An algebraic topology proof of a result from analysis A colleague of mine recently brought up the following result from real analysis:
Theorem: If $f:\mathbb{R}^2\to\mathbb{R}^2$ is continuous and $|f(x)-f(y)|\geq |x-y|$ for all $x,y$, then $f$ is onto.
It feels like there should be a slick proof of this result using the fundamental or singular homology groups but I couldn't quite find a way to make it work. Does anyone know of a nice proof using algebraic topology?
 A: Not sure if that's enough algebraic topology for you:
Since $\lvert f(x) - f(y)\rvert \geqslant \lvert x-y\rvert$, $f$ is injective, and by the invariance of domain ($\leftarrow$ algebraic topology), it is an open map. Hence $f$ is a homeomorphism between $\mathbb{R}^2$ and $U = f(\mathbb{R}^2)$. Let $g \colon U\to \mathbb{R}^2$ be the inverse. Since $g$ is Lipschitz continuous, it has a continuous extension $\hat{g}\colon \overline{U}\to \mathbb{R}^2$. This extension must still be injective: Suppose $x_1\neq x_2$ with $\hat{g}(x_1) = \hat{g}(x_2) = \xi$. There is an $x_0\in U$ with $g(x_0) = \xi$ since $g$ is surjective. Let without loss of generality $x_0 \neq x_1$. Let $V,W$ be disjoint neighbourhoods of $x_0$ resp. $x_1$. Since $g$ is open, $g(V)$ is a neighbourhood of $\xi$, and since $g$ is injective, $g(W\cap U) \cap g(V) = \varnothing$. But that means that $\xi \notin \overline{g(W\cap U)}$, and hence
$$\xi = \hat{g}(x_1) \in \hat{g}(\overline{W\cap U}) \subset \overline{\hat{g}(W\cap U)} = \overline{g(W\cap U)} \subset \mathbb{R}^2\setminus \{\xi\}$$
is a contradiction.
Since the extension $\hat{g}$ of $g$ is still injective, and $g$ is surjective, it follows that $\overline{U} = U$, and since $\mathbb{R}^n$ is connected and $U\neq\varnothing$ it follows that $U = \mathbb{R}^2$.
(Note: the same proof works in all $\mathbb{R}^n$.)
