GL_n is a topological group

Idea. There is nothing new to prove except to check the definitions: the group structure is the usual one and the topology is the subspace topology from $M_n(\mathbb{R})$.

Step 1: group. $GL_n(\mathbb{R})$ is a group under multiplication (Problem 67 gives closure: $\det(AB)\neq0$).

Step 2: topology. The subspace topology on $GL_n$ is Hausdorff (subspace of $\mathbb{R}^{n^2}$). Multiplication $GL_n\times GL_n\to GL_n$ is the restriction of the continuous multiplication $M_n\times M_n\to M_n$ (Problem 67), and inversion is continuous (Problem 68). Hence $GL_n(\mathbb{R})$ is a topological group; in fact a *Lie group*, since $GL_n$ is an open subset of $\mathbb{R}^{n^2}$ and both operations are polynomial or rational.

Step 3: the determinant. $\det(AB)=\det A\det B$ says $\det:GL_n\to\mathbb{R}^\times=\mathbb{R}\setminus\{0\}$ is a group homomorphism for multiplication on $\mathbb{R}^\times$; it is continuous (Problem 66). Its kernel is $\{A:\det A=1\}=SL_n(\mathbb{R})$.

Step 4: consequences.
- $SL_n(\mathbb{R})=\det^{-1}(\{1\})$ is a *closed* normal subgroup of $GL_n(\mathbb{R})$, being the preimage of a closed set by a continuous map.
- The cosets of $SL_n$ are the level sets $\{\det=c\}$. The map $A\mapsto\det A$ is surjective onto $\mathbb{R}^\times$ (use $\operatorname{diag}(c,1,\dots,1)$), so $GL_n/SL_n\cong\mathbb{R}^\times$.

Remark. $\mathbb{R}^\times$ has two connected components, and $\det$ is surjective and continuous. So $GL_n(\mathbb{R})$ has *at least* two components (the signs of $\det$); Part 5 shows that it has exactly two.