Can one prove the covering space of a manifold is second countable without using the fact that the fundamental group of a manifold is countable?

Question

I'm trying to prove that the covering space X of a manifold Y is a manifold of the same dimension. I was stuck on proving X is second countable, more specifically, on proving the fiber of any element in Y is at most countable.

I find a proof at Is a covering space of a manifold second countable?, but that proof is based on the countability of the fundamental group of a manifold. Could there be a more elementary approach without using this fact?

You sholud assume that $X$ is connected (which implies that also $Y$ must be connected). Otherwise there are trivial counterexamples. — Paul Frost, Nov 24 '23 at 13:55
Yes, I see that as part of the definition of a covering space. — minukesis, Nov 25 '23 at 08:10

score 1 · Answer 1 · answered Nov 24 '23 at 20:27

If you take $X$ to be the universal covering space of $Y$, its fiber over $p\in Y$ is in bijection with $\pi_1(Y,p)$. Thus, proving that this fiber is at most countable in general is no easier than proving that the fundamental group is at most countable.

Can one prove the covering space of a manifold is second countable without using the fact that the fundamental group of a manifold is countable?

1 Answers1