# Universal Covering Space of Wedge Products

Today I was studying for a qualifying exam, and I came up with the following question;

Is there a simple description in terms of the subspaces universal covers for the universal cover of a wedge product?

This question came about after calculating universal covers of the wedge of spheres ($\mathbb{S}^1 \vee\mathbb{S}^1$ and $\mathbb{S}^1 \vee\mathbb{S}^n$) and the wedge of projective space with spheres. In these cases, the universal cover looks like the cross product of the sheets of the universal covers of each space in the wedge.

For the case of wedging two spheres, we can use the fact that $\pi_{n\geq2}\left(U\right)$ is isomorphic to $\pi_{n\geq2}\left(X\right)$ for $U$ covering $X$.

I googled around a bit to try and find something, but nothing appeared.

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The universal covering of $S^1\vee S^1$ is quite different from the cross product of the universal covering spaces of the factors. – Mariano Suárez-Alvarez Jan 22 '10 at 4:59
(By the way, the wedge of two spaces is usually denoted \vee in LaTeX.) – Mariano Suárez-Alvarez Jan 22 '10 at 5:00
Mariano, I understand that it is the fractal "snowflake". I did not mean for that comment to be taken literally, however it looks somewhat like it in the sense that in every sheet of the helix, we connect another helix and at every sheet of it... etc. Perhaps I shouldn't of included that idea, it was mostly rough thinking. Thanks for pointing this out though. Also, thanks for the \vee, I was wondering about that :) – B. Bischof Jan 22 '10 at 5:07
Typically in the context of topology, $\vee$ is the wedge sum, and $A\wedge B := (A\times B)/(A\vee B)$ is called the smash product. Mariano, just for the record, the exterior product of modules is called the wedge product and uses $\wedge$. This notation has been around since the 1930s, long before the wedge sum notation. So I guess it's fair in this case to blame the topologists. – Harry Gindi Jan 22 '10 at 9:22
When our spaces are not pointed, the smash product becomes a specialized type of pushout, but I've only seen this used once in HTT, so I can't fill you in on the details, so to speak. – Harry Gindi Jan 22 '10 at 9:37

If $X$ and $Y$ are two reasonable spaces with universal covers $\tilde{X}$ and $\tilde{Y}$, there is a nice picture of the universal cover $\widetilde{X \vee Y}$ which has the combinatorial pattern of an infinite tree. The tree is bipartite with vertices labeled by the symbols $X$ and $Y$. The edges from an $X$ vertex are bijective with the fundamental group $\pi_1(X)$, and likewise for $Y$ vertices and $\pi_1(Y)$. To make $\widetilde{X \vee Y}$, replace each $X$ vertex by $\tilde{X}$ and each $Y$ vertex by $\tilde{Y}$. The base point of $X$ lifts to $|\pi_1(X)|$ points in $\tilde{X}$, and likewise for $Y$. In $\widetilde{X \vee Y}$, copies of $\tilde{X}$ are attached to copies of $\tilde{Y}$ at lifts of base points. For example, if $X = Y = \mathbb{R}P^2$, then the tree is an infinite chain and $\widetilde{X \vee Y}$ is an infinite chain of 2-spheres.
@HJRW I'm not familiar with the language of this paper and Bass-Serre theory but could you tell me whether in this graph construction the edge between two vertices corresponds to wedge sum of spaces representing them ($\tilde X$, $\tilde Y$ respectively)? Examples of covering spaces from top of my head seems to confirm this interpretation. – 108592 Feb 18 '15 at 17:33