Finding $n$-sheeted path-connected covering spaces of the wedge sum $X\vee Y$ of two spaces.

All references/definitions are from Hatcher.

Suppose I have two path-connected, locally path-connected, and semilocally simply-connected spaces $$X$$ and $$Y$$ and I want to enumerate the $$n$$-sheeted path-connected covering spaces of the wedge sum $$X\vee Y$$ up to isomorphism (considering basepoints). My question is: in general, what would the $$n$$-sheeted covering spaces of $$X\vee Y$$ look like, specifically in terms of $$X$$ and $$Y$$ (or covering spaces thereof). Assume we have concrete descriptions of what all the path-connected covering spaces of $$X$$ and $$Y$$ look like, but nothing beyond that about $$X$$ or $$Y$$ themselves. Can it even be done in this case?

My own thoughts: An $$n$$-sheeted covering space of $$X\vee Y$$ would necessary contain $$n$$ liftings $$\tilde x_0$$ of the wedge point $$x_0$$. Near each of these points $$\tilde x_0$$ would be some neighborhood that "looks like" the some neighborhood of the wedge point of $$X\vee Y$$. This makes me feel like the an $$n$$-sheeted covering space of $$X\vee Y$$ would look like some sort of chain or graph of $$X$$'s and $$Y$$'s. (Or maybe it's chains/graphs of $$n$$-sheeted covering spaces of $$X$$ and $$Y$$? Not exactly sure here.). Is my intuition correct?

Some other potential variations of the question that I am interested in (although including answers to these are by no means necessary in any answer to this question):

1. What could we say about the $$n$$-sheeted covering spaces if we only knew the fundamental groups of $$\pi_1(X)$$ and $$\pi_1(Y)$$, rather than the path-connected covering spaces of $$X$$ and $$Y$$ (beyond the fact that the fundamental groups of these covering spaces would be of index $$n$$ in $$\pi_1(X\vee Y)$$)
2. What if we only knew the $$n$$-sheeted covering spaces of $$X$$ and $$Y$$ or only the universal covers of $$X$$ and $$Y$$, instead of all path-connected covering spaces of $$X$$ and $$Y$$?
3. Among our conditions that $$X$$ and $$Y$$ be path-connected, locally path-connected, and semilocally simply-connected, which of those could we discard without any major change in the context of this problem?

Thanks!

• Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.
– Community Bot
Nov 8 '21 at 4:29
• I think the linked blog could be helpful math3ma.com/blog/a-recipe-for-the-universal-cover-of-x-y Nov 8 '21 at 7:04
• @love_sodam Thanks for that, although it only addresses how to find the universal cover of the wedge sum of spaces, and I'm interested in enumerating n-sheeted covering spaces for some n. Nov 8 '21 at 8:41
• @isaiahtx7 I remember there's some exercise problem in Hatcher which asks to find $2$ and $3$-sheeted covering space of $S^1\vee S^1$. And also in Hatcher, there is a page full of covering spaces of $S^1\vee S^1$. Locally, looks like $S^1\vee S^1$ but globally no that much. Nov 8 '21 at 16:27

I believe I have figured out the answer.

To give a "concrete description" of all the $$n$$-sheeted covering space up to isomorphism, we need to know all the $$m$$-sheeted covering spaces of $$X$$ and $$Y$$ for $$m\leq n$$. This includes $$X$$ and $$Y$$ themselves. Specifically, an $$n$$-sheeted covering space of $$X\vee Y$$ is a connected "graph" with the following properties:

1. There are $$n$$ edges
2. Given a vertex of our graph of valence $$m$$, that vertex is an $$m$$-sheeted connected covering space of either $$X$$ or $$Y$$.
3. No edge connects two covering spaces of $$X$$ or two covering spaces of $$Y$$.

Specifically, at each vertex, the edges attach at points belonging to the preimage of the basepoint $$x_0\in X\vee Y$$. We can pick any of these edges to be a basepoint of the covering space. Namely, once this graph is constructed, retract each of the lines connecting the vertices that make up the "edges" to points. For example, a graph with two vertices would become the wedge sum of two spaces.

A concrete example: consider the space $$B=\mathbb RP^2\vee T^2$$ (where $$T^2=S^1\times S^1$$ is the torus). $$B$$ has $$7$$ isomorphism classes of connected two-sheeted covering spaces. This can be seen from the fact that:

1. $$\mathbb RP^2$$ has one two-sheeted covering space, namely, $$S^2$$.
2. $$T^2$$ has three two-sheeted covering spaces, specifically, those corresponding to the subgroups $$\langle a^2,b\rangle$$, $$\langle a,b^2\rangle$$, and $$\langle a^2,ba^{-1}\rangle$$ of $$\pi_1(T^2)=\mathbb Z^2=\langle a,b\mid [a,b]\rangle$$ (these can be seen as the kernels of epimorphisms $$\mathbb Z^2\to\mathbb Z_2$$). Let's call these covering spaces $$\widetilde T_1$$, $$\widetilde T_2$$, and $$\widetilde T_3$$,

There are two possible connected graphs with $$2$$ edges, namely, that with three vertices obtained by connecting two line-segments end-to-end, and that with two vertices, each vertex of valence 2 with both edges leading to the other vertex.

In the former case, if middle vertex (of valence 2) is a two-sheeted covering space of $$\mathbb RP^2$$, of which there is only one choice, then the outermost vertices are both $$T^2$$. On the other hand, there are three possibilities for the middle vertex if it is a covering space of $$T^2$$, namely, $$\widetilde T_1$$, $$\widetilde T_2$$, and $$\widetilde T_3$$. The outermost vertices must be $$\mathbb RP^2$$. Hence, there are $$4$$ possible $$2$$-sheeted covering spaces of $$B$$ with three "vertices."

In the latter case, we have two vertices each of valence two. For the vertex that covers $$T^2$$, we have three choices, and for the vertex that covers $$\mathbb RP^2$$. we have only one choice, the sphere. In this way, we obtain $$3$$ more covering spaces of $$B$$.

I have drawn each of these graphs and labelled the vertices accordingly below. Red vertices correspond to covering spaces of the torus, while blue vertices correspond to covering spaces of $$\mathbb RP^2$$.