Suppose $f:\mathbb{R}^n\to \mathbb{R}^n$ is $C^1$,and $||f(x)-f(y)||\ge||x-y||$, is $f$ surjective? I asked this question a few days ago but haven't get an answer, so I ask again here.
Suppose $f:\mathbb{R}^n\to \mathbb{R}^n$ is $C^1$,and $||f(x)-f(y)||\ge||x-y||$, is $f$ surjective? What if $f$ is just $C^0$?I know it is clearly true when $n=1$, but when $n\ge 1$?
 A: $f$ is clearly injective. Being an injective $C^0$ map $\mathbb{R}^n \to \mathbb{R}^n$, it follows by invariance of domain that  the image is open. The argument on Hagen von Eitzen's answer implies it is also closed. Therefore, the image must be the whole $\mathbb{R}^n$.
The argument due to Hagen is:

Let $a\in\overline{\operatorname{im}(f)}$. Then a sequence $x_n$ with
  $f(x_n)\to a$ must be Cauchy, hence $x_n\to x$ with $f(x)=a$ (by
  continuity of $f$). We conclude that $\operatorname{im}(f)$ is closed.

A: Such $f$ is clearly injective and the inverse ${\operatorname{im}(f)}\to\Bbb R^n$ is continuous, hence we have a homeomorphism.
Let $a\in\overline{\operatorname{im}(f)}$.
Then a sequence $x_n$ with $f(x_n)\to a$ must be Cauchy, hence $x_n\to x$ with $f(x)=a$ (by continuity of $f$). We conclude that $\operatorname{im}(f)$ is closed.
Suppose $b\in\partial {\operatorname{im}(f)}$. Then $\operatorname{im}(f)\setminus\{b\}$ is still simply connected, but $\Bbb R^n\setminus\{f^{-1}(b)\}$ is not - contradiction.
