# Covering of the torus with simply connected open sets

I know this might sound like standard qualifying exam exercise in algebraic topology, so I apologize from start if this post might be inappropriate. The thing is, I've been doing some elementary combinatorial topology around Sperner's Lemma/Brouwer's fixed point theorem and came across the following result, which I don't really know how to prove in its full generality, since I'm kind of new to algebraic topology arguments... so I would really appreciate a full proof which I could understand by following Hatcher or something.

It goes like this:

Prove that it is impossible to cover the $n$-dimensional torus with $n$ simply-connected open sets. However, show one can do it with $n+1$.

The reason I'm asking this is related to Sperner's Lemma. The above has a cute consequence that no matter how you cover the $n$-dimensional cube with open sets, there will always be a path from one face to its opposite lying entirely in one of the open sets. As mentioned, I only managed to do this with Sperner, so I'm taking it as a non-trivial result...

PS. I know how to do first part for the $n=2$ case using van Kampen's theorem. I don't see how to do it for higher $n$'s...

-
In most algebraic topology texts these kinds of results appear once you've computed the cup product for singular cohomology of a torus. It also goes under the heading "Lusternik-Schnirelmann category". – Ryan Budney Oct 19 '12 at 22:57
Oh wow, thanks! I was actually thinking about this in a more general setting the other day. I found math.ucr.edu/~res/math246A/cuplength.pdf which practically answers my question. I guess I was hoping for a more complicated solution with Mayer-Vietoris or some induction as a friend suggested – Cosmin Pohoata Oct 19 '12 at 23:04
Well, the cup product is fairly sophisticated. Don't knock it as being uncomplicated until you dig in a little more. :) – Ryan Budney Oct 19 '12 at 23:07
FYI, there's a nice procedure that frequently gives you good upper bounds on the LS-category of a CW-complex -- the idea is to take various regular neighbourhoods of various complexes. For example, in an $n$-manifold with a CW-decomposition, your first ball will be a regular neighbourhood of a maximal tree in the dual 1-skeleton. Then for your next ball, take a regular neighbourhood of a maximal tree in the 1-skeleton. etc.... – Ryan Budney Oct 19 '12 at 23:23