Consider the operator
$$I: GL_n \to Gl_n : A \mapsto A^{-1}$$
which sends an invertible matrix to its inverse.
Identifying $Mat_n \cong \mathbb{R}^{n^2}$ through the map
$$\phi:A = (a_{ij}) \mapsto (a_{11}, \dots , a_{1n}, \dots, a_{n1}, \dots, a_{nn})$$
I want to show that $I:V = \phi(GL_n) \subseteq \mathbb{R}^{n^2} \to \mathbb{R}^{n^2}: \phi(A) \mapsto \phi(I(A)) = \phi(A^{-1})$ is a smooth map. I.e., $I \in C^{\infty}$.
The first thing that has to be shown is that $\phi(GL_n)$ is an open subset of $\mathbb{R}^{n^2}$, but this is okay because $GL_n(\mathbb{R})$ is open in $Mat_n(\mathbb{R})$ where the latter is equipped with any topology derived from a norm (because all norms on a finite dimensional vector space are equivalent), and because the map $\phi$ is a homeomorphism (to show this, we can pick convenient norms on the spaces and write a quick $\epsilon- \delta$ argument).
Is this correct so far?
How would I continue showing that $I$ is $C^\infty$? I think the formula $A^{-1} = (\det A )^{-1} \operatorname{adj}(A)$ is relevant.
It also suffices to show that all component functions $f_{ij}: V \to \mathbb{R}:(a_{11}, \dots , a_{1n}, \dots, a_{n1}, \dots, a_{nn}) \mapsto a_{ij}^{-1}$ are of class $C^\infty$ where $(a_{ij})^{-1}$ is the entry of $A^{-1}$ on place $(i,j)$.
Yes, the formula $A^{-1}=(\det A)^{-1}\operatorname{adj}(A)$ is relevant here. Each entry of $\operatorname{adj}(A)$ is a polynomial function of the entries of $A$. And, of course, $\det A$ is also a polynomial function of the entries of $A$. But polynomial functions are smooth. Therefore, inversion is smooth.