Is a covering space of a manifold second countable?

I have been told that the covering space of a manifold is again a manifold. Let $f :X\rightarrow Y$ be a covering map and $Y$ is a $n$-manifold. It is easy to show that $X$ is a locally euclidean and Hausdorff.

Now I am trying to show it is second countable, given the additional condition that for any subset $U \subset Y$, $f^{-1}(U)$ has finite or countably many components.

Now I guess that basis on $X$ consists of the precompact set of $x$ which seperates $x$ from other components of $f^{-1}(f(x))$ for every $x\in X $. I can verify this is a basis since $X$ is Hausdorff. The only problem is to show that this basis is countable. Question: Is it countable or not?

Any way to do it without alg top , i.e by general topology and manifold ?

Solution 1:

It is indeed second countable. To see this take a countable basis for $Y$.

1. If we discard any open set that's not evenly covered what remains is still a basis for $Y$ and is still countable.
2. The fundamental group of a manifold is countable and acts transitively on the fiber over any point $y \in Y$, so there are at most countably many sheets over any evenly covered subsets of $Y$.
3. Now for each basic open set in $Y$ we take the at most countably many open sets obtained by intersecting $f^{-1}(Y)$ with the sheets over $Y$. This gives us a (countable) $\times$ (at most countable) $=$ at most countable basis for $X$.