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I need to find CW complex structure in the sphere by identifying the north and south pole.
First of all I tried to visualize what it looks like and if I am not wrong gluing the poles together will result in something looking sort of like a torus but without a hole. Now from a question that another user has posed namely(Cell Structure on Sphere with Two points identified) I saw that the CW complex structure is the following attach the boundary of the one cell to the zero cell and then attach the boundary of the two cell to the zero cell. To me this structure results to a wedge of a sphere and a circle because attaching the one cell to the zero cell gives a circle and then attaching the two cell to the zero cell will result to a sphere. As such this structure will result to the wedge of a sphere and a circle. Can you please explain to me what's wrong with my reasoning and how the structure above actually results to the sphere with the poles identified??

Thanks in advance

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  • $\begingroup$ It is not homeomorphic to $S^1\vee S^2$ but I think it has the same homotopy type (intuitively, you connect north and south pole with a curve and continuously slide one end of it to a closed curve from north pole to north pole). Your reasoning is not very clear, at least not to me. What means "attaching a 0-cell to the boundary of a 1-cell", for example? The space $S^2/\{x,y\}$ has more complicated CW structure, anyway. $\endgroup$ – Peter Franek Mar 17 '15 at 22:38
  • $\begingroup$ The structure I described is supposed to result to the sphere with the poles identified and I saw this structure from a question of some other user and I think this structure actually gives the wedge of a sphere and a circle. I edited I actually meant attach the boundary of the 1 cell to the zero cell $\endgroup$ – TheGeometer Mar 17 '15 at 22:54
  • $\begingroup$ If you attach a 1-cell to a 0-cell, you get a circle. If you further attach a 2-cell to the 0-cell, you get $S^2\vee S^1$. But it is not the CW structure of $S^2$ with two points identified (only up to homotopy) $\endgroup$ – Peter Franek Mar 17 '15 at 22:55
  • $\begingroup$ Yeah I totally agree with you but some other user asked the same question and the answer to his question was the structure I described which I think is wrong. Do you happen to know the correct CW complex structure $\endgroup$ – TheGeometer Mar 17 '15 at 22:59
  • $\begingroup$ Well, it seems to me that you take the wedge of two circles (the result of a great circle on $S^2$ passing through north and south pole after shrinking them to one point) and glue on that two 2-cells (two hemispheres, "eastern" and "western") so that the boundary circles will be mapped to the $x*y\in S^1\vee S^1$ resp. $y^{-1}*x^{-1}$ where $x$ and $y$ are the free generators.. or something like that. $\endgroup$ – Peter Franek Mar 17 '15 at 23:08
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Start with a 0-cell $x$.

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

Attach a 2-cell $\sigma$ with attaching map defined on its oriented boundary circle $\partial\sigma$ as follows:

  • Subdivide $\partial\sigma$ as a concatenation of two arcs $\partial\sigma = \partial_1 \sigma * \partial_2 \sigma$,
  • Map $\partial_1 \sigma$ to $e$,
  • Map $\partial_2 \sigma$ to $\bar e$.

ADDED LATER: To see why this CW complex $X$ is homeomorphic to the sphere with north and south pole identified, this is the case where a picture is worth a large number of words, in order to convey the intuition. Yet it can be done with intuitive rigor also, with a few carefully chosen words, which I will attempt to supply.

First note that the characteristic map of the 2-cell $f : \sigma \to X$ is a surjective continuous map and is therefore a quotient map. So, $X$ can be reconstructed by starting with the 2-dimensional disc $\sigma$ and making the same identifications on $\sigma$ as are made by the map $f$. What are these identification? They are of two different types:

  1. Identify the two oriented arcs $\partial \sigma_1$, $\partial\bar\sigma_2$ to a single arc (which will be the 1-cell of $X$).
  2. Identify the two points $\partial(\partial \sigma_1) = \partial(\partial\sigma_2)$ to a single point (which will be the $0$-cell of $X$).

So, how do we see that this is the 2-sphere with the north and south poles identified? First do just identification 1: the quotient is the 2-sphere, and the arcs $\partial\sigma_1$, $\partial\sigma_2$ each map to a longitude line connecting the north and south poles. Next do identification 2: identify the north and south poles.

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    $\begingroup$ Thank you for your answer but I fail to see why this is indeed the CW structure. Can you please provide some geometric intuition why this is the case? $\endgroup$ – TheGeometer Mar 18 '15 at 1:47
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    $\begingroup$ @TheGeometer: I've tried to convey the geometric intuition. $\endgroup$ – Lee Mosher Mar 19 '15 at 2:02

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