Rotation is a homeomorphism

Question

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?

Answer 1

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 $f$.