# Covering the 2-sphere with 6 hemispheres

While reading chapter 2 of Wald's General Relativity titled "Manifolds", I stumbled upon the fact that the 2-sphere $$S^{2}$$ cannot be mapped into $$\mathbb{R}^{2}$$ in a continuous 1-1 manner. Wald then proceeds to construct a cover with 6 hemispheres and it all checks out. However, being somewhat of a newcomer to these mathematical areas, I can't help but wonder as to how to construct such a "solution". Further consulting Schutz's Geometrical methods of mathematical physics and Renteln's Manifolds, Tensors and Forms does not answer my question, they both give solved examples for which it can then be easily checked that all the requirements have been met.

Is it possible to prove (maybe somewhat of a strong word but I'll use it for lack of a better one) that $$S^{2}$$ cannot be covered with, say, 5 hemispheres or has 6 become the "standard" after failing to find a "better" solution?

P.S. I am aware of the other methods of covering the sphere, i.e. stereographic projection but my question concerns the use of hemispheres. I also apologize if this is something trivial but I honestly do not see it at this point in time.

• As Travis explains in the answer, one can cover the sphere with 4 hemispheres, it's just that the 6 hemisphere atlas is much more convenient to work with, since the charts are just the projections onto the coordinate planes. – jgon Mar 20 at 18:17

One cannot cover the sphere with $$3$$ open hemispheres: Any two distinct hemispheres leave uncovered the two points where their boundary (great) circles intersect. But these points are antipodal, so no third hemisphere can cover them both.
On the other hand, one can cover the sphere with $$4$$ open hemispheres: Inscribe a regular tetrahedron in the sphere, and place hemispheres centered at each of its vertices.