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Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be continuously differentiable. If there exists a real number $a>0$, such that $\forall x,y\in \mathbb{R}^n$, $d\big(f(x),f(y)\big)\geq a\cdot d(x,y)$, please show that $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is a $C^1$-diffeomorphism.

It's clear that $f$ is injective, and the differential $df(x)$ is invertible for any $x\in \mathbb{R}$, since $f(x+v)=f(x)+df(x)(v)+o(v)$, as $v\rightarrow 0$. Then by the inverse function theorem, we have $f$ is a locally diffeomorphism. If $f$ is surjective, then we can show that $f$ is a $C^1$-diffeomorphism.

How can I show that $f$ is surjective?

Arctic Char
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ALe0
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  • Welcome! Generally, we excpect questions here to show some effort towards a solution or some kind of research. If you already gave this problem a shot on your own, give us some idea of what you tried and where you got stuck. If you haven't yet, give it an honest effort and then come back if you're still stuck. – Alex Jones Mar 25 '21 at 14:27
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    Thanks for your reminding ! – ALe0 Mar 25 '21 at 14:45

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I hope that you are familiar with some metric topology. The strategy is to prove that the image of $f$ is both open and closed as a subset of $\mathbb{R}^n$. Since $\mathbb{R}^n$ is connected, it is the only non-empty subset of itself having these two properties.

We first show that $U = f(\mathbb{R}^n)$ is closed in $\mathbb{R}^n$. Let a sequence $y_n\in U$ converge to some $y \in \mathbb{R}^n$. Then there exist $x_n$ such that $y_n = f(x_n)$. By the assumption we have $d(x_n,x_m) \leq a\cdot d(y_n,y_m)$, but since $y_n$ converges in $\mathbb{R}^n$ it is a Cauchy sequence, and thus $x_n$ is Cauchy as well and converges to some $x\in \mathbb{R}^n$. It follows that $y = f(x)\in U$.

The openness follows from the inverse function theorem. It states that the function is invertible on a neighborhood (that is, an open subset of $\mathbb{R}^n$) of each point $f(y)$, which implicitly says that the neighborhood is contained in the image (cf. the proof of Theorem 2 here).

ajr
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  • Thanks a lot. What I wanted to do is to show that f(R^n) is clopen in R^n. The difficulty for me is the openness. The fact is that I ignored some conclusion of the inverse function theorem. – ALe0 Mar 25 '21 at 16:44
  • At first I overlooked it too. I was hoping that it can be done via some easy topological argument, but I have only found https://en.wikipedia.org/wiki/Invariance_of_domain which seems like a very high caliber for this. Perhaps the proof which I linked may be altered so to use only the fact that $|f(x) - f(y)| \geq c|x-y|$. – ajr Mar 25 '21 at 17:04