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Let $X$ be a connected topological space, and $C_{finite}$ the category of its finite coverings. Then I claim that the category $C$ of coverings of $X$ can be obtained by $C_{finite}$ taking projective limits and disjoint unions.

Since projective limits and disjoint unions commute the statement is reduced to say that $A:= \varprojlim A_i$, where each $A_i$ is a finite covering, is itself a covering space of $X$. For the proof this is my problem: Fixed a point $x\in X$, and a point $\{y_i \in A_i\} \in A$ of the fiber over $x$, then I must find an open neighborhood $x\in U \subset X$ such that it is homeomorphic to a convenient neighborhood $V$ of $\{y_i\}\in A$.

The passage is problematic because, if the limits is taken over an infinite index set $I$, then I'm not allowed anymore to restrict this neighborhood $U$ in a convenient way for each $A_i$ (Infinite intersection of opens can be not open).

There is a way to bypass this problem? Or my claim is simply wrong?

Probably in the solution it will be helpful (or necessary) to restrict to a base space $X$ which admits a universal covering, i.e. it is locally path connected and semi-locally simply connected.

Edit: I forgot to mention the following fact, which justify my claim. If $X$ admits a universal covering, then it can be obtained as projective limit of finite (normal) coverings of $X$.

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1 Answer 1

I don't see how your claim works, even in the simple case where $X$ is a circle. The universal covering space is the real line, winding infinitely often around the circle. The fiber over any point of the circle is a countably infinite discrete space. But a projective limit of finite coverings would, as far as I can see, have fibers that are projective limits of finite sets, so the fibers would be compact.

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First of all, thank you for your interest! Unfortunately It's some time I'm not working on this stuff, so I could be a bit slow in understanding things. May I ask you to elaborate some more? In specific: 1) Why the fibers of a projective limit of finite coverings are projective limits of finite sets? 2) Why projective limits of finite sets are compact? (I think you mean compact in the limit topology, correct?) Then I would remark, if I'm correct, that your answer doesn't answer the question. But it simply invalidates the justification I gave for it. Thank you again! :) –  Giovanni De Gaetano Nov 8 '12 at 10:27
    
@GiovanniDeGaetano: For question 1 in your comment, just use that projective limits commute with pullbacks, in particular with fibers. For question 2, use that projective limits (of Hausdorff spaces) are obtained as closed subsets of product spaces. So by Tychonoff's theorem, projective limits of compact spaces are compact. Finally, in your original question, the claim in the first paragraph is false; the universal cover of a circle is not obtainable as a projective limit of finite coverings. –  Andreas Blass Dec 27 '12 at 5:05

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