what is the homology group for The quotient space of $S^1 \times S^1$ obtained by identifying points in the circle $S1 \times\{x_0\} $
that differ by $\frac{2 \pi}{m}$ rotation and identifying points in the circle $\{x_0\} \times S^1 $ that
differ by $\frac{2 \pi}{n}$ rotation.
actually it is exercise 2.2.9.d from hatcher and I am really curious about its homology groups.
it will be awesome if you can show me an imagination of this space.and I really don't know what tools should I use to calculate homology groups,cellular one,simplicial one,mayer-vietoris…,please help me,it will be great if you give me guidance or hint,thank you very much.
Best Answer
Although you've got the answer by yourself, I would like to write an answer solving the problem with cellular homology, so that someone who asks the same question can find an answer here. I solved this problem a few month ago in an Algebraic Topology course as an exercise.
Proof:
Let $X = S^1\times S^1/ \sim$ be the space with the identifications:
$$(e^{2\pi i/m}z,x_0)\sim (z,x_0)$$
$$(x_0,e^{2\pi i/n}z)\sim (x_0,z)$$
Like @Berci said, you should imagine this space as a grid of $m$ and $n$ lines, i.e. there are $m$ vertical and $n$ horizontal repititions:
(OK. The picture is not the nicest one, but it's enough to induce an imagination.)
$X$ consists of one 0-cell ($x_0$ is $e_1^0$), two 1-cells ($a$ is $e_1^1$, $b$ is $e_2^1$) and one 2-cell (we all it $e_1^2$).
The attaching map identifies $x\in \partial D_1^2$ with $a^nb^ma^{-n}b^{-m}$.
This implies the cellular chain complex
$$0\to \mathbb{Z}[e_1^2]\overset{\partial_2=0}{\longrightarrow} \mathbb{Z}[a] \oplus\mathbb{Z}[b]\overset{\partial_1=0}{\longrightarrow} \mathbb{Z}[x_0]\to 0.$$
This implies
$$H_p(X) = \begin{cases} \mathbb{Z}\mbox{ for } p=0,2 \\ \mathbb{Z}^2\mbox{ for } p=1 \\ 0\mbox{ for } p>2 \end{cases}.$$
Otherwise you can just see, that the space $X$ is still a Torus (cf. remark above). So it is not surprising, that we've got the homology group of the Torus.