I want to prove that the map $f: A \to A^{-1}$ is continuous where $A$ is an $n \times n$ matrix.
The idea is to use
$A^{-1}=\frac{1}{\det(A)}A_\text{adj}$ where $(A_\text{adj})_{ij}=(-1)^{i+j} \det(X_{ji}(A))$.
Here $X_{ji}(A)$ is the $(n-1) \times (n-1)$ matrix obtained by crossing out row $j$ and column $i$ from $A$.
I have already proven the continuity of the map $A \to a_{ij}$ $\forall i,j$, and thus that $A \to \det(A)$ is continuous as a composition of continuous maps.
Now it remains to show that the map $A \to A_\text{adj}$ is continuous. For this I need to show that the function $a_{ij} \to A_\text{adj}$ is continuous.
I know that the function $a_{ij} \to (A_\text{adj})_{ij}$ is continuous, but I cannot seem to deduce why this implies that $a_{ij} \to A_\text{adj}$ is continuous.
How do I prove this last step? I know there are some threads on this topic, but I could not find an answer specific to my question.
Thanks a lot!
Edit: I think I can use that for an $m \times n$ matrix,
$\|A\|_{operator} \leq mn \max_{i,j}|a_{ij}|$.
Claim: A map $f: \Bbb R \to \Bbb R^{m \times n}$ is continuous if all of the coordinate maps $f_{ij}:\Bbb R \to \Bbb R$ defined by $f_{ij}(t) = [f(t)]_{i,j}$ are continuous.
Proof: Note that the Frobenius norm, which is defined by $$ \|A\|_F^2 = \sum_{i,j}a_{ij}^2, $$ satisfies $\|A\| \leq \|A\|_F$ (where $\|A\|$ denotes the Euclidean operator norm). With that, we find the following. Consider any $x \in \Bbb R$. Fix $\epsilon > 0$. There exists a $\delta > 0$ such that if $|x-y| < \delta$, then $|f_{ij}(x) - f_{ij}(y)| < \epsilon/\sqrt{mn}$ for all $i,j$. With that, we find that $$ \|f(x) - f(y)\|^2 < \|f(x) - f(y)\|_F^2 = \sum_{ij} |f_{ij}(x) - f_{ij}(y)|^2 < \sum_{ij} \frac{\epsilon^2}{mn} = \epsilon^2. $$ So indeed, $f$ is continuous at $x$, which was arbitrary. So $f$ is continuous.