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In Dynamical Systems and Ergodic Theory by Pollicott and Yuri, there is an easy, one dimensional, fixed point theorem:

If $T$ is a continuous map on a closed interval $J$ so that $T(J)\supseteq J$, then $T$ has a fixed point.

Here the space is mapped onto itself, in contrast to the "usual" fixed point theorems where it is mapped into itself.

Out of idle curiosity, I wonder if it's true in higher dimensions. To be concrete, let $D=\{(x,y): x^2+y^2\leq 1\}$ be the closed unit disk in $\mathbb{R}^2$. Suppose that $T$ is continuous on $D$ and $T(D)\supseteq D$, does $T$ have a fixed point?

If $T$ is one-to-one, for instance, this follows by applying Kakutani's fixed point theorem to the set valued map $x\mapsto T^{-1}(\{x\})$ on $D$. But I'm not sure whether the result holds in general.

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Upvote for "out of idle curiosity"! –  Tim Mar 1 '12 at 20:25
    
Isn't the proof the same as for Brouwer's fixed point theorem? Project $z \in D$ to its "shadow" on $\partial D$ if you put a light source at $f(z)$. This would give a continuous map $D \rightarrow \partial D$ that fixes $\partial D$ if $f$ has no fixed points. –  WimC Mar 1 '12 at 20:29
    
...no that doesn't work. It doesn't have to be continuous at $\partial D$. –  WimC Mar 1 '12 at 20:36
    
Could we perhaps map $z \in D$ to both intersection points of the line connecting $z$ and $f(z)$ on $\partial D$. That would give a map $D \rightarrow \partial D \times \partial D$ and $\partial D$ would map to a non-contractible curve on this torus. –  WimC Mar 1 '12 at 20:42
    
@WimC Thanks very much for your input. I am slower than you are, and I'm still processing the "no that doesn't work" comment. If you come up with a solution, please post it as an answer. Thanks! –  Byron Schmuland Mar 1 '12 at 20:49
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up vote 5 down vote accepted

Unfortunately, this is not true, as I convinced myself using a cork coaster not unlike these. A counterexample is given by a map that maps the upper third of the disk to the lower half, the lower third to the upper half, and the middle third to a band that connects the two outside the disk.

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You just beat me to it! –  TonyK Mar 1 '12 at 22:07
    
Very nice! Coasters are indispensible math tools. –  WimC Mar 1 '12 at 22:11
    
@joriki Thanks very much for this. I never would have thought of this on my own. –  Byron Schmuland Mar 1 '12 at 22:50
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