$\|f(x)-f(y)\|\ge c\|x-y\|$ for all $x,y\in U, c>0$-> continuously differentiable inverse function $g:f(U)\to U$ Let $U\subset\mathbb{R}^n$ open, $f:U\to \mathbb{R}^n$ continuously differentiable, $\|f(x)-f(y)\|\ge c\|x-y\|$ for all $x,y\in U, c>0$.
Why is


*

*$\det(Df(x))\neq 0$ for all $x\in U$ and

*$f\colon U\mapsto f(U)$ global invertible with inverse function g, wich is continuously differentiable ?


First of all, here If $f: U \rightarrow \mathbb{R}^n$ differentiable such that $|f(x)-f(y)| \geq c |x-y|$ for all $x,y \in U$, then $\det \mathbf{J}_f(x) \neq 0$ is a solution for 1. 
Now 2.
My ideas: and from $\|f(x)-f(y)\|\ge c\|x-y\|$ for all $x$, $y\in U$, $c>0$ follows that f is injective. Because otherwise for $x\not=y$ such that $f(x)=f(y)$ it follows that $0=\|f(x)-f(y)\|\ge c\|x-y\|>0$ which is a contradiction. 


*Clearly $f\colon U\to f(U)$ is surjective . And with the argument above $f$ onto $f(U)$ is bijective with inverse function $g$. But why is $g$ continuously differentiable ?


You need to use the inverse function theorem. It tells you that, since $Df(x)$ is invertible for each $x \in U$, there is a small neighborhood $x \in V \subset U$ such that $f|_V : V \to f(V)$ has a continuously differentiable inverse. The only possibility is that $(f|_V)^{-1} = g|_{f(V)}$, thus $g|_{f(V)}$ is continuously differentiable. But continuous differentiability is a local property, and we have the above for all $x\in U$, so we conclude that $g$ is continuously differentiable.
