Suppose that $f:\mathbb{R}^n \rightarrow \mathbb{R}^n,f\in C^1.$ If there exists a constant $a>0$,$\forall x,y\in \mathbb{R}^n,|f(x)-f(y)|\geqslant a|x-y|,$ prove that $f:\mathbb{R}^n \rightarrow \mathbb{R}^n $ is a $C^1$ diffeomorphism.
Here's what I've done:
$f\in C^1,\forall x,y\in \mathbb{R^n},|f(x)-f(y)|\geqslant a|x-y|$ indicates that $\forall v\in \mathbb{R^n},|\frac{\partial f}{\partial v}(x_0)|\geqslant a>0.$ So $\forall x_0 \in \mathbb{R^n},df(x_0)$ is invertible. Applying the inverse function theorem, for each point $x$ there exists an open set $U_x$ such that $f|_{U_x}$ is invertible and $(f|_{U_x})^{-1}\in C^1.$ Therefore we define $f^{-1}:\cup f({U_x})\rightarrow \mathbb{R^n}.$Verify that $f^{-1}$ is well-defined:$\forall z\in U_x\cap U_y,\exists U_z\subset U_x\cap U_y,$since $f|_{U_x}$ is invertible,$(f|_{U_z})^{-1}\equiv {(f|_{U_x})^{-1}}|_{U_z},$ so ${(f|_{U_x})^{-1}}|_{U_z}\equiv {(f|_{U_y})^{-1}}|_{U_z}.$
Therefore,we just need to verify that $f$ is surjective(so that $\cup f(U_x)=\mathbb{R}^n$). I got stuck here. (Maybe fixed point theorem could help?)