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I am in need of solution or tip for this question. I thank you.

Let $ p: \widetilde X \to X $ be a covering space with $ p^{-1}(x) $ finite and nonempty for all $ x \in X$. Show that $ \widetilde X$ is compact Hausdorff iff $ X$ is compact Hausdorff.

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I changed your title to be more descriptive. Using a descriptive title will attract more relevant users to your question, as opposed to a generic "I need a solution". – Ted Dec 16 '12 at 7:49
up vote 2 down vote accepted

As you had got hint from the previous answer, hope you have completed the solution. Here I want to give the idea of the proof of this statement from another point of view which does not require an open covering of $ \widetilde{X} $.(I'll show the difficult part only)

All we have to note is that a compact Hausdorff space is regular. So the base space $\ X $ is regular. Now first cover $\ X $ with evenly covered neighbourhoods {$\ U_{\alpha} $} . Using regularity condition we can get a refinement of this open covering - say {$\ V_{\beta} $} - such that closure of $\ V_{\beta} $ is inside some $\ U_{\alpha} $.

So, $\ p^{-1} (\overline{V_{\beta}})$ is disjoint union of finitely many closed sets. (Using the fact fibres are finite and $ U_{\alpha} $ is an evenly covered neighbourhood). Now $\ \overline{V_{\beta}} $ is compact so is its preimage because the covering map is local homeomorphism. Now get a finite subcover of $\ X$ from the collection {$\ V_{\beta} $} and hence preimage of the corresponding $\ \overline{V_{\beta}} $ 's will cover $ \widetilde{X} $ . So we get $ \widetilde{X} $ is union of finitely many compact sets which implies $ \widetilde{X} $ is compact.

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(Giving just hints for now; but if they’re not clear after you’ve thought about them a bit, I’m happy to expand in more detail if you’d like.)

For Hausdorffness: given two distinct points $x, y \in \tilde{X}$, consider their images $p(x)$, $p(y)$. Are these also distinct? If so, then you can separate them with open sets in $X$; use these to construct open sets separating $x$ and $y$ in $\tilde{X}$. Otherwise, if $p(x) = p(y)$, then you can use the covering space condition to find open neighbourhoods separating $x$ and $y$.

For compactness: suppose given an open covering $\mathcal{U}$ of $\tilde{X}$. Call an open set $W \subseteq X$ “$\mathcal{U}$-good” if $p^{-1}(W)$ is a disjoint union of homeomorphic copies of $W$ (as in the definition of covering space), and moreover, each of these disjoint copies is a subset of some $U \in \mathcal{U}$. Now, show that the $\mathcal{U}$-good sets form an open cover of $X$. Thus some finite set of $\mathcal{U}$-good sets covers $X$; from these, build a finite subcover of $U$.

I found both of these using the principle of “think how you could make use of the assumptions”. For the second, for instance, we start with an open cover of $\tilde{X}$, and want a subcover. We know we’ll need somehow to use compactness of $X$, so we can expect the proof may well go something like: from $U$, construct some open cover of $X$; from a finite subcover of that, build a finite subcover of $\tilde{X}$. With the first cover of $X$ I tried (the images of the sets in $\mathcal{U}$), I couldn’t get the second step to work, so I thought: how can I refine the construction so that from the finite cover of $X$, I can get back to a finite subcover of $\mathcal{U}$?

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