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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.

Thanks in advance!

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    $\begingroup$ The universal covering of $S^1\vee S^1$ is quite different from the cross product of the universal covering spaces of the factors. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 22, 2010 at 4:59
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    $\begingroup$ (By the way, the wedge of two spaces is usually denoted \vee in LaTeX.) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 22, 2010 at 5:00
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    $\begingroup$ 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 :) $\endgroup$
    – B. Bischof
    Commented Jan 22, 2010 at 5:07
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    $\begingroup$ 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. $\endgroup$
    – Harry Gindi
    Commented Jan 22, 2010 at 9:22
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    $\begingroup$ 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. $\endgroup$
    – Harry Gindi
    Commented Jan 22, 2010 at 9:37

1 Answer 1

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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.

This tree picture nicely and dramatically generalizes to Bass-Serre theory.

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  • $\begingroup$ Thanks! Another excellent answer. Also, thanks for the link, that looks very interesting. $\endgroup$
    – B. Bischof
    Commented Jan 22, 2010 at 14:08
  • $\begingroup$ For the record, this picture is often credited to Scott and Wall in their paper: Topological methods in group theory. Homological group theory (Proc. Sympos., Durham, 1977), pp. 137--203, $\endgroup$
    – HJRW
    Commented Jan 22, 2010 at 19:44
  • $\begingroup$ @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. $\endgroup$
    – Stephen Dedalus
    Commented Feb 18, 2015 at 17:33
  • $\begingroup$ @108592, yes, that's right. $\endgroup$
    – HJRW
    Commented Feb 18, 2015 at 21:00

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