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Fix $\alpha\in \mathbb{R}$. Let $X$ be the interval $[0,1]$ with points $0$ and $1$ identified. Define $f:X\rightarrow X$ such that $f(x)=x+\alpha\mod 1$.

I need to show that $f$ is a homeomorphism.

My idea is to use the fact that $X$ is compact Hausdorff, then we only need to show that it is continuous and bijective. For continuous I'd like to use the distance on $X$: $d(x,y)=\min(|x-y|,1-|x-y|)$. But then I need to calculate $d(x+\alpha \mod\ 1\ ,\ y+\alpha\mod\ 1\ )$, and I'm not so comfortable with that. For bijective, I thought of $f^{-1}(x)=x-\alpha\mod\ 1\ $ but I'm not sure that makes sense on $X$.

Does anyone have a solution to this?

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You don't really need the compact Hausdorff space argument because your $f^{-1}$ is of the same kind, hence also continuous. –  Hagen von Eitzen Mar 2 '13 at 12:10
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up vote 1 down vote accepted

This solution is not how you started to do it, and maybe it's not the shortest possible, but I think it gives some insight as to what's going on:

  1. Let $g : X \to \Bbb C$ be the function $g(x) = \exp(2\pi i x)$.
  2. $g$ is continuous (as a simple exponent and also since $g(0) = g(1)$) and injective (again this is easy to see).
  3. Like you said, $X$ is compact Hausdorff, so $g$ is a homeomorphism between $X$ and $g(X)$, which happens to be the unit circle.
  4. Rotation of the unit circle counterclockwise by $2\pi\alpha$ radians is just multiplication by a unit constant in $\Bbb C$; this is again a homeomorpism (of the unit circle and itself).
  5. After the rotation, apply $g^{-1}$; this gives your $f$.
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With the usual metric on $[0,1]$ this is an isometry. –  ncmathsadist Mar 2 '13 at 12:30
    
@ncmathsadist: Not in the way I wrote it (treating $g(X)$ as a metric subspace of $\Bbb C$), but if one defines a metric on $g(X)$ induced from the metric of $X$, you get an isometry, of course :) –  Yoni Rozenshein Mar 2 '13 at 13:12
    
True. But that is just a stretching by a constant factor of $2\pi$. –  ncmathsadist Mar 2 '13 at 13:14
    
I don't think you're right. It's more like the arcsine of the distance times $2\pi$. The regular $\Bbb C$ distance between two points on the circle is the length of the straight interval between them. The isometric metric you want is the chordal distance between them (path along the circle). The relationship between them is trigonometric and can be worked out easily from the law of sines. –  Yoni Rozenshein Mar 2 '13 at 13:19
    
You are right. I am thinking about the arcwise distance. But you still have a homeomorphis.m –  ncmathsadist Mar 2 '13 at 13:26
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