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Let X be the 2 complex obtained from $S^{1}$ with its usual cell structure by attaching two 2 cells by maps of degrees 2 and 3 , respectively. (a) Compute the homology groups of all the subcomplexes $A ⊂ X $ and the corre- sponding quotient complexes X/A . (b) Show that X is homotopy equivalent to $S^{2}$ and that the only subcomplex $A ⊂ X$ for which the quotient map X →X/A is a homotopy equivalence is the trivial subcomplex, the 0 cell.

I have calculated the homologies and these are:

Case 1 : A is 1-skeleton ,$H_0(X/A)= Z $, $H_2(X/A)= Z\bigoplus Z$ and $0$ otherwise.

Case 2: For other non-trivial proper subcomplexes ,$H_i(X/A)= Z$ for $i=0,2$ and $0$ otherwise.

But I need some help for the second part of question.

Thanks!

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What is the usual cell structure on $S^1$? I know at least two which deserve that name! –  Mariano Suárez-Alvarez Dec 10 '12 at 22:59
1  
@MarianoSuárez-Alvarez The one with 1 $0$ - cell $x_0$ and $1$ 1-cell attached via the map $f : S^0 \to x_0$ (the constant map at a point). –  user38268 Dec 10 '12 at 23:02
    
Take it 1 0-cell and 1 1-cell. –  Shraddha Srivastava Dec 10 '12 at 23:02
    
I don't think your computations are correct. For example, if you contract the 1-skelenton (which is the $S^1$ you started with) the resulting space is two spheres attached at a point. –  Mariano Suárez-Alvarez Dec 10 '12 at 23:02
    
I have done it as follows X/A has cell structure 1 0-cell and 2 2-cells.Sorry,that was a typo and thanks to point out. –  Shraddha Srivastava Dec 10 '12 at 23:11

1 Answer 1

I know this a Hatcher HW problem, so I won't give away the answer. But Allen Hatcher himself thought this problem was too hard, and gives an extra hint on this page (you have to scroll down to see it):Hatcher Additional Exercises

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well,I appreciate that you think it is a homework problem but in actual it is not.Thanks for the hint. –  Shraddha Srivastava Dec 13 '12 at 9:56

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