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.)


I'm not sure which of two spaces you mentioned you are interested in, so let me just describe for both spaces the simplest cell-structures I can think of.

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 $J\to J/N{\sim}S$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.