Four sheeted cover of $\mathbb{R}P^2$ wedged with circles

Question

Let:

$$ X_n = \underbrace{S^1 \vee S^1 \vee \dots \vee S^1}_n \vee \mathbb{R}P^2 $$

Problem says: By computing a homology group of a suitable 4-sheeted covering space of $X_n$ show that it's not homotopy equivalent to a compact connected surface for $n\geq1$.

The usual cover of $\mathbb{R}P^2$ is only two sheeted. What is the idea here?

Answer 1

I think the covering space you want is two copies of $\mathbb{R}P^2$ (Edit: should be $S^2$) which are connected by $4n$ line segments, 2 at the basepoint corresponding to each circle and 2 at the negative basepoint corresponding to each circle. If $n>0$ you can show this is homotopy equivalent to $S^2 \vee S^2 \vee (S^1 \vee \dots \vee S^1 )$ where there are $4n-1$ copies of the circle. It's cellular homology is easy to compute; in the top dimension it is $\mathbb{Z}^2$. The classification of surfaces then tells us that this can't be the top dimensional homology of any manifold (this does not immediately answer our question).

Now imagine if our base space was homotopy equivalent to a surface. Under a specific homotopy equivalence we have a bijection between subgroups of the fundamental group of each space. I think it is the case (it is for the universal cover) that the covering spaces of these two topological spaces corresponding to subgroups under this bijection are homotopy equivalent.

Collecting all of this: we know that if our space was homotopy equivalent to a surface then there is a 4 sheeted cover of that surface that is homotopy equivalent to a space with $\mathbb{Z}^2$ as its 2 dimensional homology. It is the case that a finite sheeted cover of a compact space is compact, so the cover of the surface would be another surface. However this is impossible, since its homology must be $\mathbb{Z}^2$.

Answer 2

I find covers of $S^1\vee \mathbb RP^2$ A little easier to visualize if you consider attaching a 1d handle instead of wedging the circle.

There are covers of such a space looking like strings of $S^2$’s all connected by intervals, like a pearl necklace, which have many nonhomologous 2-cells. In fact theyre equivalent to a circle wedged with many 2-spheres.