What are all the topological spaces that the $S^1$ can cover up to homeomorphism?

Question

I'm currently studying $G$-coverings for the first time and I came across an interesting question. First of all, I know that all connected coverings of $S^1:=\{z\in\mathbb{C}\mid |z|=1\}$ are of the form $S^1\to S^1$ and are given as $z\mapsto z^n$ for some integer $n$ (up to isomorphisms of covering maps). This is done by looking at the subgroups of $\pi_1(S^1,1)\cong \mathbb{Z}$ which are always of the form $n\mathbb{Z}$.

But what about the opposite? That is, given a covering map $p:S^1\to X$, or maybe even given a regular $G$-covering $p:S^1\to X$, can we say something about the space $X$ up to homeomorphism? And what if we take $\mathbb{C}^\times:=\mathbb{C}\setminus\{0\}$ instead of $S^1$? (I think this is interesting since $\mathbb{C}^\times$ and $S^1$ are homotopy-equivalent, so their fundamental groups are isomorphic). So far, I wasn't able to find any example in which $X\ncong S^1$. At the time my lecturer actually didn't know the answer to this and didn't want to think about it further as it is not important for the course. But now I'm very interested in what spaces $S^1$ (and homotopy-equivalent spaces) can cover.

Answer 1

$S^1$ can cover only itself. To see this, let $\pi:S^1\to X$ be a covering. Then $X$ is a connected topological $1$-manifold*. But it is well-known that connected there are only two connected topological $1$-manifolds upto homeomorphism! They are $S^1$ and $\mathbb{R}$. Since $S^1$ has nontrivial fundamental group, so does $X$. Thus $X$ is homeomorphic to the circle. If we identify $X$ with $S^1$, then the classification theorem you just mentioned tells us that $\pi$ is isomorphic to the power map corresponding to some integer $n$.

There is not much I can say about the classification for the spaces covered by $\mathbb{C}^\times$; perhaps you could use classifications of $2$-manifolds just as above. My apologies.

You might find this this MO thread interesting.

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(*)Indeed, every covering map is a local homeomorphism, so $X$ is locally Euclidean of dimension $1$. It is also second countable because $S^1$ is so and covering maps are open. (EDITED 2021/05/25. My initial argument for the Hausdorff condition was flawed.) It takes a little more effort to see that $X$ is Hausdorff; we can argue as follows:

• First of all, points are closed in $X$. For if $x\in X$ is a non-closed point, then neither is its fiber $\pi^{-1}(x)$ (because coverings are quotient maps) and hence we can find a point $p\in\overline{\pi^{-1}(x)}\setminus \pi^{-1}(x)$. But then each neighborhood of $p$ intersects $\pi^{-1}(x)$ infinitely many often, so that $\pi(p)$ cannot have a evenly covered neighborhood, a contradiction.
• We then observe that $\pi$ is a finite covering, because each fiber is a closed, discrete subset of the compact space $S^1$.
• It follows that the compact subsets of $X$ have compact preimages under $\pi$: If $K\subset X$ is a compact set and $\mathcal{U}$ is an open cover of $\pi^{-1}(K),$ then the fact that $\pi$ is a finite cover gives us a covering of $K$ by open sets whose preimages under $\pi$ lies in some element in $\mathcal{U}$, and then by compactness of $K$ we see that in fact finite such open sets suffices to cover $K$.
• The previous point implies that $\pi$ must be cloesd: If $C\subset S^1$ is a closed set, then $C$ is compact (for $S^1$ is compact), so by what we have just remarked $\pi^{-1}(\pi(C))$ is compact and hence closed. (Since $S^1$ is Hausdorff.) Since coverings are quotient maps, this means that $\pi(C)$ is closed.
• We may now conclude that $X$ is Hausdorff, because closed quotients of compact Hausdorff spaces are Hausdorff.