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Let $p\colon \mathbb{S}^2 \to X$, where $\mathbb{S}^2 = \{x \in \mathbb{R}^3 : \|x\| = 1\}$, be a covering projection. Is it possible to characterize the space $X$ up to a homeomorphism?

It is certainly clear that $X$ can be a 2-sphere or a projective plane. Are there any other possibilities for $X$?

pizet
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There are no other $X$.

If $p\colon Y \to X$ is a covering space and one of $X$, $Y$ is a surface, then so must the other (this is more or less immediate from the fact that $p$ is a local homeomorphism). Moreover, by compactness, if $Y$ is a closed surface, then so is $X$. Thus, if $Y = S^2$, we note that since it is simply connected and compact it can only cover closed surfaces $X$ with finite fundamental group (no covering space with infinitely many sheets is compact), and there's only two of them: $S^2$ itself and $\mathbb{R}\mathrm{P}^2$.

Ben Steffan
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  • Your first claim isn't quite accurate. In general, $X$ needn't be Hausdorff, hence not a surface. However, it's not hard to show that a space finitely covered by a Hausdorff space is Hausdorff (hence in our case a surface), so this ought to be shown first. Also, it is true that $Y$ being compact forces it to be finitely sheeted, but since it's not clear a priori that the fibers are closed, this requires an argument. – Thorgott Mar 23 '24 at 14:23
  • Fibers are of course closed, as the pre image of a closed subset. That's not what matters, though, only that fibers are discrete. The key part here is that $Y$ is a compact manifold: that's enough to show $X$ is also a compact manifold. – Steve D Mar 23 '24 at 17:20
  • @SteveD It's not clear that the fibers are preimages of closed subsets, because it's not clear a priori that points are closed in $X$ ($p$ being a local homeomorphism only implies they are locally closed). It will ultimately be true, but I don't see a way of proving that without first proving that they are finite. It's not that I don't agree that $Y$ being a compact manifold implies $X$ is a compact manifold, but the current argument is incomplete. – Thorgott Mar 23 '24 at 23:34
  • @Thorgott I don't think it's incomplete. $p$ a local homeo. implies $X$ is locally euclidean and fibers are discrete; $p$ surjective implies $X$ is compact; $Y$ compact implies fibers are finite, and that easily implies $X$ is Hausdorff, hence a manifold. – Steve D Mar 24 '24 at 03:41
  • @SteveD I think all of these implications are true, but the last two steps are subtler than one might think, precisely because the "obvious" approach (implying the fibers are closed) does not work. Similar questions have been asked before on this site and this point has always caused some errors and confusion. – Thorgott Mar 24 '24 at 12:20
  • @Thorgott but at that point we know the fibers are finite, so they're closed. The only point that needs work is showing finite fibers implies Hausdorff. – Steve D Mar 24 '24 at 12:47
  • @SteveD How do you know the fibers are finite? I have an argument like this in mind. If you have a more convenient one, please share. – Thorgott Mar 24 '24 at 13:15
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    @Thorgott if a fiber is infinite, it has a limit point, and $p$ can't be a local homeomorphism there. See for example this answer. – Steve D Mar 24 '24 at 18:26