[proofplan]
We prove openness by viewing the determinant as a continuous scalar-valued function on the finite-dimensional real [vector space](/page/Vector%20Space) $M(n,\mathbb F)$. The general linear group is exactly the preimage of the [open set](/page/Open%20Set) $\mathbb F\setminus\{0\}$ under this determinant map. Once openness is established, the manifold statement follows from the standard smooth structure on open subsets of finite-dimensional real vector spaces, together with the real dimension count for real and complex matrix spaces.
[/proofplan]
[step:Declare the real topologies and the determinant map]
Fix $n\in\mathbb N$ and $\mathbb F\in\{\mathbb R,\mathbb C\}$. Regard $M(n,\mathbb F)$ as a real vector space: for $\mathbb F=\mathbb R$ this is the vector space of real $n\times n$ matrices, and for $\mathbb F=\mathbb C$ this is the underlying real vector space of the complex vector space of complex $n\times n$ matrices.
Define the determinant map
\begin{align*}
\det_{\mathbb F}:M(n,\mathbb F)\to\mathbb F,\qquad A\mapsto \det A.
\end{align*}
The codomain $\mathbb F$ is equipped with its usual real topology, namely the usual topology on $\mathbb R$ when $\mathbb F=\mathbb R$ and the topology inherited from the real vector space identification $\mathbb C\cong\mathbb R^2$ when $\mathbb F=\mathbb C$.
[/step]
[step:Prove that the determinant map is continuous]
Let $S_n$ denote the [symmetric group](/page/Symmetric%20Group) on $\{1,\dots,n\}$. For $A=(A_{ij})\in M(n,\mathbb F)$, the determinant is given by the Leibniz formula
\begin{align*}
\det A=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^{n}A_{i,\sigma(i)}.
\end{align*}
This is a polynomial expression in the matrix entries. Therefore $\det_{\mathbb F}:M(n,\mathbb F)\to\mathbb F$ is continuous as a map between finite-dimensional real vector spaces.
[guided]
The point of writing the determinant explicitly is to reduce the topology question to a standard continuity fact about polynomials. Let $S_n$ be the symmetric group on $\{1,\dots,n\}$, and write each matrix $A\in M(n,\mathbb F)$ as $A=(A_{ij})$. The Leibniz formula gives
\begin{align*}
\det A=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^{n}A_{i,\sigma(i)}.
\end{align*}
Each factor $A_{i,\sigma(i)}$ is a coordinate function on the finite-dimensional real vector space $M(n,\mathbb F)$. Products and finite sums of continuous coordinate functions are continuous. Hence the displayed formula expresses $\det_{\mathbb F}$ as a polynomial map in real coordinates when $\mathbb F=\mathbb R$.
When $\mathbb F=\mathbb C$, each complex coordinate $A_{ij}$ is two real coordinates, its real and imaginary parts. Complex addition and multiplication are polynomial operations in these two real coordinates, so the same Leibniz formula expresses the real and imaginary parts of $\det A$ as real polynomial functions on the real vector space underlying $M(n,\mathbb C)$. Thus $\det_{\mathbb C}:M(n,\mathbb C)\to\mathbb C\cong\mathbb R^2$ is continuous.
[/guided]
[/step]
[step:Identify $GL(n,\mathbb F)$ as an open preimage]
By definition of the general linear group,
\begin{align*}
GL(n,\mathbb F)=\{A\in M(n,\mathbb F):\det A\neq 0\}=\det_{\mathbb F}^{-1}(\mathbb F\setminus\{0\}).
\end{align*}
The singleton $\{0\}$ is closed in the [Hausdorff space](/page/Hausdorff%20Space) $\mathbb F$, so $\mathbb F\setminus\{0\}$ is open. Since $\det_{\mathbb F}$ is continuous, the preimage $\det_{\mathbb F}^{-1}(\mathbb F\setminus\{0\})$ is open in $M(n,\mathbb F)$. Therefore $GL(n,\mathbb F)$ is open in $M(n,\mathbb F)$.
[/step]
[step:Compute the inherited smooth manifold dimensions]
An open subset of a finite-dimensional real vector space inherits a smooth manifold structure whose dimension is the real dimension of the ambient vector space. The map
\begin{align*}
M(n,\mathbb R)\to\mathbb R^{n^2},\qquad (A_{ij})\mapsto (A_{ij})_{1\leq i,j\leq n}
\end{align*}
is a real vector space isomorphism, so $\dim_{\mathbb R}M(n,\mathbb R)=n^2$. Hence $GL(n,\mathbb R)$ is a smooth manifold of real dimension $n^2$.
Similarly, the map
\begin{align*}
M(n,\mathbb C)\to\mathbb R^{2n^2},\qquad (A_{ij})\mapsto (\operatorname{Re}A_{ij},\operatorname{Im}A_{ij})_{1\leq i,j\leq n}
\end{align*}
is a real vector space isomorphism, so $\dim_{\mathbb R}M(n,\mathbb C)=2n^2$. Hence $GL(n,\mathbb C)$ is a smooth manifold of real dimension $2n^2$. This proves the theorem.
[/step]
