Universal Covering Space of Wedge Products - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:34:03Z http://mathoverflow.net/feeds/question/12606 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12606/universal-covering-space-of-wedge-products Universal Covering Space of Wedge Products B. Bischof 2010-01-22T04:46:55Z 2010-01-22T18:00:59Z <p>Today I was studying for a qualifying exam, and I came up with the following question;</p> <blockquote> <p>Is there a simple description in terms of the subspaces universal covers for the universal cover of a wedge product?</p> </blockquote> <p>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. </p> <p>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 $X$.</p> <p>I googled around a bit to try and find something, but nothing appeared.</p> <p>Thanks in advance!</p> http://mathoverflow.net/questions/12606/universal-covering-space-of-wedge-products/12611#12611 Answer by Greg Kuperberg for Universal Covering Space of Wedge Products Greg Kuperberg 2010-01-22T06:08:30Z 2010-01-22T18:00:59Z <p>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.</p> <p>This tree picture nicely and dramatically generalizes to <a href="http://en.wikipedia.org/wiki/Bass%E2%80%93Serre%5Ftheory" rel="nofollow">Bass-Serre theory</a>.</p>