Prove that the homeomorphisms generate action such the projection to quotient is a covering space. I need to prove that the action in the homeomorphism $\varphi:S^3\to S^3$ defined by
$$
\varphi(z_1,z_2) = \Bigl( e^{\frac{2\pi i}{n}}z_1, e^{\frac{2\pi mi}{n}}z_2 \Bigr)
$$
is a covering space on the projection to quotient.
Identify $S^3\subset\mathbb{C}^2$ with $S^3=\{(z_1,z_2)=|z_1|^2+|z_2|^2=1\}$.
Note that $\varphi^n(z_1,z_2)=(z_1,z_2)$, so $\langle \varphi \rangle$ is a cyclic group of order n, so it must be $\mathbb{Z}_n$. Then $\varphi$ is an action of $\mathbb{Z}_n$ on $S^3$.
I think I have some details to do in that explanation, but here is my real problem:
I want to use the Hatcher's 1.40 proposition that will guarantee that $\pi: S^3 \to S^3 \mathbin{/} \mathbb{Z}_n$ is a covering space, but I have to prove that

For each $z=(z_1,z_2)\in S^3$ exists a neighborhood $U_z$ such that all images of action $k(U_z)$ with $k\in\mathbb{Z}_n$ are disjoint.

If $k_1,k_2$ are actions, i think we can describe $$
k_1(z_1,z_2)=f^{k_1}(z_1,z_2), k_2(z_1,z_2)=f^{k_2}(z_1,z_2)
$$
So, how to guarantee that $k_1(U_z)=f^{k_1}(U_z)$ and $k_2(U_z)=f^{k_2}(U_z)$ are disjoints?
Obs: Sorry for my bad English. It will help if someone edited the mistakes.
 A: To apply Hatcher's Proposition 1.40, we have to verify condition $(*)$ om p.72.
Let us first observe that the action of $G = \mathbb Z_n$ is free which means that it does not have a fixed point (i.e. a point $y$ such that $g\cdot y= y$ for some $g \in G$ with $g \ne e$ = neutral element of $G$; see Hatcher p. 73). To see this, just look at $\varphi^k$ for $k = 1,\ldots,n-1$.
Now consider $y \in S^3$. The points $y_k = k(y)$ with $k = 0,\dots,n-1$, are distinct because the action is free. Since $S^3$ is Hausdorff, for each pair $(y_k,y_j)$ with $k < j$ we find disjoint open neighborhoods $U_{kj}$ of $y_k$ and $U_{jk}$ of $y_j$. Then the $U_k = \bigcap_{j \ne k}U_{kj}$ are pairwise disjoint open neighborhoods of the $y_k$ because for $k < j$ we have $U_k \subset U_{kj}$ and $U_j \subset U_{jk}$. Define
$$U = \bigcap_{k=0}^{n-1} k^{-1}(U_k) .$$
Then $k(U) \subset U_k$. Thus for $k < j$ we get $k(U) \cap j(U) \subset U_k \cap U_j = \emptyset$. This proves $(*)$.
Remark: Also look at exercise 23  on p.81.
