TITLE: Is $f$ surjective, where $|f(x)-x| \leq 2$?
QUESTION [6 upvotes]: Let $f$ be continuous and $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$.
Suppose  $|f(x)-x| \leq 2$ holds for all $x$. Is $f$ surjective?

REPLY [6 votes]: You can do this by Brouwer's fixed-point theorem. Let $a\in\Bbb R^2$ be arbitrary, and define $g:\Bbb R^2\to\Bbb R^2$ by $g(x)=x-f(x)+a$. By hypothesis, $g$ is continuous and has image contained in the disk of radius$~2$ and center$~a$. Restricting to that disk, $g$ has a fixed point $p$ in it by mentioned theorem. But $g(p)=p$ means $f(p)=a$, and since $a$ was arbitrary, $f$ is surjective.