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The exercise is the following :

Compute the Homology of the quotient space of $S^1 \times S^1$ obtained by identifying points in the circle $S^1\times \{x_0\}$ that differ by $2\pi/m$ rotation and identifying points in the circle $\{x_0\}\times S^1$ that differ by $2\pi/n$ rotation.

Now I thought I had come up with a solution for this exercise, but after looking for the answer online I found my result was wrong however I don't understand why.

My idea was to consider $S^1\times S^1$ and the sets $A_n$ and $A_m$ which are the points that differ by $2\pi/n$ and $2\pi/m$ respectively.

Now if we consider $\tilde X:=(S^1\times S^1)/ A_m$, then our desired space will be $X=\tilde X/A_n$.

Now first one needs to compute the homology of $\tilde X$, but since $A_m\rightarrow S^1\times S^1$ is a cofibration we have that $H_k(S^1\times S^1, A_m)\cong \tilde H(\tilde X)$. Using exact sequence of homology relative to a pair we find out a short exact sequence of the form

$0\rightarrow \mathbb{Z}\bigoplus \mathbb{Z}\rightarrow H_1(\tilde X)\rightarrow \mathbb{Z}^{m-1}\rightarrow 0$

which is a split short exact sequence , and so $H_1(\tilde X)\cong \mathbb{Z}^{m+1}$. Now doing something analogous for $\tilde X$ and $A_n$ I got that $H_1(X)=\mathbb{Z}^{n+m}$. However this result does not agree with what I found , but I am not sure what is wrong with me reasoning. Any help is appreciated. Thanks in advance.

Someone
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    You're not collapsing a subspace to a point, you're identifying points in a space. – John Palmieri Jun 19 '21 at 21:51
  • I am confusing something , since if we identify those $m$ points together say , the ones that are $2\pi/m$ in $S^1\times {x_0}$ apart, how is it different from doing $(S^1\times S^1)/A_m$ @JohnPalmieri – Someone Jun 19 '21 at 21:59
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    You should be doing infinitely many identifications, not just $m$ of them. Write a point in $S^1$ as an angle $\theta$. Then you want to identify $\theta$ with $\theta + 2\pi/m$ for each $\theta$. – John Palmieri Jun 19 '21 at 22:05
  • Or another way of saying this: you want to glue the arc of $S^1$ with angles $[0, 2\pi/m)$ to the arc with angles $[2\pi/m, 4\pi/m)$, etc. – John Palmieri Jun 19 '21 at 22:06
  • Oh I see , I was confusing the space , I thought we just wanted those points. Alright thanks. @JohnPalmieri – Someone Jun 19 '21 at 22:06

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