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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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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. –  Haskell Curry Dec 29 '12 at 15:50
    
Also check this recent question. –  user1551 Dec 29 '12 at 16:18

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

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Yes. But I originally meant $R^d$ instead of $R$. –  Ash GX Dec 31 '12 at 10:53

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