# Applying a cell structure to $S^2$ with north and south pole identified.

This question may be a bit nit-picky but I want to get a better handle on constructing cell structures on particular spaces.

Identifying the north and south pole of $S^2$ is the same as attaching an arc between the two poles.

I want to place a cell structure on this space.

In Hatcher, a cell structure is placed on this space by viewing the two poles as our $0$-cells, the arc between the two poles as a $1$-cell and the rest of $S^2$ as a $2$-cell.

I am not sure how this is a valid cell structure.

In general we construct $S^2$ by attaching the boundary, $S^1$, of a $2$-cell to a $0$-cell. However in this case we have two $0$-cells.

How do we attach $\textbf{one}$ $2$-cell so that $S^2$ contains both $0$-cells as poles?

(I can think of ways to do this using three $1$-cells and two $2$-cells but not using only one $2$-cell and one $1$-cell.)

For $S^2\cup I$, a sphere with an arc outside of $S^2$ which connects the north and the south pole, we have two $\text{$0$-cells}$ $N$ and $S$, two $1$-cells, the arc $I$ and another arc $J\subset S^2$ from $N$ to $S$, and a $2$-cell.
There is a homeomorphism \begin{align} D^2 &\to S^2\cap\{(x,y,z)\mid y\ge 0\} \\ (x,z) &\mapsto \left(x,\sqrt{1-x^2-y^2},z\right) \end{align} and then there's a map \begin{align} S^2\cap\{(x,y,z)\mid y\ge 0\} &\to S^2 \\ \pmatrix{ \sin\theta \sin\phi & \sin\theta \cos\phi & \cos \theta \\ } &\mapsto \pmatrix{ \sin\theta \sin2\phi & \sin\theta \cos2\phi & \cos \theta \\ } \end{align} The composite wraps the disk around the $2$-sphere such that its boundary becomes the arc $J$.
For $S^2/N{\sim}S$, we simply shrink $I$ to a point, so that $N$ and $S$ become a single $0$-cell. The result is still a CW complex, and the attaching map $\partial D^2\to J$ is just composed with the quotient map $J\to J/N{\sim}S$.