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This is an exercise in Allen Hatcher's Algebraic Topology book(page 53, 7)

Let X be the quotient space of $S^2$ obtained by identifying the north and south poles to a single point.Put a cell complex structure on X and use this to compute $\pi_1(X)$.

How to cut X to see how many 0-cells,1-cells and 2-cells.I use a way to cut X from up to down through the identified point,so each half is constructed by 3 0-cells,4 1-cells and 1 2-cell.Am I right?I'm a little suspicious.

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  • $\begingroup$ The fundamental group should be isomorphic to $\mathbb{Z}$. $\endgroup$ – Kerry Apr 11 '12 at 5:34
  • $\begingroup$ One can also show that $X \cong S^2 \vee S^1.$ $\endgroup$ – Ehsan M. Kermani Apr 11 '12 at 5:43
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Hint: you can also think about this space in terms of the torus $T^2$. The torus has a standard CW decomposition with one $0$-cell, two $1$-cells, and one $2$-cell. The space in question is homeomorphic to taking the quotient of the torus by collapsing one of the $1$-cells. This should suggest to you how to give a CW decomposition of this space with one cell in each dimension $1,2,3$

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  • $\begingroup$ So the CW decomposition of X is with one 0-cell,one 1-cell $a$,and one 2-cell.The 2-cell is attached along $aa^{-1}$,so $\pi_1(X)=Z$.Am I right?Thank you! $\endgroup$ – Jiangnan Yu Apr 11 '12 at 13:13
  • $\begingroup$ That looks good to me :) There is one generator $a$, and one trivial relation $aa^{-1}$ $\endgroup$ – William Apr 11 '12 at 15:54

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