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Suppose $f: \mathbb{R}^n \to \mathbb{R}^n$ is continuous and for some $\lambda>0$, we have $\|f(x)-f(y)\|\geq \lambda \|x-y\|$ for all $x,y\in\mathbb{R}^{n}$. Is $f$ surjective?

I can show the image is closed by showing it contains its limit points. So, if I can show it is also open, then I am done (I want to avoid using the Invariance of Domain theorem).

Obviously, $f$ is one-to-one. By considering the the inverse map defined from $E=f(\mathbb{R}^n)$ back to $\mathbb{R}^n$ we get the following equivalent formulation of the problem:

Suppose $E \subset \mathbb{R}^n$ is closed and $f\colon E \to \mathbb{R}^n$ is Lipschitz, one-to-one, and onto. Is $E$ necessarily equal to all of $\mathbb{R}^n$?

INQUISITOR
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    Maybe this might help? https://math.stackexchange.com/q/956753/27978 – copper.hat Feb 12 '21 at 06:05
  • Well then, is there a way to show onto without using the Invariance of domain theorem? – INQUISITOR Feb 12 '21 at 20:15
  • Naive question: if $f:E\to\mathbb{R}^n$ is Lipschitz, one-to-one, and onto then isn't it true that $f^{-1}(\mathbb{R}^n)$ is open (continuous pre-image of open set) and also closed (continuous pre-image of closed set). Then you can conclude that $E=\mathbb{R}^n$ by connectedness? – Jürgen Sukumaran Mar 28 '21 at 19:07
  • That only tells you that $f^{-1}(\mathbb{R}^n)$ is relatively open in $E$. – INQUISITOR Mar 28 '21 at 19:20
  • I see. I think you can show the set $E$ is unbounded. Let $S$ be an unbounded closed subset of $\mathbb{R}^n$, then $f^{-1}(S)$ is closed in $E$. Take a sequence $s_k\in S$ with $|s_k|\to\infty$, if $E$ is bounded then there is a convergent sequence (technically subsequence but consider relabelling WLOG) $x_k\in E$ such that $f(x_k)=s_k$. Then $|f(x_\infty)|=\lim |f(x_k)| = \lim |s_k| = \infty$. – Jürgen Sukumaran Mar 29 '21 at 09:34

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