2 edited body; edited title

A $\mathbb{Z}^2$-action \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$. The proof (in a paper I'm refereeing) is clear but somewhat laborious. It would be nice to either have a succinct proof, or a reference, rather than a detailed proof from first principles. Has anyone seen this before? Thanks! 1 # Free$\mathbb{Z}^2$-actions match at some point I have in front of me a proof of this lemma: If$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!