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Does there exist a homeomorphism of the unit disk with two holes $$\left\{(x,y):x^2+y^2 \le 1\right\} \setminus \left (\left \{(x,y):\left(x+ \frac 1 2 \right)^2+y^2 < \frac 1 {10} \right \} \bigcup \left \{(x,y):\left(x-\frac 1 2 \right)^2+y^2 < \frac 1 {10}\right\}\right)$$ on itself without any fixed point?

user64494
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  • Try finding a homeomorphism to a simpler space, and solve the problem there. –  Oct 07 '14 at 10:35

1 Answers1

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Yes. Your space is homeomorphic to the standard unit sphere $S^2\subseteq\mathbb R^3$ with three open disks removed. These disks can be chosen so that the rotation of $\mathbb R^3$ by $\frac{2\pi}3$ around some axis cyclically permutes them. This rotation, however, has two fixed points, where the axis of rotation intersects $S^2$. Compose it with a reflection across the plane perpendicular to your axis and you are done. (The disks can be chosen so that this reflection preserves them.)

Dejan Govc
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    Is it true that any orientation preserving homeomorphism does have a fixedpoint? I think so – Hagen von Eitzen Oct 07 '14 at 11:45
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    @ronno: But if said fixed point is in the holes, then it's not a fixed point of the homeomorphism of the space of interest. –  Oct 07 '14 at 21:38
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    Incidentally, fixed point free orientation preserving examples do not exist. Just use the Lefschetz number. – Moishe Kohan Oct 11 '14 at 00:56