Compactness and Covering Spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:26:40Z http://mathoverflow.net/feeds/question/32111 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32111/compactness-and-covering-spaces Compactness and Covering Spaces Eric Haengel 2010-07-16T04:27:51Z 2010-07-16T11:38:35Z <p>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.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/32111/compactness-and-covering-spaces/32112#32112 Answer by Dick Palais for Compactness and Covering Spaces Dick Palais 2010-07-16T04:59:38Z 2010-07-16T04:59:38Z <p>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.</p> http://mathoverflow.net/questions/32111/compactness-and-covering-spaces/32113#32113 Answer by Greg Kuperberg for Compactness and Covering Spaces Greg Kuperberg 2010-07-16T05:13:14Z 2010-07-16T05:13:14Z <p>A direct argument without the use of nets:</p> <p>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 $\mathcal{D}'$.)</p> http://mathoverflow.net/questions/32111/compactness-and-covering-spaces/32154#32154 Answer by Georges Elencwajg for Compactness and Covering Spaces Georges Elencwajg 2010-07-16T11:26:38Z 2010-07-16T11:38:35Z <p>Dear Eric, here is a Bourbaki-style proof.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p><strong>NB</strong> I have used Bourbaki's definition "universally closed" for proper. As I said, this <em>implies</em> 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. </p>