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Let $f:[0,1] \to [0,1]$ be a homeomorphism with $f(0)=0$ and $f(1)=1$. If $f$ is not the identity map, is it true that $f^n \neq f$ for all integers $n>1$?

Edit: By $f^n$ I mean $f$ iterated $n$ times.

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  • $\begingroup$ Does $f^n$ mean to the power $n,$ or iterated $n$ times? I ask because if it means power then you get $f^{n-1}=1$ which implies $f$ is $0$ or $1$ at each $x$ (can't happen if $f$ continuous). $\endgroup$
    – coffeemath
    Apr 28 '16 at 3:37
  • $\begingroup$ After thinking about it some more, here is my loose reasoning for the case $n=2$. If $f$ is not the identity, there is a point $x$ with $f(x) \neq x$. $f$ is surjective, so there exists $y$ with $f(y)=x$. Then $f(f(y))=f(x)$. If $f^2=f$, then $f(f(x))=f(x)$ and $f(f(y))=f(y)$. This is a contradiction because $f$ is injective. Does that sound reasonable? $\endgroup$
    – G Pace
    Apr 28 '16 at 3:42
  • $\begingroup$ @coffeemath I meant $f$ iterated n times, I realize now that was unclear. $\endgroup$
    – G Pace
    Apr 28 '16 at 3:43
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Hint: By the Intermediate Value Theorem, $f$ must be order preserving (strictly increasing).

If $n>1$ and $f^n=f$, then you have $f^{n-2}\circ f=f^{n-1}=\iota$; thus the identity is the composition of two order preserving functions one of which is $f$. Now can you prove $f=\iota$?

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