How to find covering space of wedge sum? Consider the wedge sum of 2 CW complices $X\vee Y$. Let $p:C\rightarrow X\vee Y$ be a m-sheeted covering map. Then what do we know in general about the space $C$? Is that true $C$ is also a wedge of CW complices in some natural sense?
It seems that $C$ should be a wedge sum of several CW complices and each of them is a covering space of $X$ or $Y$. But I don't know if exhausted.
In general, is there any method to find all the covering space of a given index of $X\vee Y$?
 A: For me it's simpler to think about a mild variation of the wedge sum where instead of just identifying two points we insert a path between them. Then we can describe the covering spaces in quite a nice way: they are "graphs of spaces" where the "edges" correspond to lifts of the path connecting the wedge points and the "vertices" are covering spaces of the spaces we're wedging. The rule for how to connect "vertices" up by "edges" is that if a given "vertex" is an $n$-fold cover then it contains $n$ lifts of its wedge point which means it must have "degree" $n$ in this "graph." (There is also some auxiliary data needed to uniquely determine a covering map.) Furthermore this covering space is an $m$-fold cover iff there are $m$ "edges" total.
It's much easier to explain this with pictures. You can find some in this blog post where $X = B C_2$ and $Y = B C_3$, so that the wedge sum is the classifying space of the modular group $\Gamma = PSL_2(\mathbb{Z}) \cong C_2 \ast C_3$.
This observation, together with the classification of covering spaces and the Seifert-van Kampen theorem, implies the Kurosh subgroup theorem classifying subgroups of free products, which is a generalization of the classical fact that subgroups of free groups are free.
I think the story should be the same for the ordinary wedge sum of CW complexes (but probably not more general spaces); we should be able to shrink the "edge" without affecting the covering spaces. But I'm not 100% confident.
