Is this action of $\mathbb{Z}/n\mathbb{Z}$ a covering space action?

Question

I define $Z=\{(w,z)\in(\mathbb{C}\setminus\{0\})^2\,\colon w^n=z\}$ for some fixed $n\in\mathbb{N}$. I define an action of $\mathbb{Z}_n$ on $Z$ by $k(w,z)=(we^{\frac{2\pi i k}{n}},z)$. I want to show that $\mathbb{Z}_n$ is a covering space action, in the sense Hather defines it at page 72. I know that $\mathbb{Z}_n$ is finite, and acts freely on $Z$ and $Z$ is Hausdorff. Is this enough to have a covering space action? I see where freeness of the action and Hausdorffness of the space come in, but where do I need finiteness of $G$?

Answer 1

On p.72 of Hatcher you find the condition $(*)$ assuring that you get a covering space action. On p.73 Hatcher remarks that $(*)$ implies that the action is free and gives an example showing that the converse is false. Moreover he says "An exercise at the end of the section is to show that for actions on Hausdorff spaces, freeness plus proper discontinuity implies condition $(*)$. Note that proper discontinuity is automatic for actions by a finite group."

This should answer your question.

Let me remark that $Z$ is a copy of $\mathbb{C}^\times = \mathbb{C}\setminus\{0\}$. See my answer to Finding the fundamental group of a subspace of $\mathbb{R}^4$. Under this identification your action correponds to $k(w)=we^{\frac{2\pi i k}{n}}$ which makes things a little bit easier.

Let us determine what $\mathbb{C}^\times /\mathbb Z _n$ is.

Let $p : \mathbb{C}^\times \to \mathbb{C}^\times /\mathbb Z _n$ denote the the quotient map and $\mu_n : \mathbb{C}^\times \to \mathbb{C}^\times, \mu_n(z) = z^n$. It is well-known (and easy to verify) that $\mu_n$ is an $n$-fold covering map. Obviously we have $\mu_n(z) = \mu_n(z')$ iff $z'= \rho z$ with an $n$-th unit root $\rho$. Hence $\mu_n(z) = \mu_n(z')$ iff $z,z'$ belong to the same equivalence class in $\mathbb{C}^\times /\mathbb Z _n$. This shows that $\mu_n$ induces a continuous bijection $h_n : \mathbb{C}^\times /\mathbb Z _n \to \mathbb{C}^\times$ such that $h_n \circ p = \mu_n$. But now $\mu_n$ is a local homeomorphism, hence an open map and therefore a quotient map. We have $h_n^{-1} \circ \mu_n = p$ which implies that $h_n^{-1}$ is continuous. That is, $h_n$ is a homeomorphism.