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?


  • $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Nov 8 '21 at 4:29
  • $\begingroup$ I think the linked blog could be helpful math3ma.com/blog/a-recipe-for-the-universal-cover-of-x-y $\endgroup$
    – love_sodam
    Nov 8 '21 at 7:04
  • $\begingroup$ @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. $\endgroup$
    – isaiahtx7
    Nov 8 '21 at 8:41
  • $\begingroup$ @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. $\endgroup$
    – love_sodam
    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 $\mathbb RP^2$.

Two-sheeted covering spaces of RP^2 v T^2


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