# All regular covering projections of wedge product of $\mathbb{RP}^3$ with $S^1$

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 $\mathbb{Z}_2\ast\mathbb{Z}=<x,y|x^2=1>$ should also come in pairs (one where there can be any number of $x$s in a word, and one where there must be an even number of $x$s). Am I on the right track? If there's some mechanical process for finding all the regular coverings from the universal covering I must have missed it. Thanks. If I didn't make anything clear please say so and I'll try to explain better or draw a picture or something.

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I'm not sure what the intention was behind the question, but I very much doubt that there's any simple way to enumerate the covers of $\mathbb{RP}^3 \lor S^1$.
If $G$ is any finite group, then $G$ is isomorphic to a quotient of $\mathbb{Z}_2\ast\mathbb{Z}$ if and only if $G$ can be generated by two elements, one of which squares to the identity. Every such $G$ corresponds to a finite-sheeted regular cover of $\mathbb{RP}^3 \lor S^1$, whose structure resembles the Cayley graph of $G$ with respect to this generating set.
For example, the pairs of circles "bridged" by copies of $\mathbb{RP}^3$ that you mentioned correspond to the groups $\mathbb{Z}_2\times \mathbb{Z}_n$ (if the circles are oriented in the same direction), or the dihedral groups of order $2n$ (if the circles are oriented in opposite directions).
Among other things, the class of groups $G$ described above includes cyclic groups, groups of the form $\mathbb{Z}_2\times\mathbb{Z}_n$, the dihedral groups $D_{2n}$, the symmetric groups $S_n$, the alternating groups $A_n$, the group $\mathrm{PSL}(2,p)$ for any prime $p$, and so forth. According to GAP, there are 344 finite groups with order up to 100 in this class, so I don't think there's any realistic chance of a simple enumeration.