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A homeomorphism f is said to be orientation reversing if for any $x<y<z$ we have $f(z)<f(y)<f(x)$. Show that every orientation reversing homeomorphism of the real line has a fixed point.

This is a question on my assignment sheet (not for credit) that I've been thinking about for days but not made any progress on. I feel like this will be easy to answer once I find the trick, any hints would be great!

SpaghettiMonsterMan
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1 Answers1

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If $f(0)=0$, we are done.

If $f(0)>0$, then $f(f(0))<f(0)$, and if $f(0)<0$, then $f(f(0))>f(0)$. In both cases, we have poitns $x_1,x_2$ with $f(x_1)<x_1$ and $f(x_2)>x_2$. Then the Intermediate Value Theorem tells us that the continuous function $x\mapsto f(x)-x$ has a zero between $x_1$ and $x_2$, i.e., $f$ has a fixed point.

Hagen von Eitzen
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  • So we don't need homeomorphism, just continuous. – lhf Oct 13 '18 at 18:46
  • @ihf I'm not sure what your comment means, but the hypothesis that it $f$ is an orientation reversing homeomorphism has been used in this proof, i.e. in the case $f(0)>0$ we may conclude $f(f(0))<f(0)$. This proof would break down if you only assumed $f$ was continuous. – Lee Mosher Oct 13 '18 at 20:32
  • @LeeMosher, I meant, does $f$ need to be a bijection? With continuous inverse? Sure, orientation reversing is essential. – lhf Oct 14 '18 at 10:56
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    What should be the meaning of "orientation reversing" for a map which is not a homeomorphism? – Paul Frost Oct 14 '18 at 12:13
  • @PaulFrost, $x \mapsto \arctan(-x)$ is an orientation reversing map $\mathbb R \to \mathbb R$ that is not a homeomorphism (though it is a homeomorphism with its image). – lhf Oct 15 '18 at 14:39
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    Okay. The only assumption you need is that $f$ is strictly decreasing. In this case $f(\mathbb{R})$ is an open interval (bounded or unbounded), and $f$ is an orientation reversing homeomorphism from $\mathbb{R}$ to $f(\mathbb{R})$. – Paul Frost Oct 15 '18 at 14:58