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Timeline for Compactness and Covering Spaces

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Jul 16, 2010 at 15:48 comment added BMann 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 $Y$.
Jul 16, 2010 at 11:26 answer added Georges Elencwajg timeline score: 1
Jul 16, 2010 at 5:13 answer added Greg Kuperberg timeline score: 12
Jul 16, 2010 at 5:12 comment added Yemon Choi @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 $U$?
Jul 16, 2010 at 4:59 answer added Dick Palais timeline score: 4
Jul 16, 2010 at 4:57 comment added Dylan Moreland If I recall correctly, you don't need Hausdorff.
Jul 16, 2010 at 4:50 comment added Tyler Lawson @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.)
Jul 16, 2010 at 4:31 comment added Yemon Choi 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?
Jul 16, 2010 at 4:27 history asked Eric Haengel CC BY-SA 2.5