Let $U \subseteq \mathbb{R}^n$ be open and let $f: U \to \mathbb{R}^n$ be continuous differentiable and one-to-one. Let $Df(x)$ be invertible for all $x \in U$.
I have already shown that the set $V:=f(U)$ is open and that there exists a inverse function $g: V \to U$.
Currently I'm struggling to show thath $g$ is continuous differentiable with $Dg(y) = [Df(g(y))]^{-1}$ for all $y \in V$. It seems like I have to apply the inverse function theorem here which would give me exactly that statment.
The inverse function theorem states:
$U \subset \mathbb{R}^n$ open, $f: U \to \mathbb{R}^n$ continuous differentiable, $a \in U, b:=f(a)$ with $Df(a) \in \mathbb{R}^{n \times n} $ invertible Then:
$\exists V \subset U$ open neighborhood of $a$
$\exists \tilde{V}$ open neighborhood of $b$ with $f:V \to \tilde{V}$ bijective and $g=f^{-1}:\tilde{V} \to V$ continuous differentiable with $Dg(f(a))= [Df(a)]^{-1}$ respectively $Dg(b) = [Df^{-1}(b)]^{-1}$.
I do not see how the existence of $g:V \to U$ and the fact that $f(U)=V$ is open should lead to the continous differentiability of $g$ using this theorem.
I think I figured it out. If someone got an alternative proof or thinks my proof is wrong feel free to post your idea or to edit this answer.
Let $a \in U, b:=f(a)$ Using the inverse function theorem we find an open neighborhood $W \subset U$ of $a$ and an open neighborhood $\tilde{W}$ of $b$ with $f: W \to \tilde{W}$ bijective and $g=f^{-1}:\tilde{W} \to W$ continuous differentiable with $Dg(f(a))=[Df(a)]^{-1}$ respectivly $Dg(b) = [Df(g(b))]^{-1}$.
Since $a \in U$ was arbitrary and $V=f(U)$ is open, $g$ is continuous differentiable on $V$ with $Dg(b) = [Df(g(b)]^{-1}$ for all $b$ in $V$. $\Box$