Proof

Observe that under the identification $Mat_{n \times n}(k) \simeq k^{(n^2)}$ matrix multiplication is a polynomial function

Similarly matrix inversion is a rational function. Now rational functions are continuous on their domain of definition, and since a real matrix is invertible previsely if its determinant is non-vanishing, the domain of definition for matrix inversion is precisely $GL(n,k) \subset Mat_{n \times n}(k)$.

Proof

On the one had, the universal property of the mapping space (this prop.) gives that the inclusion

is a continuous function for $GL(n,\mathbb{R})$ equipped with the Euclidean metric topology, because this is the adjunct of the defining continuous action map

This implies that the Euclidean metric topology on $GL(n,\mathbb{R})$ is equal to or finer than the subspace topology coming from $Map(\mathbb{R}^n, \mathbb{R}^n)$.

We conclude by showing that it is also equal to or coarser, together this then implies the claims.

Since we are speaking about a subspace topology, we may consider the open subsets of the ambient Euclidean space $Mat_{n \times n}(\mathbb{R}) \simeq \mathbb{R}^{(n^2)}$. Observe that a neighborhood base of a linear map or matrix $A$ consists of sets of the form

for $\epsilon \in (0,\infty)$.

But this is also a base element for the compact-open topology, namely

where $K_i \coloneqq \{e_i\}$ is a singleton and $V_i \coloneqq B^\circ_{A e^i}(\epsilon)$ is the open ball of radius $\epsilon$ around $A e^i$.

Proof

First observe that $GL(1,k) = k \setminus \{0\}$ has this property:

1. $\mathbb{C} \setminus \{0\}$ is path-connected,

2. $\mathbb{R} \setminus \{0\} = (-\infty,0) \sqcup (0,\infty)$ is not path connected.

Now for the general case:

1. For $k = \mathbb{C}$: every invertible complex matrix is diagonalizable by a sequence of elementary matrix operations (this prop.). Each of these is clearly path-connected to the identity. Finally the subspace of invertible diagonal matrices is the product topological space $\underset{ \{1, \cdots, n\} }{\prod} (\mathbb{C} \setminus \{0\})$ and hence connected (by this prop., since each factor space is).

2. For $k = \mathbb{R}$: the determinant function is a continuous function $GL(n,k) \to \mathbb{R} \setminus \{0\}$, and since the codomain is not path connected, the domain cannot be either.

Proof

Observe that

is an open subspace, since it is the pre-image under the determinant function (which is a polynomial and hence continuous, as in the proof of lemma ) of the of the open subspace $k \setminus \{0\} \subset k$.

As an open subspace of Euclidean space, $GL(n,k)$ is not compact, by the Heine-Borel theorem.

As Euclidean space is Hausdorff, and since every subspace of a Hausdorff space is again Hausdorff, so $Gl(n,k)$ is Hausdorff.

Similarly, as Euclidean space is locally compact and since an open subspace of a locally compact space is again locally compact, it follows that $GL(n,k)$ is locally compact.

From this it follows that $GL(n,k)$ is paracompact, since locally compact topological groups are paracompact (this prop.).