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Let $f,g$ be homeomorphisms of the topological space $\mathbb{I}\times\mathbb{I}$ such that both of $f,g$ have more than one fixed point. Must $fg$ have more than one fixed point ?

I tried looking for counterexamples for some time but I didn't find.

Thank you

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2 Answers

up vote 9 down vote accepted

I'm assuming that by $\mathbb{I}$ you mean a unit interval.

Flip the unit square across one diagonal, and then across the other. Each of those maps is an isometry, so definitely a homeomorphism, and each one has infinitely many fixed points. However, the composition of the two maps only has one fixed point, namely the point at the center of the square.

Does that work?

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Yes. Surely works :) –  Amr Sep 1 '13 at 3:37
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For simplicity, let us assume that $\mathbb I=[-1,1]$. Consider the following maps $$g(x,y)=(-x,y)\ \text{and} \ f(x,y)=(x,-y)$$ Both of them have more than one fixed point (actually uncountably many ) but $(0,0)$ is the only fixed point of $f\circ g$.

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