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Let ${\mathcal C}$ be the class of topological spaces which carry a CW-structure (note that I do not want to fix some particular CW-structure).

Is it true that for a covering map $E\stackrel{f}{\to} B$ with $E\in{\mathcal C}$ we have $B\in{\mathcal C}$, too?

It is true that the total space of a covering lies in ${\mathcal C}$ if the base space does, but the reverse implication is not clear to me.

Edit

As Algori pointed out, the quotient space is not even Hausdorff in general. What about finite regular coverings, i.e. those which come from a free action of a finite group on the total space? Is it true then that the quotient space carries a CW-structure, too?

I'm interested in that because this would imply that given a free group action of a finite group on a "nice" space like a CW-complex, one can always choose a CW-structure with respect to which $G$ just permutes cells. Then the corresponding cellular complex would be a (possibly nice) complex of ${\mathbb Z}G$-modules (for example, if the space was a sphere, then this procedures can be used to construct a periodic ${\mathbb Z}G$-resolution of the trivial module ${\mathbb Z}$, showing that the group has to have periodic invariants like homology and cohomology; in this particular case, however, things behave well as the quotient space ${\mathbb S}^n/G$ is still a compact manifold).

Thank you.

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is there a counterexample where $B$ is hausdorff? –  Martin Brandenburg Jan 13 '10 at 16:03

2 Answers 2

up vote 9 down vote accepted

Ok, here is a simple example when this fails: let $X$ be $\mathbf{R}^2$ minus the origin and consider the automorphism of $X$ given by $(x,y)\mapsto (2x,y/2)$. This automorphism generates a group $G$. The quotient map $X\to X/G$ is a covering, but $X/G$ is not Hausdorff.

This answers the question as it is stated, but maybe you wanted $B$ to be Hausdorff (in which case I don't know what to do).

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This example can be found on wikipedia: en.wikipedia.org/wiki/Covering_space. (That's a good thing; I just wanted to point it out.) It's interesting -- the fact that the quotient by a free, properly discontinuous group action can lose the Hausdorff property is something I had never thought of before. Does this phenomenon show up naturally in algebraic or analytic geometry? –  Pete L. Clark Jan 13 '10 at 3:18
    
Pete -- thanks for the reference! Honestly, I can't see any direct analogies with algebraic geometry, only some fairly vague ones, when the quotient turns out to be worse than expected (e.g an algebraic space instead of a scheme etc). One of the problems is that I don't see how to translate "properly discontinuous" into algebraic/categorical language. –  algori Jan 13 '10 at 4:57
    
Thank you, algori. That's interesting. –  Hanno Becker Jan 13 '10 at 7:06
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There are different uses for the phrase "properly discontinuous G-action" but if you take the definition that "for any compact subset K only finitely many images of K intersect itself", then the above counterexample disappears. Under this definition, if G acts freely and properly discontinuously on a manifold X, then the quotient X/G is a (Hausdorff) manifold and the projection is a covering map. –  Igor Belegradek Jan 19 '10 at 16:16

This is not an answer it might help.

  1. I tried searching on "$G$-CW-complexes" or "equivariant-CW-complexes", and this old paper popped up. There it is mentioned that in the case of free actions of finite groups the notions of $G$-CW-complex and $G$-cell complex coincide (see Remark 1.12). In other words, if you have a cell complex $X$ on which a finite group $G$ acting freely by permuting cells, then $X$ is automatically a $G$-CW-complex.

  2. On the other hand, I am skeptical that any free action of a finite group on a cell complex can be made cellular. For example, it is a hard theorem of Kirby-Siebenmann that any compact topological manifold has a CW-structure. If what you want is true, then this hard theorem would be easy for spherical space forms, because they are covered by spheres which are smoothable, hence have CW-structures.

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