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If a homeomorphism $f:R\rightarrow R$ satisfies $f^2=1$, prove that it has at least one fix point.

What if we set $f^n=1$ instead of $f^2=1$?

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    Also check http://math.stackexchange.com/questions/233246/periodic-orbits/233252#2332522012-12-29
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    I think that it is better to replace $ 1 $ by $ \text{id} $. I initially mistook $ 1 $ for the constant function that takes the value $ 1 $ everywhere.2012-12-29
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    Also check [this recent question](http://math.stackexchange.com/questions/263753/fixed-point-in-a-continuous-map).2012-12-29

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HINT: The sets $U=\{x\in\Bbb R:f(x)>x\}$ and $V=\{x\in\Bbb R:f(x) are open and disjoint, $f[U]=V$, and $f[V]=U$.

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    Yes. But I originally meant $R^d$ instead of $R$.2012-12-31