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This is a problem from Bredon's 'Topology and Geometry,' it's not homework, just something I'm doing for general learning.

The problem is as follows:

Let $K = \{ (x,y) \in \mathbb R^2| x^2 + y^2 \leq 16, (x-2)^2 + y^2 \geq 1, (x+2)^2 + y^2 \geq 1 \}$. This is just a disk with two smaller disks removed from the interior. If $f: K \to K$ is a homeomorphism with no fixed points, show that $f$ must cyclically permute the three boundary components, and reverse orientations. Construct an example of such a map.

For this problem, I thought I had worked out a map with the desired properties that realized this space as a kind of 'pair of pants' type object, but when I want to check with a more careful sketch, it turns out my map actually was not orientation reversing, and also had fixed points(!). My new ideas is to take the pair of pants, and try turning them inside out, permuting the holes, and then putting it back down in the plane, where a leg is now the torso, the torso a leg, and the second leg is now the first.

I believe this map will do what I want, but since I can't write it down explicitly, it seems hopeless to check if it has fixed points or not.

Actually, now with some more thinking, I'm worried this map may have fixed points too, in the 'crotch' of the pants. I know that once I produce a map, I can check if it's fixed point free or not with the Lefschetz-Hopf trace formula, if I can give a description of the induced map on homology (or just chain groups, by a theorem in Bredon). Since my map sends generators of the chain group to generators, we can actually just check that this has Lefschetz number different from $0$, and so it does have a fixed point. So that ideas is dead in the water.

OK, so I'm rather stuck on this problem, even after some new thoughts while writing this up. Maybe I should shoot for the general case first, and see if that's more insightful. Either way, some helpful hints would be appreciated.

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View $K$ as homeomorphic to a sphere with three identical circular holes punched in it at equdistant points along the equator.

Mirror everything across the equatorial plane and then turn the sphere by 120° around the axis.

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    Comment rescinded. I misunderstood your answer. Let me think on it for a moment. I think this will work though. Now I will have to think about how to pass to the general case. Thank you!2017-02-24