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


*

*If we discard any open set that's not evenly covered what remains is still a basis for $Y$ and is still countable.

*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 $Y$.

*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 $X$.

A: You don't need any algebraic topology for that!
Of course it is not true in general, since you can always cover $Y$ by $Y\times F$, where $F$ is an uncountable discrete space. But you have strong assumptions that exclude such pathology.
I'm also posting this answer to provide some details (in the last paragraph) that I find important and that were skipped in the brief explanation provided by Jim.
For each $p\in Y$ take an evenly covered neighbourhood. Since we work with a manifold, fix a smaller neighbourhood $B(p)$ diffeo with a unit ball in $\mathbb R^n$ and a family of its subneighbourhoods diffeo with balls forming a local basis at $p$. Now, take a basis $\mathcal B'$ of $X$ consisting of all those balls. By this question and the assumption that there is some countable basis $\mathcal V$ for $X$ you can choose a countable subbasis of $\mathcal B'$ - let's call it $\mathcal B$.
Now - for each $B\in \mathcal B$ we have $f^{-1}(B)\simeq B\times F_{B}$ for some discrete space $F_{B}$. Note that $F_{B}$ is at most countable because components of $B\times F_{B}$ are exactly of the form $B\times \{g\}$, where $g\in F_{B}$.
Our basis of $Y$ will be 
$$\tau=\bigcup_{B\in \mathcal B} \left\{B\times \{g\}\ | \ g\in F_{B} \right\}.$$
It is clearly countable as a countable union of at most countable sets.
Why is it a basis? We know that $f$ is a local homeomorphism, so it preserves local bases. Did we utilise it well?
That's where our balls (coming from the assumption that we deal with a manifold, not a random second countable space!) are used. Having $p\in B_0\in \mathcal B$ and a sheet $B_0 \times \{g\}$ of its inverse image, we know that the local basis $\mathcal B_p^{B_0}$ at $p$ (restricted to subsets of $B_0$) is "transported" to $B_0 \times \{g\}$ (by that I mean that one of the sheets [that form $\tau$] over any set from $\mathcal B_p^{B_0}$ must lie inside $B_0 \times \{g\}$, because the representation as a product with a discrete space is the same for the subset). It doesn't have to be true in the case of not connected neighbourhoods!
