n. 3) Real projective spaces RP2n+1 and RP2n ∨ S2n+1 are not homotopy equivalent by comparing their cohomology rings."> n. 3) Real projective spaces RP2n+1 and RP2n ∨ S2n+1 are not homotopy equivalent by comparing their cohomology rings.">
Final 2014 Sol
Final 2014 Sol
Final 2014 Sol
Spring 2014
1.Show that every map f : S k+l S k S l , k, l > 0, induces the
(S k S l )
trivial homomorphism of reduced cohomology groups f : H
(S k+l ).
H
SOLUTION: The cohomology ring H (S m ) = Z[]/(2 ) with || =
m. Since the groups H r (S m ) are free for all r and m, the cohomology
ring H (S k S l ) is isomorphic to Z[]/(2 ) Z[]/( 2 ). Since k, l <
k + l, we obtain H k (S k+l ) = 0 = H l (S k+l ). Hence f () = 0 = f ().
Thus, f is zero homomorphism in dimensions > 0.
2. Show that there is no map f : CP m CP n with m > n inducing
a nontrivial homomorphism f : H 2 (CP n ) H 2 (CP m ).
SOLUTION: Note that H (CP r ) = Z[r ]/(rr+1 ) with |r | = 2. If
f (n ) 6= 0, then f (n ) = km with k 6= 0. Then we would have a
contradiction:
n+1
0 = f (nn+1 ) = f (n )n+1 = k n+1 m
6= 0.
5. Show that a compact manifold does not retract onto its boundary.
SOLUTION: Let M be a compact n-manifold with boundary. We
consider the homology exact sequence of the pair (M, M ) for Z2 coefficients:
Hn1 (M ; Z2 )
Hn (M ; Z2 ) Hn (M, M ; Z2 ) Hn1 (M ; Z2 )
Ccomp X,mN
is exact since the tensor product is a right exact functor. For each
there is a commutative diagram:
(lim G ) F1 (lim G ) F2 (lim G ) A 0
x
x
x
G F1
x
G F0
x
G A
x
G F1
G F0
G A
Hence there is the limit diagram which is also commutative:
0.
Contradiction.
9. Let p : X M2 be a covering map that corresponds to a
subgroup generated by the element a1 b1 in 1 (M2 ) =< a1 , a2 , b1 , b2 |
[a1 , b1 ][a2 b2 ] = 1 >.
(a) Compute H1 (X) and H2 (X); (b) Compute H 1 (X) and H 2 (X).
SOLUTION: The element a1 b1 generates a cyclic subgroup of 1 (M2 )
of infinite order. Hence 1 (X) = Z. Therefore, H1 (X) = Z. Since
p1 (x0 ) is closed subset homeomorphic to Z (i.e. discrete infinite), the
space X is not compact. Since X is a non-closed 2-manifold, H2 (X) =
0.
The cohomology groups H (X) can be obtained from the Universal
Coefficient Theorem. They are H 1 (X) = Z and H 2 (X) = 0.
10. Prove that if a closed orientable manifold M of dimension 2k
has Hk1 (M ; Z) torsion free, then Hk (M ; Z) is also torsion free.
SOLUTION: By the Poincare Duality Hk (M ) = H k (M ). By the
Universal Coefficient theorem applied to a space with finitely generated
homology, H k (M ) takes the free part from Hk (M ) and the torsion part
from Hk1 (M ). Since the torsion part T Hk1 (M ) = 0. The result
follows.
REMARK. The fact that the groups Hi (M ) are finitely generated for
a closed n-manifolds which do not admit simplicial complex structure
can be obtained as follows. One can show that the topological (covering) dimension of M is n (see Munkers). Then M can be embedded
to S N for large enough N (Munkres). Then by the Alexander Duality and the fact stated in the Problem 4 it follows that all the groups
H j (M ) are countable. This contradicts to the following Proposition
proven in class: If Hn (X) is not finitely generated then either H n (X)
or H n+1 (X) is uncountable.