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Can anyone please provide the cell structure of $S^2 \times S^1$? I know that there are one cell in each dimension from 0 to 3 but I am not sure about the attaching maps.

Thanks in advance.

Paladin
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    Related: http://math.stackexchange.com/questions/100170/cartesian-product-of-two-cw-complexes – Arthur Jul 31 '14 at 12:32

2 Answers2

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If $S^2$ has $e_0$ and $e_2$ as a $0$-cell and $2$-cell with attaching map $\varphi$ taking $S^1$ (the boundary of $D^2$) to $e_0$ , and $S^1$ a $0$-cell $f_0$ and a $1$-cell $f_1$ with attaching map $\phi$, then the cell structure of $X= S^2 \times S^1$ is:

$X^0 = e_0\times f_0$ as a $0$-cell,

$X^1= e_0 \times f_1$ (a $1$-cell) attached to the $0$-cell with attaching map $1 \times \phi$,

$X^2= e_2 \times f_0$ (a $2$-cell) attached to $X^1$ via $\varphi \times 1$

$X= e_2 \times f_1$ (a $3$-cell) attached to $X^2$ via $\varphi \times \phi$

inquisitor
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$S^3=S^1\wedge S^2=S^1\times S^2/S^1\vee S^2$. In other words, $S^1\times S^2 = S^1\vee S^2 \cup_\varphi e^3$, where $\varphi$ is the attaching map $$\partial e^3=(\partial e^1\times e^2)\cup(e^1\times\partial e^2)\to S^1\vee S^2,$$ which sends $\partial e^2$ and $\partial e^1$ to the base point of $S^1\vee S^2$.

Quang Hoang
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