Charts needed for an atlas

I just read this question link and asked myself, if there is any easy way to decide how many charts you actually need to cover a given compact manifold in $\mathbb{R}^3$, maybe at least in this special situation there is an easier answer to the question. I mean, the answer accepted in the cited question seems to use advanced techniques.

What I would be interested in is to understand why some manifolds like the torus or the real projective space cannot have $2$ charts. I mean, $1$ chart is of course wrong by compactness and the same argument tells us that if we have two charts, then the intersection of the open sets on the manifold cannot be empty, but what is the problem with two charts on these manifolds. What is it that distinguishes them from the sphere let's say? Maybe it is even possible to extend this argument further for general manifolds in $\mathbb{R}^3$?

You are looking for the smallest number of open contractible sets needed to cover $M.$ This is so well studied, that it has a name: *the Lyusternik-Shnirelman category of $M$." For references, you can look at the nice paper by Gomez-Larranaga, Heil, and Gonzalez-Acuna, or just look at the Wikipedia article.
• yes, this is also the answer in the link that I referred to, but if I want to calculate this for a submanifold in $\mathbb{R}^3$ are there any ways to do this somehow directly? Like how can I see that this number is three for a torus and not just two? I mean there should be simpler arguments available in $\mathbb{R}^3$ and that is what I am looking for. Is there an easy way to do this? – Physicist 2.0 Apr 20 '15 at 23:28
• Submanifolds of $R^3$ are orientable surfaces. For such, the sum of betti numbers is $2g+2,$ where $g$ is the number of handles. For the relationship between this and the Lyusternik-Shnirelman category, see the wikipedia article. – Igor Rivin Apr 20 '15 at 23:40
Proposition If a compact connected topological surface $S$ is covered by two connected open sets $U$ and $V$ with $U$ homeomorphic to a disk, then $U \cap V$ is connected.
If we assume also that $V$ is homeomorphic to a disk, we get $H_1(U)\simeq H_1(V)\simeq H_0^\sharp(U)\simeq H_0^\sharp(V) \simeq 0$ and using the Mayer-Vietoris sequence, $H_1(U \cup V) \simeq H_0^\sharp(U\cap V) \simeq 0$. Thus the Euler characteristic is 2 and the surface is a sphere.
Proof of proposition Identify $U$ to the open unit disk in the complex plane and look at the subset $C_n$ of $U$ corresponding to $\{z \mid 1-1/n < |z| < 1\}$. The intersection of the closures of the $C_n$ in $S$ is included in $V$, so by compactness for some n, (the closure of) $C_n$ is in $V$. So there is a corona both in $U$ and $V$. Given any point in $U \cap V$, we can connect it to a given point in $V\backslash U$ with a path in $V$. Looking at that path in the chart $U$, we see that it must meet the corona before leaving $U$. Hence any point in $U\cap V$ can be connected in $U\cap V$ to the corona.