TITLE: Does a continuous map $f$ from the $n$-ball $B$ into $R^n$ such that $B\subset f(B)$ have a fixed point?
QUESTION [6 upvotes]: If $f$ is a continuous map from the $n$-ball $B$ into itself, the Brouwer fixed point theorem guarantees a fixed point. What if we assume that $f$ maps $B$ into all $R^n$, and $f(B)$ contains $B$? For the one-dimensional case it is still true that such $f$ has a fixed point. Does this property also hold for higher dimensions?

REPLY [5 votes]: (Really just a comment on Nik Weaver's answer, but too long for a comment)
The same idea with a different (more easily expressible) parametrization:  Use polar coordinates, and 
consider the "square"  $D:=\{(r,\psi):  \frac12 \le r \le 1, 0\le \psi \le \alpha\}$, with some small angle $\alpha$.

For $0\le\psi \le\frac14 \alpha$, map $(r,\psi)$ to $(r,\frac12\alpha + 2\psi)$, covering the region between $\frac12\alpha$ and $\alpha$. 
For $\frac34 \alpha\le\psi\le \alpha$, map $(r,\psi)$ to $(r,2*(\psi-\frac34\alpha))$, covering the region between $0$ and $\frac12\alpha$. 
For $\frac 14 \alpha\le \psi\le \frac34\alpha$, map $(r,\psi)$ to $(r, c(\psi-\frac14\alpha)+d)$, with $d:=\alpha$ and $c\cdot\frac12\alpha+d=2\pi$.