Is it true that the orbit space of a free finite group action on a CW-complex is also a CW-complex? Suppose a finite group G acts freely and continuously on an n-dimensional CW-complex X. Then can we conclude that the orbit space of this action is still an n-dimensional CW-complex? (or homotopy equivalent to an n-dimensional CW-complex?)  In particular, we do not assume G acts cellularly on X.
 A: The 3-sphere 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 3-sphere. 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 3-ball to get a CW structure on $S^3$ with the Alexander horned sphere being the boundary of the closure of a 3-cell, which is a contradiction to the other question. 
A: 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 CW-complex. This follows from Prop. 1.15 and Ex. 1.17(2) of tom Dieck: Transformation Groups
A: Lemma If $X$ is a countable locally finite CW-complex and $G$ acts freely and properly discontinuously on $X$, then $X/G$ is homotopy equivalent to a CW-complex.
Proof Any metrizable ANR is homotopy equivalent to a CW-complex
(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 CW-complex, even the case when $X$ is a PL manifold is unclear. The difficulty is that it seems unknown which topological manifolds are homeomorphic to CW-complexes (Kirby-Siebenmann 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 CW-complexes but whose finite covers
are PL. 
A: 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 one-dimensional CW complex with continuous action of a finite group $G$, then $X/G$ is again a one-dimensional CW complex (up to homeomorphism, not just homotopy equivalence).
