Suppose a finite group G acts freely and continuously on an ndimensional CWcomplex X. Then can we conclude that the orbit space of this action is still an ndimensional CWcomplex? (or homotopy equivalent to an ndimensional CWcomplex?) In particular, we do not assume G acts cellularly on X.

$\begingroup$ Perhaps you can perform repeated subdivisions on the cell structure of $X$ to arrive at the cellular case? I'm not claiming this is always possible, but it may be. $\endgroup$ – Mark Grant Oct 15 '12 at 14:10

3$\begingroup$ Here is a suggestion for proving that $X/G$ is homotopy equivalent to a CWcomplex in a special case when the complex $X$ is countable and locally finite. Any metrizable ANR is homotopy equivalent to a CWcomplex. If the complex $X$ is countable and locally finite, then it is a metrizable separable ANR, and I suspect metrizability and separability are inherited by $X/G$. Now a metrizable separable space that is locally an ANR is globally an ANR, so $X/G$ would then be a metrizable ANR. $\endgroup$ – Igor Belegradek Oct 15 '12 at 14:28

1$\begingroup$ In seeing whether $X/G$ is homeomorphic to a CWcomplex, even the case when $X$ is a smooth manifold is unclear. Indeed, if the $G$action is nonsmoothable, then $X/G$ would only be a topological manifold, and in general it is unclear to me whether $X/G$ is homeomorphic to a CWcomplex. (I think the existence of a CW structure on a topological manifold is unknown in dimension 4 and also for noncompact manifolds in higher dimensions, at least the proof in KirbySiebenmann's book is for compact case only). $\endgroup$ – Igor Belegradek Oct 15 '12 at 14:35

$\begingroup$ I would see if their is some sort of Borel construction. At the very least, you can take the singular chains on $X$, make the group action free with a simplicial Borel construction, then take the geometric realization of that. After this process, you will be in possession of a CW complex with the right weak homotopy type. I suspect that if you began with a CW complex, you will have a homotopy equivalence. $\endgroup$ – Spice the Bird Oct 15 '12 at 15:27

1$\begingroup$ @Igor: I think you should write your comments as an answer. Note that if $X$ is metrizable, so is $X/G$ (by averaging the distance function on $X$ under the group action: Sum of distance functions is again a distance function). Local finiteness passes to the quotient, separability too. $\endgroup$ – Misha Oct 15 '12 at 18:07
Lemma If $X$ is a countable locally finite CWcomplex and $G$ acts freely and properly discontinuously on $X$, then $X/G$ is homotopy equivalent to a CWcomplex.
Proof Any metrizable ANR is homotopy equivalent to a CWcomplex (I am not sure who proved it first but see Theorem 3.6.1 here. Since $X$ is countable and locally finite, it is a metrizable separable ANR. As Misha remarks in comments averaging the metric over the group action implies that $X/G$ is metrizable. Also a countable dense subset of $X$ projects to a countable dense subset of $X/G$. Finally, if a metrizable separable space is locally ANR, it is an ANR (see Borsuk's "Theorey of Retracts", Corollary 10.4, Chapter IV). It follows that $X/G$ is a metrizable ANR as desired.
Remark In seeing whether $X/G$ is homeomorphic to a CWcomplex, even the case when $X$ is a PL manifold is unclear. The difficulty is that it seems unknown which topological manifolds are homeomorphic to CWcomplexes (KirbySiebenmann prove this for compact manifolds of dimension $\ge 6$ (or maybe $\ge 5$?, but certainly not $4$). So there might exist manifolds not homeomorphic to CWcomplexes but whose finite covers are PL.

$\begingroup$ I wonder when $G$ is finite and the CWcomplex $X$ is of dimension $n$, can we choose the CWcomplex homotopic to the orbit space $X/G$ to be $n$dimensional too? $\endgroup$ – Li Yu Oct 16 '12 at 2:48

$\begingroup$ @Igor: I do not think you need compactness for this. The point is that you can exhaust an open topological manifold $X$ by an increasing sequence of codimension $0$ compact submanifolds with boundary $X_i$. Let $B_i=\partial X_i$. Then, inductively, the handle structure extends from the collar $B_i\times I\subset X_{i+1}$ to the rest of $X_{i+1}$ (all the existence results are relative). Incidentally, the handle decomposition of KirbySiebenmann works in dimensions $\ge 6$; it is extended to dimension $5$ by Frank Quinn in "Ends of MapsIII". $\endgroup$ – Misha Oct 16 '12 at 5:00
The 3sphere gives an example of an action with fixed points. If one takes the solid Alexander horned sphere, then Bing proved that its double is homeomorphic to the 3sphere. So the quotient of the involution acting on $S^3$ is the solid Alexander horned sphere. However, the solid horned sphere is not homeomorphic to a CW complex. This follows from the answer to this question on the Alexander horned sphere. If the solid Alexander horned sphere were a CW complex, then one could attach the exterior 3ball to get a CW structure on $S^3$ with the Alexander horned sphere being the boundary of the closure of a 3cell, which is a contradiction to the other question.

$\begingroup$ This action is not free, as requested, the subspace of fixed points is Alexander's horned sphere (the common boundary). $\endgroup$ – Fernando Muro Oct 15 '12 at 21:10

$\begingroup$ Ian, does not your involution have fixed points? $\endgroup$ – Igor Belegradek Oct 15 '12 at 21:11

$\begingroup$ I also missed the assumption that the action was supposed to be free. However, the question is still a good one without that assumption and Ian's answer is instructive so I hope he leaves it up. $\endgroup$ – Neil Strickland Oct 15 '12 at 21:24

$\begingroup$ I missed the freeness assumption, but as Neil suggests I'll leave it up. $\endgroup$ – Ian Agol Oct 15 '12 at 21:46
If $G$ (finite or more generally discrete) acts cellularly on $X$, i.e.
 if $\sigma$ is an open cell of $X$ then $g\sigma$ is again an open cell in $X$ for all $g \in G$
 if $g \in G$ fixes an open cell $\sigma$ (i.e. $g\sigma=\sigma$), then it fixes $\sigma$ pointwise (i.e. $gx=x$ for all $x \in \sigma$)
then $X/G$ is a CWcomplex. This follows from Prop. 1.15 and Ex. 1.17(2) of tom Dieck: Transformation Groups

2$\begingroup$ What I want to know is exactly the case when G does not acts cellularly! $\endgroup$ – Li Yu Oct 15 '12 at 9:15

2$\begingroup$ Yes, but when he answered this, you had not made that clear $\endgroup$ – David White Oct 15 '12 at 13:38

1$\begingroup$ I am very sorry. I thought the cellularly action case is easy. So I did not mention it at the beginning. But after your answer, I realize I should emphasize where the difficulty of the question lies. $\endgroup$ – Li Yu Oct 15 '12 at 13:56

1$\begingroup$ @David White: your comment seems unfair. If you look in the edits, the first version of the question did not assume the action was cellular; it said "Suppose a finite group G acts freely and continuously on a CWcomplex X. Then can we conclude that the orbit space of this action is still a CWcomplex? (or homotopy equivalent to a CWcomplex?)". In my view this is clearly stated, and it was Ralph who misread the question; why blame the OP? $\endgroup$ – Igor Belegradek Oct 15 '12 at 16:18

1$\begingroup$ @Igor: 1) Right, the question in its original version makes no assumption on cellularity. Hence there are two case: a) cellular action b) noncellular action. My answer obviously treats case a). So how do you conclude I misread the question ? 2) If one is only interested in case b) then it's good style to point it out, as mentioned by David.  Anyway, now it's clear and I'm curious about the answer of this interesting problem. $\endgroup$ – Ralph Oct 15 '12 at 17:45
This is not really an answer, but a comment about an interesting special case. Suppose that $G$ acts smoothly on $S^2$. By averaging we can choose a $G$invariant Riemannian metric. This gives $S^2$ a conformal structure, making it a Riemann surface. Any Riemann surface homeomorphic to $S^2$ is conformally equivalent to the standard Riemann sphere. Thus, we can reduce to the case where $G$ acts on $\mathbb{C}\cup\{\infty\}$ by conformal and anticonformal maps, which must have the form $z\mapsto (az+b)/(cz+d)$ or $z\mapsto (a\overline{z}+b)/(c\overline{z}+d)$. I think it even works out here that the quotient $(\mathbb{C}\cup\{\infty\})/G$ is always either a sphere or a disc. Thus, one cannot get any local pathology in this context. This contrasts with other settings where smooth functions can generate topological pathology: for example, any closed subset of $\mathbb{R}^n$, however fractal, can be expressed as the zero set of a smooth function $f\colon\mathbb{R}^n\to\mathbb{R}$.
Along somewhat similar lines, I think one can show that when $X$ is a onedimensional CW complex with continuous action of a finite group $G$, then $X/G$ is again a onedimensional CW complex (up to homeomorphism, not just homotopy equivalence).

$\begingroup$ Neil: Even more, if $G$ acts topologically freely and properly discontinuously on a topological manifold $X$ of dimension $\ne 4$ then the quotient $X/G$ is homeomorphic to a CWcomplex. The same applies if $X$ is merely a simplicial complex of dimension $\le 3$ and $G$ acts topologically. $\endgroup$ – Misha Oct 15 '12 at 22:04

$\begingroup$ Misha: is your $X$ compact? If not, would you give a reference for the first claim, say in higher dimensions? $\endgroup$ – Igor Belegradek Oct 15 '12 at 22:42