My question is related to this one here but is different in that I am wondering about the CW structure on such a space. I am trying to put a CW structure on $S^2/S^0$ and I think that we have $1$ 0 -cell, $1$ 1 - cell and $1$ 2-cell. My $1$ - cell is just some path connecting the north pole to the south pole. However then I run into trouble because by Van - Kampen the fundamental group of $S^2/S^0$ is zero and not $\Bbb{Z}$ as stipulated in the link above.

Question: What is wrong with this cell structure on $S^2/S^0$? It seems to get $\Bbb{Z}$ I would need it to have $2$ 1 -cells but how is this obvious from the definition of $S^2/S^0$?

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    $\begingroup$ You have to attach the $1$-cell before the $2$-cell. So, if you have only one $0$-cell, there is no north (or south) pole to which to attach the $1$-cell. $\endgroup$ – Joe Johnson 126 Nov 7 '12 at 2:16
  • $\begingroup$ Start with two $0$-cells. Then use three $1$-cells. $\endgroup$ – Joe Johnson 126 Nov 7 '12 at 2:17
  • $\begingroup$ @JoeJohnson126 With regards to your first comment, why can't I start with one $0$ - cell? Because in the quotient space there is only one $0$ - cell no? $\endgroup$ – user38268 Nov 7 '12 at 3:04
  • $\begingroup$ @BenjamLim: You may be able to start with one $0$-cell. But, your $2$-cells have to be attached along the $1$-skeleton. With only one $0$-cell, your $1$-skeleton will be $S^1$. $\endgroup$ – Joe Johnson 126 Nov 7 '12 at 5:30
  • $\begingroup$ @JoeJohnson126 This is the thing right now that makes me confused because then by Van Kampen I have that $\pi_1(X) = \pi_1(S^1)/N$ where $X$ is my space now and $N$ is some normal subgroup in $\pi_1(S^1)$. I get that $N = \Bbb{Z}$ and so how do I get $\pi_1(X) = \Bbb{Z}$? $\endgroup$ – user38268 Nov 7 '12 at 6:51

Take the CW-decomposition to be one 0-cell and one 1-cell and one 2-cell, where the boundary of the 2-cell maps onto the 0-cell and the boundary of the 1-cell also maps onto the 0-cell. Then by van-Kampen on $X=A\cup B$, with $A$ slightly containing more than the 2-cell and $B$ slightly containing more than the 1-cell, we have $A\cap B\simeq\lbrace\text{0-cell}\rbrace$ and $\pi_1A\cong\pi_1S^2=0$ and $\pi_1B\cong\pi_1S^1=\mathbb{Z}$ and hence $\pi_1X\cong\mathbb{Z}$.

The problem with your given cell structure is that you haven't actually told us the attaching maps... for instance, where is your 1-cell attaching to? By definition, it must attach to cells of dimension $<1$, i.e. to your given 0-cell.

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    $\begingroup$ I don't understand why this cell decomposition is giving us $S^2/S^0$. It seems to me it's giving us $S^2 \vee S^1$, with the basepoint of the wedge being the $0$-cell. $\endgroup$ – Ben Blum-Smith Feb 20 '17 at 23:28
  • $\begingroup$ The two spaces are homotopy-equivalent. Via a homotopy, the wedged $S^1$ can be "opened up" into an interval $[0,1]$ where one of the endpoints runs away from the other endpoint (the original wedge point) along $S^2$. Then you can "shrink" the interval, which contracts the two endpoints together and hence squishes the $S^2$ at two points. $\endgroup$ – Chris Gerig Feb 20 '17 at 23:32
  • $\begingroup$ Thanks - so this cell complex is adequate for computing $\pi_1$. $\endgroup$ – Ben Blum-Smith Feb 22 '17 at 4:53

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