.

LECTURE 16
Designing homology groups and homology with

coefficients
• Designing cell complexes

We announced last time that cell complexes enable us to design spaces with
prescribed homology groups. We are going to prove this result today.
We will need a construction on spaces that we have already used in special
cases.
Recall that the wedge X ∨ Y of two pointed spaces (X,x) and (Y,y) is
defined as the quotient of X ⊔ Y modulo x ∼ y, i.e., the disjoint union with x and
y identified. We can think of X ∨ Y glued together at the joint point [x] = [y].
This generalizes the wedge of spheres that we have seen before. This construction
generalizes to infinite wedges.
If each point xα is a deformation retract of a neighborhood Uα in Xα , then
the wedge satisfies a formula for reduced homology that we are used to for the
homology of disjoint unions:
H̃∗ ( Xα ) ∼
_ M
= H̃∗ (Xα ).
α α
Now we can prove the following result:
Theorem: Moore spaces

Let A∗ be any graded abelian group with An = 0 for n < 0. Then there
exists a cell complex X with H̃∗ (X) = A∗ .
Proof: Let us start with just a single abelian group A. By choosing generators
for A, we can define a surjective homomorphism
F0 → A
151
152
CHAPTER 16. DESIGNING HOMOLOGY GROUPS AND HOMOLOGY WITH COEFFICIENTS
from a free abelian group F0 . The kernel of this homomorphism, denoted by
F1 , is also free, since it is a subgroup of a free abelian group.
We write J0 for a minimal set of generators of F0 such that we have a surjection
F0 → A and J1 for a minimal set of generators of F1 .
For n ≥ 1, we define Xn to be
_
Xn := Sαn .
α∈J0
The nth homology of Xn is Hn (Xn ) = Z[J0 ].

Now we are going to define an attaching map
a
f: Sβn → Xn
β∈J1
by specifying it on each summand Sβn .

In F0 , we can write the generator β of F1 as a linear combination of the
generators of F0
s
X
β= ni α i .
i=1
We can reproduce this relation in topology. For, let

s
_
Sn → Sαni
i=1
be the map obtained by pinching s − 1 circles on S n to points. The effect of this

map in homology is to send the generator in Hn (S n ) to the s-tuple of generators
in Hn (Sαni ):
s
_ s
M
n
Hn (S ) → Hn ( Sαni ) = Hn (Sαni ), 1 7→ (1, . . . ,1).
i=1 i=1
For each i, we choose a map Sαni → Sαni of degree ni .

The map on the summand Sβn is now defined as the composite
s
_ _
Sβn → Sαni → Sαn .
i=1 α
153
Taking the disjoint union of all these maps as attaching maps, we get a cell
complex X whose cellular chain complex looks like
0 → F1 → F0 → 0
with F0 in dimension n and F1 in dimension n + 1, and whose homology is
(
A for q = n
H̃q (X) =
0 for q ̸= n.
We write M (A, n) for the CW -complex produced this way and call it a Moore
space of type A and n.
Finally, for a graded abelian group A∗ as in the theorem, we define X to
be the wedge of all the M (An , n). QED
Moore spaces are not functorial

It is important to note that the construction of Moore spaces cannot be
turned into a functor Ab → hoTop. This might surprise at first glance.
For given a homomorphism g : A → B we can construct a continuous map
γ : M (A, n) → M (B, n) such that Hn (γ) = g.
However, this construction depends on the various choices we make.
That means that for homomorphisms
g1 g2
A−
→ B, and B −
→C
we cannot guarantee that γ2 ◦ γ1 is the same map as the one we would have
constructed by starting with g2 ◦ g1 : A → C directly.
Despite this caveat, we have witnessed an important phenomenon that still

motivates a l ot of exciting research:
From Algebra to Topology

The proof demonstrates a common phenomenon in Algebraic Topology.
Whereas our initial goal was to translate topology into algebra, now we
went in the opposite direction. Starting with an algebraic structure we
modelled a space whose homology reproduces the algebra. Since
being an abelian group is not the only the algebraic structure and homology
not the only invariant out there, we can imagine that the above theorem is
only a first glance at the makings of a huge mathematical industry.
154
The proof also shows why Topology is particularly well suited for this
endeavour. The gluing construction we used for building cell complexes
is unique for topological spaces. Requiring any additional structure
usually stops us from producing cell complexes.
For example, there are no cell complexes of smooth manifolds or
algebraic varieties. Nevertheless, there are some inventive procedures to
remedy this defect...
• Homology with coefficients

Now it is time to move on and to develop new algebraic invariants which add to
the information we get from singular homology, or possibly simplify computations.
Recall that homology produces abelian groups. As nice as abelian groups are,
it would be good to have additional structure, for example as vector spaces over
a field. So one might wonder if there is a version of singular homology with
values in the category of vector spaces over a field, or more generally the
category of modules over a ring.
Actually, if R is a ring (with unit and commutative), there is an obvious
candidate for such a theory: We define
Sn (X; R) := RSingn (X)
to be the free R-module over the set Singn (X) of n-simplices. What we have
done so far, was the special case R = Z.
Now we can use the face maps and the same formula we had before for defining
a boundary operator
X XX
(26) ∂n : Sn (X; R) → Sn−1 (X; R), rj σj 7→ (−1)i rj (σj ◦ ϕni ).
j j i
which is now a homomorphism of R-modules. The same calculations as before

yield ∂ ◦ ∂ = 0.
Now we can form the homology as usual
Ker (∂n : Sn (X; R) → Sn−1 (X; R))
Hn (X; R) := .
Im (∂n+1 : Sn+1 (X; R) → Sn (X; R))
For each n ≥ 0, Hn (X; R) is an R-module and is called the singular homology
of X with coefficients in R.
155
More generally, if M is any abelian group, we can form the tensor product
M
Sn (X; M ) = Sn (X) ⊗Z M = M
σ∈Singn (X)
( )
X
= mj σj : σj ∈ Singn (X), mj ∈ M .
j
The boundary operator is defined as before by
∂nM : Sn (X; M ) → Sn−1 (X; M ), ∂nM = ∂n ⊗ 1.
More explicitly, ∂nM is given by the formula in (26) with rj s replaced with mj s.
Since ∂ ◦ ∂ = 0, we get ∂ M ◦ ∂ M = 0.
Homoglogy with coefficients

For a pair of spaces (X, A), we define singular homology of (X,A) with
coefficients in M Hn (X,A; M ) to be the nth homology of the chain com-
plex
S∗ (X; M )
S∗ (X,A; M ) := .
S∗ (A; M )
Homology with coefficients is functorial: That is, a map of pairs
f : (X,A) → (Y,B)
induces, by composing simplices with f , a homomorphism
f∗ : Hn (X,A; M ) → Hn (Y,B; M ) for all n ≥ 0
which we denote just by f∗ to keep the notation simple. Moreover, we have

(g ◦ f )∗ = g∗ ◦ f∗ .
Note if M = R is a ring, this is the same definition as above, and for M = Z we
recover Hn (X,A; Z) = Hn (X,A). We will often refer to these groups as integral
homology groups.
If M is an R-module, then the groups Hn (X,A; M ) have the additional struc-
ture as an R-module itself.
156
Eilenberg-Steenrod Axioms are satisfied

Singular homology with coefficients in M satisfies the Eilenberg-Steenrod
axioms with the only modification
(
M for n = 0
Hn (pt; M ) =
0 for n > 0.
Since everything we proved for singular homology was based on these proper-
ties, we can transfer basically all our work to homology with coefficients.
Let us point out two crucial facts:
• The calculations for spheres can be transfered and we get
(
n M for k = n
H̃k (S ; M ) =
0 otherwise.
• If X is a cell complex, there is a ceullular chain complex

M
C∗ (X; M ) with Cn (X; M ) = M
en
α
where the sum is taken over the n-cells of X. As for M = Z, the nth
homology of C∗ (X; M ) is isomorphic to Hn (X; M ).
The reduced homology groups H̃n (X; M ) with coefficients in M are defined
as the homoogy groups of the augmented chain complex
ϵ
. . . → S1 (X; M ) → S0 (X; M ) →
− M →0
P P
where ϵ is the homomorphism which sends j mj σj to j mj ∈ M .
For a homomorphism of groups φ : M → N there is an induced morphism
of chain complexes S∗ (X,A; M ) → S∗ (X,A; N ) which induces a homomorphism
in homology
φ∗ : H∗ (X,A; M ) → H∗ (X,A; N ).
This homomorphism is compatible with f∗ for maps of pairs and with long exact
sequences of pairs.
For the calculations using the cellular chain complex, we need to check the
following lemma:
157
Lemma: Degrees with coefficients

Let f : S n → S n be a map of degree k. Then f∗ : H̃n (S n ; M ) → H̃n (S n ; M )
is given by multiplication with k, where k denotes the image of k in M .
Proof: Let φ : Z → M be a homomorphism of groups (0Z 7→ 0M ) which sends

1 ∈ Z to an element m ∈ M . Then the assertion follows from the commutativity
of the diagram
f∗
Z Hn (S n ; Z) / Hn (S n ; Z) Z
φ φ∗ φ∗ φ

M Hn (S n ; M ) / Hn (S n ; M ) M.
f∗
That the outer diagram commutes follows from the way we compute the homology
groups of S n with coefficients Z and M via the Eilenberg-Steenrod axioms. QED
• Why coefficients?
The coefficients that are most often used are the fields Fp , for a prime number
p, and the field Q, R and sometimes C.
In order to get an idea of what happens when we use different coefficients, let
us look at the homology of RPn for R = F2 . We use the cellular chain complex
which looks like this
0 → F2 → F2 → . . . → F2 → F2 → 0.
We showed that the differentials alternated between multiplication by 2 and 0.
But in F2 , 2 = 0 which means that all differentials vanish and we get
(
F2 for 0 ≤ k ≤ n
Hk (RPn ; F2 ) =
0 otherwise.
We learn from this example that

• The calculation of F2 -homology can be particularly easy, and it might
see more nontrivial groups than integral homology.
• Nevertheless, F2 -homology is often sufficient to distinguish between triv-

ial and nontrivial spaces or maps.
158
The situation is quite different if we take R to be any field of characteristic
different from two. Then the cellular chain complex of RPn looks like (with
the left-hand copy of R in dimension n)
∼
= 0 ∼
= ∼
= 0
0→R−
→R→
− R−
→ ... −
→R→
− R→0
for n even, and
0 ∼
= 0 ∼
= 0
0→R→
− R−
→R→
− ... −
→R→
− R→0
for n odd.
Thus, for n even, we get
(
R for k = 0
Hk (RPn ; R) =
0 otherwise
and, for n odd,

R for k = 0

n
Hk (RP ; R) = R for k = n

0 otherwise.
In other words, away from 2, real projective n-space looks for R-homology
like a point if n is even and like an n-sphere if n is odd.
This teaches us already that different coefficients can tell quite different
stories.
This notwithstanding one might wonder whether integral homology is the
finest invariant and all other homologies are just coarser variations. This is not
the case, and it is indeed possible that homology with coefficients detects
more than integral homology. Let us look at an example:
Example: When Z/m-homology sees more

Let X = M (Z/m,n) be a Moore space we constructed in the previous
lecture: We start with an n-sphere S n and form X by attaching an n + 1-
dimensional cell to it via a map f : S n → S n of degree m
X = S n ∪f Dn+1 .
Let
q : X → X/S n ≈ S n+1
159
be the quotient map. It induces a trivial homomorphism in reduced

integral homology. For, the only nontrivial homology occurs in degrees n
and n + 1 where we have
q∗
→ H̃n+1 (X/S n ; Z)
H̃n+1 (X; Z) = 0 −
and
q∗
→ H̃n+1 (X/S n ; Z) = H̃n (S n+1 ; Z) = 0.
H̃n (X; Z) −
Hence integral homology cannot distinguish between the quotient map
and a constant map.
However, Z/m-homology does see the difference between q and a
constant map.
For, the Z/m-cellular chain complex of X is
m
0 → Z/m −
→ Z/m → 0
with copies of Z/m in dimensions n + 1 and n. Thus, the Z/m-homology
of X is 
Z/m if k = n + 1

H̃k (X; Z/m) = Z/m if k = n

0 otherwise.
The long exact sequence of the pair (X,S n ) in dimension n + 1 then yields
q∗
→ H̃n+1 (X/S n ; Z/m).
0 = H̃n+1 (S n ; Z/m) → H̃n+1 (X; Z/m) −
Since the left-hand group is 0, exactness implies that q∗ is injective and
hence nontrivial, since both H̃n+1 (X; Z/m) and H̃n+1 (X/S n ; Z/m) are
isomorphic to Z/m.
This example demonstrates that homology groups with coefficients are similar,
but often a bit different than integral homology groups. This raises the question
how different they can be. More generally, we could ask:
Question
Given an R-module M H∗ (X; R), what can we deduce about H∗ (X; M )?
For example, let M be the Z-module Z/m. One might wonder if Hn (X; Z/m)
is just the quotient Hn (X; Z)/mHn (X; Z), since the latter is isomorphic to the
tensor product Hn (X; Z) ⊗ Z/m.
160
But we have to be careful. For, we do have a short exact sequence of chain
complexes
m
0 → S∗ (X; Z) −
→ S∗ (X; Z) → S∗ (X; Z/m) → 0.
Such a short exact sequence induces a long exact sequence of the respective
homology groups a part of which looks like
m m
Hn (X; Z) −
→ Hn (X; Z) → Hn (X; Z/m) → Hn−1 (X; Z) −
→ Hn−1 (X; Z).
Using the exactness of this sequence yields a short exact sequence
(27)
0 → Hn (X; Z)/mHn (X; Z) → Hn (X; Z/m) → m-Torsion(Hn−1 (X; Z)) → 0
where m-Torsion(Hn−1 (X; Z)) denotes the m-torsion, i.e., the kernel of the map
m
Hn−1 (X; Z) −
→ Hn−1 (X; Z) given by multiplication by m.
In fact, the short exact sequence (27) provides a tool to determine Hn (X; Z/m)
when we know both Hn (X; Z) and Hn−1 (X; Z). However, in general, we will
need a more sophisticated method to understand the relationship of H∗ (X; Z)⊗
M and H∗ (X; M ).
As a first generalization, we have the following result:
Long exact sequence of coefficients

Assume we have a short exact sequence of abelian groups
0 → M ′ → M → M ′′ → 0.
For any pair of spaces (X,A), there is an induced short exact sequence of
chain complexes
0 → S∗ (X,A; M ′ ) → S∗ (X,A; M ) → S∗ (X,A; M ′′ ) → 0.
Such a short exact sequence induces a long exact sequence
··· / Hn+1 (X,A; M ′′ )
∂
r
Hn (X,A; M ′ ) / Hn (X,A; M ) / Hn (X,A; M ′′ )
∂
r
Hn−1 (X,A; M ′ ) / ···
LECTURE 17
Tensor products, Tor and the Universal Coefficient

Theorem
Our goal for this lecture is to prove the Universal Coefficient Theorem
for singular homology with coefficients. This will require some preparations in
homological algebra. For some this will be a review. Though to keep everybody
on board, this is what we have to do.
We will not treat the most general cases, but rather focus on the main ideas.
Any text book in homological algebra will provide more general results.
• Tensor products
Let A and B be abelian groups. We would like to combine A and B into just
one object, denoted A ⊗ B, in such a way that having a bilinear homomorphism
f: A×B →C
is the same as having a homomorphism from A ⊗ B into C.
That f is bilinear means
f (a1 + a2 ,b) = f (a1 ,b) + f (a2 ,b)
f (a,b1 + b2 ) = f (a,b1 ) + f (a,b2 ).
We can achieve this by brute force.
Tensor product
For, we can construct A ⊗ B as the quotient of the free abelian group
generated by the set A × B modulo the subgroup generated by {(a + a′ ,b) −
(a,b) − (a′ ,b)} and {(a,b + b′ ) − (a,b) − (a,b′ )} for all a,a′ ∈ A and b,b′ ∈ B.
We denote the equivalence class of (a,b) in A ⊗ B by a ⊗ b.
We call A ⊗ B the tensor product of A and B.
Let us collect some immediate observations:

161
CHAPTER
162 17. TENSOR PRODUCTS, TOR AND THE UNIVERSAL COEFFICIENT THEOREM
• For any a ∈ A, b ∈ B and any integer n ∈ Z, the relations imply
n(a ⊗ b) = (na) ⊗ b = a ⊗ (nb).
• The abelian group A ⊗ B is generated by elements a ⊗ b with a ∈ A and
b ∈ B.
• P
Elements in the abelian group A⊗B are finite sums of equivalence classes
m
i=1 ni (ai ⊗ bi ).
• The tensor product is symmetric up to isomorphism with isomorphism
given by
m m
∼
=
X X
A⊗B − → B ⊗ A, ni ai ⊗ bi 7→ n i b i ⊗ ai .
i=1 i=1
• The tensor product is associative up to isomorphism:

A ⊗ (B ⊗ C) ∼ = (A ⊗ B) ⊗ C.
• For homomorphisms f : A → A′ and g : B → B ′ , there is an induced
homomorphism
f ⊗ g : A ⊗ B → A′ ⊗ B ′ , (f ⊗ g)(a ⊗ b) = f (a) ⊗ g(b).
• The tensor product has the desired universal property:
∼
=
Hombilinear (A × B,C) −
→ HomAb (A ⊗ B,C),
i.e., if we have a bilinear map A × B → C, then there is a unique (up to
isomorphism) dotted map which makes the diagram commutative
q
A×B / A⊗B
%
C.
• The universal property of the tensor product implies that we have an
isomorphism
Aα ) ⊗ B ∼
M M
( = (Aα ⊗ B).
α α
Now it is time to see some examples:

• For every abelian group A, we have isomorphisms
A⊗Z∼ =A∼ =Z⊗A
which sends a ⊗ n 7→ na and inverse a 7→ a ⊗ 1.
• For every abelian group A and every m, we have an isomorphism
A ⊗ Z/m ∼ = A/mA, a ⊗ [n] 7→ [an] ∈ A/mA.
163
• If M is an abelian group and X a space, we can form the tensor product

Sn (X; M ) := Sn (X) ⊗ M ∼
M
= M
σ∈Singn (X)
( )
X
= mj σj : σj ∈ Singn (X), mj ∈ M .
j
There is a boundary operator defined by

∂nM : Sn (X; M ) → Sn−1 (X; M ), ∂nM = ∂n ⊗ 1.
This turns S∗ (X; M ) into a chain complex. The homology of this complex
is the homology of X with coefficients in M .
Tensor products are great. Except for the following:

• Tor functor
Suppose we have an abelian group M and a surjective homomorphism of
abelian groups
B ↠ C.
Then we can check by looking at the generators that
B⊗M ↠C ⊗M
is also surjective.
More generally, we can show that tensor products preserve cokernels:
Lemma: Tensor products preserve cokernels

Let M be an abelian group. Suppose we have an exact sequence
i j
A→
− B→
− C → 0.
Then taking the tensor product − ⊗ M yields an exact sequence
i⊗1 j⊗1
A ⊗ M −−→ B ⊗ M −−→ C ⊗ M → 0
where 1 denotes the identity map on M . In other words, the functor − ⊗ M
is right exact and preserves cokernels.
Proof: We are going to show that − ⊗ M preserves cokernels. This is in fact

equivalent to the other statements.
CHAPTER
Let f : B ⊗ M → Q be a homomorphism. We need to show that there is a
unique factorization as indicated by the dotted arrow in the diagram
i⊗1 j⊗1
A⊗M / B⊗M / C ⊗M / 0
f
% y
0
Q.
By the universal property and the fact that C × M generates C ⊗ M , this is
equivalent to a unique factorization of the diagram of bilinear maps
i×1 j×1
A×M / B×M / C ×M / 0
F
% y
0
Q.
But now we only need to find an appropriate extension C → Q the existence of
which is implied by assumption. □
However, suppose we have an injective homomorphism
A ,→ B.
Then it is in general not the case that
A⊗M →B⊗M
is injective.
2
For example, take the map Z → − Z given by multiplication by 2. It is clearly
injective. But if we tensor with Z/2, we get the map
2=0
Z/2 −−→ Z/2
which is not injective.
Thus, tensor products do not preserve exact sequences, in general.
We would like to remedy this defect. And, in fact, the tensor product is not
so far from being exact. For, if M is a free abelian group, then the functor
M ⊗ − is exact, i.e., it preserves all exact sequences.
We can see this as follows: Assume M is the free abelian group on the set
S. Then M ⊗ N = ⊕S N , since tensoring distributes over direct sums, as we
remarked above.
To exploit this fact we make use of the following observation:
165
Lemma: Direct sums of exact sequences

If Mi′ → Mi → Mi′′ is exact for every i ∈ I, then
M M M
Mi′ → Mi → Mi′′
i i i
is exact.
Proof: The composition is zero and if (xi )i is sent to 0 in i Mi′′ , then each
L
xi must be sent to 0 in Mi′′ . Hence each xi comes from some x′i , and hence (xi )i
comes from (x′i )i . We just need to remember to choose x′i = 0 whenever xi = 0.
□
Now let A be any abelian group. As we did in the previous lecture, we choose
a free abelian group F0 mapping surjectively onto A
F0 ↠ A.
The kernel F1 of this map is also free abelian as a subgroup of a free abelian
group. Hence we get an exact sequence of the form
0 → F1 ,→ F0 ↠ A.
Free resolutions
Such an exact sequence with F1 and F0 free abelian groups, is called a free
resolution of A of length two.
(Note that the fact that we can always choose such a free resolution of length
two is particular to the case of abelian groups, i.e., Z-modules. For R-modules
over other rings, one might only be able to find projective resolutions of higher
length. The fact that Z is a principal ideal domain, a PID, does the trick.)
For any abelian group M , tensoring these maps with M yields an exact se-
quence
F1 ⊗ M → F0 ⊗ M → A ⊗ M → 0.
The kernel of the left-hand map is not necessarily zero, though.

This leads to the following important definition:
CHAPTER
Definition: Tor
The kernel of the map A ⊗ F1 → A ⊗ F0 is called Tor(A,M ). Hence by
definition we have an exact sequence
0 → Tor(A, M ) → A ⊗ F1 → A ⊗ F0 → A ⊗ M → 0.
This group measures how far − ⊗ M is from being exact.
Note that if we replace abelian groups with R-modules over other rings than
Z and take tensor products over R, we might have to consider higher Tor-terms.
Hence we should really write TorZ1 (A, M ) for Tor(A, M ). But we are going to
keep things simple and focus on the idea rather than general technicalities.
It is again time to see some examples:
• If M is a free abelian group, then Tor(A,M ) = 0 for any abelian group
A. That follows from the lemma above.
• Let M = Z/m. Then we can take F0 = F1 = Z and

m
Z−
→ Z → Z/m → 0
as a free resolution of Z/m. For an abelian group A, the sequence defining
Tor looks like
1⊗m
0 → Tor(A,Z/m) → A ⊗ Z −−→ A ⊗ Z → A ⊗ Z/m → 0.
Since we know A ⊗ Z/m = A/mA, we get
Tor(A,Z/m) = Ker (m : A → A) = m-torsion in A.
Hence Tor(A,Z/m) is the subgroup of m-torsion elements in A.
• For a concrete case, let us try to calculate Tor(Z/4, Z/6). We use the
free resolution
6
Z→
− Z → Z/6 → 0.
Tensoring with Z/4 yields
2
Z/4 →
− Z/4 → Z/4 ⊗ Z/6 → 0
2
where we use 6 = 2 in Z/4. The kernel of Z/4 →
− Z/4 is Z/2. Thus
Tor(Z/4, Z/6) = Z/2.
• More generally, we get
Tor(Z/n, Z/m) = Z/ gcd(n, m)
167
where gcd(n, m) denotes the greatest common divisor of n and m.

The last three examples explain the name Tor.
We should hold our breath for a moment and check a couple of things. For
example, that Tor does not depend on the choice of free resolution, that it is a
functor etc. So let us get to work:
Lemma: Lifting resolutions

i p
Let f : M → N be a homomorphism and 0 → E1 → − E0 → − M and 0 →
j q
F1 → − F0 →
− N be free resolutions. Then we can lift f to a chain map
f∗ : E∗ → F∗ , i.e, to a commutative diagram
p
0 / E1
i / E0 / M / 0
f1 f0 f
j q
0 / F1 / F0 / N / 0.
Moreover, this lift is unique up to chain homotopy, i.e., for another lift f∗′
of f , there is a chain homotopy h between f∗ and f∗′ :
0 / E1 / E0 / 0
h
f1′ f1 f0′ f0
~
0 / F1 / F0 / 0.
Proof: • Since E0 is a free abelian group, we know there is some set S of

generators such that E0 = ZS. Now we can map the elements in S to M via the
p f q
map E0 → − M , and further to N via M → − N . Since F0 →− N is surjective, we
can choose lifts in F0 of the elements in f (p(S)). Since a homomorphism on a
free abelian group is determined by the image of the generators, we can extend
this process to get a homomorphism
f0
E0 −
→ F0 such that f ◦ p = q ◦ f0 .
Now can define f1 to be the restriction of f0 to the kernel of p which is E1 by

definition. This yields the desired commutative diagram.
CHAPTER
• Now let f0′ and f1′ be another choice of maps which lift f . The differences
g0 := f0 − f0′ and g1 := f1 − f1′ are then maps which lift f − f = 0 : M → N :
p
0 / E1
i / E0 / M / 0
g1 g0 0
j q
0 / F1 / F0 / N / 0.
Since the diagram commutes, we get q ◦ g0 = p ◦ 0 = 0. Therefore, the universal
property of kernels implies that we can lift g0 to a map h : E0 → F1 such that
j ◦ h = g0 :
p
0 / E1
i / E0 / M / 0
h
g1 g0 0
~ j q
0 / F1 / F0 / N / 0.
Moreover, since E1 is the kernel of i, we must have h ◦ i = g1 . Thus h is a

chain homotopy between f∗ and f∗′ (the next map E1 → 0 being trivial). □
With this result at hand we can finally prove:
Corollary: Tor is independent of resolutions

Tor is independent of the choice of a free resolution: For any free resolution
i
0 → E1 → − E0 → M of M , there is a unique isomorphism
∼
=
Ker (i ⊗ 1) −
→ Tor(A,M ).
Proof: We just apply the previous result to the identity of M to get that,
with whatever resolution we calculate Tor, there is an isomorphism between any
two different ways. And this isomorphism is unique by the theorem on chain
homotopies and their induced maps on homologies. □
There are other properties of Tor the proof of which we are going to omit:
• Tor is functorial: For any homomorphisms of abelian groups A → A′
and M → M ′ , there is a homomorphism
Tor(A,M ) → Tor(A′ ,M ′ ).
• Tor is symmetric, i.e., Tor(A,M ) ∼

= Tor(M,A).
• If M is free, then Tor(A,M ) = 0 for any abelian group A.

169
• L
Since the direct sum of free resolutions of each Ai is a free resolution of
i Ai , we know that Tor commutes with direct sums:
Ai ,M ) ∼
M M
Tor( = Tor(Ai ,M ),
i i
• Let T (M ) be the subgroup of torsion elements of M . Then

Tor(A,M ) ∼
= Tor(A,T (M ))
for any abelian group A.
Now we can prove the main result in this story:
Theorem: Universal Coefficient Theorem

Let C∗ be a chain complex of free abelian groups and let M be an abelian
group. Then there are natural short exact sequences
0 → Hn (C∗ ) ⊗ M → Hn (C∗ ⊗ M ) → Tor(Hn−1 (C∗ ),M ) → 0
for all n. These sequences split, but the splitting is not natural.
Proof: We write Zn for the kernel and Bn−1 for the image of the differential
d : Cn → Cn−1 . Since Cn and Cn−1 are free, both Zn and Bn−1 are free as well.
Together with the differential in C∗ , this yields a morphism of short exact
sequences
dn
0 / Zn / Cn / Bn−1 / 0
dn dn dn−1
dn−1
0 / Zn−1 / Cn−1 / Bn−2 / 0.
By definition of Zn and Bn , the restriction of the differentials to these groups

vanish. This implies that (Z∗ ,d) and (B∗ ,d) are chain complexes (with trivial
differentials).
Hence we get a short exact sequence of chain complexes
(28) 0 → Z∗ → C∗ → B∗−1 → 0.
Since all groups in these chain complexes are free, tensoring with M yields
again a short exact sequence of chain complexes
0 → Z∗ ⊗ M → C∗ ⊗ M → B∗−1 ⊗ M → 0.
CHAPTER
This can be checked as in the above lemma on direct sums of exact sequences.
Since the differentials in Z∗ and B∗ are trivial, the associated long exact
sequence in homology looks like
∂
n ∂n−1
· · · → Bn ⊗ M −→ Zn ⊗ M → Hn (C∗ ⊗ M ) → Bn−1 ⊗ M −−−→ Zn−1 ⊗ M → · · ·
∂
n
The connecting homomorphism Bn ⊗ M −→ Zn ⊗ M in this sequence is in ⊗ 1,
where in : Bn ,→ Zn denotes the inclusion and 1 denotes the identity on M . This
can be easily checked using the definition of the connecting homomorphism.
A long exact sequence can always be cut into short exact sequences of
the form
0 → Coker(in ⊗ 1) → Hn (C∗ ⊗ M ) → Ker (in−1 ⊗ 1) → 0.
Since the tensor product preserves cokernels, the cokernel on the left-
hand side is just
Coker(in ⊗ 1) = Coker(in ) ⊗ M = Zn /Bn ⊗ M = Hn (C∗ ) ⊗ M.
For Ker (in−1 ⊗ 1), we observe that

in−1
Bn−1 −−→ Zn−1 → Hn−1 (C∗ ) → 0
is a free resolution of Hn−1 (C∗ ). Hence after tensoring with M we get an exact
sequence
in−1 ⊗1
0 → Ker (in−1 ⊗ 1) → Bn−1 ⊗ M −−−−→ Zn−1 ⊗ M → Hn−1 (C∗ ) ⊗ M → 0.
Thus, since Tor is independent of the chosen free resolution,

Ker (in−1 ⊗ 1) = Tor(Hn−1 (C∗ ),M ).
Finally, to obtain the asserted splitting we use that subgroups of free abelian
groups are free. That implies that sequence (28) splits and we have
Cn ∼
= Zn ⊕ Bn−1 .
Tensoring with M yields
Cn ⊗ M ∼
= (Zn ⊗ M ) ⊕ (Bn−1 ⊗ M ).
Now one has to work a little bit more to get that this induces a direct sum
decomposition in homology. We skip this here. □
Since the singular chain complex S∗ (X,A) is an example of a chain complex
of free abelian groups, the theorem implies:
171
Corollary: UCT for singular homology

For each pair of spaces (X,A) there are split short exact sequences
0 → Hn (X,A) ⊗ M → Hn (X,A; M ) → Tor(Hn−1 (X,A),M ) → 0
for all n, and these sequences are natural with respect to maps of pairs
(X,A) → (Y,B).
One of the goals of introducing coefficients is to simplify calculations. The

simplest case is often when we consider a field as coefficients. For example, the
finite fields Fp or the rational numbers Q. The UCT tells how we can recover
integral homology from these pieces. We will figure out how this works in the
exercises.
Since we put so much work into defining Tor, let us mention another important
theorem. It tells us how the homology of the product of two spaces depends on
the homology of the individual spaces. For that relation is not as straight forward
as one might hope:
Künneth Theorem
For any pair of spaces X and Y and every n, there is a split short exact
sequence
M M
0→ (Hp (X) ⊗ Hq (Y )) → Hn (X × Y ) → Tor(Hp (X), Hq (Y )) → 0.
p+q=n p+q=n−1
This sequence is natural in X and Y . But the splitting is not natural.
The maps Hp (X) → Hn (X × Y ) and Hq (Y ) → Hn (X × Y ) arise from the

cross product construction on singular chains. We will not have time to discuss
this in class though.
LECTURE 18
Singular cohomology
We are going to define a new algebraic invariant, called singular cohomol-

ogy. At first glance it might look almost the same as homology, but we will see
that there is a striking difference between homology and cohomology. For,
singular cohomology allows us to define an additional algebraic structure: mul-
tiplication.
As a motivation, we start with the following familiar situation. Recall that
2
R a path γ : [a,b] → R and
in calculus, we learn to calculate path integrals. Given
a 1-form pdx + qdy. Then we can form the integral γ pdx + qdy, and we learned
all kinds of things about it.
In particular, we can consider taking the integral as a map
Z
γ 7→ pdx + qdy ∈ R.
γ
Since any path γ can be reparametrized to a 1-simplex, we can think of taking

the integral of a given 1-form over a path as a map
Sing1 (R2 ) → R.
This map captures certain geometric and topological information. It is an

important example of a 1-cochain, a concept we are now going to define.
Definition: Singular cochains

Let X be a topological space and let M be an abelian group. An n-cochain
on X with values in M is a function
Singn (X) → M.
We turn the set
S n (X; M ) := Map(Singn (X),M )
of n-cochains into a group by defining c + c′ to be the function which sends
σ to c(σ) + c′ (σ).
173
174 CHAPTER 18. SINGULAR COHOMOLOGY
• As an example, let us look at the case M = Z. Then an n-cochain on X
is just a function which assigns to any n-simplex σ : ∆n → X a number
in Z.
We know that simplices in different dimensions are connected via the face
maps. As for chains, the face maps induce an operator between cochains in dif-
ferent dimensions. But note that, for cochains, the degree will increase instead
of decrease.
Definition: Coboundaries
The coboundary operator
δ : S n (X; M ) → S n+1 (X; M )δ(c)(σ) = c(∂σ)
is defined as follows:
Given an n-cochain c and an n + 1-simplex σ : ∆n+1 → X. Then we define
the n + 1-cochain δ(c) as
n+1
X
δ n (c)(σ) = c(∂n+1 (σ)) = (−1)i c(σ ◦ ϕn+1
i )
i=0
where ϕn+1
i is the ith face map. This defines δ n (c) as a function on
Singn+1 (X).
• For an example, let us look again at the case m = Z. We learned that

an n-cochain on X is a function which assigns to any n-simplex σ : ∆n → X a
number in Z. In order to be an n-cocycle, the numbers assigned to the boundary
of an n + 1-simplex cancel out (with the sign convention).
To get more concrete, let c ∈ S 1 (X; Z) be a 1-cochain. Let σ : ∆2 → X be a
2-simplex. Then, for c to be a cocycle, we need that the numbers it assigns to
the faces of σ cancel out in the sense that
c(d0 σ) − c(d1 σ) + c(d2 σ) = 0.
• Let us have another look at the example from calculus we started with.
A function
f : R2 → R
is a 0-cochain on R2 with values in R. For it assigns to each zero-simplex, i.e.,

a point x ∈ R2 , a real number f (x).
175
Then the 1-cochain δ(f ) is the function which assigns to a (smooth) path γ
the number f (γ(1)) − f (γ(0)):
δ(f ) : γ 7→ f (γ(1)) − f (γ(0)).
By Green’s Theorem, this is also the value of the integral

Z Z
fx dx + fy dy = df
γ γ
which is the integral of the 1-form df along γ.

Hence the cochain complex, while it looks very much like homology, also has a
natural connection to calculus. In fact, there is some justification for saying
that cochains and cohomology are more natural notions than chains and
homology.
Back to the general case. The coboundary operator turns S ∗ (X; M ) into a
cochain complex, since δ ◦ δ = 0 which follows from our previous calculation.
For, given an n + 1-simplex σ and an n − 1-cochain c, we get
(δ n ◦ δ n−1 (c))(σ) = (δ n c)(∂n (σ)) = c(∂n ◦ ∂n+1 (σ)) = 0.
An equivalent way to obtain this complex, is to look at homomorphisms

of abelian groups from Sn (X) to M , i.e., we have
S n (X; M ) = HomAb (Sn (X),M ).
The coboundary operator is just the homomorphism induced by the boundary
operator on chains:
δ = Hom(∂,M ) : HomAb (Sn (X),M ) → HomAb (Sn+1 (X),M ), c 7→ c ◦ ∂.
In other words, δ = ∂ ∗ equals the pullback along ∂.
The subgroup given as the kernel of δ n is denoted by
Z n (X; M ) = Ker (δ : S n (X; M ) → S n+1 (X; M ))
and called the group of n-cocylces of X.
The image of δ n−1 is called the group of n-coboundaries of X and is denoted
by
B n (X; M ) = Im (δ n−1 : S n−1 (X; M ) → S n (X; M )).
Since δ ◦ δ = 0, we have
B n (X; M ) ⊆ Z n (X; M ).
In other words, every coboundary is a cocycle.
Definition: Singular cohomology

Let X be a topological space and let M be an abelian group. The nth
singular cohomology group of X is defined as the nth cohomology group
of the cochain complex S ∗ (X; M ), i.e.,
Z n (X; M )
H n (X; M ) = n .
B (X; M )
Integrals over forms yield elements in cohomology with coefficients in R.

This is in fact the origin of cohomology theory and is connected to de Rham
cohomology. Though as natural as de Rham cohomology is, it has the drawback
that we have to stick to coefficients in R.
This demonstrates why it might be smart to take the detour via singular
simplices and taking maps in chains. For we gain the flexibility to study
singular cohomology with coefficients in an arbitrary abelian group.
As a first example, let us try to understand H 0 (X; M ).
Cohomology in dimension zero
A 0-cochain is a function
c : Sing0 (X) → M.
Since Sing0 (X) is just the underlying set of X, a 0-cochain corresponds to

just an arbitrary, that is not necessarily continuous, function
f : X → M.
Now what does it mean for such a function to be a cocycle? To figure this out
we need to calculate δ(f ). Since δ(f ) is defined on 1-simplices, let σ : ∆1 → X
be a 1-simplex on X. The effect of δ(f ) is to evaulate f on the boundary of
σ:
δ(f )(σ) = f (∂σ) = f (σ(e0 )) − f (σ(e1 )).
Since this expression must be 0 for every 1-simplex, we deduce that f is a

cocycle if and only if it is constant on the path-components of X.
If we denote by π0 (X) the set of path-components, then we have shown:
177
H 0 (X; M ) = Map(π0 (X),M ).
Cohomology of a point
If X is just a point, then Singn (pt) consists just of the constant map for
each n. Hence an n-cochain c ∈ S n (pt; M ) is completely determined by its
value mc on the constant map, and therefore Hom(Singn (pt),M ) ∼ = M for all n.
The coboundary operator takes c ∈ Hom(Singn (pt),M ) to the alternating sum
n+1
X n+1
X
i
δ(c) = (−1) c(constant map ◦ ϕni ) = (−1)i mc .
i=0 i=0
Hence the coboundary is trivial if n is even and the identity if n is odd.

The cochain complex therefore looks like
0 1 0 1 0
0→M →
− M→
− M→
− M→
− M→
− ···
Thus the cohomology of a point is given by

(
M if n = 0
H n (pt; M ) =
0 else.
Now let us see what else we know about singular cohomology.

Properties of singular cohomology
Fix an abelian group M . Singular cohomology has the following properties:
• Cohomology is contravariant, i.e., a continuous map f : X → Y
induces a homomorphism
f ∗ : S ∗ (Y ; M ) → S ∗ (X; M ).
This map works as follows: Let c ∈ S n (Y ; M ) be an n-cochain on Y .
Then f ∗ c is the map which assigns to n-simplex σ : ∆n → X, the value
σ f
(f ∗ c)(σ) = c(f ◦ σ) = c(∆n →
− X→
− Y ).
Since f ∗ is in fact a map of cochain complexes (which is defined in
analogy to maps of chain complexes), this induces a homomorphism on
cohomology
f ∗ : H ∗ (Y ; M ) → H ∗ (X; M ).
This assignment is functorial, i.e., the identity map is sent to the
identity homomorphism of cochains and if g : Y → Z is another map,
then
(g ◦ f )∗ = f ∗ ◦ g ∗ .
• In our calculus example, the contravariance corresponds to re-

striction of a form to an open subspace.
Why cohomology?
At first glance it seems like cohomology and homology are the same
guys, just wrapped up in slightly different cloths and reversing the
arrows. In fact, this is kind of true as we will see in the next lecture.
However, there is also a striking difference which is due to the
innocent looking fact that cohomology is contravariant. We are
going to exploit this fact as follows:
Assume that R is a ring, and let
∆
X−
→ X × X, x 7→ (x,x)
be the diagonal map. It induces a homomorphism in cohomology
∆∗
H ∗ (X × X; R) −→ H ∗ (X; R).
Now we only need to construct a suitable map H p (X; R) ⊗
H q (X; R) → H p+q (X L
× X; R) to get a multiplication on the di-
∗
rect sum H (X; R) = q H q (X; R):
∆∗
H p (X; R) ⊗ H q (X; R) → H p+q (X × X; R) −→ H p+q (X; R).
It will still take some effort to make this idea work. Nevertheless,
this gives us a first idea of how contravariance can be useful.
• Cohomology is homotopy-invariant, i.e., if the maps f and g are

homotopic f ≃ g, then they induce the same map in cohomology
f ∗ = g∗.
In fact, the proof we used for homology dualizes to cohomology.
For, recall that a homotopy between f and g induces a chain homo-
topy between h between the maps f∗ and g∗ on singular chain complexes.
Now we use that the singular cochain complex is the value of the singular
chain complex under the functor Hom(−,M ).
179
Then Hom(h,M ) is a homotopy between the maps of cochain com-

plexes
f ∗ = Hom(f∗ ,M ) and g ∗ = Hom(g∗ ,M ).
For the relation
f ∗ − g∗ = h ◦ ∂ + ∂ ◦ h
implies the relation
f ∗ − g ∗ = δ ◦ Hom(h,M ) + Hom(h,M ) ◦ δ.
• If A is a subspace of X, then there are also relative cohomology

groups. Let i : A ,→ X denote the inclusion map. We consider the
cochain complex
i∗
→ S ∗ (A; M ))
S ∗ (X,A; M ) = Ker (S ∗ (X; M ) −
consisting of those maps Singn (X) → M which vanish on the subset
Singn (A).
The nth relative cohomology group is defined as the cohomology
of this cochain complex
H n (X,A; M ) = H n (S ∗ (X,A; M )).
By definition, there is a short exact sequence
0 → S ∗ (X,A; M ) → S ∗ (X; M ) → S ∗ (A; M ) → 0
which induces a long exact sequence of the cohomology groups of
the complexes in the same way as this was the case for chain complexes
and homology:
∂n
· · · → H n (X,A; M ) → H n (X; M ) → H n (A; M ) −→ H n+1 (X,A; M ) → · · ·
• There is also a reduced version of cohomology.

P P ϵ : S0 (X; M ) → M
Let
be the augmentation map sending i mi σi to i mi ∈ M . Since we
know ∂0 ◦ ϵ = 0, we observe that applying the functor Hom(−, M ) yields
the augmented singular cochain complex
ϵ∗ δ0 δ1
→ S 0 (X; M ) −
0→M − → S 1 (X; M ) −
→ ···
The reduced cohomology of X with coefficients in M is the cohomol-
ogy of the augmented singular cochain complex.
• Cohomology satisfies Excision, i.e., if Z ⊂ A ⊂ X with Z̄ ⊂ A◦ ,
then the inclusion k : (X − Z, A − Z) ,→ (X, A) induces an isomorphism
∼
=
k ∗ : H ∗ (X, A; M ) −
→ H ∗ (X − Z, A − Z; M ).
• Cohomology sends sums to products, i.e.,
H ∗ ( Xα ; M ) ∼
a Y
= H ∗ (Xα ; M ).
α α
• Cohomology has Mayer-Vietoris sequences, i.e., if {A,B} is a

cover of X, then, for every n, there are connecting homomorphisms
d which fit into a long exact sequence
i∗
 
 A 
∗ ∗ ∗
h i
d −i B jA jB d
− H n (X; M ) −−−−→ H n (A; M ) ⊕ H n (B; M ) −−−−−−→ H n (A ∩ B; M ) →
··· → − H n+1 (X; M ) → · · ·
Note that the maps go in the other direction and the degree of
the connecting homomorphism increases.
Here we used the inclusion maps
jA
A ∩ _ B / A _
jB iA

/
B iB
X.
To prove, for example, the statement about Mayer-Vietoris sequences,

let us go back to the proof in homology.
Let A = {A,B} be our cover. We used a short exact sequence of chain
complexes
0 → S∗ (A ∩ B) → S∗ (A) ⊕ S∗ (B) → S∗A (X) → 0
where S∗A (X) denotes the A-small chains.

We would like to turn this into an exact sequence in cohomology. As we
have learned last time, not all functors preserve exactness.
And, in fact, Hom(−,M ) is unfortunately no exception. We will study
the behaviour of Hom next time, but for the present purpose we observe a fact
which saves our day.
181
For, the singular chain complexes involved in the above short exact sequence
consist of free abelian groups in each dimension. And exactness is indeed pre-
served by Hom for such sequences.
More precisely, we would like to use the following fact:
Lemma: Exactness of Hom-functor on free complexes

Let M be an abelian group and let
0 → A∗ → B∗ → C∗ → 0
be an exact sequence of chain complexes of free abelian groups. Then the
induced sequence of cochain complexes
0 → Hom(C∗ ,M ) → Hom(B∗ ,M ) → Hom(A∗ ,M ) → 0
is exact.
Proof: By definition of exactness for sequences of complexes, it suffices to

show the assertion for a short exact sequence of free abelian groups.
The key is that any short exact sequence of free abelian groups splits. The
splitting induces a splitting on the induced sequence of Hom-groups.
More concretely, let
i p
0→A→
− B→
− C→0
be a short exact sequence of free abelian groups.
Since C is free and p is surjective, there is a dotted lift in the solid diagram
?B
s
p

C C
which makes the diagram commute, i.e., p ◦ s = 1C .
This implies
s∗ ◦ p∗ = 1Hom(C,M ) ,
and hence s∗ is a section of p∗ in
p∗ i∗
0 → Hom(C, M ) −
→ Hom(B, M ) −
→ Hom(A, M ) → 0.
This implies that Hom(B,M ) = Hom(A,M ) ⊕ Hom(C,M ) and that i∗ is sur-

jective. That the sequence is exact at Hom(B,M ) is now obvious as well. QED
Warning: Note that if A = Z[S] is a free abelian group, then
M Y
Hom(A,M ) = Hom( Z,M ) = Hom(Z,M )
S S
which might be an uncountable product. This leads to the annoying fact
that Hom(A, M ) is not a free abelian group, in general.
Back to the proof of the MVS, with this fact at hand we get an induced
short exact sequence of cochain complexes
0 → Hom(S∗A (X), M ) → Hom(S∗ (A), M ) ⊕ Hom(S∗ (B), M ) → Hom(S∗ (A ∩ B), M ) → 0
where we also use that Hom commutes with direct sums.
Again, such a short exact sequence of cochain complexes induces a long
exact sequence of cohomology groups.
The final step of the proof is that we need to check that the induced map
of cochain complexes
Hom(S∗ (X),M ) → Hom(S∗A (X),M )
induces an isomorphism in cohomology.
In fact, this follows from the Small Chain Theorem and the following fact:
Proposition: From isos in homology to isos in cohomol-

ogy
Let C∗ and D∗ be two chain complexes of free abelian groups.
φ
Assume that there is a map C∗ −
→ D∗ which induces an isomorphism in
homology
∼
=
φ∗ : H∗ (C∗ ) −
→ H∗ (D∗ ).
Then, for any abelian group M , the map φ
φ∗ : H ∗ (D; M ) → H ∗ (C; M )
induces an isomorphism in cohomology with coefficients in M as well.
Here we wrote H ∗ (C; M ) = H ∗ (Hom(C∗ ,M )) and H ∗ (D; M ) =
H ∗ (Hom(D∗ ,M )) for the cohomology of the induced cochain complexes.
We are going to deduce this result from the Universal Coefficient Theorem
in cohomology which we will prove in the next lecture. Roughly speaking, it
will tell us how homology and cohomology are related.
As a first approach, we observe the following phenomenon.
183
The Kronecker pairing

Let M be an abelian group. For a chain complex C∗ and cochain complex
C ∗ := Hom(C∗ ,M ) there is a natural pairing given by evaluating a cochain on
chains:
⟨−,−⟩ : C n ⊗ Cn → M, (φ,a) 7→ ⟨φ,a⟩ := φ(a).
This is called the Kronecker pairing.
The boundary and coboundaries are compatible with this pairing, i.e.,
⟨δφ,a⟩ = δ(φ)(a) = φ(∂(a)) = ⟨φ,∂a⟩.
Lemma: Kronecker pairing

The Kronecker pairing induces a well-defined pairing on the level of coho-
mology and homology, i.e., we get an induced pairing
⟨−,−⟩ : H n (C ∗ ) ⊗ Hn (C∗ ) → M.
Proof: Let φ be a cocycle, i.e., δφ = 0. Then we get

⟨φ, a + ∂b⟩ = ⟨φ, a⟩ + ⟨φ, ∂b⟩ = ⟨φ, a⟩ + ⟨δφ, b⟩ = ⟨φ, a⟩.
Thus, the map ⟨φ, −⟩ descends to homology if φ is a cocycle.

It remains to check that this map vanishes if φ is a coboundary. So assume
φ = δψ and a is a cycle, i.e., ∂a = 0. Then we get
⟨φ, a⟩ = ⟨δψ, a⟩ = ⟨ψ, ∂a⟩ = 0.
This show that the pairing is well-defined on H n (C ∗ ) and Hn (C∗ ). □
Kronecker homomorphism
Thus, applied to the integral singular chain complex and the cochain com-
plex with coefficients in M , this pairing yields a natural homomorphism
κ : H n (X; M ) → Hom(Hn (X),M ),
which sends the class [c] of a cocycle to the homomorphism κ([c]) defined
by
κ([c]) : Hn (X) → M, [σ] 7→ ⟨c, σ⟩ = c(σ).
This leads to the important question:

From homology to cohomology?

If we know the singular homology of a space, what can we deduce about its
cohomology?
More concretely, what can we say about the map κ? Is κ injective? Is κ
surjective?
LECTURE 19
Ext and the Universal Coefficient Theorem for cohomology
In the previous lecture, we introuced the singular cochain complex and defined
singular chomology. Along the way we ran into some exact sequences to which
applied the Hom-functor. In particular, we constructed the Kronecker map
κ : H n (X; M ) → Hom(Hn (X),M ).
Our goal for this lecture is to study the Hom-functor in more detail and to
prove the Universal Coefficient Theorem for singular cohomology which will
tell us that κ is surjective. However, κ is not injective in general, but the UCT
will tell us what the kernel is.
Again, for some this will be a review of known results in homological algebra.
Nevertheless, those who have not seen this before, should get a chance to catch
up.
We will again focus on the main ideas.
Let M be an abelian group. We would like to understand the effect of the

functor Hom(−,M ) on exact sequences.
Before we start, note that Hom is not symmetric in general, i.e., Hom(A,M )
and Hom(M,A) might be very different indeed. For example,
Hom(Z,Z/n) ∼
= Z/n, but Hom(Z/n,Z) = 0.
Our next observation tells us that Hom is left-exact:
Lemma: Hom is left-exact

(a) Let M be an abelian group. Suppose we have an exact sequence
i j
A→
− B→
− C → 0.
185
186
CHAPTER 19. EXT AND THE UNIVERSAL COEFFICIENT THEOREM FOR COHOMOLOGY
Then applying Hom(−,M ) yields an exact sequence

j∗ i∗
0 → Hom(C, M ) −
→ Hom(B, M ) −
→ Hom(A, M ).
In other words, the functor Hom(−, M ) is left-exact and sends cokernels
to kernels.
(b) Similarly, applying Hom(M,−) to an exact sequence of the form
i j
0→A→
− B→
− C
yields an exact sequence
∗ i j∗
0 → Hom(M, A) −
→ Hom(M, B) −
→ Hom(M, C).
In other words, the functor Hom(M, −) is left-exact and sends kernels to
kernels.
Proof: (a) To show that j ∗ is injective, assume that γ ∈ Hom(C, M ) satis-

fies j ∗ (γ) = 0. That means
j ∗ (γ)(b) = (γ ◦ j)(b) = γ(j(b)) = 0 for all b ∈ B.
But j is surjective, and hence every element in C is of the form j(b) for some
b ∈ B. Hence γ = 0 is the trivial homomorphism.
The composition i∗ ◦ j ∗ is clearly 0, since j ◦ i = 0 by assumption. Thus
Im (j ∗ ) ⊆ Ker (i∗ ).
Now if β ∈ Hom(B,M ) is in Ker (i∗ ), then
0 = i∗ (β)(a) = β(i(a)) for all a ∈ A.
In other words, β is trivial on the image of i and hence factors as
β : B → B/Im (i) → M.
But B/Im (i) ∼= C, since the intial sequence was exact. Hence β is the composition
j γ
of a map B →
− C→ − M for some γ ∈ Hom(C,M ). Thus, β ∈ Im (j ∗ ).
(b) The proof is of course similar. To show that i∗ is injective, let α ∈
Hom(M,A) be a map such that i∗ (α) = 0. That means
i∗ (α(m)) = i(α(m)) = 0 for all m ∈ M.
Since i is injective, this implies α(m) = 0 for all m ∈ M , and hence α = 0.
The composition j∗ ◦ i∗ is clearly 0, since j ◦ i = 0 by assumption. Thus
Im (i∗ ) ⊆ Ker (j∗ ).
187
If β ∈ Hom(M,B) is in Ker (j∗ ), then

0 = j∗ (β)(m) = j(β(m)) for all m ∈ M.
In other words, β(m) ∈ Ker (j) for all m ∈ M . Since Ker (j) = Im (i), we get
β(m) ∈ Im (i) for all m ∈ M . Hence β factors as
α i
β: M −
→A→
− B
for some α ∈ Hom(M,A). Thus, β ∈ Im (i∗ ). QED
However, suppose we have an injective homomorphism
A ,→ B.
Then it is in general not the case that the induced map
Hom(B,M ) → Hom(A,M )
is surjective.
2
For example, take the map Z →
− Z given by multiplication by 2. It is clearly
injective. But if we apply Hom(−,Z/2), we get the map
2=0
Hom(Z,Z/2) ∼
= Z/2 −−→ Z/2 ∼
= Hom(Z,Z/2)
which is not surjective.
We would like to remedy this defect. And we can already guess how this can
be achieved. As we have seen in the previous lecture, Hom(−,M ) is not so far
from being exact. For, if we apply Hom(−,M ) to a short exact sequence of
free abelian groups, then the induced sequence is still short exact.
So let A be an abelian group and let us choose a free resolution of A as in a
previous lecture
0 → F1 ,→ F0 ↠ A.
Applying Hom(−,M ) to this equence yields an exact sequence

0 → Hom(A,M ) → Hom(F0 ,M ) → Hom(F1 ,M ).
The right-hand map is not necessarily surjective, or in other words, the cok-
ernel of the right-hand map is not necessarily zero.
This leads to the following important definition:
188
Definition: Ext
The cokernel of the map Hom(F0 ,M ) → Hom(F1 ,M ) is called Ext(A,M ).
Hence by definition we have an exact sequence
0 → Hom(A,M ) → Hom(F0 ,M ) → Hom(F1 ,M ) → Ext(A,M ) → 0.
Roughly speaking, the group Ext(A,−) measures how far Hom(A,−) is from
being exact.
Let us calculate some examples:

• Let A = Z/p. Then we can take F0 = F1 = Z and
p
Z→
− Z → Z/p → 0
as a free resolution of Z/p. For an abelian group M , the sequence defining
Ext looks like
p
0 → Hom(Z/p,M ) → Hom(Z,M ) →
− Hom(Z,M ) → Ext(Z/p,M ) → 0.
Since Hom(Z,M ) = M , this sequence equals
p
0 → p-torsion in M → M →
− M → Ext(Z/p,M ) → 0.
Thus
p
Ext(Z/p,M ) = Coker(M →
− M ) = M/pM.
• For a concrete case, let us calculate Ext(Z/2, Z/2). We use the free
resolution
2
Z→
− Z → Z/2 → 0.
Applying Hom(−,Z/2) yields
2
0 → Hom(Z/2,Z/2) → Hom(Z,Z/2) →
− Hom(Z,Z/2).
This sequence is isomorphic to
2=0
0 → Z/2 → Z/2 −−→ Z/2.
Since 2 = 0 in Z/2, the second map is trivial. Hence the cokernel of
this map is just Z/2. Thus
Ext(Z/2, Z/2) = Z/2.
• More generally, one can show
Ext(Z/n,Z/m) = Z/ gcd(n, m)
where gcd(n, m) denotes the greatest common divisor of n and m.
189
Now we should study Ext in more detail. As a first step we show that it can
be viewed as a cohomology group:
Lemma: Ext and Hom as cohomology groups

j
Let A and M be abelian groups and 0 → F1 → − F0 → A → 0 be a free
resolution of A. Consider the cochain complex Hom(F∗ ,M ) given by
j∗
0 → Hom(F0 , M ) −
→ Hom(F1 , M ) → 0
with Hom(F0 , M ) in dimension zero and Hom(F1 , M ) in dimension one.
Then we have
H 0 (Hom(F∗ , M )) = Hom(A, M ) and H 1 (Hom(F∗ , M )) = Ext(A, M ).
Proof: By definition, Ext(A, M ) is the cokernel of j ∗ . Since the differential

out of Hom(F1 , M ) is trivial, the first cohomology is just
H 1 (Hom(F∗ , M )) = Hom(F1 , M )/Im (j ∗ ) = Coker(j ∗ ) = Ext(A, M ).
For H 0 we remember that the augmented sequence

j∗
0 → Hom(A, M ) → Hom(F0 ,M ) −
→ Hom(F1 , M )
is exact.
Hence Hom(A,M ) is isomorphic to its image in Hom(F0 ,M ) which is, by
exactness of the sequence, the kernel of j ∗ . But this kernel is the cohomology
group of Hom(F∗ ,M ) in dimension 0:
H 0 (Hom(F∗ ,M )) = Ker (j ∗ ) = Hom(A,M ). □
We should check that Ext does not depend on the choice of a free resolution.
To do this, we are going to apply the lemma we proved for the Tor-case which
states that maps can be lifted to resolutions and any two lifts are chain homotopic
in a suitable sense.
Proposition: Ext is independent of resolutions

i
Ext is independent of the choice of a free resolution: If 0 → E1 →
− E0 → A
j
and 0 → F1 → − F0 → A are two free resolutions of A, there is a unique
isomorphism
∼
=
Coker(Hom(i,M )) −
→ Coker(Hom(j,M )).
190
Proof: We know from the result on lifting resolutions that we can lift the
identity map on A to a map of resolutions
0 / E1
i / E0 / A / 0
f1 f0
j
0 / F1 / F0 / A / 0.
The other way around we get a lift
0 / E1
i / E0 / A / 0
g1 g0
j
0 / F1 / F0 / A / 0.
We write E∗ for the complex 0 → E1 → E0 → 0 and F∗ for the complex

0 → F1 → F0 → 0.
Composition yields maps f∗ ◦ g∗ : E∗ → E∗ and g∗ ◦ f∗ : F∗ → F∗ which lift
the identity map on A. But since the identity maps on E∗ and F∗ , respectively,
also lift the identity on A, the lemma of a previous lecture implies that there is
a chain homotopy hE between f∗ ◦ g∗ and 1E∗ and a chain homotopy hF between
g∗ ◦ f∗ and 1F∗ .
Now we apply Hom(−,M ). Then hE induces a cochain homotopy Hom(hE ,M )
0 / Hom(E0 ,M ) / Hom(E1 ,M ) / 0
h∗
1Hom(E0 ,M ) g0∗ ◦f0∗ g1∗ ◦f1∗ 1Hom(E1 ,M )
v
0 / Hom(E0 ,M ) / Hom(E1 ,M ) / 0.
between
Hom(f∗ ◦ g∗ ,M ) = Hom(g∗ ,M ) ◦ Hom(f∗ ,M ) and Hom(1E∗ ,M ) = 1Hom(E∗ ,M ) .
Whereas hF induces a cochain homotopy Hom(hF ,M )

0 / Hom(F0 ,M ) / Hom(F1 ,M ) / 0
h∗
1Hom(F0 ,M ) f0∗ ◦g0∗ f1∗ ◦g1∗ 1Hom(F1 ,M )
v
0 / Hom(F0 ,M ) / Hom(F1 ,M ) / 0.
between
Hom(g∗ ◦ f∗ ,M ) = Hom(f∗ ,M ) ◦ Hom(g∗ ,M ) and Hom(1F∗ ,M ) = 1Hom(F∗ ,M ) .
191
Thus, the maps induced by the compositions on cohomology are equal to

the respective identity maps. In other words, the induced maps f ∗ and g ∗ on
cohomology are mutual inverses to each other.
Moreover, since the chain homotopy type of f∗ and g∗ is unique by the lemma
of the lecture on Tor, they induce in fact a unique isomorphism
∼
=
Coker(Hom(i,M )) = H 1 (Hom(E∗ ,M ) −
→ H 1 (Hom(F∗ ,M ) = Coker(Hom(j,M )).
QED
Lemma: Induced exact sequence

Let M be an abelian group and assume we have a short exact sequence of
abelian groups
i p
0→A→
− B→
− C → 0.
Then there is an associated long exact sequence
0 / Hom(C,M ) / Hom(B,M ) / Hom(A,M )
s
Ext(C,M ) / Ext(B,M ) / Ext(A,M ) / 0.
Proof: Let 0 → E1 → E0 → A → 0 be a free resolution of A, and 0 → F1 →

F0 → C → 0 be a free resolution of C. This data gives us a free resolution of B
by forming direct sums:
0 → E1 ⊕ F1 → E0 ⊕ F0 → B → 0.
By the result of the previous lecture, we can lift the maps in the short exact
sequence to maps of resolutions
0 / E1 / E1 ⊕ F1 / F1 / 0

0 / E0 / E0 ⊕ F0 / F0 / 0

0 / A / B / C / 0

0 0 0
192
The horizontal sequences are short exact, since the middle term is a direct
sum of the other terms. Hence we get a short exact sequence of chain complexes
0 → E∗ → E∗ ⊕ F∗ → F∗ → 0.
Since all three complexes consist of free abelian groups, applying Hom(−,M )
yields a short exact sequence of cochain complexes
0 → Hom(F∗ ,M ) → Hom(E∗ ⊕ F∗ ,M ) → Hom(E∗ ,M ) → 0.
By taking cohomology of these cochain complexes, we get an induced long

exact sequence of the associated cohomology groups. This is the desired exact
sequence together with the identification of H 1 with Ext and H 0 with Hom of
the previous lemma. □
This lemma also gives a hint to where the name Ext comes from:
Ext and extensions

• We can think of a short exact sequence of abelian groups
0→A→B→M →0
as an extension of M by A. We can then say that two extensions are
equivalent if they fit into an isomorphism of short exact sequences
0 / A / B / M / 0
∼
=

0 /A / B′ /M / 0.
• Note that we can always construct a trivial extension by taking the direct
sum of A and M :
(1,0)
0 → A −−→ A ⊕ M → M → 0.
Recall that we say that such a sequence splits.
• The group Ext(A,M ) measures how far extensions of M by A can be from
being from the trivial extension. For, we have
Ext(A,M ) = 0 ⇐⇒ every extension of M by A splits.
Proof: Given an extension, applying Hom(−,M ) yields an exact sequence
Hom(B,M ) → Hom(M,M ) → Ext(A,M ).
1
Thus the identity map M → − M lifts to a map B → M if Ext(A,M ) = 0.
But that is equivalent to that the initial short exact sequence splits. QED
193
• Now one can show in general that Ext(A,M ) is in bijection with the set
of all equivalence classes of extensions of M by A.
• For example, we computed Ext(Z/2,Z/2) = Z/2. The trivial element in
Ext corresponds to the trivial extension
0 → Z/2 → Z/2 ⊕ Z/2 → Z/2 → 0
whereas the non-trivial element corresponds to the extension
2
0 → Z/2 →
− Z/4 → Z/2 → 0.
We summarize some further properties of Ext:

• Ext is functorial: For any homomorphisms of abelian groups A → A′
and M → M ′ , there are homomorphisms
Ext(A′ ,M ) → Ext(A,M ) and Ext(A,M ) → Ext(A,M ′ ).
This follows from the lemma on liftings of resolutions.
• If A is free, then Ext(A,M ) = 0 for any abelian group A. This follows

1
from the fact that 0 → A →
− A → 0 is a free resolution of A.
• Ext commutes with finite direct sums, i.e.,

Ext(A1 ⊕ A2 ,M ) ∼
= Ext(A1 ,M ) ⊕ Ext(A2 ,M ).
This follows from the fact that the direct sum of free resolutions of each
A1 and A2 is a free resolution of A1 ⊕ A2 .
• Let A be a finitely generated abelian group and let T (A) denote its
torsion subgroup. Since Ext(Z/m,Z) = Z/m, the structure theorem for
finitely generated abelian groups and the previous two points imply that
Ext(A,Z) ∼
= T (A).
Now we prove the main result which connects homology and cohomology and
answers the question we raised last time about the Kronecker map κ:
Theorem: Universal Coefficient Theorem

Let C∗ be a chain complex of free abelian groups and let M be an abelian
group. We write C ∗ = Hom(C∗ ,M ) for the induced cochain complex.
Then there are natural short exact sequences
κ
0 → Ext(Hn−1 (C∗ ),M ) → H n (C ∗ ) →
− Hom(Hn (C∗ ),M ) → 0
194
for all n. These sequences split, but the splitting is not natural.
The proof builds on the same ideas as for the UCT in homology. But let us
do it anyway to get more practice.
Proof: • We write Zn for the kernel and Bn−1 for the image of the differ-
ential d : Cn → Cn−1 . Since Cn and Cn−1 are free, both Zn and Bn−1 are free as
well.
By definition of Zn and Bn , the restriction of the differentials to these groups
vanish. This implies that (Z∗ ,d) and (B∗ ,d) are chain complexes (with trivial
differentials).
Hence we get a short exact sequence of chain complexes
d
(29) 0 → Z∗ → C∗ →
− B∗−1 → 0.
• Since all groups in these chain complexes are free, applying the functor
Hom(−,M ) yields again a short exact sequence of cochain complexes
0 → Hom(B∗−1 ,M ) → Hom(C∗ ,M ) → Hom(Z∗ ,M ) → 0.
This follows from the lemma we proved in the previous lecture.
• Since the differentials in Z∗ and B∗ are trivial, the nth cohomology of
Hom(B∗−1 ,M ) is just Hom(Bn−1 ,M ), and the nth cohomology of Hom(Z∗ ,M ) is
just Hom(Zn ,M ).
Hence the long exact sequence in cohomology associated to the short exact
sequence (29) looks like
∂ d∗ i∗ ∂
· · · → Hom(Zn−1 ,M ) → → H n (Hom(C∗ ,M )) −
− Hom(Bn−1 ,M ) − → Hom(Zn ,M ) →
− Hom(Bn ,M ) → · · ·
∂
• The connecting homomorphism Hom(Zn ,M ) → − Hom(Bn ,M ) in this
sequence is i∗n = Hom(in ,M ), where in : Bn ,→ Zn denotes the inclusion. For, the
connecting homomorphism is defined as follows. Consider the maps
Hom(Cn ,M ) / Hom(Zn ,M )
δ

Hom(Bn ,M ) / Hom(Cn+1 ,M ).
δ
A preimage of φ ∈ Hom(Zn ,M ) is any map ψ : Cn → M which restricts to Zn .

Such a preimage exists since the upper horizontal map is surjective. Then ψ is
195
mapped to ψ ◦ d ∈ Hom(Cn+1 ,M ) by δ. Since every boundary is a cycle, we have

ψ ◦ d = φ ◦ d.
Now it remains to find a map φ̃ : Bn → M such that
ψ ◦ d = φ ◦ d = φ̃ ◦ d.
There is a canonical candidate for φ̃, namely the restriciton of φ to Bn . This is
exactly i∗n (φ).
16 M
φ FO
ψ
Zn / CO n φ̃ φ
d
in
/ Bn /
d
Cn+1 Zn
• A long exact sequence can always be cut into short exact sequences of
the from
0 → Coker(Hom(in−1 ,M )) → Hn (C ∗ ) → Ker (Hom(in ,M )) → 0.
Since the functor Hom(−,M ) sends cokernels to kernels, the kernel on the
right-hand side is just
Ker (Hom(in ,M )) = Hom(Coker(in ),M ) = Hom(Zn /Bn ,M ) = Hom(Hn (C∗ ),M ).
For the cokernel on the left-hand side, we use that

in−1
Bn−1 −−→ Zn−1 → Hn−1 (C∗ ) → 0
is a free resolution of Hn−1 (C∗ ).
Hence, after applying Hom(−,M ), we get an exact sequence
i∗n−1
0 → Hom(Hn−1 (C∗ ),M ) → Hom(Bn−1 ,M ) −−→ Hom(Zn−1 ,M ) → Coker(Hom(in−1 ,M )) → 0.
Thus, since Ext(−,M ) is independent of the chosen free resolution,

Coker(Hom(in−1 ,M )) = Ext(Hn−1 (C∗ ),M ).
Finally, to obtain the asserted splitting we use that subgroups of free abelian
groups are free. That implies that sequence (29) splits and we have
Cn ∼
= Zn ⊕ Bn−1 .
196
Applying Hom(−,M ) yields
Hom(Cn ,M ) ∼
= Hom(Zn ,M ) ⊕ Hom(Bn−1 ,M ).
Now one has to work a little bit more to get that this induces a direct sum
decomposition in homology.
It remains to check that the right-hand map in the theorem is in fact the
previously defined map κ. We leave this as an exercise. QED
Now we can prove the result we claimed in the previous lecture:
Corollary: From isos in homology to isos in cohomology

Let C∗ and D∗ be two chain complexes of free abelian groups. Let M be
an abelian group.
φ
Assume that there is a map C∗ −
→ D∗ which induces an isomorphism in
homology
∼
=
φ∗ : H∗ (C∗ ) −
→ H∗ (D∗ ).
Then this map also induces an isomorphism in cohomology with coeffi-
cients in M
∼
=
φ∗ : H ∗ (D∗ ) −
→ H ∗ (C ∗ ).
Proof: Since the construction of the long exact sequence we used in the proof
of the theorem is functorial, we see that φ induces a commutative diagram
0 / Ext(Hn−1 (C∗ ),M ) / H n (C ∗ ) / Hom(Hn (C∗ ),M ) / 0

(φ∗ )∗ φ∗ (φ∗ )∗

0 / Ext(Hn−1 (D∗ ),M ) / H n (D∗ ) / Hom(Hn (D∗ ),M ) / 0
The assumption that φ∗ induces an isomorphism implies that the two outer
vertical maps are isomomorphisms. The Five-Lemma implies that the middle
vertical map φ∗ is an isomorphism as well. QED
Our previous oberservations about Ext and torsion subgroups together with
the theorem imply:
197
Corollary: Computing cohomology from homology

Assume that the homology groups Hn (C∗ ) and Hn−1 (C∗ ) of the chain com-
plex are finitely generated. Let Tn ⊆ Hn (C∗ ) and Tn−1 ⊆ Hn−1 (C∗ ) denote
the torsion subgroups. Then we can calculate the integral cohomology of
C ∗ = Hom(C∗ ,Z) by
H n (C ∗ ; Z) ∼
= (Hn (C∗ )/Tn ) ⊕ Tn−1 .
Since the singular chain complex S∗ (X,A) is an example of a chain complex

of free abelian groups, the theorem implies:
Corollary: UCT for singular cohomology

For each pair of spaces (X,A) there are split short exact sequences
0 → Ext(Hn−1 (X,A),M ) → H n (X,A; M ) → Hom(Hn (X,A),M ) → 0
for all n, and these sequences are natural with respect to maps of pairs
(X,A) → (Y,B).
As a final remark, we mention that there are versions of Ext for the category of
R-modules for any ring. The corresponding Ext-groups ExtR (M,N ) will depend
on the ring R as well as on the modules M and N . Moreover, there might be
non-trivial higher Ext-groups ExtiR (M,N ) for i ≥ 2, in general.
But the theory is very similar to the case of abelian groups, i.e., Z-modules,
as long as R is a principal ideal domain (PID). For, then submodules of free R-
modules are still free over R (which is not true in general). Hence free resolutions
of length two exist, and higher Ext groups vanish also in this case.
For example, fields are examples of PIDs. However, note that, for example,
Ext(Z/2,Z/2) = Ext1Z (Z/2,Z/2) = Z/2 whereas Ext1Z/2 (Z/2,Z/2) = 0. Hence the
base rings matter.
LECTURE 20
Cup products in cohomology
We are now going to define the additional algebraic structure on coho-

mology that we promised earlier: multiplication.
There are many different ways to define a product structure in cohomology.
As always, each of these ways has its advantages and disadvantages. We will take
a direct path to the construction. This has the advantage to get a product right
away. The price we are going to pay is that we will have to work harder for some
results later. Note also that, even though we emphasized the importance of the
diagonal map in a previous lecture, this will not become clear from our direct
approach today. Though it matters nevertheless. :)
What we do take advantage of and which would not work for singular chains
is that a cochain is by definition a map to a ring. So we can multiply images of
cochains. Hence we could try to multiply cochains pointwise. We will just need
to figure out the images of which points we need to multiply.
We need to assume that we work with coefficients in a ring R. We will
always assume that R is commutative and that there is a neutral element 1 for
multiplication (even though not all arguments require all these assumptions).
Our main examples will be, of course, Z, Z/n, Q.
Definition: Cup products

For cochains φ ∈ S p (X; R) and ψ ∈ S q (X; R), we define the cup product
φ ∪ ψ ∈ S p+q (X; R) to be the cochain whose value on the p + q-simplex
σ : ∆p+q → X
(φ ∪ ψ)(σ) = φ(σ|[e0 ,...,ep ] )ψ(σ|[ep ,...,ep+q ] )
where the product is taken in R (here it comes already quite handy that we
work with coefficients in a ring).
Note: The symbol σ|[e0 ,...,ep ] refers to the restriction of σ to the front face
of ∆p+q
σ
σ|[e0 ,...,ep ] : ∆p ,→ ∆p+q →
− X, σ|[e0 ,...,ep ] (t0 , . . . , tp ) = σ(t0 , . . . , tp , 0, . . . , 0).
199
200 CHAPTER 20. CUP PRODUCTS IN COHOMOLOGY
Similarly, the symbol σ|[ep ,...,ep+q ] refers to the restriction of σ to the back
face of ∆p+q
σ
σ|[ep ,...,ep+q ] : ∆q ,→ ∆p+q →
− X, σ|[ep ,...,ep+q ] (t0 , . . . , tq ) = σ(0, . . . , 0, t0 , . . . , tq ).
We would can think of this construction as evaluating φ on the front face of

σ, evaluating ψ on the back face of σ, and then taking the product of the two
results.
To make sure that this construction yields something meaningful on the level
of cohomology we need to check a couple of things.
Lemma: Cup products and coboundaries

For cochains φ ∈ S p (X; R) and ψ ∈ S q (X; R), we have
δ(φ ∪ ψ) = δφ ∪ ψ + (−1)p φ ∪ δψ.
For the next proof and the remaining lecture, recall that the notation êi means
that the vertex ei is omitted.
Proof: By definition, for a simplex σ ∈ ∆p+q+1 → X, we have
δ(φ ∪ ψ)(σ) = (φ ∪ ψ)(∂σ)

p+q+1
!
X
i
= (φ ∪ ψ) (−1) σ|[e0 ,...,êi ,...,ep+q+1 ]
i=0
p+1
X
= (−1)i φ(σ|[e0 ,...,êi ,...,ep ] )ψ(σ|[ep+1 ,...,ep+q+1 ] )
i=0
p+q+1
X
+ (−1)i φ(σ|[e0 ,...,ep ] )ψ(σ|[ep ,...,êi ,...,ep+q+1 ] )
i=p
where the split into the two sums is justified by the fact that the last term of the
first sum is exactly (−1)-times the first term of the second sum.
Now it remains to observe that these two sums are exactly the definition of
(δφ ∪ ψ)(σ) and (−1)p (φ ∪ δψ)(σ). □
We would like this construction to descend to cohomology. Therefore, we
need to check:
201
• Assume that φ and ψ are cocycles, i.e., δφ = 0 and δψ = 0. Then φ ∪ ψ

is a cocycle, since
δ(φ ∪ ψ) = δφ ∪ ψ ± φ ∪ δψ = 0 ± 0 = 0.
• Assume that φ is a cocycle, i.e., δφ = 0, and ψ is a coboundary, i.e.,
there is a cochain ψ ′ with ψ = δψ ′ . Then φ ∪ ψ is a coboundary, since
δ(φ ∪ ψ ′ ) = δφ ∪ ψ ± φ ∪ δψ ′
= 0 ± φ ∪ ψ.
In other words, φ ∪ ψ is the image of ±φ ∪ ψ ′ under δ.
• Similarly, we can show that φ∪ψ is a coboundary if φ is a coboundary
and ψ is a cocycle.
Thus we have shown:
Cup product in cohomology

For any p and q, the cup product defines a map on cohomology groups
∪
H p (X; R) × H q (X; R) −
→ H p+q (X; R).
As we can easily check by evaluating on a simplex:

• The product is associative, i.e.,
(φ ∪ ψ) ∪ ξ = φ ∪ (ψ ∪ ξ).
• The product is distributive, i.e.,
φ ∪ (ψ + ξ) = φ ∪ ψ + φ ∪ ξ.
• The 0-cocycle ϵ ∈ H 0 (X; R) defined by taking value 1 for every 0-simplex
is a neutral element, i.e.,
ϵ ∪ φ = φ = φ ∪ ϵ for all φ ∈ H p (X; R).
Before we address commutativity, let us first check how the cup product be-
haves under induced homomorphisms:
Proposition: Cup products are natural

Let f : X → Y be a continuous map and let f ∗ : H p+q (Y ; R) → H p+q (X; R)
be the induced homomorphism. Then
f ∗ (φ ∪ ψ) = f ∗ φ ∪ f ∗ ψ
for all φ ∈ H p (Y ; R) and ψ ∈ H q (Y ; R).
Proof: We can check this formula already on the level of cochains. For, given
a simplex σ : ∆p+q → X, we get
(f ∗ φ ∪ f ∗ ψ)(σ) = f ∗ φ(σ|[e0 ,...,ep ] )f ∗ ψ(σ|[ep ,...,ep+q ] )
= φ(f ◦ σ|[e0 ,...,ep ] )ψ(f ◦ σ|[ep ,...,ep+q ] )
= (φ ∪ ψ)(f ◦ σ)
= f ∗ (φ ∪ ψ)(σ).
Now we are going to address the remaining natural property of multiplication:

commutativity. It will turn out that the cup product is not exactly symmetric.
This is annoying, but so is life sometimes. However, it is very close to being
symmetric. For the next result, recall that we assume that R itself is commutative.
Theorem: Cup products are graded commutative

For any classes φ ∈ H p (X; R) and ψ ∈ H q (X; R), we have
φ ∪ ψ = (−1)pq (ψ ∪ φ).
The proof of this result will require some efforts. Before we think about it,
let us collect some consequences of this theorem and of the construction of
the cup product in general.
• Many cup products are trivial just for degree reasons. For classes φ ∈
H p (X; R) and ψ ∈ H q (X; R) with p + q such that H p+q (X; R) = 0, then
φ ∪ ψ = 0 no matter what.
• This can happen for example if X is a finite cell complex.
• If φ ∈ H p (X; R) and p is odd, then
2
φ2 = (−1)p φ2 = −φ2 .
Therefore, 2φ2 = 0 in H 2p (X; R).
If R is torsion-free or if R is a field of characteristic different
from 2, this implies
φ2 = 0.
Proof for a special case: In order to find a strategy for the proof of the
theorem, let us look at a special case. So let [φ], [ψ] ∈ H 1 (X; R), and let σ : ∆2 →
X be a 2-simplex.
203
The respective cup products are then determined by their effect on a 2-simplex
σ : ∆2 → X:
(φ ∪ ψ)(σ) = φ(σ|[e0 ,e1 ] )ψ(σ|[e1 ,e2 ] ).
and
(ψ ∪ φ)(σ) = ψ(σ|[e0 ,e1 ] )φ(σ|[e1 ,e2 ] )
= φ(σ|[e1 ,e2 ] )ψ(σ|[e0 ,e1 ] )
where we use that R is commutative.
Hence in order to show that these two expressions are related, we would like
to reshuffle the vertices. As a first attempt we are going to reverse the order of
all vertices, i.e., we replace σ with σ̄ defined by
σ(ei ) = σ(e2−i ).
We will also use the notation

σ|[e2 ,e1 ,e0 ] = σ̄
which expresses
σ̄(t0 ,t1 ,t2 ) = σ(t2 ,t1 ,t0 ).
Recall that, a long time ago, we showed that reversing the order of vertices
on a 1-simplex corresponds, at least up to boundaries, multiplying the simplex
with (−1).
Hence we should consider inserting a sign as well. So we define maps
ρ1 : S1 (X) → S1 (X) and ρ2 : S2 (X) → S2 (X)
both defined by sending a simplex σ to −σ̄.
Surprisingly, the comparison of the two cup products after taking pullbacks
along the ρs becomes easier. For,
(ρ∗1 φ ∪ ρ∗1 ψ)(σ) = φ(−σ|[e1 ,e0 ] )ψ(−σ|[e2 ,e1 ] )
= φ(σ|[e1 ,e0 ] )ψ(σ|[e2 ,e1 ] )
and
(ρ∗2 (ψ ∪ φ))(σ) = −ψ(σ|[e2 ,e1 ] )φ(σ|[e1 ,e0 ] )
= −φ(σ|[e1 ,e0 ] )ψ(σ|[e2 ,e1 ] )
using that R is commutative.
Hence we get
ρ∗1 φ ∪ ρ∗1 ψ = −ρ∗2 (ψ ∪ φ).
In other words, up to ρ∗1 and ρ∗2 we have shown the desired equality.
Now we remember that we are still on the level of cochains. The theorem is
about an equality of cohomology classes. Hence all we need to show is that ρ∗1
and ρ∗2 will vanish once we pass to cohomology.
This leads to the idea to show that ρ1 and ρ2 are part of a chain map which
is chain homotopic to the identity. So let us try to do this.
First, we want that ρ1 and ρ2 commute with the boundary operator:
(ρ1 ◦ ∂)(σ) = ρ(σ|[e1 ,e2 ] − σ|[e0 ,e2 ] + σ|[e0 ,e1 ] )
= −σ|[e2 ,e1 ] + σ|[e2 ,e0 ] − σ|[e1 ,e0 ] )
= ∂(−σ|[e2 ,e1 ,e0 ] )
= (∂ ◦ ρ2 )(σ).
Now we would like to construct a chain homotopy between ρ and the identity
chain map.
The idea is to interpolate between the identity and ρ by permuting the vertices
one after the other until the order is completely reversed. Then we sum up all
these maps. Along the way we need to introduce some signs.
Before we can define maps, we need to recall the prism operator we used to
construct a chain homotopy which showed that singular homology is homotopy
invariant.
These were maps
pni : ∆n+1 → ∆n × [0,1]
determined by
(
(ek ,0) if 0 ≤ k ≤ i
pni (ek ) =
(ek−1 ,1) if k > i.
Let us write e0k := (ek ,0) and e1k := (ek ,1). Given an n-simplex σ, we would
like to compose it with pni and also permute vertices.
Consider the permutation of simplices
s
∆n+1 −
→i
∆n+1 , (e0 , . . . ,en+1 ) 7→ (e0 , . . . ,ei ,en+1 , . . . ,ei+1 ).
205
To simplify the notation, we are going to write

σ|[e0 ,...,ei ,en ,...,ei ] : ∆n+1 → X
for the n + 1-simplex defined by the composition of si with
pn pr σ
∆n+1 −→
i
∆n × [0,1] −
→ ∆n →
− X.
Now we define three maps

h0 : S0 (X) → S1 (X), σ 7→ σ|[e0 ,e0 ] ,
for n = 0,
h1 : S1 (X) → S2 (X), σ 7→ −σ|[e0 ,e1 ,e0 ] − σ|[e0 ,e1 ,e1 ] ,
for n = 1, and
h2 : S2 (X) → S3 (X), σ 7→ −σ|[e0 ,e2 ,e1 ,e0 ] + σ|[e0 ,e1 ,e2 ,e1 ] + σ|[e0 ,e1 ,e2 ,e2 ]
for n = 2.
(You will see that it does not matter so much how these maps are defined.
It is just important that we have some consistent way of moving from Sn (X) to
Sn+1 (X).)
For a 1-simplex σ : ∆1 → X, we compute
(∂ ◦ h1 )(σ) = ∂(−σ|[e0 ,e1 ,e0 ] − σ|[e0 ,e1 ,e1 ] )
= −(σ|[e1 ,e0 ] − σ|[e0 ,e0 ] + σ|[e0 ,e1 ] )
− (σ|[e1 ,e1 ] − σ|[e0 ,e1 ] + σ|[e0 ,e1 ] )
and
(h0 ◦ ∂)(σ) = h0 (σ|[e1 ] − σ|[e0 ] )
= σ|[e1 ,e1 ] − σ|[e0 ,e0 ] .
Taking these terms together we get

(∂ ◦ h1 )(σ) + (h0 ◦ ∂)(σ) = −σ|[e1 ,e0 ] + σ|[e0 ,e0 ] − σ|[e0 ,e1 ]
− σ|[e1 ,e1 ] + σ|[e0 ,e1 ] − σ|[e0 ,e1 ]
+ σ|[e1 ,e1 ] − σ|[e0 ,e0 ]
= −σ|[e1 ,e0 ] − σ|[e0 ,e1 ]
= ρ(σ) − σ.
Thus, we have shown the homotopy relation
∂ ◦ h1 + h0 ◦ ∂ = ρ1 − id.
Similarly, for a 2-simplex σ : ∆2 → X, we calculate
(∂ ◦ h2 )(σ) = ∂(−σ|[e0 ,e2 ,e1 ,e0 ] − σ|[e0 ,e1 ,e2 ,e1 ] + σ|[e0 ,e1 ,e2 ,e2 ] )
= −(σ|[e2 ,e1 ,e0 ] − σ|[e0 ,e1 ,e0 ] + σ|[e0 ,e2 ,e0 ] − σ|[e0 ,e2 ,e1 ] )
+ (σ|[e1 ,e2 ,e1 ] − σ|[e0 ,e2 ,e1 ] + σ|[e0 ,e1 ,e1 ] − σ|[e0 ,e1 ,e2 ] )
+ (σ|[e1 ,e2 ,e2 ] − σ|[e0 ,e2 ,e2 ] + σ|[e0 ,e1 ,e2 ] − σ|[e0 ,e1 ,e2 ] )
= −σ|[e2 ,e1 ,e0 ] + σ|[e0 ,e1 ,e0 ] − σ|[e0 ,e2 ,e0 ] + σ|[e1 ,e2 ,e1 ] + σ|[e0 ,e1 ,e1 ]
+ σ|[e1 ,e2 ,e2 ] − σ|[e0 ,e2 ,e2 ] − σ|[e0 ,e1 ,e2 ]
and
(h1 ◦ ∂)(σ) = h1 (σ|[e1 ,e2 ] − σ|[e0 ,e2 ] + σ|[e0 ,e1 ] )
= −σ|[e1 ,e2 ,e1 ] − σ|[e1 ,e2 ,e2 ] + σ|[e0 ,e2 ,e0 ] + σ|[e0 ,e2 ,e2 ] − σ|[e0 ,e1 ,e0 ] − σ|[e0 ,e1 ,e1 ] .
This gives
(∂ ◦ h2 + h1 ◦ ∂)(σ) = ρ2 (σ) − σ.
Thus, we have again shown the homotopy relation

∂ ◦ h2 + h1 ◦ ∂ = ρ2 − id.
This indicates that ρ1 and ρ2 are part of a chain map which is chain homotopic
to the identity.
To prove the general case, we adapt this strategy we developed for n = 1.

Proof of the theorem: When we evaluate the two cup products on a simplex
σ : ∆p+q → X, they differ only by a permutation of the vertices of σ. The idea
of the proof consists of
• choosing a nice permutation which simplifies notation and computations,
• and then to construct a chain homotopy between the resulting cup prod-
uct and the identity.
Now let us get to work:
• For an n-simplex σ, let σ̄ be the n-simplex obtained by composing it first
with the linear transformation which reverses the order of the vertices.
In other words,
σ̄(ei ) = σ(en−i ) for all i = 0, . . . ,n.
207
or
σ̄(t0 , . . . , tn ) = σ(tn , . . . , t0 ).
We will also use the notation

σ|[en ,...,e0 ] = σ̄.
For this will make it easier to combine it with the restriction to the n − 1-
dimensional faces of ∆n .
• Since the reversal of the vertices is the product of n + (n − 1) + · · · + 1 =

n(n + 1)/2 many transpositions, our test case motivates the definition of the
homomorphism
n(n+1)
ρn : Sn (X) → Sn (X), ρn (σ) = (−1) 2 σ̄.
n(n+1)
To simplify the notation we will write ϵn := (−1) 2 .
• We claim that ρ is a map of chain complexes which is chain homotopic

to the identity map. Assuming that the claim is true we can finish the proof
of the theorem as follows.
For σ : ∆p+q → X, we can then calculate
(ρ∗ φ ∪ ρ∗ ψ)(σ) = φ(ϵp σ|[ep ,...,e0 ] )ψ(ϵq σ|[ep+q ,...,ep ] )
= ϵp ϵq φ(σ|[ep ,...,e0 ] )ψ(σ|[ep+q ,...,ep ] )
and
(ρ∗ (ψ ∪ φ))(σ) = ϵp+q ψ(σ|[ep+q ,...,ep ] )φ(σ|[ep ,...,e0 ] ).
Now we observe
(p + q)(p + q + 1) p2 + 2pq + q 2 + p + q
=
2 2
p(p + 1) q(q + 1) 2pq
= + +
2 2 2
p(p + 1) q(q + 1)
= + + pq.
2 2
Thus
ϵp+q = (−1)pq ϵp ϵq .
We conclude from these two computations
ρ∗ φ ∪ ρ∗ ψ = (−1)pq ρ∗ (ψ ∪ φ).
Now we use that ρ is chain homotopic to the identity. That implies that
when we pass to cohomology classes, ρ∗ is the identity and we obtain the desired
identity
φ ∪ ψ = (−1)pq (ψ ∪ φ).
Now we are going to prove the claims we made:

• ρ is a chain map.
We need to show that ∂ ◦ ρ = ρ ◦ ∂. For an n-simplex σ we calculate the
effects of the two maps:
n
X
(ρ ◦ ∂)(σ) = ρ( (−1)i σ|[e0 ,...,êi ,...,en ] )
i=0
n
X
= ϵn−1 (−1)i σ|[en ,...,êi ,...,e0 ]
i=0
n
X
= ϵn−1 (−1)n−i σ|[en ,...,ên−i ,...,e0 ] by changing the order of summation
i=0
n
X
= ϵn−1 (−1)i−n σ|[en ,...,ên−i ,...,e0 ] using (−1)j = (−1)−j
i=0
n
X
= ϵn−1 (−1) n
(−1)i σ|[en ,...,ên−i ,...,e0 ] again using (−1)n = (−1)−n
i=0
n
X
= ϵn (−1)i σ|[en ,...,ên−i ,...,e0 ]
i=0
= ∂(ϵn σ|[en ,...,e0 ] )
= (∂ ◦ ρ)(σ)
where we used the identity ϵn = (−1)n ϵn−1 .
• There is a chain homotopy between ρ and the identity.
We are going to use again the notation we introduced for the special case
above. The idea for the chain homotopy is to interpolate between ρ which reverses
the order of all vertices and the identity by, step by step, reversing the order up
209
to some vertex while the others remain fixed. Then we throw in some signs to
make things work.
We define homomorphisms hn for each n by
hn : Sn (X) → Sn+1 (X)
X
σ 7→ (−1)i ϵn−i σ|[e0 ,...,ei ,en ,...,ei ] .
i=0
Now we can show by calculating ∂ ◦hn and hn−1 ◦∂ that h is a chain homotopy,
i.e., we have
∂ ◦ hn + hn−1 ◦ ∂ = ρ − id.
We have
X
(∂ ◦ hn )(σ) = ∂( (−1)i ϵn−i σ|[e0 ,...,ei ,en ,...,ei ] )
i=0
X
= (−1)i (−1)j ϵn−i σ|[e0 ,...,êj ,...,ei ,en ,...,ei ]
j≤i
X
+ (−1)i (−1)i+1+n−j ϵn−i σ|[e0 ,...,ei ,en ,...,êj ,...,ei ] .
j≥i
The overlap of the summation indices is necessary. For, at j = i, only the two
sums together yield all the summands we need:
X
ϵn σ|[en ,...,e0 ] + ϵn−i σ|[e0 ,...,ei−1 ,en ,...,ei ]
i>0
X
n+i+1
+ (−1) ϵn−i σ|[e0 ,...,ei ,en ,...,ei+1 ] − σ|[e0 ,...,en ] .
i<n
Now we observe that the two sums in the last expression cancel out, since if
we replace i by i − 1 in the second sum turns the sign into
(−1)n+i ϵn−i+1 = −ϵn−i .
Hence, for j = i, what remains is exactly

ϵn σ|[en ,...,e0 ] − σ|[e0 ,...,en ] = ρ(σ) − σ.
Hence it suffices to show that the terms with j ̸= i in (∂ ◦ hn )(σ) cancel out
with (hn−1 ◦ ∂)(σ). So we calculate
X
(hn−1 ◦ ∂)(σ) = hn−1 ( (−1)j σ|[e0 ,...,êj ,...,en ] )
j=0
X
= (−1)i−1 (−1)j ϵn−i σ|[e0 ,...,êj ,...,ei ,en ,...,ei ]
j<i
X
+ (−1)i (−1)j ϵn−i−1 σ|[e0 ,...,ei ,en ,...,êj ,...,ei ] .
j>i
Since ϵn−i = (−1)n−i ϵn−i−1 , the two sums cancel with the two corresponding
sums in (∂ ◦ hn )(σ). Hence h is a chain homotopy between ρ and the identity.
□
LECTURE 21
Applications of cup products in cohomology
We are going to see some examples where we calculate or apply multiplicative

structures on cohomology. But we start with a couple of facts we forgot to
mention last time.
Relative cup products
Let (X,A) be a pair of spaces. The formula which specifies the cup product
by its effect on a simplex
(φ ∪ ψ)(σ) = φ(σ|[e0 ,...,ep ] )ψ(σ|[ep ,...,ep+q ] )
extends to relative cohomology.

For, if σ : ∆p+q → X has image in A, then so does any restriction of σ. Thus,
if either φ or ψ vanishes on chains with image in A, then so does φ ∪ ψ.
Hence we get relative cup product maps
H p (X; R) × H q (X,A; R) → H p+q (X,A; R)

H p (X,A; R) × H q (X; R) → H p+q (X,A; R)
H p (X,A; R) × H q (X,A; R) → H p+q (X,A; R).
More generally, assume we have two open subsets A and B of X. Then the
formula for φ ∪ ψ on cochains implies that cup product yields a map
S p (X,A; R) × S q (X,B; R) → S p+q (X,A + B; R)
where S n (X,A+B; R) denotes the subgroup of S n (X; R) of cochains which vanish

on sums of chains in A and chains in B.
The natural inclusion
S n (X,A ∪ B; R) ,→ S n (X,A + B; R)
211
212 CHAPTER 21. APPLICATIONS OF CUP PRODUCTS IN COHOMOLOGY
induces an isomorphism in cohomology. For we have a map of long exact coho-
mology sequences
H n (A ∪ B) / H n (X) / H n (X,A ∪ B) / H n+1 (A ∪ B) / H n+1 (X)

H n (A + B) / H n (X) / H n (X,A + B) / H n+1 (A + B) / H n+1 (X)
where we omit the coefficients. The small chain theorem and ∼
our results on
=
cohomology of free chain complexes imply that H n (A ∪ B; R) −
→ H n (A + B; R)
is an isomorphism for every n. Thus, the Five-Lemma implies that
∼
=
H n (X,A ∪ B; R) −
→ H n (X,A + B; R)
is an isomorphism as well.
Thus composition with this isomorphism gives a cup product map
H p (X,A; R) × H q (X,B; R) → H p+q (X,A ∪ B; R).
Now one can check that all the formulae we proved for the cup product also
hold for the relative cup products.
Cohomology ring
All we are going to say now also works for relative cohomology. But to keep
things simple, we just describe the absolute case.
We will now often drop the symbol ∪ to denote the cup product and just
write
αβ = α ∪ β.
The cohomology ring of a space X is the defined as

M
H ∗ (X; R) = H n (X; R)
n
as the direct sum of all cohomology groups. Note that, while the symbol ∗
previously often indicated that something holds for an arbitrary degree, we now
use it to denote the direct sum over all degrees.
The product of two sums is defined as
X X X
( αi )( βj ) = α i βj .
i j i,j
213
This turns H ∗ (X; R) into a ring with unit, i.e., multiplication is associative,
there is a multiplicatively neutral element 1, and addition and multiplication
satisfy the distributive law.
We consider the cohomological degree n in H n (X; R) as a grading of H ∗ (X; R).
If an element α is in H p (X; R) we call p the degree of α and denote it also by |α|.
Since multiplication respects this grading in the sense that it defines a map
H p (X; R) × H q (X; R) → H p+q (X,A; R),
we call H ∗ (X; R) a graded ring.
Moreover, as we have shown with a lot of effort last time, the multiplication
is commutative up to a sign which depends on the grading:
αβ = (−1)|α||β| βα.
Hence H ∗ (X; R) a graded commutative ring.
Moreover, there is an obvious scalar multiplication by elements in R which
turns H ∗ (X; R) into a graded R-algebra.
Finally, if f : X → Y is a continuous map, then the induced map on coho-
mology
f ∗ : H ∗ (Y ; R) → H ∗ (X; R)
is a homomorphism of graded R-algebras.
Now we should determine some ring structures and see what they can tell us.
As a first, though disappointing, example, let us note that the product in
the cohomology of a sphere S n (with n ≥ 1) is boring, since H 0 (S n ; R) is just R
and the product on H n (S n ; R) is trivial for reasons of degrees:
H n (S n ; R) × H n (S n ; R) → H 2n (S n ; R) = 0.
So let us move on to more interesting cases.
Cohomology ring of the torus

Even though the cohomology ring of S 1 was boring, the cohomology ring of
the product T = S 1 × S 1 , i.e., of the torus, is not. Let us assume R = Z.
We computed the homology of T using its structure as a cell complex with
one 0-cell, two 1-cells, and one 2-cell.
The cellular chain complex has the form
2 d 1 d
0→Z−
→ Z⊕Z−
→ Z→0
where d1 (a,b) = a + b and d2 (s) = (s, − s) (the attaching map of the 2-cell to the
two 1-cells was aba−1 b−1 ). This yields the homology of T .
We can then apply the UCT to deduce that the singular cohomology of T is
given by

Z
 if i = 0
i
H (T ; Z) = Z ⊕ Z if i = 1

Z if i = 2
and H i (T ; Z) = 0 for i > 2.

Let α and β be generators of H 1 (T ; Z). We could obtain them for example as
the dual of the basis {a,b} of H1 (T ; Z) and the isomorphism of the UCT:
H 1 (T ; Z) = Hom(H1 (T ; Z),Z).
Being a dual basis means, in particular,
α(a) = ⟨α,a⟩ = 1, α(b) = ⟨α,b⟩ = 0, β(a) = ⟨β,b⟩ = 0, β(b) = ⟨β,b⟩ = 1
where the funny brackets denote the Kronecker pairing we had defined earlier.
Since multiplication is graded commutative, we have
2α2 = 0 = 2β 2 .
Since Z is torsion-free, this implies
α2 = 0 = β 2 .
Now we would like to understand the product αβ. Therefore, we need to

evaluate it on a generator of H2 (T ; Z). Such a generator is given by the 2-chain
σ − τ , where σ and τ are the 2-simplices indicated in the picture (that this is a
generator needs to be checked; we just accept this for the moment):
215
It is a cycle, since
∂(σ − τ ) = ∂(σ) − ∂(τ ) = b − d + a − (a − d + b) = 0
where d denotes the diagonal.
Now we can calculate
(α ∪ β)(σ − τ ) = α(σ|[e0 ,e1 ] )β(σ|[e1 ,e2 ] ) − α(τ|[e0 ,e1 ] )β(τ|[e1 ,e2 ] )
= α(a)β(b) − α(b)β(a)
= 1 − 0 = 1.
Thus, since H 2 (T ; Z) = Hom(H2 (T ; Z),Z) by the UCT, we see that αβ is a

generator of H 2 (T ; Z).
Hence we can conclude that the cohomology ring of the torus is the ring with
generators α and β and relations
H ∗ (T ; Z) = Z{α,β}/⟨α2 = 0 = β 2 , αβ = −βα⟩.
Another way to formulate this is to say that H ∗ (T ; Z) is the exterior algebra

over Z with generators α and β:
H ∗ (T ; Z) = ΛZ [α,β].
In general, the exterior algebra ΛR [α1 , . . . ,αn ] over a commutative ring R with
unit is defined as the free R-module with generators αi1 · · · αik for i1 < · · · < ik
with associative and distributive multiplication defined by the rules
αi αj = −αj αi if i ̸= j, and αi2 = 0.
Setting Λ0 = R, ΛR [α1 , . . . ,αn ] becomes a graded commutative ring with odd
degrees for the αi s and unit 1 ∈ R.
For the n-torus T n = S 1 × · · · × S 1 , defined as the n-fold product of S 1 , we
then get
H ∗ (T n ; Z) = ΛZ [α1 , . . . ,αn ].
Cohomology of projective spaces

The cohomology rings of projective spaces are truncated polynomial algebras:
Cohomology rings of RPn and CPn

• For every n ≥ 1 and F2 -coefficients, we have an isomorphism of graded
rings
H ∗ (RPn ; F2 ) ∼
= F2 [x]/(xn+1 ), and H ∗ (RP∞ ; F2 ) ∼
= F2 [x]
with |x| = 1.
• For every n ≥ 1 and integral coefficients, we have an isomorphism of
graded rings
H ∗ (CPn ; Z) ∼
= Z[y]/(y n+1 ), and H ∗ (CP∞ ; Z) ∼
= Z[y]
with |y| = 2.
The proof of this result requires some efforts. We will postpone its proof and
rather see some consequences of it.
Cup products detect more
Consider the wedge of spheres S 2 ∨ S 4 . We know that its homology is given
by
H̃∗ (S 2 ∨ S 4 ; Z) = H̃∗ (S 2 ; Z) ⊕ H̃∗ (S 4 ; Z).
In other words,
(
Z if i = 0, 2, 4
Hi (S 2 ∨ S 4 ; Z) =
0 else.
Hence the homologies of CP2 and S 2 ∨ S 4 are the same. Since all the groups
are free, this also implies that the cohomology groups of the two spaces are the
same. Thus, neither homology nor cohomology groups can distinguish between
these two spaces.
The cup product, however, can.
217
For, we know that the square of a generator in H n (S n ; Z) is zero, since

H (S n ; Z) = 0. Thus
2n
H ∗ (S n ; Z) = Z[t]/(t2 = 0) with |t| = n,
and hence we have a generator s ∈ H 2 (S 2 ∨ S 4 ; Z) with s2 = 0 and a generator

t ∈ H 2 (S 2 ∨ S 4 ; Z) with t2 = 0.
If there was an isomorphism of graded Z-algebras
H̃ ∗ (CP2 ; Z) ∼
= H̃ ∗ (S 2 ∨ S 4 ; Z)
it would have to send the generator y ∈ H 2 (CP2 ; Z) to the generator s ∈ H 2 (S 2 ∨

S 4 ; Z). But y 2 ̸= 0 in H 4 (CP2 ; Z), whereas s2 = 0 in H 4 (S 2 ∨ S 4 ; Z).
Thus, such an isomorphism of graded rings cannot exist.
Thus, the cup product structures show that there does not exist a homotopy
equivalence between CP2 and S 2 ∨ S 4 , something our previous invariants could
not prove.
Hopf maps
As an important application of what we just learned, we consider the following
situation.
Many problems can be reduced to checking whether a map is null-homotopic,
i.e., homotopic to a constant map, or not.
Given a map f : X → Y , we can form the mapping cone Cf = CX ∪f Y
(which we introduced in the exercises). It is the pushout of the diagram
X × {1} / CX
f

Y
i / Cf .
If f is homotopic to a constant map, then the diagram is equivalent to the diagram
pt / CX/(X × {1})

Y
i / SX ∨ Y
where we use that CX/(X × {1}) is the suspension SX of X.
Thus, if f is null-homotopic, then there is a homotopy equivalence
≃
Cf −
→ SX ∨ Y.
Let us look at an example. Let

η : S 3 → CP1 ≈ S 2 , x 7→ [Cx] = {λx ∈ C2 : λ ∈ C}
be the complex Hopf map which sends a point x ∈ S 3 ⊂ C2 to the complex
line in C2 which passes through x.
This is exactly the map which attaches the 4-cell to CP1 ≈ S 2 in the cell
structure of CP2 . The mapping cone Cη of η is CP2 , since the cone of S 3 is just
D4 :
CS 3 = (S 3 × [0,1])/(X × {0}) ≈ D4
and hence
Cη = CS 3 ∪η S 2 ≈ D4 ∪η S 2 ≈ D4 ∪η CP1 ≈ CP2 .
Now we use that we showed in the exercises that the suspension of S 3 is

homeomorphic to S 4 . Thus, if η was null-homotopic, then the argument above
would imply
≃
CP2 ≈ Cη −
→ S 2 ∨ S 4.
But we just showed that such a homotopy equivalence cannot exits. Thus, η
is not null-homotopic.
More Hopf maps

Note that there is also a quaternionic Hopf map
ν : S 7 → S 4,
and an octonionic Hopf map
σ : S 15 → S 8 .
They are constructed in the same way as η by replacing C with the quater-
nions H and the octonions O, respectively. There are corresponding projec-
tive spaces HPn and OPn with HP1 ≈ S 4 and OP1 ≈ S 8 , and polynomial
rings as cohomology rings:
H ∗ (HP2 ; Z) = Z[z]/(z 3 ), |z| = 4, and H ∗ (OP2 ; Z) = Z[w]/(w3 ), |w| = 8.
The homotopy classes of η, ν and σ
[η] ∈ π3 (S 2 ), [ν] ∈ π7 (S 4 ), [σ] ∈ π15 (S 8 )
219
play a crucial role in the stable homotopy category.
Is there a multiplication on Rn ?
For the next application, we are going to assume one more result, namely that
the cohomology ring of the product of RPn × RPn is given
H ∗ (RPn × RPn ; Z/2) ∼
= F2 [α1 ,α2 ]/(α1n+1 ,α2n+1 ).
This implies the following algebraic fact:
Theorem: Multiplication on Rn
Assume there is a R-bilinear map
µ : Rn × Rn → Rn
such that µ(x, y) = 0 implies x = 0 or y = 0.
Then n must be a power of 2.
In fact, n must be 1, 2, 4 or 8. In all these dimensions we have such multipli-

cations by identifying
R2 ∼
= C, R4 ∼
= H, R8 ∼
= O.
But to show that there are no other such algebra structures on Rn is a much
harder task. The only known proofs of this fact are using algebraic topology!
In fact, for showing this we need to study the famous Hopf Invariant One-
Problem. This is beyond the scope of this lecture. So let us be modest and just
prove the result stated above.
Proof: • Since µ is linear in both variables, it induces a continuous map
µ̄ : RPn−1 × RPn−1 → RPn−1 .
Then µ̄ induces a homomorphism of cohomology rings which has the form

µ̄∗ : F2 [α]/(αn ) → F2 [α1 ,α2 ]/(α1n ,α2n ).
• Since µ does not have a zero-divisor, the restriction of µ to Rn × {a} for

any a ∈ Rn is an isomorphism. Hence the restriction of µ̄ to RPn−1 × {y} for any
point y ∈ RPn−1 is a homeomorphism.
This implies that the composite
µ̄
RPn−1 → RPn−1 × {y} ,→ RPn−1 × RPn−1 →
− RPn−1
is a homeomorphism as well. Hence the induced homomorphism of cohomology
rings must send α to α.
Repeating this argument for {y} × RPn−1 , we see that the image of α under
µ̄∗ must be
µ̄∗ (α) = α1 + α2 .
Since both rings are polynomial algebras, µ̄∗ is completely determined by this
identity.
• Since αn = 0, we must have µ̄∗ (α)n = 0, i.e.,
X n
n
(α1 + α2 ) = α1k α2n−k = 0.
k
k
The sum on the right-hand side can only be zero if all the coefficients of the
monomials α1k α2n−k vanish for 0 < k < n. Since we are working over F2 , this
means that all the numbers nk for 0 < k < n must be even.
To prove this fact is equivalent to proving the following claim about the poly-
nomial ring F2 [x]:
• Claim: In F2 [x], we have
(1 + x)n = 1 + xn ⇐⇒ n is a power of 2.
First, if n is a power of 2, then the equation (a + b)2 = a2 + b2 modulo 2 shows

the if part:
r r−1 2 r−2 r
(1 + x)2 = (1 + x2 )2 = (1 + x2 )2 = · · · = 1 + x2 in F2 [x].
For the other direction, write n as

n = 2r m with m odd and m > 1.
Then
rm r r
(1 + x)n = (1 + x)2 = (1 + x2 )m = 1 + mx2 + . . . + xn ̸= 1 + xn in F2 [x]
since m is odd. □
LECTURE 22
Poincaré duality and intersection form
We are going to meet an important class of topological spaces and study one of
their fundamental cohomological properties. This lecture will be short of proofs,
but rather aims to see an important theorem and structures at work.
Manifolds
We start with defining an important class of spaces.
Definition: Topological manifolds

A n-dimensional topological manifold is a Hausdorff space in which each
point has an open neighborhood which is homeomorphic to Rn .
In this lecture, the word manifold will always mean a topological manifold.
You know many examples of manifolds, most notably Rn itself, any open
subset of Rn , n-spheres S n , tori, Klein bottle, projective spaces. Even though the
definition does not refer to this information, any manifold M can be embedded
in some RN for some large N (which depends on M ).
Though it is a crucial point that N and n can and usually are different. For
example, S 2 is a subset of R3 , but each point on S 2 has a neighborhood which
looks like a plane, i.e., is homeomorphic to R2 .
221
222 CHAPTER 22. POINCARÉ DUALITY AND INTERSECTION FORM
There are many reasons why manifolds are important. One of them is that
we understand and can study them locally, while they can be very complicated
globally.
Poincaré duality
In this lecture, all homology and cohomology groups will be with F2 -coefficients.
Recall that there is a pairing
⟨−,−⟩
H k (X; F2 ) ⊗ Hk (X; F2 ) −−−→ F2
defined by evaluating a cocycle φ on a cycle σ which is an element in F2 .

We are going to study the consequences of the following famous fact:
223
Theorem: Poincaré duality mod 2

Let M be a compact topological manifold of dimension n. Then there
exists a unique class [M ] ∈ Hn (M ; F2 ), called the fundamental class of
M , such that, for every p ≥ 0, the pairing
∪ ⟨−,[M ]⟩
H p (M ; F2 ) ⊗ H n−p (M ; F2 ) −
→ H n (M ; F2 ) −−−−→ F2
is perfect.
That the pairing is perfect means that the adjoint map

⟨a∪−,[M ]⟩
H p (X; F2 ) −−−−−−→ Hom(H n−p (X; F2 ),F2 ), a 7→ ⟨a ∪ −,[M ]⟩
is an isomorphism.
Here are some first consequences of this theorem:
• Since cohomology vanishes in negative dimensions, we must have H p (X; F2 ) =
0 for p > n as well.
• Since M is assumed to be compact, we know that π0 (M ), the set of
connected components of M , is finite. Moreover, we once showed that
H 0 (M ; F2 ) equals Map(π0 (M ),F2 ). Hence we get
H n (M ; F2 ) = Hom(H 0 (M ; F2 ),F2 ) = Hom(Map(π0 (M ),F2 ),F2 ) = F2 [π0 (M )].
• A vector space admitting a perfect pairing is finite-dimensional. Hence

H p (M ; F2 ) is finite-dimensional for all p.
There is a version of the Universal Coefficient Theorem with F2 -coefficients.

Since F2 is a field, it implies that there is an isomorphism
Hom(H n−p (M ; F2 ),F2 ) ∼

= Hn−p (M ; F2 ).
(Note that we formulated the UCT with the roles of homology and cohomology
reversed. But, since the map arose from the Kronecker pairing, we can also
produce the claimed version of the UCT. As mentioned in the intro to this lecture,
we rush through some points for the sake of telling a good story.)
Composition with the above pairing yields an isomorphism
∼ ∼
H p (X; F2 )
= / Hom(H n−p (M ; F2 ),F2 ) o
=
Hn−p (M ; F2 ).
2
∼
=
Definition: Poincaré duals

Homology and cohomology corresponding to each other under the dotted
isomorphism are said to be Poincaré dual to each other.
Intersection pairing
Combining this isomorphism for different dimensions, we can write the cup
product pairing in cohomology as a pairing in homology (where we drop the
coefficients which are still F2 )
Hp (M ) ⊗ Hq (M )
⋔ / Hp+q−n (M )
∼
= ∼
=

∪ /
H n−p (M ) ⊗ H n−q (M ) H 2n−p−q (M ).
The top map is called the intersection pairing in homology.

Here is how we should think about it:
• Let α ∈ Hp (M ) and β ∈ Hq (M ) be homology classes.
• Represent them, if possible, as the image of fundamental classes of sub-

manifolds of M . That means that there are submanifolds Y and Z in M
of dimensions p and q, respectively, such that
α = i∗ [Y ] and β = j∗ [Z]
where i∗ : Hp (Y ) → Hp (M ) and j∗ : Hq (Z) → Hq (M ) are the homomor-

phisms induced by the inclusions i : Y ,→ M and j : Z ,→ M .
• Move them a bit if necessary to make them intersect transversally.
• Then their intersection is a submanifold of dimension p + q − n and its

image will represent the homology class α ⋔ β.
Let us look at an example:

225
Example: Intersection on a torus

Let M = T 2 = S 1 × S 1 be the two-dimensional torus. We know
H 1 (M ) = F2 ⟨a,b⟩
with a2 = 0 = b2 , and H 2 (M ) is generated by ab = ba.
The Poincaré duals α and β of a and b are represented by cycles which
wrap around one or the other factor circle of M .
The cycles α and β can be made to intersect in a single point. This reflects
the equation
⟨a ∪ b, [M ]⟩ = 1.
But this equation also tells us that α and β can only be moved in such a
way that they intersect in an odd number of points.
The fact that a2 = 0 reflects that the fact that its Poincaré dual α can be
moved so as not to intersect itself.
Intersection form
Let us look at a particular case of Poincaré duality. Let us assume that M
is even-dimensional, say of dimension n = 2p. Then Poincaré duality implies
that we have a symmetric bilinear form on the F2 -vector space H p (M ):
H p (M ) ⊗F2 H p (M ) → H 2p (M ) ∼
= F2 .
As we just observed, this can be interpetred as a bilinear form on homology

Hp (M ). Evaluating this form can be viewed as describing (modulo 2) the number
of points where two p-cycles intersect, after they have been moved in general
position, i.e., a position where they intersect transversally.
Definition: Intersection form

This form Hp (M ) ⊗ Hp (M ) → F2 is called intersection form and will be
denoted
α · β := ⟨a ∪ b, [M ]⟩
where a and b are Poincaré dual to α and β, respectively.
Let us consider two examples:

• For the sphere S 2 , the first homology is trivial, and so is the intersection
form on S 2 .
• In the example of the torus, the intersection form can be described
in terms of the basis α and β by the matrix (since any such form looks
like (v,w) 7→ v T Hw)

0 1
H= .
1 0
Such a form is called hyperbolic.
Apparently, it would good to know a bit more about such forms. We are
going to review what we need to know about them now and then get back to the
application in topology in the next lecture.
A digression on symmetric bilinear forms

We need to have a brief look at such forms.
So let V be a finite-dimensional vector space over F2 together with a nonde-
generate symmetric bilinear form. Such a form restricts to any subspace W of V ,
but the restricted form may be degenerate. But any subspace has an orthogonal
complement
W ⊥ = {v ∈ V : v · w = 0 for all w ∈ W }.
Then we have the following lemma:
Lemma
The restriction of a nondegenerate symmetric bilinear form on V to a sub-
space W is nondegenerate if and only if W ∩ W ⊥ = 0.
227
In this case, the restriction to W ⊥ is also nondegenerate and the splitting

V ∼
= W ⊕ W⊥
respects the forms.
We can use this lemma to inductively decompose all finite-dimensional sym-

metric bilinear forms:
• If there is a vector v ∈ V with v · v = 1, then it generates a nondegen-
erate subspace, i.e., a subspace on which the restriction of the form is
nondegenerate, and
V = ⟨v⟩ ⊕ ⟨v⟩⊥
where ⟨v⟩ denotes the subspace generated by v.
• Continue to split off one-dimensional subspaces until we reach a nonde-
generate symmetric bilinear form such that v · v = 0 for all vectors.
• Unless we ended up with zero space, we can pick a nonzero vector v.
Since the form is nondegenerate, there must be a vector w such that
v · w = 1.
• The two vectors v and w generate a hyperbolic subspace, i.e., one on
which the form is represented by the matrix

0 1
H= .
1 0
• Split this space off, and continue the process.

This procedure shows:
Proposition: Classification of nondegenerate forms

Any finite-dimensional nondegenerate symmetric bilinear form over F2 splits
as an orthogonal sum of forms with matrices

0 1
I = (1) and H = .
1 0
We are going to continue the study of forms and get back to topology in the
next lecture.
LECTURE 23
Classification of surfaces
We will first continue the study of bilinear forms, and then use this knowl-
dege to classify all compact connected surfaces, i.e., compact connected two-
dimensional manifolds. Then we are going to contemplate a bit more on Poincaré
duality. In this lecture, all vector spaces, homology and cohomology groups will
be over F2 .
The monoid of nondegenerate symmetric bilinear forms

Last time we showed:
Proposition: Classification of nondegenerate forms

Any finite-dimensional nondegenerate symmetric bilinear form over F2 splits
as an orthogonal sum of forms with matrices

0 1
I = (1) and H = .
1 0
Now let Bil be the set of isomorphism classes of nondegenerate symmetric

bilinear forms over F2 . This is a commutative monoid under the operation of
taking orthogonal direct sums (that means it is like a group except that there no
inverses).
Since any such form corresponds to a matrix, we can identify Bil also with the
set of invertible symmetric matrices modulo the equivalence relation
of similarity:
• Two matrices A and B are called similar, denoted A ∼ B, if B = P AP T
for some invertible matrix P .
• Every form corresponds to a matrix A determined by v · w = v T Aw.
• Assume we have given two vector spaces V1 and V2 with nondegenerate
symmetric bilinear forms which are represented by matrices A1 and A2 ,
229
230 CHAPTER 23. CLASSIFICATION OF SURFACES
∼
=
respectively. Then there is an isomorphism φ : V −
→ W such that
φ(v ·V w) = φ(v) ·W φ(w),
if and only if A1 and A2 are similar.
Hence, in order to understand Bil we can aim to understand invertible ma-
trices modulo similarity. Here is a crucial fact:
Lemma
Over F2 we have the similarity
   
0 1 0 1 0 0
1 0 0 ∼ 0 1 0 .
0 0 1 0 0 1
Proof: The assertion is equivalent to saying there is an invertible matrix P

such that
 
0 1 0
1 0 0 = P P T .
0 0 1
 
1 1 0
This is the case for P = 1 0 1 where we need to remember that we work
1 1 1
over F2 . QED
Since neither nI nor mH are similar to other matrices, I + H = 3I is the only
relation. As a consequence we get:
Bilinear forms via generators and relations

The commutative monoid Bil is generated by I and H modulo the relation
I + H = 3I.
Now we are going to apply this knowledge to the intersection form.
Intersection form
Let us look at a particular case of Poincaré duality. Let us assume that M is
even-dimensional, say of dimension n = 2p. Then Poincaré duality defines a a
231
symmetric bilinear form on the F2 -vector space H p (M ):
H p (M ) ⊗F2 H p (M ) → H p (M ).
As we observed last time, this can be interpetred as a bilinear form on ho-

mology Hp (M ). Recall that evaluating this form can be viewed as describing
(modulo 2) the number of points where two p-cycles intersect, after they have
been moved in general position, i.e., a position where they intersect transversally.
Intersection form
For a compact manifold of dimension n = 2p, the intersection pairing
Hp (M ; F2 ) ⊗ Hp (M ; F2 ) → F2 , α · β := ⟨a ∪ b, [M ]⟩
defines a nondegenerate symmetric bilinear form on Hp (M ; F2 ), called the
intersection form. Here a and b are Poincaré dual to α and β, respectively.
• We have seen the example of the torus for which the intersection form
is hyperbolic, i.e., can be described in terms of the basis α and β by
the matrix

0 1
H= .
1 0
• For another example, take M = RP2 . We know H1 (RP2 ) = F2 . More-

over, RP2 can be viewed as a Möbius band with a disk glued along
the boundary. On the Möbius band, there is a nontrivial intersection.
Hence the intersection form is nontrivial and therefore given by I
according to our classification, since on a one-dimensional space there
are only two options. As a consequence we see that in whatever way
we try to move the boundary of the Möbius band in RP2 , it will always
intersect itself in an odd number of points.
Note that the open Möbius band itself is a two-dimensional manifold, but
it is not compact. While the closed Möbius is compact, it is not a manifold
according to the definition we stated last time. Though it is a manifold with
boundary. The story is different if we allow boundaries.
Connected sums
There is an interesting geometric operation on manifolds which produces
new ones out of old:
Given two compact connected manifolds M1 and M2 both of dimension n.
Then we can
• cut out a small open n-dimensional disk Dn of each one, and
• sew them together along the resulting boundary spheres S n−1 , i.e., iden-
tify the boundaries via a homeomorphism.
• The resulting space is called the connected sum of M1 and M2 and
is denoted by M1 #M2 . Note M1 #M2 is a connected compact n-
dimensional manifold.
Let us see two examples:
• There is not much happening if we take S 2 #S 2 as it is homeomorphic
to S 2 :
233
• But we get a new surface for T 2 #T 2 :
Lemma: Homology of connected sums

There is an isomorphism
Hi (M1 #M2 ) ∼
= Hi (M1 ) ⊕ Hi (M2 ) for all 0 < i < n.
Proof: We start with the pair (M1 #M2 , S n−1 ). Since M1 and M2 are man-
ifolds of dimension n, there is an open neighborhood around S n−1 in M1 #M2
which retracts onto S n−1 . Thus, by a result we showed some time ago when we
discussed cell complexes and wedge sums, we know
H∗ (M1 #M2 , S n−1 ) ∼
= H∗ ((M1 #M2 )/S n−1 ,pt) ∼
= H̃∗ (M1 ∨ M2 ).
Now we consider the long exact sequence of the pair (M1 #M2 , S n−1 ):
· · · → H̃i (S n−1 ) → Hi (M1 #M2 ) → Hi (M1 #M2 , S n−1 ) → H̃i−1 (S n−1 ) → · · ·
Since only H̃n−1 (S n−1 ) is nonzero, we deduce

H̃i (M1 #M2 ) ∼
= Hi (M1 #M2 , S n−1 ) ∼
= H̃∗ (M1 ∨ M2 ) for all i < n − 1.
Hence, for 0 < i < n − 1, we have

Hi (M1 #M2 ) ∼
= Hi (M1 ) ⊕ Hi (M2 ).
The remaining part of the long exact sequence is then

0 → Hn (M1 #M2 ) → Hn (M1 ∨ M2 ) → Hn−1 (S n−1 ) → Hn−1 (M1 #M2 ) → Hn−1 (M1 ∨ M2 ) → 0
where the zeros on both ends are explained by the vanishing of the corresponding
homologies of S n−1 .
Since fundamental classes are natural, the map
∼
=
(30) → Hn (M1 ∨ M2 ) → Hn−1 (S n−1 )
Hn (M1 ) ⊕ Hn (M2 ) −
sends the fundamental classes of both M1 and M2 to the fundamental class of
S n−1 . Thus, this map is surjective and we deduce from the exactness of the
sequence that
Hn−1 (M1 #M2 ) ∼= Hn−1 (M1 ) ⊕ Hn−1 (M2 ).
We also see that Hn (M1 #M2 ) is the kernel of the map in (30).
QED
Lemma: Connected sums and intersction forms

Assume both M1 and M2 are of dimension n = 2p. Then the isomorphism
∼
=
Hp (M1 #M2 ) −
→ Hp (M1 ) ⊕ Hp (M2 )
is compatible with the intersection form.
Proof: Fundamental classes are natural in the sense that the homomorphism
∼
= ∼
=
Hn (M1 #M2 ) −
→ Hn (M1 ∨ M2 ) −
→ Hn (M1 ) ⊕ Hn (M2 ), [M1 #M2 ] 7→ [M1 ] + [M2 ]
sends the fundamental class of [M1 #M2 ] to the sum of the fundamental classes
of M1 and M2 .
Moreover, the cup product is natural so that we get a commutative diagram
∪ ⟨−,[M1 #M2 ]⟩
H p (M1 #M2 ) ⊗ H p (M1 #M2 ) / H 2p (M1 #M2 ) / F2
⟨−,[M1 ]⟩+⟨−,[M2 ]⟩
∪ / /
H p (M1 ) ⊗ H p (M1 ) ⊕ H p (M2 ) ⊗ H p (M2 ) H 2p (M1 ) ⊕ H 2p (M2 ) F2 .
Now it remains to translate this into the intersection pairing in homology which
proves the claim. QED
Classification of surfaces
Motivated by the examples of the torus and real projective plane we are going
to focus now on the case n = 2, i.e., two-dimensional manifolds which we are
235
going to call surfaces. In fact, we are going to study compact surfaces. In

this case we have an intersection form on H1 (M ).
We write Surf for the set of homeomorphism classes of compact con-

nected surfaces. The connected sum operation provides it with the structure
of a commutative monoid. The neutral element being S 2 , since S 2 #Σ ≈ Σ
for any surface Σ.
There is the following important result:
Theorem: Classification of surfaces

Associating the intersection form to a surface defines an isomorphism of
commutative monoids
∼
=
Surf −
→ Bil.
This theorem is great because it gives us a complete algebraic classifi-

cation of a class of geometric objects. This is one reason why algebraic
topology is so useful.
Actually, we are not finished with proving the theorem yet. Our examples
show us that T 2 corresponds to H and RP2 corresponds to I. And S 2 is sent to
the neutral element.
It remains to show the relation (and that this is the only relation)
(31) T 2 #RP2 ∼
= RP2 #RP2 #RP2 .
One way to do this is to triangulate the surfaces involved. This requires

too much geometric thinking for us today.
Instead, we make the following observation. We have not defined an ori-
entation, but assuming we know what that means it is a surprising fact that
even though we never assumed anything about orientations and worked with
F2 -coefficients, the theorem tells us what the orientable surfaces look like.
For, the orientable surfaces correspond to the forms gH where g is the
genus of the surface. This follows from the facts that T 2 is orientable whereas
RP2 is not.
In other words, any compact connected orientable surface Σ of genus g is
homeomorphic to the connected sum of g tori
Σ∼ = T 2 # · · · #T 2 .
Since S 2 is also an orientable surface, we allow g = 0 for this case too.
The real projective plane is not orientable. Therefore, any surface which is
homeomorphic to a connected sum of at least one copy of RP2 is not orientable.
In fact, the classification theorem and the relation (31) tell us that such a surface
is actually homeomorphic to a connected sum of copies of just RP2 s.
Quadratic refinement and the Kervaire invariant

Recall that away from characteristic 2 there is a bijection between quadratic
forms and symmetric bilinear forms. However, since we are working over
F2 , we can ask whether there is a quadratic refinement q of the intersection
form such that
q(x + y) = q(x) + q(y) + x · y.
For such a refinement to exist requires x · x = 0 for all x ∈ H1 (M ; F2 ), since
0 = q(2x) = q(x) + q(x) + x · x = x · x.
Hence we can only expect such a refinement on a sum of tori, i.e., on an
orientable surface.
The existence of a quadratic refinement is an additional structure associated
with the intersection form. Geometrically, it corresponds to a trivialization
of the normal bundle of an embedding into an RN for some N sufficiently
large. Such a trivialization is called a framing. There is an invariant
for quadratic forms in characteristic two, called the Arf invariant. In the
case of a surface, or more generally a manifold of dimension 4k + 2 (the only
dimension where interesting things happen for this invariant), this invariant
is called the Kervaire invariant. This invariant is a measure for if we can
do certain surgery manoeuvres on a manifold or not. Kervaire and Milnor
used this invariant to study the differentiable structures on spheres.
But there were certain dimensions they could not completely explain. To
settle the missing dimensions remained an open problem for about 60 years
237
until Mike Hill, Mike Hopkins, and Douglas Ravenel finally solved the
mystery (almost completely as there is one dimension left, it is 126) in a
groundbreaking work in 2009 (published in 2016) using highly sophisticated
methods in equivariant stable homotopy theory.
LECTURE 24
More on Poincaré duality
We continue our discussion of Poincaré duality. First we see two applications

of the theorem with F2 coefficients. Then we will discuss what we need to do
for other coefficients. This will lead to an important concept, orientations of
manifolds, and an important algebraic structure, the cap product.
Dualities reflect fundamental properties

Poincaré duality is extremely interesting, since it reflects a deep symmetry
in the homology and cohomology groups on manifolds. For, the cohomology
in dimension p determines the homology in dimension n−p. This symmetry
has many consequences which make the study of manifolds particularly
interesting.
Duality theorems arise in many areas of mathematics and always reflect
deep and interesting structures.
We start with an application of Poincaré duality modulo 2.
Applications of Poincaré duality with F2 -coefficients

Recall the important theorem:
Theorem: Poincaré duality mod 2

Let M be a connected compact manifold of dimension n. Then there
exists a unique class [M ] ∈ Hn (M ; F2 ), called the fundamental class of
M , such that, for every p ≥ 0, the pairing
∪ ⟨−,[M ]⟩
H p (M ; F2 ) ⊗ H n−p (M ; F2 ) −
→ H n (M ; F2 ) −−−−→ F2
is perfect.
239
240 CHAPTER 24. MORE ON POINCARÉ DUALITY
Since real projective space is a compact connected n-dimensional manifold,
Poincaré duality applies. And, in fact, we can use this result to deduce the algebra
structure on the cohomology of real projective space:
Corollary: Cohomology of RPn

Let x be the nonzero element in H 1 (RPn ; F2 ). Then xk is the nonzero
element of H k (RPn ; F2 ) for k = 2, . . . , n.
Thus H ∗ (RPn ; F2 ) is the truncated polynomial algebra
H ∗ (RPn ; F2 ) ∼
= F2 [x]/(xn+1 )
generated by x in degree 1 and truncated by setting xn+1 = 0.
Moreover, H ∗ (RP∞ ; F2 ) is a polynomial algebra
H ∗ (RPn ; F2 ) ∼
= F2 [x]
generated by x in degree 1.
Proof: The proof is by induction on n.

By the construction of the cell structure on RPn , we know that the inclusion
jk : RPk ,→ RPk+1 is a map of cell complexes which induces an isomorphism
∼
=
H i (RPk ; F2 ) −
→ H i (RPk+1 ; F2 ) for all i = 0, . . . ,k,
which sends the nonzero element x ∈ H 1 (RPk ; F2 ) to the nonzero element in
H 1 (RPk+1 ; F2 ) which we therefore also denote by x.
Hence, assuming xk is the nonzero element in H k (RPk ; F2 ), it suffices to show
that x ∪ xk is nonzero in H k+1 (RPk+1 ; F2 ).
By Poincaré duality, the pairing
∪ ⟨−,[RPk+1 ]⟩
H 1 (RPk+1 ; F2 ) ⊗ H k (RPk+1 ; F2 ) −
→ H k+1 (RPk+1 ; F2 ) −−−−−−−→ F2
is perfect. Since x and xk are nonzero by assumption, this implies x ∪ xk = xk+1
is nonzero as well.
For RPn , we know that H i (RPn F2 ) = 0 for i > n, since there are no cells in
dimensions bigger than n. Thus xn+1 = 0.
For RP∞ we just continue the induction process. □
As an application of this calculation, we are going to prove another famous

theorem, the Borsuk-Ulam Theorem.
241
Lemma
Let f : RPm → RPn be a continuous map which induces a nontrivial map
f∗ ̸= 0 : H1 (RPm ; F2 ) → H1 (RPn ; F2 )
Then m ≤ n.
Proof: Since H 1 (X; F2 ) ∼

= Hom(H1 (X; F2 )), the assumption implies that the
induced map in cohomology
f ∗ : H 1 (RPn ; F2 ) → H 1 (RPm ; F2 )
is nontrivial as well.
Let x ̸= 0 be the nonzero element in H 1 (RPn ; F2 ). Then f ∗ (x) ̸= 0 is nonzero
in H 1 (RPm ; F2 ). By the calculation of the H ∗ (RPm ; F2 ), we have
0 ̸= (f ∗ (x))m = f ∗ (xm ).
Thus, xm ̸= 0 in H m (RPn ; F2 ) which implies m ≤ n. □
Lemma: Paths between antipodal points

Let p ∈ S n and let σ : ∆1 → S n be a 1-simplex on S n which connects p and
its antipodal point −p in S n , i.e., σ(e0 ) = p and σ(e1 ) = −p. Let
π : S n → RPn
be the quotient map.
Then π∗ (σ) = π ◦σ is a cycle on RPn which represents a nonzero element
in H1 (RPn ; F2 ).
Proof: First, that π∗ (σ) is a cycle on RPn just follows from the fact
[π(σ(e0 ))] = [−π(σ(e0 ))] = [π(σ(e1 ))] in RPn .
It remains to show that it is not a boundary.
Recall that there is a cell structure on S n with skeleta
S 0 ⊂ S 1 ⊂ · · · ⊂ S n−1 ⊂ S n .
By symmetry, we can assume that p and −p are the points of S 0 .
• For n = 1, we have a homeomorphism RP1 ≈ S 1 (for example, one could
use the stereographic projection). Since σ connects p and −p on S 1 , there is an
integer k such that σ walks around S 1 (k + 1/2)-many times. Thus π∗ (σ) walks
around RP1 (2k + 1)-many times, i.e., an odd number of times.
Now recall that we showed
∼
=
→ H1 (S 1 ; Z), m 7→ (z 7→ z m )
Z−
where we use the identification π(S 1 ) = H1 (S 1 ; Z) that we showed in the ex-
ercises. This implies that with F2 -coefficients, even numbers correspond to 0 in
H1 (RP1 ; F2 ) and odd numbers correspond to the nonzero element in H1 (RP1 ; F2 ).
Thus, the image of π∗ (σ) in H1 (RP1 ; F2 ) is nonzero.
• For n > 1, we first choose a path τ on S 1 ⊂ S n which connects p and −p on
S . By the previous case, we know [π∗ (τ )] ̸= 0 in H1 (RP1 ; F2 ). The inclusion
1
map RP1 ,→ RPn induces an isomorphism

∼
=
H1 (RP1 ; F2 ) −
→ H1 (RPn ; F2 ).
Hence [π∗ (τ )] ̸= 0 in H1 (RPn ; F2 ) as well.

But for n > 1, the difference σ − τ is a boundary, since it is homotopic to a
constant map. This implies
[π∗ (σ)] = [π∗ (τ )] ̸= 0 in H1 (RPn ; F2 ).
Lemma: No antipodal maps

For any n, there is no continuous map f : S n+1 → S n with
f (−p) = −f (p) for all p ∈ S n+1 .
Proof: Assume there was such a map f . Since f (−p) = f (−p) for all p, f
induces a map
f¯: RPn+1 → RPn
which fits into a commutative diagram
f
S n+1 / Sn
π n+1 πn

RPn+1 / RPn .
f¯
243
Now we take take a 1-simplex σ which connects two antipodal points on S n+1 .
Its image f∗ (σ) = f ◦ σ is then a 1-simplex which connects two antipodal points
on S n , since f (−p) = f (−p).
By the previous lemma, π∗n (f∗ (σ)) ̸= 0 in H1 (RPn ; F2 ). Thus
f¯∗ (π n+1 (σ)) = π n (f∗ (σ)) ̸= 0.
∗ ∗
In other words,
f¯∗ ̸= 0 : H1 (RPn+1 ; F2 ) → H1 (RPn ; F2 )
is nontrivial. By the other lemma, this is not possible. Hence f cannot exist.
□
The Borsuk-Ulam Theorem

Let g : S n → Rn be a continuous map. Then there is a point p ∈ S n with
g(p) = g(−p).
Proof: If there is no such point, we can define a continuous map

g(p) − g(−p)
f : S n → S n−1 , p 7→ .
|g(p) − g(−p)|
But this map satisfies
f (−p) = −f (p).
This contradicts the previous lemma. □
Orientation and fundamental classes
We now leave the world of F2 -coefficients and contemplate on what we need
for a Poincaré duality theorem with other coefficients. Since we will only sketch
the main ideas anyway, we will just look at Z-coefficients.
We start with the following observation on the homology groups of a manifold

at a point:
Lemma: Local homology on manifolds

Let M be an n-dimensional topological manifold. For any point x ∈ M ,
there is an isomorphism
Hn (M,M − {x}; Z) ∼
= Z.
and Hi (M,M − {x}; R) = 0 for all i ̸= n.
Proof: Since M is a manifold, there is an open neighborhood U around x in

M such that U ∼= Rn . We set Z = M − U and apply excision to get
Hi (M,M − {x}; Z) ∼
= Hi (M − Z,(M − {x}) − Z); Z) (by excision)
= Hi (U,U − {x}; Z)
∼
= Hi (Rn ,Rn − {0}; Z)
∼
= Hi−1 (Rn − {0}; Z) (by homotopy invariance and long ex. seq.)
∼
= Hi−1 (S n−1 ; Z) (by homotopy invariance).
This implies
(
Z if i = n
Hi (M,M − {x}; Z) ∼
=
0 else.
Local orientation
The group Hn (M,M − {x}; Z) is often called the local homology of M
at x. It is an infinite cyclic group and therefore has two generators.
A choice of a generator µx ∈ Hn (M,M − {x}; Z) is a local orientation of
M at x.
For every point x ∈ M , we can choose such a generator. Note that such a
choice was not necessary in F2 , since there is only one generator. That makes
F2 -coefficients quite special.
245
The natural question is how all these choices are related. In other words,
is it possible to choose these generators in a compatible way?
More precisely, let x and y be two points in M which both lie in some subset
U ⊂ M . The inclusions ix : {x} ,→ M and iy : {y} ,→ M induce maps
ix∗ iy∗
Hn (M, M − {x}; Z) o Hn (M, M − U ; Z) / Hn (M, M − {y}; Z).
A class µU ∈ Hn (M,M − U ; Z) which maps to generators in Hn (M, M −

{x}; Z) and Hn (M, M − {y}; Z). Such an µU would define local orientations
µx := ix∗ (µU ) and µy := iy∗ (µU ) at x and y, respectively. We call such an element
µU a fundamental class at U .
Around every point in M there is a little neighborhood U with a fundamental
class at U . The crucial question is: how large can we choose such a U ? Ideally,
we would like to be able to choose U = M such that Hn (M,M − U ) = Hn (M ).
Unfortunately, this is not always possible. This leads to an important concept:
Orientations
Let M be a compact connected n-dimensional manifold.
• An orientation of M is a function x 7→ µx , where µx ∈ Hn (M, M −
{x}; Z) is a generator, which satisfies the following condition:
At any point x ∈ M , there is a neighborhood U around x and
an element µU ∈ Hn (M, M − U ; Z) such that iy∗ (µU ) = µy for all
y ∈ U.
• If such an orientation exists, we say that M is orientable.
If M is orientable, then there are exactly two orientations. If M is ori-

entable, and we have chosen an orientation, then we say that M is oriented.
We can reformulate this in terms of a particular class in homology, the funda-

mental class. The following statement is both a definition and proposition. We
skip the proof, since we only have time for a rough sketch of the story.
Fundamental classes and orientability

Let M be a compact connected n-dimensional manifold.
• A fundamental class of M is an element µ ∈ Hn (M ; Z) such that,

for every point x ∈ M , the image of µ under the map
Hn (M ; Z) → Hn (M,M − {x}; Z)
induced by the inclusion (M, ∅) ,→ (M, M − {x}) is a generator.
• M is orientable if and only if M has a fundamental class.
• M is orientable if and only if Hn (M ; Z) = Z.
For example, RP2n is not orientable, since H2n (RP2n ; Z) = 0. Whereas

RP2n+1 is orientable with H2n+1 (RP2n+1 ; Z) = Z.
Spheres and tori are orientable. The Klein bottle is not orientable.
The cap product

There is another type of product that has elements in both cohomology and
homology and has a homology class as output. Actually, there are several other
such products. But that is a story for another day.
Cap products are defined for arbitrary spaces. So we leave the world of man-
ifolds for a moment and get back to it afterwards. Again we only discuss Z-
coefficients, but everything works for any ring R as coefficients.
Definition: Cap product

Let X be any space. The cap product is defined to be the Z-bilinear map
∩
S q (X) × Sp (X) −
→ Sp−q (X)
defined by sending a q-cochain φ ∈ S q (X) and a p-simplex σ : ∆p → X to
the p − q-chain
φ ∩ σ := φ(σ|[e0 ,...,eq ] )σ|[eq ,...,ep ] .
If p < q, then the cap product is defined to be 0.
After checking the relation

∂(φ ∩ σ) = φ ∩ (∂σ)
we see that the cap product descends to a Z-linear map on cohomology and
homology
∩
H q (X) ⊗ Hp (X) −
→ Hp−q (X).
247
Given a continuous map f : X → Y , there is the following formula which

expresses the naturality of the cap product:
φ ∩ f∗ (σ) = f∗ (f ∗ φ ∩ σ).
The cap product is important for us, since (one form of) Poincaré duality can
be formulated by saying that the cap product with the fundamental class is an
isomorphism:
Poincaré duality
Let M be a compact n-dimensional oriented manifold. Let [M ] ∈ Hn (M ; Z)
be its fundamental class. Then taking the cap product with [M ] yields an
isomorphism
∼
=
D : H p (M ; Z) −
→ Hn−p (M ; Z), φ 7→ φ ∩ [M ].
Note that there are many different ways to formulate Poincaré duality. In
particular, there is also the stronger statement in terms of perfect pairings on
cohomology groups that we have seen in the mod 2-case.
The idea of the proof of this theorem is to study the case of open subsets of
Rn first. Then we use that every point in M has an open neighborhood which is
homeomorphic to an open subset in Rn . Since M is compact, we only need to take
finitely many such open neighborhoods to cover M . The Mayer-Vietoris sequence
then allows to patch the overlapping open subsets together. Unfortunately, there
are some technical difficulties to take care of along the way, e.g., that certain
diagrams actually commute.
Bibliography
[1] Christian Baer, Algebraic Topology, Lecture Notes, Universität Potsdam, Winter Term
2016/17.
[2] Glen E. Bredon, Topology and geometry, Graduate Texts in Mathematics, vol. 139,
Springer-Verlag, New York, 1993. MR1224675
[3] William Fulton, Algebraic topology, Graduate Texts in Mathematics, vol. 153, Springer-
Verlag, New York, 1995. A first course. MR1343250
[4] Allen Hatcher, Algebraic topology, Cambridge University Press, Cambridge, 2002.
MR1867354
[5] J. P. May, A concise course in algebraic topology, Chicago Lectures in Mathematics, Uni-
versity of Chicago Press, Chicago, IL, 1999. MR1702278
[6] Haynes Miller, Lectures on Algebraic Topology, Lecture Notes, MIT, Fall 2016.
[7] John W. Milnor and James D. Stasheff, Characteristic classes, Princeton University Press,
Princeton, N. J.; University of Tokyo Press, Tokyo, 1974. Annals of Mathematics Studies,
No. 76. MR0440554
[8] James R. Munkres, Topology, Prentice Hall, Inc., Upper Saddle River, NJ, 2000. Second
edition of [ MR0464128]. MR3728284
[9] , Elements of algebraic topology, Addison-Wesley Publishing Company, Menlo Park,
CA, 1984. MR755006
[10] Christoph Schweigert, Algebraic Topology, Lecture Notes, Universität Hamburg, Summer
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Springer-Verlag, New York, 1994. An introduction to algebraic topology. MR1254439
249
APPENDIX A
Exercises
1. Exercises after Lecture 2

1 Let f : X → Y be a continuous map between topological spaces X and
Y.
a) Let K ⊆ X be compact. Show that f (K) ⊆ Y is compact.
b) Give an example of a map f and a compact subset K ⊆ Y such
that f −1 (K) ⊆ X is not compact.
2 Draw a picture of S 2 as a cell complex with six 0-cells, twelve 1-cells and
eight 2-cells.
3 Show that the stereographic projection

(
x
̸ 1
y=
ϕ : S 1 → R ∪ {∞}, (x,y) 7→ 1−y
∞ y=1
defines a homeomorphism from S 1 to the one-point compactification R̂ =
R ∪ {∞} of R.
4 a) Let X and Y be topological spaces. Show that homotopy defines an

equivalence relation on the set C(X,Y ) of continuous maps X → Y .
b) Show that being homotopy equivalent defines an equivalence relation
on topological spaces.
5 a) Show that S 1 is a strong deformation retract of D2 \ {0}.

b) Show that D2 \ {0} is not contractible.
251
252 CHAPTER A. EXERCISES
1 Let f ∈ C((X,A),(Y,B)) be a map of pairs.
a) Show that, for every n ≥ 0, f induces a homomorphism Hn (X,A) →
Hn (Y,B).
b) Show that the connecting homomorphisms fit into a commuative
diagram
Hn (f )
Hn (X,A) / Hn (Y,B)
∂ ∂

Hn−1 (A) / Hn−1 (B).
Hn−1 (f|A )
2 Let X be a nonempty topological space. Recall that if ω is a path on

X, i.e., a continuous map ω : [0,1] → X, then we define an associated
1-simplex σω by
σω (t0 ,t1 ) := ω(1 − t0 ) = ω(t1 ) for t0 + t1 = 1, 0 ≤ t0 ,t1 ≤ 1.
a) Show that if ω is a constant path, then σω is a boundary.

b) Let γ1 and γ2 be paths in X, and let γ := γ1 ∗ γ2 be the path given
by first walking along γ1 and then walking along γ2 , i.e., the map
(
γ1 (2t) for 0 ≤ t ≤ 21
γ = γ1 ∗ γ2 : [0,1] → X, t 7→
γ2 (2t − 1) for 21 ≤ t ≤ 1.
Show that the 1-chain σγ − σγ1 − σγ2 is a boundary.

2. EXERCISES AFTER LECTURE 6 253
For the next two exercises, recall that, given a topological space X and a
subspace A ⊂ X, A is called a retract of X if there is a retraction ρ : X → A,
i.e., a continuous map ρ : X → A with ρ|A = idA . Moreover, we can consider ρ
ρ
also as a map X → X via the inclusion X → − A ⊂ X. If ρ is then in addition
homotopic to the identity of X, then A is called a deformation retract of X.
3 For every n ≥ 2, show that S n−1 is not a deformation retract of the unit
disk Dn .
4 Show that if A is a retract of X then the map Hn (i) : Hn (A) → Hn (X)

induced by the inclusion i : A ⊂ X is injective.
5 In this bonus exercise we show that the additivity axiom is needed only
for infinite disjoint unions:
For two topological spaces X and Y , let iX : X ,→ X⊔Y and iY : Y ,→
X ⊔ Y be the inclusions into the disjoint union of X and Y . Without
referring to the additivity axiom show that the remaining Eilenberg-
Steenrod axioms imply that the induced map
Hn (iX ) ⊕ Hn (iY ) : Hn (X) ⊕ Hn (Y ) → Hn (X ⊔ Y )
is an isomorphism for every n. (Hint: You may want to apply the long
exact sequence and excision with U = X ⊂ X ⊔ Y .)
Recall the definition from Lecture 7: For n ≥ 1, let f : S n → S n be a con-

tinuous map. The degree of f , denoted by deg(f ), is the integer determined by
Hn (f )([σ]) = deg(f ) · [σ] for a generator [σ] ∈ Hn (S n ) ∼
= Z.
Since we know Hn+1 (Dn+1 , S n ) ∼ = Z, we can apply the same definition also to
selfmaps of the pair (Dn+1 , S n ): Let f : (Dn+1 , S n ) → (Dn+1 , S n ) be a continuous
map of pairs. The degree of f , again denoted by deg(f ), is the integer determined
by Hn+1 (f )([σ]) = deg(f ) · [σ] for a generator [σ] ∈ Hn+1 (Dn+1 , S n ) ∼ = Z.
1 Show that the degree has the following properties:

a) The identity has degree 1, i.e., deg(id) = 1.
b) The degree of a constant map is 0.
c) If f,g : S n → S n are two continuous maps, then deg(f ◦ g) =
deg(f ) deg(g).
d) If f0 and f1 are homotopic, then deg(f0 ) = deg(f1 ).
e) If f : S n → S n is a homotopy equivalence, then deg(f ) = ±1.
f ) For f : (Dn+1 , S n ) → (Dn+1 , S n ), let f|S n denote the restriction of f
to S n . Then deg(f ) = deg(f|S n ).
2 Let a : S n → S n be the antipodal map, i.e.,

a : (x0 ,x1 , . . . ,xn ) 7→ (−x0 , − x1 , . . . , − xn ).
a) Show deg(a) = (−1)n+1 .

(Hint: Use the result from Lecture 7 on the degree of a reflection.)
b) For n even, show that the antipodal map is not homotopic to the
identity on S n .
(Hint: Use what you have just learned in the previous exercises.)
3 a) If f : S n → S n is a continuous map without fixed points, i.e., f (x) ̸=

x for all x ∈ S n , then deg(f ) = (−1)n+1 .
(Hint: Show that f is homotopic to the antipodal map.)
b) If f : S n → S n is a continuous map without an antipodal point, i.e.,
f (x) ̸= −x for all x ∈ S n , then deg(f ) = 1.
(Hint: Show that f is homotopic to the identity map.)
c) If n is even and f : S n → S n is any continuous map, show that there
is a point x ∈ S n with f (x) = ±x.
(Hint: Apply the previous observations.)
n n n+1
A vector field on S is a continuous map v : S → R with x ⊥ v(x) for all
x ∈ S n (x and v(x) are orthogonal to each other).
4 Prove the following theorem: The n-dimensional sphere S n admits a

vector field v without zeros, i.e., v(x) ̸= 0 for all x ∈ S n , if and only if n
is odd.
In particular, every vector field on S 2 must have a zero. This is often
rephrased as: you cannot comb a hairy ball without leaving a bald spot.
(Hint: If n even, show that the assumption v(x) ̸= 0 would allow to
define a homotopy between the identity and the antipodal map. When
you write down the homotopy make sure the image lies on S n .)
1 Let X be a space, A ⊂ X be a subspace and j : (X,∅) ,→ (X,A) be the
inclusion map. Suppose A is contractible.
a) Show that the natural homomorphism Hn (j) : Hn (X) → Hn (X,A)
is an isomorphism for all n ≥ 2.
b) Show that Hn (j) is an isomorphism for all n ≥ 1 if A and X are
path-connected.
c) For n ≥ 1, let p ∈ S n be a point. Show that S n \ {p} is contractible.
d) For two distinct points p1 , p2 ∈ S n , is S n \ {p1 ,p2 } contractible?
2 Let f : S n → S n be a continuous map. If f is not surjective, then

deg(f ) = 0.
3 Our goal in this exercise is to construct a surjective map f : S 1 → S 1

with deg(f ) = 0.
a) Start with a map
(
e−is if s ∈ [0,π)
g : S 1 → S 1 , eis 7→
eis if s ∈ [π,2π).
Show that g has degree 0.

b) Compose g with a another map such that the composition becomes
a surjective map f : S 1 → S 1 of degree 0.
4 Let f : S n → S n be a continuous map with deg(f ) = 0. Show that there

must exist points x,y ∈ S n with f (x) = x and f (y) = −y.
5 With this exercise we would like to refresh our memory on real projective
spaces and connect it to questions on the existence of fixed points.
Recall from Lecture 2 that the real projective space RPk is defined
to be the quotient of Rk+1 \ {0} under the equivalence relation x ∼ λx
for λ ∈ R \ {0}. The topology on RPk is the quotient topology.
a) Show that any invertible R-linear map F : Rk+1 → Rk+1 induces a
continuous map f : RPk → RPk .
b) Show that for any invertible R-linear map F : Rk+1 → Rk+1 with an
eigenvector, the induced map f : RPk → RPk has a fixed point.
c) Show that any continuous map f : RP2n → RP2n that is induced by
an invertible R-linear map F : R2n+1 → R2n+1 has a fixed point.
2n−1 2n−1
d) Show that there are continuous maps f : RP → RP without
fixed points.
6 Let p(z) = z n + an−1 z n−1 + · · · + a1 z + a0 be a polynomial of degree

n ≥ 1 with coefficients in C. The goal of this exercise is to prove the
Fundamental Theorem Algebra, i.e., we would like to show that there is
a z ∈ C with p(z) = 0.
We are going to show this as wollows:
Consider p as a map C → C. Assume that p had no root. Then
we can define a new map
p(z)
p̂ : S 1 → S 1 , z 7→ .
|p(z)|
We are going to show that this assumption leads to a contradiction.
a) Show that p̂ is homotopic to a constant map. What is the degree of
p̂?
b) Show that the map
tn p( zt )
H : S 1 × (0,1] → S 1 , (z,t) 7→
|tn p( zt )|
can be continuously extended to a map S 1 ×[0,1], i.e., analyze H(z,t)

for t → 0. What is the degree of p̂?
c) Deduce that p must have a root, i.e., there must be a z ∈ C with
p(z) = 0.
7 In this exercise we continue our study of the Fundamental Theorem Al-

gebra. Our goal is to connect the degree and the multiplicity of a root
of a polynomial.
a) Let f : S 1 → S 1 be a continuous map. Show that if f can be ex-
tended to a map on D2 , i.e., if there is a continuous map F : D2 → S 1
such that F|S 1 = f , then deg(f ) = 0.
Now let p(z) = z n + an−1 z n−1 + · · · + a1 z + a0 be a polynomial of
degree n ≥ 1 with coefficients in C.
b) Assume that p has no root z with |z| ≤ 1. Then we can define the
map
p(z)
p̂ : S 1 → S 1 , z 7→ .
|p(z)|
Show that the degree of p̂ is 0.
c) Assume that p has exactly one root z0 with |z0 | < 1 and no root z
with |z| = 1. Then we can define the map
p(z)
p̂ : S 1 → S 1 , z 7→ .
|p(z)|
Show that the degree of p̂ equals the multiplicity of the root z0 ,
i.e., deg(p̂) = m where m ≥ 0 is the unique number such that
p(z) = (z − z0 )m q(z) with q(z0 ) ̸= 0.
Finally, we switch perspectives a bit. We know that the polynomial p
satisfies lim|z→∞| |p(z)| = ∞. Hence we can extend the map p : C → C to
a map p : S 2 → S 2 where we think of S 2 as the one-point-compactification
of C ∼= R2 . We are going to use the following fact: Let fm : S 2 → S 2 , z 7→
z m . The effect of H2 (fm ) as a selfmap of H2 (S 2 ) and as a selfmap of
H2 (S 2 , S 2 \ {0}) is given by multiplication by m.
d) Let zi be a root of p. Show that the local degree deg(p|zi ) of p at zi
is equal to the multiplicity of zi as a root of p.
5. EXERCISES ON THE HUREWICZ MAP 259
5. Exercises on the Hurewicz map
This will be a guided tour to the fundamental group and the Hurewicz ho-
momorphism. You can read about this topic in almost every textbook. But you
could also take the time to solve the following exercises and enjoy the fun of
developing the maths on your own.
Note that we already seen some of the following problems in previous exercises.
But feel free to do them again. :)
We fix the following notation:
Let X be a nonempty topological space and let x0 be a point in X. We write
I = [0,1] for the unit interval and ∂I = {0,1}.
We denote by
Ω(X,x0 ) := {γ ∈ C([0,1],X) : γ(0) = x0 = γ(1)}
the set of continuous loops based at x0 .

We are going to call two loops γ1 and γ2 based at x0 are homotopic relative
to ∂I if there is a continuous map

h(s,0) = γ1 (s) for all s

h : [0,1] × [0,1] → X wich satisfies h(s,1) = γ2 (s) for all s

h(0,t) = x = h(1,t) for all t.
0
On this exercise set, we will always use the word loop to denote a loop based
at x0 and say that two loops are homotopic when they are homotopic relative to
∂I.
Let γ1 and γ2 be two loops based at x0 . We define the loop γ1 ∗ γ2 to be
the loop given by first walking along γ1 and then walking along γ2 with doubled
speed, i.e., the map
(
γ1 (2s) for 0 ≤ s ≤ 21
γ1 ∗ γ2 : [0,1] → X, s 7→
γ2 (2s − 1) for 12 ≤ s ≤ 1.
Our first goal is to prove the following theorem:

The fundamental group

The set of equivalence classes of loops modulo homotopy
π1 (X,x0 ) := Ω(X,x0 )/ ≃
becomes a group with group operation
[γ1 ] · [γ2 ] := [γ1 ∗ γ2 ].
The class of the constant loop ϵx0 is the neutral element. The inverse of [γ]
is [γ̄], where γ̄ denotes the loop in reverse direction s 7→ γ(1 − s).
The group π1 (X,x0 ) is called the fundamental group of X at x0 .
1 In this exercise we are going to show that π1 (X,x0 ) together with the
above described operation is a group.
Let γ, γ ′ , ξ, ξ ′ , ζ denote loops based at x0 . Let ϵx0 with ϵx0 (t) = x0
for all t be the constant loop at x0 . Recall that we use the notation
γ ≃ γ ′ to say that the two loops γ and γ ′ are homotopic relative to ∂I.
a) Show that if γ ≃ γ ′ and ξ ≃ ξ ′ , then γ ∗ ξ ≃ γ ′ ∗ ξ ′ .
Let φ : [0,1] → [0,1] be a continuous map with φ(0) = 0 and φ(1) = 1.

The composition γ ◦ φ is called a reparametrization of γ.
b) Show that φ is homotopic to the identity map of [0,1]. Deduce that
γ ◦ φ is homotopic to γ.
c) Choose appropriate reparametrizations of the paths involved to show
ϵx0 ∗ γ ≃ γ ≃ γ ∗ ϵx0 .
d) Choose a φ corresponding to the following picture to show that
γ ∗ (ξ ∗ ζ) is a reparametrization of (γ ∗ ξ) ∗ ζ by φ. Conclude that
(γ ∗ ξ) ∗ ζ ≃ γ ∗ (ξ ∗ ζ).
e) Show that γ ∗ γ̄ ≃ ϵx0 ≃ γ̄ ∗ γ by writing down precise formulae for

the following picture.
Our next goal is to construct a homomorphim π1 (X,x0 ) → H1 (X).
2 Recall that if γ is a loop on X we define an associated 1-simplex σγ by
σγ (1 − t,t) := γ(t) for 0 ≤ t ≤ 1.
Note that if γ is a loop, then σγ is a 1-cycle.

A brief reminder before we start. For solving the following prob-
lems remember that if you want to construct a 2-simplex with a certain
boundary, you need to define a map on all of ∆2 and not just its bound-
ary. Omitting to describe the map on all of ∆2 would make the tasks
trivial. Now let us get to work:
a) Show that if γ = ϵx0 is the constant loop at x0 , then σγ is a boundary.
b) Show that the 1-chain σγ1 + σγ2 − σγ1 ∗γ2 is a boundary.
c) Show that if γ1 and γ2 are homotopic loops, then σγ1 − σγ2 is a
boundary.
(Hint: For a homotopy h between γ1 and γ2 , think of I × I as a
square. Then you can either collapse it to a triangle or divide it
along the diagonal to get two triangles. This will give you a way to
construct 2-simplices out of h.)
d) Show that the 1-chain σγ + σγ̄ is a boundary.
e) Conclude that the map
ϕ : π1 (X,x0 ) → H1 (X), [γ] 7→ [σγ ]
is a homomorphism of groups. It is called the Hurewicz homomor-

phism.
The fundamental group π1 (X,x0 ) is in general not abelian. Since H1 (X) is by

definition an abelian group, the homomorphism φ factors through the maximal
abelian quotient of π1 (X,x0 ). Reall that this quotient is defined as follows:
For a group G, the commutator subgroup [G,G] of G is the smallest subgroup
of G containing all commutators [g,h] = [ghg −1 h−1 ] for all g,h ∈ G. Note that
[G,G] is a normal subgroup. The quotient Gab := G/[G,G] is the maximal abelian
quotient of G and is called the abelianization of G.
The abelianization has the following universal property: Let q : G → G/[G,G]
be the quotient map. If H is an abelian and ηG → H a homomorphism of groups,
then there is a unique homomorphism of abelian groups ηab : Gab → H such that
the following diagram commutes
η
G /
=H
q
ηab

Gab .
3 Assume that X is path-connected. We are going to show that the induced

homomorphism
ϕab : π1 (X,x0 )ab → H1 (X),
which is also called the Hurewicz homomorphism, is an isomorphism.

We are going to construct an inverse ψ of ϕab as follows:
For any x ∈ X, we choose a continuous path λx from x0 to x. If
x = x0 , then we choose λx0 to be the constant path at x0 .
Let σ : ∆1 → X be a 1-chain in X. Denote the associated path in X
by
γσ : [0,1] → X, t 7→ σ(1 − t,t).

Then we define a loop
ψ̂(σ) : [0,1] → X, t 7→ λσ(e0 ) ∗ γσ ∗ λ̄σ(e1 ) .
We extend this definition Z-linearly to obtain homomorphism of

abelian groups
ψ̂ : S1 (X) → π1 (X,x0 )ab .
a) As a preparation show the following: Let β : ∆2 → X be a 2-simplex.
Let αi be the path corresponding to the ith face β ◦ ϕi of β. Show
that the loop α2 ∗ α0 ∗ ᾱ1 based at y0 := β(e0 ) is homotopic to the
constant loop ϵy0 at y0 .
Note: If you do not want to derive formulae, draw a picture to
convince yourself that the statement makes sense and describe in
words why it is true.
b) Show that ψ̂ sends the group B1 (X) of 1-boundaries to the neutral
element 1 ∈ π1 (X,x0 )ab .
(Hint: Use that π1 (X,x0 )ab is abelian and that the loop given by
walking along the boundary of a 2-simplex is homotopic to the con-
stant loop.)
c) Conclude that ψ̂ induces a homomorphism of abelian groups
ψ : H1 (X) → π1 (X,x0 )ab .
d) Show that if γ is a loop, then ψ(ϕab ([γ])) = [γ].
e) Let σ be a 1-simplex. Show that ϕab (ψ([σ])) = [σ + κσ(e0 ) − κσ(e1 ) ],
where κy denotes the constant 1-simplex with value y.
f ) Show that, if c is a 1-cycle, then ϕab (ψ([c])) = [c].
g) Conclude that ψ is an inverse of ϕab and hence that ϕab is an iso-
morphism.
1 In this exercise we give another proof of the exactness of the Mayer-
Vietoris sequence. We start with an algebraic lemma which provides
good practice in diagram chasing and then we use this result to deduce
the MVS from the Excision Axiom.
a) Assume we have a map of long exact sequences
i′n a′n b′n
(32) ··· / Kn′ / L′n / Mn′ / ′
Kn−1 / ···
fn gn ∼
= kn fn−1

··· / Kn / Ln / Mn / Kn−1 / ···
in an bn
∼
=
such that kn : Mn′ −
→ Mn is an isomorhism for every n. For each n,
we define the homomorphism ∂n to be
a k−1 b′
n
∂n : Ln −→ → Mn′ −→
n
Mn −− n ′
Kn−1 .
Show that the sequence
 
f
 n 
′
h i
′
−i n ′
in gn ∂n ′
· · · → Kn −−−−→ Kn ⊕ Ln −−−−−−→ Ln −→ Kn−1 → ···
is exact.
b) Let {A,B} be a cover of X. Apply the previous algebraic observation
to the long exact sequences of the pairs (X,A) and (B,A∩B) and use
the excision isomorphism to deduce the Mayer-Vietoris sequence.
2 Let A and B be two disjoint closed subsets of R2 .

a) Show that there is an isomorphism
H1 (R2 − (A ∪ B)) ∼= H1 (R2 − A) ⊕ H1 (R2 − B).
Recall that a path-component of a space X is a maximal path-
connected subspace (where the ordering is given by inclusion). For ex-
ample, if X is path-connected itself, then it has one path-component. If
X is the disjoint union of two path-connected spaces U and V , then U
and V are the path-components of X.
b) Show that the number of path-components of R2 − (A ∪ B) is one
less than the sum of the numbers of path-components of R2 − A and
R2 − B.
Definition: Mapping cylinder

Let f : X → Y be a continuous map. The mapping cylinder of f is
defined to be te quotient space
Mf := (X × [0,1] ⊔ Y )/((x,0) ∼ f (x)).
The mapping cylinder fits into a commutative diagram

f
X /
>Y
f1 g
Mf
where f1 maps x to (x,1) and g maps (x,t) to f (x) for all x ∈ X and t ∈ [0,1]
and y ∈ Y to y.
3 a) Show that the inclusion i : Y ,→ Mf is a deformation retract and g

is a deformation retraction.
b) We can construct the Möbius band M := Mf as the mapping
cylinder of the map
f : S 1 → S 1 , z 7→ z 2 (S 1 ⊂ C).
Determine the homology of the Möbius band.
c) For n ≥ 1 and m ∈ Z, let Mf be the mapping cylinder of a map
f : S n → S n with deg(f ) = m.
Show that Hn (f1 ) is given by multiplication with m.
d) For n ≥ 1 and m ≥ 2, let Mf be the mapping cylinder of a map
f : S n → S n with deg(f ) = m.
Show that X = S n is not a weak retract of Mf .
4 We can consider the real projective plane RP2 as a two dimensional

disk D2 with a Möbius band M attached at its boundary. Writing A =
D2 and B = M , we have A ∩ B ≃ S 1 . Calculate the homology groups
of RP2 .
On this exercise set we are going to explore some additional, important topics.
We will use them later in the lectures.
We start with an application of the excision property:
1 Let U ⊂ Rn and V ⊂ Rm be nonempty open subsets. Show that if there

≈
is a homeomoprhism φ : U −
→ V , then we must have n = m.
(Hint: Take a point x ∈ U and compare Hn (Rn ,Rn − {x}) and
Hn (U,U − {x}).)
Now we are going to introduce a slight modification of homology:
Definition: Reduced homology

Let X be a nonempty topological space. We define the reduced homology
of X to be the homology of the augmented complex of singular chains
∂ 2 ∂ 1 ϵ̃
· · · S2 (X) −
→ S1 (X) −
→ S0 (X) →
− Z→0
P P
where ϵ̃( i ni σi ) := i ni (and Z is placed in degree −1). Recall that we
checked before (in Lecture 4) that ϵ̃ ◦ ∂1 = 0. Hence the above sequence is a
chain complex. Moreover, the construction of the augmented chain complex
is functorial.
The nth reduced homology group of X is denoted by H̃n (X).
Reduced homology does not convey any new information, but is convenient
for stating things. It also helps focusing on the important information, since it
disregards the contribution in H0 (X) which comes from a single point.
Here are some basic properties of reduced homology:
2 Let X be a nonempty topological space. Show that reduced homology

satisfies the following properties:
a) H̃0 (X) = Ker (ϵ : H0 (X) → H0 (pt)).
b) H̃n (X) = Hn (X) for all n ≥ 1.
c) If X is path-connected, then H̃0 (X) = 0.
d) For any point x ∈ X, Hn (X) ∼ = H̃n (X) ⊕ Hn ({x}) and H̃n (X) ∼
=
Hn (X,{x}) for all n ≥ 0.
Now we move on towards an important construction on spaces, the suspension.
It does not look spectacular, but will prove extremely useful and important later
on in our studies of Algebraic Topology:
Definition: Suspension of a space

Let X be a topological space. The suspension of X is defined to be the
quotient space
(X × [0,1])/((x1 ,0) ∼ (x2 ,0) and (x1 ,1) ∼ (x2 ,1) for all x1 ,x2 ∈ X).
In other words, SX is constructed by taking a cylinder over X and then
collapsing all points X × {0} to a point p0 and all points X × {1} to a point
p1 . The topology on SX is the quotient topology.
For any continuous map f : X → Y , there is an induced continuous map

S(f ) : SX → SY, [x,s] 7→ [f (x),s].
3 Our goal in this exercise is to understand SX a bit better and to show

that there are isomorphisms
H̃n+1 (SX) ∼
= H̃n (X) for all n ≥ 0
(Note that reduced homology makes it much easier to state this re-
sult. For, without reduced homology we would have to write H1 (SX) ∼ =
Ker (H0 (X) → H0 (pt)) for n = 0.)
a) Show that SX is path-connected and hence H0 (SX) ∼ = Z.
b) Show that SX − {p1 } is contractible.
c) Show that SX − {p0 , p1 } is homotopy equivalent to X.
d) Use the Mayer-Vietoris sequence to determine H̃n+1 (SX) for all n ≥
0.
A crucial example is the suspension of the sphere.

4 a) Show that the suspension SS n−1 of the (n − 1)-sphere is homeomor-

phic to the n-sphere.
b) For n ≥ 1, let f : S n → S n be a continuous map and let S(f ) : SS n →
SS n be the induced map on suspensions. By using either SS n ≈
S n+1 or Hn+1 (SS n ) ∼= Hn (S n ) show that Hn+1 (S(f )) is given by
multiplication by an integer which we denote by deg(S(f )). Show
that deg(S(f )) = deg(f ).
c) For n ≥ 1, show that, for any given k ∈ Z, there is a map f : S n →
S n with deg(f ) = k.
1 Show that the two different cell structures on S n we discussed in the
lecture lead to cellular chain complexes which have the same homology
groups.
2 Show the statement of the lecture that the isomorphism between the ho-
mology of the cellular chain complex is functorial in the following sense:
Let f : X → Y be a cellular (or filtration-preserving) map between cell
complexes, i.e., f (Xn ) ⊆ Yn for all n. Show that f induces a homomor-
phism of cellular chain complexes C∗ (f ) : C∗ (X) → C∗ (Y ) which fits into
a commutative diagram
H∗ (C∗ (f ))
H∗ (C∗ (X)) / H∗ (C∗ (X))
∼
= ∼
=

H∗ (X) / H∗ (Y ).
H∗ (f )
3 Let X be a cell complex and A a subcomplex. Show that the quotient

X/A inherits a cell structure such that the quotient map q : X → X/A
is cellular.
4 Consider S 1 with its standard cell structure, i.e. one 0-cell e0 and one
1-cell e1 . Let X be a cell complex obtained from S 1 by attaching two
2-cells e21 and e22 to S 1 by maps f2 and f3 of degree 2 and 3, respectively.
We may express this construction as
X = S 1 ∪f2 e21 ∪f3 e22 .
a) Determine all the subcomplexes of X.

b) Determine the cellular chain complex of X and compute the homol-
ogy of X.
c) For each subcomplex Y of X, compute the homology of Y and of
the quotient space X/Y .
d) As a more challenging task show that the only subcomplex Y of
q
X for which X → − X/Y is a homotopy equivalence is the trivial
subcomplex consisting only of the 0-cell.
(Hint: Study the effect of H2 (q).)
Note that one can nevertheless show that X is homotopy equivalent
to S 2 . But we are lacking some results in homotopy theory to prove
this.
For the next exercise, note that if X and Y are cell complexes, then X × Y
is a cell complex with cells the products enα,X × em n
β,Y where eα,X ranges over the
m
cells of X and eβ,Y ranges over the cells of Y .
5 Show that the Euler characteristic has the following properties:

a) If X and Y are finite cell complexes, then
χ(X × Y ) = χ(X)χ(Y ).
b) Assume the finite cell complex X is the union of the two union of
two subcomplexes A and B. Then
χ(X) = χ(A) + χ(B) − χ(A ∩ B).
We start with proving some properties about the Tor-functor that we men-
tioned in the lecture.
1 Let M be an abelian group.

a) Let A be an abelian group and 0 → F1 → F0 → M → 0 be a free
resolution of M . Consider the chain complex K∗ given by
(33) 0 → A ⊗ F1 → A ⊗ F0 → 0
with A ⊗ F1 in dimension one and A ⊗ F0 in dimension zero.
Show that H1 (K∗ ) = Tor(A,M ) and H0 (K∗ ) = A ⊗ M .
b) Show that for any short exact sequence of abelian groups
0→A→B→C→0
there is an associated long exact sequence
0 → Tor(A,M ) → Tor(B,M ) → Tor(C,M ) → A ⊗ M → B ⊗ M → C ⊗ M → 0.
c) For any abelian group A, show that Tor(A,M ) = 0 if A or M is a
free abelian group.
d) Show that, for any abelian group A, Tor is symmetric:
Tor(A,M ) ∼
= Tor(M,A).
(Note: It takes a wile to show this without any additional tools from
homological algebra. It is important though that you think about
what you have to do to prove the statement and that you try. The
arguments used in the lecture are useful here, too.)
e) For any abelian group A, show that Tor(A,M ) = 0 if A or M is
torsion-free, i.e., the subgroup of torsion elements vanishes.
(Hint: You may want to use the fact that a finitely generated
torsion-free abelian group is free.)
f ) Let A be an abelian group and let T (A) denote the subgroup of
torsion elements in A. Show Tor(A,M ) = Tor(T (A),M ).
(Hint: A/T (A) is torsion-free.)
Definition: Mapping cone

For a space X, the cone over X is defined as the quotient space
CX := (X × [0,1])/(X × {0}).
Let f : X → Y be a continuous map. The mapping cone of f is defined

to be te quotient space
Cf := (CX ⊔ Y )/((x,1) ∼ f (x)) = Y ∪f CX.
2 Let f : X → Y be a continuous map, and let M be an abelian group.

a) Show that the homology of the mapping cone of f fits into a long
exact sequence
f∗ i∗
· · · → H̃n+1 (Cf ; M ) → H̃n (X; M ) −
→ H̃n (Y ; M ) −
→ H̃n (Cf ; M ) → H̃n−1 (X; M ) → · · ·
(Hint: Relate Cf to the mapping cylinder Mf from a previous exer-
cise set.)
b) Show that f induces an isomorphism in homology with coefficients
in M if and only if H̃∗ (Cf ; M ) = 0.
The first part of the next exercise requires some familiarity with Tor beyond
the discussion of the lecture. But you should think about it anyway and definitely
note the statement.
3 Let X and Y be topological spaces.

a) Show that H̃∗ (X; Z) = 0 if and only if H∗ (X; Q) = 0 and H∗ (X; Fp ) =
0 for all primes p.
(Hint: Use the UCT with Fp -coefficients to control the torsion part
and with Q-coefficients to control the torsion-free part. Then apply
suitable points of the first exercise.)
b) Show that a map f : X → Y induces an isomorphism in integral
homology if and only if it induces an isomorphism in homology
with rational coefficients and in homology with Fp -coefficients for
all primes p.
4 Let X be a finite cell complex, and let Fp be a field with p elements. Show
that the Euler characteristic χ(X) can be computed by the formula
X
χ(X) = dimFp (−1)i Hi (X; Fp ).
i
In other words, χ(X) is the alternating sum of the dimensions of the

Fp -vector spaces Hi (X; Fp ).
(Hint: Use the UCT.)
5 Use the Künneth Theorem of the lecture to show:

a) The homology of the product X × S k is satisfies
Hn (X × S k ) ∼
= Hn (X) ⊕ Hn−k (X).
b) The homology of the n-torus T n defined as the n-fold product T n =
S 1 × . . . × S 1 is given by
n
= Z( i ) .
H (T n ) ∼
i
1 Let M be an abelian group. Let X be a cell complex and let Xn denote
the n-skeleton of X. We set
C ∗ (X; R) := H n (Xn ,Xn−1 ; M ).
We would like to turn this into a cochain complex. We define the
differential
dn : C n (X; M ) → C n+1 (X; M )
as the composite
dn /
C n (X; M ) = H n (Xn , Xn−1 ; M ) H n+1 (Xn+1 ,Xn ; M ) = C n+1 (X; M )
4
jn * ∂n
H n (Xn ; M )
where ∂ n is the connecting homomorphism in the long exact sequence of
cohomology groups of pairs and j n is the homomorphism induced by the
inclusion (Xn ,∅) ,→ (Xn ,Xn−1 ). Define the cellular cochain complex
of X with coefficients with M to be the cochain complex (C ∗ (X; M ),d∗ ).
Note that the cup product defines a product on the cellular cochain
complex.
a) Show that C ∗ (X; M ) is in fact a complex, i.e., dn ◦ dn−1 = 0.
b) Show that C ∗ (X; M ) is isomorphic to the cochain complex Hom(C∗ (X),M )
where C∗ (X) is the cellular chain complex of X.
(Hint: Remember the Kronecker map κ.)
c) Use the UCT for cohomology and the isomorphism between Hn (X)
and Hn (C∗ (X)) to show
H n (X; M ) ∼
= H n (C ∗ (X; M )).
Note that the isomorphism we produce this way is not functorial.
2 Let X = M (Z/m,n) be a Moore space constructed by starting with an

n-sphere S n and then forming X by attaching an n + 1-dimensional cell
to it via a map f : S n → S n of degree m
X = S n ∪f Dn+1 .
Let
q : X → X/S n ≈ S n+1
be the quotient map.
276 CHAPTER 1. EXERCISES
Recall that we showed that q induces a trivial map on H̃i (−; Z) for
all i.
a) Show H n+1 (X; Z/m) ∼ = Z/m and that H n+1 (q; Z/m) is nontrivial.
(Hint: Use the UCT for cohomology.)
b) Use the previous example to show that the splitting in the UCT for
cohomology cannot be functorial.
(Hint: You need to show that a certain square induced by the UCT
does not commute.)
3 Show that if a map g : RPn → RPm induces a nontrivial homomorphism

g ∗ : H 1 (RPm ; Z/2) → H 1 (RPn ; Z/2),
then n ≥ m.
4 Show that there does not exist a homotopy equivalence between RP3 and
RP2 ∨ S 3 .
The next exercise is a bit more challenging.
5 Let X be the cell complex obtained by attaching a 3-cell to CP2 via a

map
S 2 → S 2 = CP1 ⊂ CP2
of degree p. Let Y = M (Z/p,2) ∨ S 4 where M (Z/p,2) is a Moore space.
We observe that the cell complexes X and Y have the same 2-skeleton,
but the 4-cell is attached via different maps.
a) Show that X and Y have isomorphic cohomology rings with Z-
coefficients.
b) Show that the cohomology rings of X and Y with Z/p-coefficients
are not isomorphic.

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Algebraic Topology

What this class is about:

We can then describe topology as the science which studies properties of

Typical question in topology

Let us look at a familiar example. Fix two natural numbers n < m.

Algebraic Topology in a nutshell

In particular, this implies: if we find values of an invariant that differ for X

This is a really deep result!

Here is another example of the use of an algebraic invariant:

To calculate the number of edges e we observe that every pentagon has 5

To calculate the number of vertices v we observe again that every pentagon

Now we apply Euler’s formula:

“Euler characteristic of the surface”

The rough initiating idea for our algebraic invariants

Cell complexes and homotopy

Our main goal today is to introduce cell complexes as an important type of

The set of open sets in Rn

is a subset of all subsets of Rn and has the following properties:

Definition: Topological spaces

TY := {V ⊂ Y : there is a U ∈ TX such that V = U ∩ X}.

Definition: Open neighborhoods

Definition: Continuous maps

Examples of continuous maps include:

Some examples are:

Example: A bijection which is not a homeomorphism

We know that f is bijective and continuous from Calculus and Trigonometry

Here is an extremely important property a subset in a topological space can

• By the Theorem of Heine-Borel, a subset Z ⊂ Rn is compact if and

Simple examples of connected spaces are given by intervals in R.

Definition: Hausdorff spaces

Every subspace of RN (with the relative topology) is a Hausdorff space. More-

Lemma: Closed in compact implies compact

Lemma: Continuous images of compact sets are com-

As a consequence we can deduce a useful criterion for when continuous bijec-

Lemma: Continuous bijection from compact to Haus-

Proof: Since f is a bijection, there is a set-theoretic inverse map which we

Homeomorphisms preserve topological properties

We will remind ourselves of many other important topological properties along

Definition: Product topology

Definition: Disjoint unions or sums of spaces

Another important construction for producing new topological spaces is to

commutes, the map f is continuous iff f¯ is continuous.

• Take a subset X ⊂ Rn and consider it with the induced topology as a

Real projective space

There is also a complex version:

Complex projective space

Projective spaces play an important role in geometry and topology. We will

onto the upper hemisphere ρ : D2 → S 2 by

Definition: One-point compactification

Examples of one-point compactifications are spheres. For S n is the one-

Sphere as a quotient revisited

large. If c ∈ ∂Dn is a boundary point, then ρ(c) = ∞. Since an → c, the

Definition: Cell complexes

with the weak topology, i.e., a set A ⊂ X is open (or closed) if

We have already seen some examples of cell complexes:

Finally, we attach an open 2-cell e2 ⊂ D2 via the attaching map

• Actually, every compact smooth manifold can be turned into a fi-

What makes topology unique

Definition: Euler characteristic for cell complexes

For example, the Euler characteristic of S n is