Fixed-point free action of $\mathbb{Z}/p\mathbb{Z}$ on a finite CW complex Problem 2a on this old qualifying exam asks the following:

Suppose $X$ is a finite CW complex and $X$ admits a fixed-point free
  action of $G:= \mathbb{Z}/p\mathbb{Z}$ for some prime $p$. Prove that
  $p$ divides $\chi(X)$.

We can show this using the Lefschetz fixed point theorem. If $\sigma \in G$ is a generator, then $\sigma_{\star}$ on $H_k (X, \mathbb{Q})$ satisfies $\sigma_{\star}^{p} = \text{Id}$, so all its eigenvalues are $p$th roots of unity. It follows that the characteristic polynomial of $\sigma_{\star}$ is of the form $$p(T) = (T-1)^{C_k} (T^{p-1} + \cdots + T + 1)^{D_k}$$ and hence $\text{Tr}(\sigma_{\star} : H_k) = C_k - D_k$. However, we also have $$\text{Tr}(\text{Id} : H_k) = \dim(H_k) = C_k + (p-1)D_k$$ so it follows that $\chi(X) = \tau(\text{Id}) \equiv \tau(\sigma) \pmod{p}$, where $\tau(f)$ denotes the Lefschetz number of $f$. It follows that if $\sigma$ is fixed-point free, then $p$ divides $\chi(X)$.
This proof is just fine, but what I really want to do is note that $X \to X/G$ is a degree $p$ covering map and call it a day. The result cited here lets us conclude this way if we know that $X/G$ is of "bounded finite type," i.e. all homology groups are finitely generated and vanish in sufficiently high degree.
Question: Under the hypotheses of the problem, can we conclude $X/G$ is of bounded finite type? Why or why not?
 A: If we have a compact metric space with a free $G$-action we can assume there is a nonzero lower bound to the distance from $x$ to $gx$ for all $x$ in $X$ and all $g \neq 1$ in $G$. That means that $X$ and $X/G$ are locally homeomorphic. For a compact space $Y$, I think of the finite-type property you want as "essentially" coming from good local behavior of $Y$. More precisely, the basic hypothesis which implies it is for $Y$ to have a finite "contractible covering", i.e. a covering by open sets of which every nonempty finite intersection is contractible. (For that implies that $Y$ is homotopy-equivalent to the finite polyhedron which is the nerve of the covering.) That is always true for a smooth manifold (for one can choose a Riemannian metric and take a covering by small geodesic balls). (But I suspect it need not be the case for a topological manifold, which shows that one needs the good behavior to be in some sense "uniform".) A stronger property is to have arbitrarily fine contractible coverings, which is also true for a smooth (or PL) manifolds. If your space $X$ has that stronger property then I think it is inherited by $X/G$, because one can choose the sets in $X$ too small ever to meet their $G$-translates.
But does a finite CW complex have such coverings? I thought at first yes, but then felt a bit cautious—the attaching maps can be very badly behaved.
As somewhat of an aside, some folks from the distant past who have greatly impacted algebraic topology, hated to be called algebraic topologists, because they wanted to think that they studied "natural" rather than "man-made" objects. A CW complex, in contrast to an algebraic variety or a smooth manifold, would be their archetype of something invented by homotopy theorists which does not occur in nature. They really did not like CW complexes (or topological manifolds), and some have to some extent inherited their prejudice—at any rate, I think of spaces with fine contractible coverings as a nicer class than CW complexes. (But a modern person would point out that the two categories are alternative model categories for the same homotopy category.)
Update: No sooner had I posted my answer than it struck me that I had been carelessly wrong in saying that the property of having fine contractible coverings is inherited by the orbit space. I do not see why that should be so.
In fact it made me think more carefully about your clever-if-indirect argument, which had convinced me at first. The essential point is your assertion that when one has a map $f: X \to X$, where $X$ is a finite CW complex, then if $f$ has no fixed points its Lefschetz number must be zero. That is certainly easy to prove if, say, $f$ is a finite polyhedron, for one can make a sufficiently small deformation of $f$ to a map which is simplicial (for a subdivision of the triangulation) and still have no fixed points. But it is not clear to me how one would prove it for a CW complex, where, as far as I can see, deforming $f$ to a cellular map could introduce fixed points. As with my own incorrect argument, the trouble is that the group action may look dreadful from the point of view of the cell decomposition.
On the other hand, if one looks at the matter in terms of the Leray spectral sequence, either for the map $X \to X/G$ or, better, for an arbitrary open cover of $X/G$ which is sufficiently fine for the covering map to be trivial over each of its open sets, then the fact that the Euler number of $X$ is divisible by $p$ seems completely obvious. But then one is using Čech cohomology in an essential way, and not using any CW hypothesis—but one would need to know that $X/G$ can be covered by a finite number of small open sets which, even if they are not contractible, have finite-dimensional Čech cohomology.
Very puzzling, I have no clue on how to move forward.
