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Let p : Y -> X be an n-sheeted covering map, where X and Y are topological spaces. If X is compact, prove that Y is compact.

I realize that this seems like a very simple problem, but I want to stress the lack of assumptions on X and Y. For example, this is very easy to prove if we can assume that X and Y are metrizable, for sequential compactness is then equivalent to compactness and it is easy to lift sequential compactness from X to Y.

I asked three people in person this question and all of them immediately made the assumption that X and Y are metrizable, so I feel like I should put in this warning here that they are not.

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Are you assuming X and Y are also Hausdorff? If so, then I can't see what goes wrong with the natural approach: take an open cover of Y, push it down to an open cover of $X$ (because $p$ is surjective it will be open) take a finite subcover downstairs and lift it up with multiplicity $n$ to a finite subcover upstairs. What have I missed? – Yemon Choi Jul 16 '10 at 4:31
@Yemon: What does "lift it up with multiplicity n" mean? How do you choose sets from the original cover to cover your new cover? (And I apologize for that sentence.) – Tyler Lawson Jul 16 '10 at 4:50
If I recall correctly, you don't need Hausdorff. – Dylan Moreland Jul 16 '10 at 4:57
@Tyler: good point, I was being over hasty. Seeing as my topology is rusty: by an n-sheeted covering, do we mean that (a) p is a quotient map of topological spaces; (b) each point $x\in X$ has an open neighbourhood $U$ suchthat $p^{-1}U)$ is the disjoint union of $n$ open sets, each of which is mapped homeomorphically onto $U$? – Yemon Choi Jul 16 '10 at 5:12
I think a fix of the proof above goes as follows: Let $U_\gamma$ be a cover of $Y$. Choose a refinement $V_\alpha$ of this cover so that each $V_\alpha$ is small enough to be homeomorphic to its image under the covering map. It suffices to show that $V_\alpha$ has a finite subcover. Since the sets $p(V_\alpha)$ form an open cover of $X$, they have a finite subcover, $p(V_\alpha)_\beta$, each of which has $n$ lifts. The lifts of this subcover provide a finite subcover of $Y$. – BMann Jul 16 '10 at 15:48

A direct argument without the use of nets:

Let $\mathcal{C}$ be an open cover of $Y$. For each $p \in X$, choose an open set $p \in U \subseteq X$ such that $Y$ is trivial over $U$, and such that each lift of $U$ is contained in some element of $\mathcal{C}$. This is an open cover $\mathcal{D}$ of $X$, which has a finite subcover $\mathcal{D}'$ since $X$ is compact. The lift of $\mathcal{D}'$ to $Y$ is also a finite cover, as well as a cover that refines $\mathcal{C}$. Thus $\mathcal{C}$ must have a finite subcover. (The fact that $Y$ is a finite cover is used twice, first to make each $U$, second to lift $\mathcal{D}'$.)

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Can you enlighten as to what is the closest statement that may be true for infinite sheeted covers? I understand that for a cover downstairs to lift to a cover upstairs one needs the number of sheets to be finite but why does the definition of $U$ depend on finiteness of the number of sheets? Take a locally trivializing neighbourhood of $p$ and look at some intersection of its pre-image with some open set in $C$ and then project it down. Doesn't that give the kind of $U$ one needs? May be I am missing something obvious! – Anirbit Jul 16 '10 at 9:57
@Anirbit: infinite intersections of open sets are not open. Because in the end you want a finite subcover from $\mathcal{C}$, so on each "sheet" you need to pick an open set in $\mathcal{C}$ containing $U$. – Willie Wong Jul 16 '10 at 15:20
Ah..yes. I was mistaken to think that taking intersection on only 1 sheet should be enough. To define the kind of $U$ that Greg wanted one needs to take intersection on all the sheets and that will mess up. – Anirbit Jul 16 '10 at 17:35

Well, the obvious argument that any sequence has a convergent subsequence that your three friends used for the metrizeable case generalizes easily to show that any net has a convergent subnet in the general case.

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Nice! maybe this could be used as an example for… – Yemon Choi Jul 16 '10 at 5:15

Dear Eric, here is a Bourbaki-style proof.

Recall that a continuous map $f: Y\to X$ is called proper by Bourbaki if, for all spaces $Z$, the map $f\times 1_Z: Y \times Z\to X \times Z$ is closed. For example the trivial finite covering $X\times \{ 1,\ldots n \}\to X$ is proper.

Now, your $X$ is covered by opens $X_\iota \subset X$ such that the restricted/corestricted maps $f_{X_\iota }:f^{-1} (X_\iota) \to X_\iota $ are trivial finite coverings, hence are proper by the example above. We deduce that the original covering $f:Y\to X$ is proper: this follows easily from the definition of "proper" and (if a reference is needed) is proved in Bourbaki's General Topology, Chapter 1, §10, Proposition 3.

But a proper map has the property that the inverse image of a quasi-compact subset of the target (in our case all of $X$) is quasi-compact (ibid., Proposition 6). Hence $Y$ is quasi-compact if $X$ is.

NB I have used Bourbaki's definition "universally closed" for proper. As I said, this implies that inverse images of quasi-compact subsets are quasi-compact.This last property is often taken as the definition of proper. For locally compact spaces, both definitions coincide.

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