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In http://en.wikipedia.org/wiki/Manifold_(mathematics)#Construction, it says that 6 charts can be used to make an atlas for a sphere. But the text shows that you have a chart for the northern hemisphere, and you can make a similar chart for the southern hemisphere. Hence, these two charts cover the entire sphere.

What am I doing wrong?

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    Those two charts don't cover the great circle on the equator. So you need four more charts to fully cover that circle. You can cover the sphere with just two charts using [stereographic projection](http://en.wikipedia.org/wiki/Stereographic_projection).2011-07-30

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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).

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    @Charles That makes sense as well.2011-08-02