Hatcher's exercise 28, chapter 4 I'm trying to solve the exercise number 28 from the 4th chapter of Hatcher's Algebraic Topology.

Show that the group $ \mathbb{Z}_p \times \mathbb{Z}_p $ with $p$ prime cannot act freely on any sphere $S^n$, by filling in details of the following argument. Such an action would define a covering space $S^n \to M $ with $M$ a closed manifold. When n > 1 , build a $K( \mathbb{Z}_p \times \mathbb{Z}_p , 1)$ from $M$ by attaching a single $(n + 1)$-cell and then cells of higher dimension. Deduce that $H^{n + 1} (K(\mathbb{Z}_p \times \mathbb{Z}_p , 1); \mathbb{Z}_p )$ is $\mathbb{Z}_p$ or $0$ , a contradiction. (The case $n = 1$ is more
elementary.)

I found online this solution (the Proposition $0.0.2$ of the paper I linked). The argument is quite clear (and it is more or less what I was already thinking while trying to solve the exercise), but I am stuck with the followiung doubt:
Once they build the $K( \mathbb{Z}_p \times \mathbb{Z}_p , 1)$ as suggested in the exercise, to conclude they computed the cellular cohomlogy with $\mathbb{Z}_p$ coefficients and used the fact that the space had a single $(n+1)$-cell to conclude that the $(n+1)$-cohomology group must be either $\mathbb{Z}_p$ or $0$. My problem is that I don't know that the space $M$ is a $CW$-complex: I know that I attched cells, but I didn't have a cell complex to begin with. So how can I compute cell cohomology if there isn't a cell ctructure? Or does the orbit space inherit a cell structure from the covering? If this is the case, could you please explain me how or give me a refernce?
Alternatively, could any of you give me a different solution?
 A: It is not relevant whether or not $M$ is a CW complex. (You will not be able to prove it: there are actions of $\Bbb Z/2$ on $S^4$ whose quotients are non-smoothable manifolds homotopy equivalent but not homeomorphic to $\Bbb{RP}^4$, and I believe it is unknown whether these manifolds have cell structures.)
The point is that you have a space $M$ with homotopy groups $\Bbb Z/p \times \Bbb Z/p$ in degree 1, zero in degrees $1 < k < n$, and $\Bbb Z$ in degree $n$.
You may now build a space $X = K(\Bbb Z/p \times \Bbb Z/p, 1)$ by attaching cells to $M$ in order of their dimension. No, $X$ need not be a CW complex. But this structure is that of a `relative CW complex'.
You can still calculate the cohomology of $X$ perfectly well using knowledge of the homology of $M$ and iterated use of the Mayer-Vietoris sequence (which is all cellular homology actually amounts to). If you like, you can assemble this into a chain complex where $C_k(X)$ is free abelian on the k-cells of $X$ for $k>n$ and $C_n(X) = H_n(M)$.
