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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
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
@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

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

share|cite|improve this answer
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

Yo can find an idea for the solution here:, page 26.

share|cite|improve this answer

Your Answer


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.