Let $S:=T^2\setminus{\rm int}(D^2)$ be the torus minus an open disk. Must every continuous function $f:S\to S$ have a fixed point?

This is essentially the conclusion of the Brouwer fixed point theorem. However, that theorem only applies to spaces homeomorphic to $D^n$, and so I can't directly apply it to this question. On the other hand, I can't come up with a counterexample either.

  • $\begingroup$ Wait, what exactly $T^2$ is for you and how do you subtract $D^2$ from $T^2$? $\endgroup$ – freakish Mar 2 '17 at 0:07
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    $\begingroup$ @freakish The torus… $\endgroup$ – Akiva Weinberger Mar 2 '17 at 0:08
  • $\begingroup$ To embed $D^2$ in $T^2$, choose any point and consider the set of things within a distance of $\epsilon$ from that point (for small enough $\epsilon$) $\endgroup$ – Akiva Weinberger Mar 2 '17 at 0:08

Let's express $T^2$ by letting $H \subset \mathbb{R}^2$ be a regular hexagon centered at the origin, and then gluing opposite sides of $H$ by translations.

Let $q : H \to T^2$ be the quotient map of this gluing.

Let $D^2 \subset \mathbb{R}^2$ be a small round open disc centered at the origin and contained in the interior of $H$, and so we may regard $D^2$ as embedded in $T^2$ by the map $q$. Then, as you say, we remove the interior of $D^2$ to obtain the surface $S$.

Now let $R : H \to H$ be a rotation of angle $2 \pi / 6$. The map $R$ induces a homeomorphism $S \mapsto S$ having no fixed points.

  • $\begingroup$ Ah, nice. I thought of doing a similar thing for a square instead of a hexagon, but that didn't work (since the corners, which get quotiented to a point in the quotient map, get mapped to themselves). I didn't think of doing hexagons; this seems to work ('cause now there are two corners after quotienting). $\endgroup$ – Akiva Weinberger Mar 2 '17 at 2:44

Let's sit your torus $T^2$ inside $\Bbb{C} \times \Bbb{C}$ as the subset $S^1 \times S^1$, where $S^1 = \{ z \mid |z|= 1\}$ and let's sit your disk $D^2$ so that it doesn't meet the circle $\{(z, 1) \mid z \in S^1\}$. Then the function $(z, w) \mapsto (iz,1)$ has no fixed points on any subset of $T^2$.

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    $\begingroup$ But does that example restrict to a function $f : S \to S$? $\endgroup$ – Lee Mosher Mar 2 '17 at 1:18
  • $\begingroup$ Yes, it does, subject to a small (and obvious) refinement I have made to my answer. $\endgroup$ – Rob Arthan Mar 2 '17 at 2:20
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    $\begingroup$ How could any point of the form $(z,0)$ possibly be on the torus? $0\notin S^1$. (Remember that the disk in the question is meant to be a subset of the torus.) $\endgroup$ – Akiva Weinberger Mar 2 '17 at 2:34
  • $\begingroup$ I have fixed the typo. $\endgroup$ – Rob Arthan Mar 2 '17 at 2:56
  • $\begingroup$ Oh, I misread, sorry! I thought it said $(z,w)\mapsto(iz,w)$ (a rotation). I see now that this is a rotation combined with a projection. Yes, this should work. I apologize! $\endgroup$ – Akiva Weinberger Mar 2 '17 at 3:42

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