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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 ?

Arctic Char
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2 Answers2

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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 $X$.
Jim
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  • Is there any way to do it without using algebraic top? – student user Feb 28 '13 at 07:26
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    I can't say for certain that there isn't, but I have no idea how you would get a handle on the cardinality of the fiber of a covering map without some sort of high powered machinery. – Jim Feb 28 '13 at 07:31
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    @Jim The argument with transitive action if deeply false. The covering space doesn't have to be path-connected. Moreover, it is clearly said in the question that every inverse image is at most countable, so the fiber is of course at most countable. – savick01 Aug 10 '13 at 12:48
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    @savick01: It is quite natural to assume $X$ to be connected... Otherwise there are obvious counterexamples. So caring about the non-connected cases would be deeply pointless! ;-) – Sam Aug 10 '13 at 13:02
  • @Sam I can't think of any counterexamples provided that inverse images have at most countably many components. – savick01 Aug 10 '13 at 13:05
  • @savick01: There are none if you add this hypothesis. What I meant is that the hypothesis is unnecessarily strong. As Jim's argument shows, as long as $X$ is connected, we are fine. So the only thing that might go wrong (without this hypothesis) is that $X$ has uncountably many components. And spaces with this property are quite silly... – Sam Aug 10 '13 at 13:11
  • @Sam Right, they are. But the thing is that the OP is apparently not advanced enough to know the fact that fundamental group is countable for manifolds - that's why their teacher gave them the extra assumption. – savick01 Aug 10 '13 at 13:16
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    @savick01: I take covering spaces to be connected by definition. – Jim Aug 11 '13 at 01:55
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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!

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