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.



Is it true that the orbit space of a free finite group action on a CW-complex is also a CW-complex?
**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](http://image.diku.dk/aasa/oldpage/aasa.pdf). 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. 


