Lower bound for the size of an atlas This question came up in a graduate-level class on differential topology I'm currently taking; the instructor couldn't come up with an answer off the top of her head and while I'm very new to the subject, it struck me as the kind of thing for which there ought to already be some results.
Let $M$ be a (not-necessarily smooth) manifold. Is there any information on the least number of pairs $(U_i, \phi_i)$ where $U_i$ is an open subset of $M$ and $\phi_i: U_i\rightarrow \mathbb{R}^n$ is a homeomorphism needed to produce an atlas for $M$? Or, perhaps a better way of phrasing the question is, is there any information on a smallest open cover $\{U_i\}$ of $M$ with the property that all the covering sets are homeomorphic to $\mathbb{R}^n$?
Some examples I can work out pretty easily on my own are spheres of any dimension (where compactness of the sphere proves that 1 chart is not sufficient and the stereographic projection proves that 2 charts are) the torus (which appears to require 4 charts (although I don't have a proof for this and if a counterexample exists I'd love to hear about it)), and the surface of genus 2 (which appears to require 6 but I'm less sure of this because I can't quite construct the charts).
[Edit: I overlooked the obvious fact that an orientable surface with an embedding in 3-space can be covered by exactly two charts with no effort. The math overflow link in the comments provides many wonderful references, but I'm still wondering if there are others that may have been missed there.
Edit to the edit: In fact, the above observation I made above is incorrect, because as Anthony Carapetis points out in the comments, those charts are not contractible in general. However, the comments agree that 3 charts suffice for any surface, and armed with the suggestion I have been able to imagine such atlases for surfaces of low genus. A reference has not been produced yet, although in my spare time I'm still searching for one.] 
If anyone is familiar with any reading materials that may shed some light on this question and any answers--partial or otherwise--which may already exist, I'd be pretty happy to have a reference. Also, I hope the algebraic topology tag is appropriate; I assume if there is anything like an answer that it can be stated in algebro-topological terms.
 A: I remember hearing that if a closed manifold $M$ can be covered by $U, V$ both homeomorphic to $\mathbb{R}^{n}$, then it is homotopy equivalent to a sphere (and hence it is topologically a sphere by Poincare conjecture, which is much more difficult). No proof was given, but I will try to give one. It uses a hefty dose of elementary algebraic topology. 
Let $dim(M) \geq 2$, as the other case is trivial. 
Clearly $M$ is path-connected and we want to show simply-connectedness. Fix a homeomorphism $\phi: \mathbb{R}^{n} \rightarrow U$ and observe that by standard compactness arguments $V$ must contain $\phi(\mathbb{R}^{n} \setminus D^{n} _{k})$, where $D^{n} _{k} = \{ x \in \mathbb{R}^{n} \ | \ | x | < k \}$. Let $x, y \in U \cap V$ and $p \in V \setminus U$. A beginning of any path $x \rightarrow p$ (in $V$) will give a path entirely contained in $U$ such that its end does not lie in $\phi(D^{n} _{k})$ and the same is true for $y$. Since $U \setminus \phi(D^{n} _{k})$ is path-connected because of $n \geq 2$, there must also exist a path $x \rightarrow y$. This shows that $U \cap V$ is path-connected and by applying van Kampen theorem we get the desired result, namely $\pi _{1}(M) = 0$. 
Observe that by standard argument considering the isomorphisms $H^{k}(M, U) \rightarrow H^{k}(M)$, $H^{k}(M, V) \rightarrow H^{k}(M)$ for $k \neq 0$ and the relative cup product, all higher cup products must vanish. Since $M$ is simply-connected, it is orientable and Poincare duality applies to show that cup-product gives a perfect pairing between $H^{k}(M), H^{n-k}(M)$ modulo torsion. As all products vanish, this means that $H^{k}(M)$ are all torsion for $k \neq 0, n$. But they cannot contain $p$-torsion for an analogous argument over $\mathbb{Z}_{p}$ and so they must be zero. 
Now the map $M \rightarrow K(\mathbb{Z}, n)$ representing the generator of $H^{n}(M, \mathbb{Z})$ can be pushed down to the $(n+1)$-skeleton of $K(\mathbb{Z}, n)$ which happens to be the $n$-sphere. This gives a map $M \rightarrow S^{n}$ that is an isomorphism on $H^{n}$ and by universal coefficient theorem must be also a homology isomorphism. But both spaces are simply-connected and by dual Whitehead theorem we conclude that $M \simeq S^{n}$.
