algebraic-topologycovering-spacesgroup-theory
I was trying to find all connected covers of $\mathbb{RP}^2 \vee \mathbb{RP}^2$.
In that regard I got the universal cover which is a disjoint wedge of countably infinite spheres.
Also I got covers corresponding to each cyclic subgroup of $\mathbb{Z}_2*\mathbb{Z}_2$.
How do I find covers corresponding to other subgroups if there exists any?
I believe that this is $\mathbf{R}P^2$ with an infinite (in one direction) string of spheres attached to it. I can think of a couple ways to arrive at this. Here's one, which requires knowledge of the universal cover:
Let $U$ be the universal cover of $X$, which as you say is an infinite (in two directions) string of spheres. If $Y$ is a cover of $X$ with fundamental group $\mathbf{Z}/2\mathbf{Z}$, then $U$ is also the universal cover of $Y$. The covering $U \to Y$ has degree $2$ and is normal/regular, so there is some subgroup, isomorphic to $\mathbf{Z}/2\mathbf{Z}$, of the deck transformations of $U \to X$ such that $U/G \approx Y$. The obvious thing to try (and I think these are the only elements of order two) is the deck transformation which is the antipodal map on one fixed sphere, and which flips the rest of the chain around that sphere. I think this quotient works.
I hesitate to write about the alternative I have in mind, since it feels less justifiable, but maybe it will help: Since a covering of $X$ restricts to a covering of $\mathbf{R}P^2$, you know that you must stick a bunch of spheres and projective planes together. Sometimes it helps me to draw some dots along a vertical line, representing the points of the cover lying above the point where the $\mathbf{R}P^2$ meet; on the left side are the things mapping to the first copy of $\mathbf{R}P^2$. Then on either side of a particular dot I can place a copy of $\mathbf{R}P^2$ or half of a sphere, which must be made whole at another point. If you play around with this for a while, I think it should yield the answer as well.
$\widetilde{\Bbb{RP}^2 \vee T^2}$ is going to look like a tree with vertices corresponding to either $S^2$ or $\Bbb R^2$ and edges corresponding to one-point union of the two spaces corresponding to the vertices it joins.
The tree is a colored tree, with vertices colored by blue and red, each blue vertex adjacent only to red vertices and each red vertex adjacent only to blue vertices. Neighborhood of a red vertex consist of $\Bbb Z/2$-many vertices and neighborhood of a blue vertex consist of $\Bbb Z^2$-many vertices. This is because the wedge point $x$ in $\Bbb{RP}^2 \vee T^2$ lifts to $\Bbb Z/2$-many points in each $S^2$, and $\Bbb Z^2$-many points in each $\Bbb R^2$. Replacing each red vertex by an $S^2$, each blue vertex by an $\Bbb R^2$ and each edge by one-point union of the two vertex spaces gives me the desired universal cover.
Here is a picture of the part of the graph. While there are infinitely many red vertices adjacent to blue vertices, only finitely many are drawn for obvious reasons and the existence of the rest are dotted. As we see, the graph is a tree with vertex set partitioned into two colors and valence of blue vertices is $|\Bbb Z^2|$ and valence of red vertices is $2$.
Thus, ultimately, the space $\widetilde{\Bbb{RP}^2 \vee T^2}$ is iterative one-point-union of infinitely many $S^2$'s and $\Bbb{R}^2$'s, with each $S^2$ wedged with two $\Bbb R^2$'s, and each $\Bbb R^2$ wedged with $\Bbb Z^2$-many $S^2$'s.
$\text{Explanation}$: To see this, note that $\Bbb R^2$ is the universal cover of $T^2$, hence $\Bbb R^2 \bigvee_{\Bbb Z^2} \Bbb{RP}^2$ ($\Bbb R^2$ with a copy of projective plane attached at each integer lattice) covers $\Bbb{RP}^2 \vee T^2$. Now $S^2$ is the universal cover of $\Bbb{RP}^2$, so you can similarly "unwrap" one of the projective planes from $\Bbb Z^2$-many of them to get the cover $\Bbb R^2 \bigvee_{\Bbb Z^2 - (0, 0)} \Bbb{RP}^2 \vee (S^2 \vee \Bbb R^2 \bigvee_{\Bbb Z^2 - (0, 0)} \Bbb{RP}^2)$. Covering all of the wedged $\Bbb{RP}^2$'s likewise, one will end up with the cover $\Bbb{R}^2 \bigvee_{\Bbb Z^2} (S^2 \vee \Bbb R^2 \bigvee_{\Bbb Z^2} \Bbb{RP}^2)$. "Unwrapping" iteratively in this process will give you a tree-like structure, entirely consisting of $S^2$ and $\Bbb R^2$, hence simply connected and thus a universal cover of your space.
$\text{Remark}$: The reason that you get a much nicer thing for $\Bbb{RP}^2 \vee \Bbb{RP}^2$ is that your tree consists of vertices corresponding only to $S^2$ and the wedge point lifts only to 2 points in each $S^2$. This implies for every $S^2$-vertex, there are only two $S^2$-vertices adjacent to it in the graph, so globally it looks like an infinite string of $S^2$'s, each two of them touching at a point. Note that the graph is still a tree, with each vertex being of valence $2$.
The presence of a space (i.e., $T^2$) with infinite fundamental group ($\pi_1(T^2) \cong \Bbb Z^2$) makes things worse.
Best Answer
Let $\pi_1(RP^2\vee RP^2)=\langle a,b\mid a^2=b^2=1\rangle$. By viewing this as the infinite dihedral group, you get the following kind of subgroups: