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?
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2 Answers
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Edit: I'm retaining my original answer below, but here is a simpler formula which builds on Goldstern's modification of my answer.
Work in the complex plane. Let $$A = \{z \in \mathbb{C}: 1 \leq |z| \leq 2\mbox{ and }0 \leq {\rm arg}(z) \leq \pi\}.$$ This set is homeomorphic to the unit ball, so for topological questions like the existence of fixed points the change is inessential.
Define $f: A \to \mathbb{C}$ by $f(z) = iz^2$. Then the image under $f$ of the part of $A$ with argument between $0$ and $\pi/4$ is $\{z: 1 \leq |z| \leq 4$ and $\pi/2 \leq {\rm arg}(z) \leq \pi\}$, and the image of the part of $A$ with argument between $3\pi/4$ and $\pi$ is $\{z: 1 \leq |z| \leq 4$ and $0 \leq {\rm arg}(z) \leq \pi/2\}$. The rest of $A$ maps into the lower half-plane. So the image of $A$ contains $A$, and it is obvious from looking at arguments that there are no fixed points.
(The alternative formula $f(z) = iz^2/|z|$ would eliminate the radial stretching, if one prefers this.)
Original answer below.
This fails for $n = 2$. For simplicity consider the unit square $C = [0,1]^2$ instead of $B$. Then define $f$ by first setting $f(x,y) = (x + 1/2, y)$ for $(x,y) \in C$ with $x \leq 1/2$ --- this shifts the left half of the square onto the right half, no fixed points there. Define $f(x,y) = (2x - 3/2, y)$ for $(x,y) \in C$ with $x \geq 3/4$ --- this takes the right one-fourth of $C$ onto the left half, again no fixed points. Then the middle strip $[1/2, 3/4] \times [0,1]$ can be stretched around over the top of the square to complete the definition of a continuous map. I have a hard time describing that last step without drawing a picture, is it clear?
answered Jul 13, 2015 at 21:54
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$\begingroup$ You may be able to use something like (z+c)^n for a well chosen complex c and integer n for the map, and show that (z+c)^n - z "has large roots". Gerhard "Not Ready For Such Complexity" Paseman, 2015.07.13 $\endgroup$ Commented Jul 13, 2015 at 22:49
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$\begingroup$ Yes, it is clear, thanks. I guess when you do the stretching and over-the-top construction, you take care to get agreement with the previous definition, over the top boundary of the square. $\endgroup$ Commented Jul 14, 2015 at 19:30
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$\begingroup$ Very nice example. I don't think this is general knowledge. I would like your permission to quote the example (including your autorship) in a work done by a student on Sharkovskii's theorem. $\endgroup$ Commented Jul 23, 2015 at 3:57
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$\begingroup$ The new answer is even nicer than mine. $\endgroup$ Commented Jul 29, 2015 at 9:50
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(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 $c\cdot\frac12\alpha+d=2\pi$.
