I am trying to figure out the $CW$ complex structure on a sphere with the north and south pole identified. I've been told the structure is the following

Start with a $0$-cell $x$.

Attach an oriented $1$-cell $e$ with both endpoints identified to $x$.

Attach a $2$-cell $σ$ with attaching map defined on its oriented boundary circle $∂σ$ as follows:

Subdivide $∂σ$ as a concatenation of two arcs $∂σ=∂1σ∗∂2σ$,

Map $∂1σ$ to $e$,

Map $∂2σ$ to $e¯$.
The problem is that I don't understand why or if this is indeed the $CW$ structure. First of all I don't know what $e¯$ means. I assume is the inverse of $e$. So does the above say we attach the half circle of the boundary of our disk to the oriented circle (of the 1 skeleton) and then the second half but in opposite orientation?. I don't understand how this attachment is possible. I don't even know what happens in the simple scenario of say we want to attach the circular part of the boundary of a half disk to a circle? This really confuses me... how can we attach something smaller to something bigger?
Can anyone explain me whats going on here

Thanks in advance

  • $\begingroup$ Pls try to use MathJax! $\endgroup$ – aGer Mar 18 '15 at 14:18
  • $\begingroup$ Instead of asking a new question so soon, you should wait for Lee Mosher to reply to your comment on your last question. $\endgroup$ – Michael Albanese Mar 18 '15 at 14:49
  • $\begingroup$ If you have a CW structure for the sphere which has at least two $0$-cells, then by identifying two distinct $0$-cells, this gives you an induced CW structure on the space in your question. In this case, you can build a CW structure for the sphere from two $0$-cells, a $1$-cell between them, and then a $2$-cell which is 'zipped up' from north to south and then glued onto the $1$-cell in the obvious way. $\endgroup$ – Dan Rust Mar 18 '15 at 16:37

Note carefully: It is homotopy equivalent to $S^2\vee S^1$.


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