I'm studying for a topology exam. One question in an exam paper from a previous year is:
Starting with the universal cover, describe all the regular covering projections of $\mathbb{RP^3}\vee S^1$.
The universal covers of $\mathbb{RP^3}$ and $S^1$ are $S^3$ and $\mathbb{R}$ respectively. I believe the universal cover of $\mathbb{RP^3}\vee S^1$ is the "tree fractal" space consisting of many (countable) copies of $S^3$ joined to each other by edges, where each copy of $S^3$ has two opposite points, each attached to two edges, and there's no loops. Sorry, my description isn't particularly clear, but if you imagine taking the valency-4 infinite tree graph that is the universal cover of $S^1\vee S^1$ and replacing each vertex with a copy of $S^3$ such that the four incoming edges attach to two opposite points on the copy of $S^3$, two edges to each point, that's what I mean. The projection maps each copy of $S^3$ to $\mathbb{RP}^3$ and each edge to $S^1$ like you'd expect.
So the fundamental group of $\mathbb{RP}^3\vee S^1$ is the free product $\mathbb{Z}_2\ast\mathbb{Z}$, but I don't think the question intends me to explicitly calculate all normal subgroups of this group since that is kind of outwith the scope of the course and I wouldn't really know where to start. (If there is an easy way please let me know.) So I just sort of thought about what simplifications of the universal cover would work, which seems to be what the question is suggesting I do. Anyway, I came up with two families of regular coverings:
- Obviously you can take the real line and attach infinitely many copies of $\mathbb{RP}^3$ along it. Or if you instead connect the line back to itself and take a finite number of copies, you get a covering which is a circle with $n\ge 1$ copies of $\mathbb{RP}^3$ attached along it.
The second family is the obvious counterpart to these where you have $S^3$ instead of $\mathbb{RP^3}$:
- Two parallel copies of the real line that are "bridged" by infinitely many copies of $S^3$ between them. Or, two concentric circles, bridged by $n\ge 1$ copies of $S^3$.
This is all that I can come up with but it seems a bit hand-wavy. I don't know if I could prove that this is all the regular coverings. (Actually, I don't know whether this is required.) But I'm inclined to believe it is, since intuitively it seems like the nontrivial normal subgroups of \mathbb{Z}_2\ast\mathbb{Z}=