Sphere construction from cells

Question

http://en.wikipedia.org/wiki/Morse_theory

In the link, there is this statement 'The number of critical points of index $\gamma$ of $f : M → \mathbb{R}$ is equal to the number of $γ$ cells in the CW structure on $M$ obtained from "climbing" $f$.'. When I let $X=S^2$, it seems like not true. We know that in sphere, there are two nondegenerate critical points, each of them has index $0$ and $2$ respectively because one is global minimum and another one is global maximum. From the statement, we should have one $0$-cell and one $2$-cell. But I don't see how we can construct a sphere by just using these two cells. Can anyone help me?

Answer 1

A sphere can be constructed by gluing the boundary of a $2$-cell to a $0$-cell (a point).

This isn't too hard to visualize by itself but if you imagine climbing your sphere and applying Morse theory you can actually see what the gluing would look like. Your $2$-cell gets bent inwards until it's boundary is brought together to be glued together at a point.

Answer 2

Intuitively, if $D$ is a closed disk, then identifying the boundary points of $D$ to a point gives a sphere.

If you prefer formulas, send the point $(r\cos\theta, r\sin\theta)$ of $D$ to the point $$ \bigl(\sin(\pi r)\cos\theta, \sin(\pi r)\sin\theta, -\cos(\pi r)\bigr) $$ of the unit sphere in $\mathbf{R}^{3}$. This mapping sends each boundary point of $D$ to the north pole $N = (0, 0, 1)$, and is easily seen to be continuous, and bijective in the open disk.