why not just 2 charts to make atlas for sphere?

Solution 1:

The northern hemisphere and southern hemisphere don't cover the entire sphere. The sphere is $$\mathbb{S}^2=\{(x,y,z)\in\mathbb{R}^3\mid x^2+y^2+z^2=1\}.$$

The northern and southern hemispheres are, respectively, $$\mathbb{S}_N^2=\{(x,y,z)\in\mathbb{S}^2\mid z>0\},\qquad\mathbb{S}_S^2=\{(x,y,z)\in\mathbb{S}^2\mid z<0\}.$$ These miss the equator $\{(x,y,z)\in\mathbb{S}^2\mid z=0\}$. Adding "east" and "west" hemispheres $$\mathbb{S}_W^2=\{(x,y,z)\in\mathbb{S}^2\mid x>0\},\qquad\mathbb{S}_E^2=\{(x,y,z)\in\mathbb{S}^2\mid x<0\}$$ still doesn't get everything: we are missing the points on the equator $(0,1,0)$ and $(0,-1,0)$. Finally, adding the last two hemispheres (east and west, only rotated 90 degrees) covers the entire sphere.

This raises the question, why are we defining our hemispheres with $>$ and $<$? Perhaps we could instead use $\leq $ and $\geq$, and this would let us cover the sphere with two hemispheres?

The answer is that a chart of a manifold needs to be a homeomorphism between an open subset of the manifold with an open subset of $\mathbb{R}^n$. The sets $$\{(x,y,z)\in\mathbb{S}^2\mid z\geq 0\},\qquad\{(x,y,z)\in\mathbb{S}^2\mid z\leq 0\}$$ are not open in the topology of $\mathbb{S}^2$ (which is the subspace topology inherited from $\mathbb{R}^3$). So we can't use them as coordinate neighborhoods in the manifold structure of $\mathbb{S}^2$.

It does warrant mentioning, however, that we can cover the sphere using only two charts, via stereographic projection. The two open subsets of $\mathbb{S}^2$ acting as our coordinate domains are $$\mathbb{S}^2-\{(0,0,1)\},\qquad\mathbb{S}^2-\{(0,0,-1)\}$$ and for each, we project a line from the removed point to the plane, which one can check gives a continuous map. It is a tedious (but important) exercise to demonstrate that the smooth structure determined by stereographic projection is the same as that of the hemispheres (i.e., they are compatible atlases).