Every two $n$-sheeted Coverings of $S^1$ Are Isomorphic In Prove that two $n$-sheeted covering space of $S^{1}$ are isomorphic. it was asked whether the Blaschke product and $z^n$ are isomorphic. I am wondering are these two coverings the only $n$-sheeted cover of $S^1$? If not, does every two coverings are isomorphic? I have difficulty to find the homeomorphism to prove every two coverings are isomorphic. Thank you.
 A: For $n\geq 1$, $\mathbf S^1$ has (up to isomorphism) only one (regular) covering of index $n$. This is because $\pi_1\left(\mathbf S^1\right)\simeq\mathbf Z$ has only one (normal) subgroup of index $n$, namely $n\mathbf Z$.
A: 
Theorem 1: Given any connected, locally path-connected,
semi-locally simply
connected
pointed space $(X,x)$ and any subgroup $H$ of $\pi_1(X,x)$ there
exists a connected covering $p\colon (\widetilde X,\widetilde{x})\to
 (X,x)$ such that $p_*\pi_1(\widetilde X,\widetilde x)=H$. Moreover, if
$q\colon (\widetilde Y,\widetilde{y})\to (X,x)$ is another connected
covering such that $q_*\pi_1(\widetilde Y,\widetilde y)=H$, then there
is a homeomorphism $\varphi\colon (\widetilde Y, \widetilde y)\to
(\widetilde X, \widetilde x)$ so that $p\circ \varphi=q$.

For proof, see Proposition 1.36. on page 66 of Hatcher's Algebraic Topology.

Theorem 2: If $p\colon (\widetilde X,\widetilde x)\to (X,x)$ is covering such that $\widetilde X$ is path-connected, then cardinality
of $p^{-1}(x)$ is the same as index of the subgroup
$p_*\pi_1(\widetilde X,\widetilde x)$ in $\pi_1(X,x)$.

For proof, see Proposition 1.32. on page 61 of Hatcher's Algebraic Topology.

Since $\Bbb S^1$ is a connected manifold, it is locally path-connected, semi-locally simply connected. Now, $\pi_1(\Bbb S^1,1)=\Bbb Z$ and the only subgroup of $\Bbb Z$ with index $n$ is $n\Bbb Z$ for any integer $n\geq 1$. Also, $p:\Bbb S^1\ni z\longmapsto z^n\in \Bbb S^1$ is an $n$-fold covering for integer $n\geq 1$ and if $q\colon Y\to \Bbb S^1$ is any other connected $n$-fold cover of $\Bbb S^1$ then there is a homeomorphism $\varphi\colon Y\to \Bbb S^1$ such that $\big(\varphi(y)\big)^n=q(y)$ for all $y\in Y$.
Note that the covering $\Bbb S^1\ni z\longmapsto z^{-n}\in \Bbb S^1$ is related to the $\Bbb S^1\ni z\longmapsto z^{n}\in \Bbb S^1$ via the homeomorphism $\Bbb S^1\ni z\longmapsto z^{-1}\in \Bbb S^1$. In other words, up to homeomorphism, there is exactly one $n$-fold connected covering map of $\Bbb S^1$.
