Question

I have in front of me a proof of this lemma:

If $f$ and $g$ are free $\mathbb{Z}_2$-actions on $S^1$, then $f(x)=g(x)$ for some $x \in S^1$.

A $\mathbb{Z}_2$-action on the unit circle $S^1$ is a homeomorphism $f \;:\; S^1 \rightarrow S^1$ such that $f(f(x))=x$ for all $x \in S^1$; and $f$ is free if $f(x) \neq x$ for all $x \in S^1$.

The proof (in a paper I'm refereeing) is clear but somewhat laborious. It would be nice to either have a succinct proof, or a reference, rather than a detailed proof from first principles. Has anyone seen this before? Thanks!

Answer 1

we can clearly assume that $f(z)=-z$ in the standard metric on $S^1\subset \mathbb C$ (as we can assume that $f$ is isometric with respect to some Riemannnian metric on $S^1$). Then $g(z)=z\cdot e^{i\alpha(z)}$ with $0<\alpha(z)<2\pi$. If $\alpha(z_0)<\pi$ then $\alpha(g(z_0))>\pi$ and there is a point $z_1$ with $\alpha(z_1)=\pi$ by the intermediate value theorem.