Prove that any connected cover of $S^1$ of finite degree is homeomorphic to $S^1.$

Question

Assume that any connected and compact $1$-manifold is homeomorphic to either the unit circle $S^1$ or the unit interval $[0, 1].$ Prove that any connected cover of $S^1$ of finite degree is homeomorphic to $S^1.$

I don't have any idea how to proceed? Could anyone share some ideas?

Thanks for investing your valuable time on my question.

Answer 1

Let $p\colon E\to B$ be a covering map of finite degree with $B$ compact. I claim that also $E$ is compact. Let $\{U_\alpha\}_{\alpha \in A}$ be an open cover of $E$ and for each point of $b\in B$ choose a trivializing neighborhood $V_b$ such that the sheets of $p^{-1}(V_b)$ are contained in some member of $\{U_\alpha\}_{\alpha\in A}$ (here we use that $p$ is of finite degree). By compactness of $B$ there are $b_1,\ldots, b_n$ such that $B\subset \bigcup_{i=1}^n V_{b_i}$. Now again by finiteness of the covering $p^{-1}(V_{b_i})$ is covered by finitely many elements of $\{U_\alpha\}_{\alpha \in A}$, hence also $E = \bigcup_{i=1}^n p^{-1}(V_{b_i})$.

Now apply this to $B=S^1$ and use that in this case $E$ must be a $1$-dimensional manifold since $p$ is a local homeomorphism and $E$ cannot be homeomorphic to $[0,1]$, since $[0,1]$ cannot cover $S^1$.

Answer 2

We can resolve this problem without any manifold theory. Just for the interest; I realise your original question focuses on the manifold perspective. We can use covering space theory instead.

Take a locally and globally path connected space $X$, which is also semilocally simply connected. Let $\upsilon:(U,u_0)\to(X,x_0)$ be a universal covering map, where $U$ is a suitable simply connected space. It can be shown that all covers of $X$ are isomorphic to quotients $U\times F/m_R\times\varrho$ where $F$ is a discrete space, $\varrho$ is a given right action of $\pi_1(X;x_0)$ on $F$ and $m_R$ is a certain canonical right action of $\pi_1(X;x_0)$ on $U$. The connected covers are precisely those for which $\varrho$ is transitive. Moreover, the resulting cover is $|F|$-sheeted. This is covered in reasonable detail by Hatcher.

In particular take $X=S^1,U=\Bbb R$ and some connected finite $n$-sheeted cover of $S^1$. It will be isomorphic to a quotient $\Bbb R\times[n]$. Identifying $\pi_1(S^1,1)\cong\Bbb Z$ in the canonical way, the action $m_R$ in this instance is $k\cdot x=x+k$. If $\varrho$ is a transitive action of $\Bbb Z$ on $[n]=\{1,2,\cdots,n\}$, it is generated by some transitive permutation $\tau$ of $[n]$ and these are precisely the $n$-cycles. Up to a relabelling of the integers, any $n$-sheeted connected cover of $S^1$ is then isomorphic to some quotient of $\Bbb R\times[n]$ which identifies $(x,j)\sim(x',j')$ iff. $x'-x$ is an integer and $x'-x=j'-j\bmod n$. Clearly we can choose a single 'stick' $\Bbb R$ as a set of representatives, whereupon two elements are equal iff. they differ by a multiple of $n$. It follows any connected $n$-sheeted cover of $S^1$ is isomorphic to the cover $\Bbb R/n\Bbb Z\cong S^1\to S^1,z\mapsto z^n$.