Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
The fundamental group should be isomorphic to $\mathbb{Z}$. –  Kerry Apr 11 '12 at 5:34
    
One can also show that $X \cong S^2 \vee S^1.$ –  Ehsan M. Kermani Apr 11 '12 at 5:43

1 Answer 1

up vote 3 down vote accepted

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 $1,2,3$

share|improve this answer
    
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! –  Jiangnan Yu Apr 11 '12 at 13:13
    
That looks good to me :) There is one generator $a$, and one trivial relation $aa^{-1}$ –  you Apr 11 '12 at 15:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.