# Is a covering space of a manifold always a manifold

Assume $M$ is a manifold and $q : E \to M$ is a covering map. I have been told a few times that a covering space of a manifold is again a manifold. Indeed, it is easy to verify that $E$ is both Hausdorff and locally euclidean. I am worried about whether $E$ needs to be second countable.

If $E$ is not connected, then it does not need to be second countable. Take the standard covering map $\coprod_{i \in \mathbb{R}} M \to M$.

Question: If $E$ is connected, then why is it second countable?

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Because of Poincaré-Volterra's theorem.
The best reference is, as usual, Forster's great Lectures on Riemann surfaces , Lemma 23.2, page 186.
A less self-contained but more general version (surprise, surprise...) is to be found in Bourbaki's General Topology : it is the very last result of Chapter 1.

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Here is a direct link to page 186 (we edited simultaneously and I don't want to interfere again). – t.b. May 27 '12 at 11:49
Dear @t.b., you are not interfering but being helpful to the community. Believe it or not, I was wondering why you hadn't answered with the same reference and was regretting that we hadn't had any contact for a long time. Glad this is no longer true! – Georges Elencwajg May 27 '12 at 12:03
Dear Georges, Thank you for your kind words! Yes, it is a pity we didn't have much interaction over the past few months. To clear up a possible slight misunderstanding, I was talking about this -- our edits happened within a few seconds. I couldn't have pointed to a good reference for this result (I didn't even know it had a name) and I just followed your reference and edited the link so as to point to the desired page. Best wishes, Theo – t.b. May 27 '12 at 12:22
Thanks Georges! I will check it out this afternoon! – DBr May 27 '12 at 21:06

In the case that the manifold has a PL structure, I believe there's an elementary proof as follows.

Since $E$ is connected, the degree of the covering is the size of the fundamental group of $M$.

Also, the fundamental group of $M$ must be countable, as follows. $M$ is homeomorphic to a locally finite simplicial complex. The fundamental group comes from paths in the 1-skeleton of that complex, considered as a graph. Since the simplicial complex is locally finite, there are only countably many such paths, and therefore the fundamental group is countable.

So the degree of the covering is countable, and we can lift a countable basis of $M$ to get a countable basis of $E$. Therefore $E$ is second countable.

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