Is $f$ surjective, where $|f(x)-x| \leq 2$? 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?
 A: A homotopy proof (outline.) Not sure how to do it with only analysis.
Assume $x_0\in\mathbb R^2$ is not in the range of $f$. consider $S=\{x\in\mathbb R^2: \left|x-x_0\right|=3\}$ be the circle of radius $3$ around $x_0$. 
Then $f:S\to\mathbb R^2\setminus\{x_0\}$ is homotopic with the inclusion map $S\to\mathbb R^2\setminus\{x_0\}$, but the map: $h(x,t) = tx + (1-t)f(x)$, which never hits $x_0$ by the rule $|f(x)-x|\leq 2$.
But the function $f$ extends to the ball $B=\{x:|x-x_0|\leq 3\}$, which means that $f_{|S}$ retracts to a constant function as a map to $\mathbb R^2\setminus\{x_0\}$. But the inclusion map on $S$ does not retract to a point, which is a contradiction.
So our assumption is false.
Basically, this is a property that is a multi-dimensional intermediate value theorem. If a continuous $f:\mathbb R^2\to\mathbb R^2$ sends some circle to a path that "winds" a non-zero total number of times around some point $y_0$, then $y_0$ is in the range of the function. There is an $n$-dimensional version of this, too. I'm just not sure how to even define this concept of "wind" without some homotopy.
A: 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.
