Computing singular homology groups of quotient space I want to compute the homology groups of $X$, the quotient of $S^2 \times S^1$ by the relation $(x,z) \sim (-x,-z)$. I've already computed the homology groups of $S^2 \times S^1$ using Mayer-Vietoris (partitioning the sphere in two fattened hemispheres).
First, I've tried to see what $X$ is homeomorphic to without success. Then, I divided the sphere in a different way (I'll attach a picture ASAP), and showed that the deformation retracts induce deformation retracts in the quotient, and proceeded as in with $S^2 \times S^1$, but this time with different groups given by the deformation retracts (2 torus and 1 circle, instead of 2 circles and 1 torus). I got $H_0(X)=H_1(X)=\mathbb{Z}$.
I can't get any further, and I suspect I'm missing something that might make it a lot easier.
 A: Method 1: Group action
Note that $X=(S^2\times S^1)/\Bbb Z_2$, where the action of $\Bbb Z_2$ is generated by $(x,z)\mapsto (-x,-z)$. We can also rewrite $X$ as a quotient space of $S^2\times\Bbb R$ using the fact that $S^1\approx \Bbb R/\Bbb Z$, then the equivalence relations are $(x,y)\sim(x,y+n)$ for $n\in\Bbb Z$ (given by the circle) and $(x,y)\sim (-x,y+1/2)$. Combining these two pieces, we see that the equivalence relation is defined by $(x,y)\sim((-1)^nx,y+n/2)$. Therefore, $X=(S^2\times\Bbb R)/\Bbb Z$, where the action of $\Bbb Z$ is generated by $(x,y)\mapsto(-x,y+1/2)$.
Note that this group action is free, so we have $\pi_1(X)\cong\Bbb Z$. According to Hurewicz, $H_1(X;\Bbb Z)\cong \Bbb Z$ as well. Since this $\Bbb Z_2$-action on $S^2\times S^1$ is orientation reversing, $X$ is non-orientable, so $H_3(X;\Bbb Z)\cong 0$ and the torsion of $H_2(X;\Bbb Z)$ is $\Bbb Z_2$.
To determine the rank of $H_2$, we use the fact that the quotient map $q:S^2\times S^1\to X$ is a double cover, so $2\chi(X)=\chi(S^2\times S^1)=0\implies \chi(X)=0$. On the other hand,
$$\chi(X)=b_0-b_1+b_2-b_3=1-1+b_2-0=b_2$$
where $b_i$'s are betti numbers. This shwos that the rank of $H_2$ is $0$.
In conclusion,
$$H_k(X;\Bbb Z)\cong\begin{cases}\Bbb Z &  k=0,1\\ \Bbb Z_2&k=2\\ 0 &\text{ otherwise}\end{cases}$$

Method 2: Cellular homology
The cell structure of $S^2$:

*

*two $0$-cells $e_N^0, e_S^0$ (one at the north pole and another one at the south pole);

*two $1$-cells $e_\alpha^1$ (oriented from the south pole to north pole)  $ e_\beta^1$ (oriented from the north pole to the south pole);

*two $2$-cells $e_\alpha^2$, $e_\beta^2$.

The cell structure of $S^1$ is constructed in the same way as $S^2$.

*

*two $0$-cells $f_N^0, f_S^0$;

*two $1$-cells $f_\alpha^1, f_\beta^1$.

The reason to choose such non-standard cellular decomposition with 2-fold symmetry is that we want to obtain another cell structure when we descends to the quotient space $X$.
The cell structure $S^2\times S^1$ should be obvious from the construction above, i.e., four $3$-cells, eight $2$-cells, eight $1$-cells, and four $0$-cells. Now, consider the equivalence relation $(x,z)\sim (-x,-z)$, we see that cells are collapsed in pairs. For instance, $e_\alpha^1\times f_N^0$ is identified with $e_\beta^1 \times f_S^0$, so we have the following cellular chain complex for $X$.
$$0\to\Bbb Z^2\overset{\partial_3}{\to}\Bbb Z^4\overset{\partial_2}{\to}\Bbb Z^4\overset{\partial_1}\to\Bbb Z^2\to 0$$
Consider the one $e_\alpha^1\times f_N^0$, its image under the boundary map is $\partial_1(e_\alpha^1)\times f_N^0=(e_N^1-e_S^1)\times f_N^0$. We don't have to consider its "partner" $e_\beta^1\times f_S^0$ because they are identified in the quotient space. Similarly, one can argue that $\partial_1(e_\beta^1\times f_N^0)=(e_S^0-e_N^0)\times f_N^0=-\partial_1(e_\alpha^1\times f_N^0)$ and that $\partial_1(e_N^0\times f_\alpha^1)=e_N^0\times(f_N^0-f_S^0)=-\partial_1(e_N^0\times f_\beta^0)$. This shows that

*

*$\ker(\partial_1)=\langle(e_\alpha^1+e_\beta^1)\times f_N^0, e_N^0\times (f_\alpha^1+f_\beta^1),e_N^0\times f_\alpha^1+e_\beta^1\times f_N^0\rangle$

*$\operatorname{im}(\partial_1)=\langle(e_N^0-e_S^0)\times f_N^0\rangle$. (compute $e_\alpha^1\times f_N^0, e_\beta^1\times f_N^0, e_N^0\times f_\alpha^1, e_N^0\times f_\beta^1$)

Similar arguments can be repeated for $\partial_2$ and $\partial_3$ using the formula of cellular boundary map. Note that we need to refer back to the equivalence relation to see the relationship between $\ker\partial_{k}$ and $\operatorname{im}\partial_{k+1}$.

*

*$\ker(\partial_2)=\langle e_\alpha^2\times (f_N^0-f_S^0), (e_\alpha^1+e_\beta^1)\times f_\alpha^1\rangle$; (compute $e_\alpha^2\times f_N^0, e_\alpha^2\times f_S^0, e_\alpha^1\times f_\alpha^1, e_\beta^1\times f_\alpha^1$)

*$\operatorname{im}(\partial_2)=\langle(e_\alpha^1+e_\beta^1)\times f_N^0, (e_N^0-e_S^0)\times f_\alpha^1+e_\alpha^1\times (f_N^0-f_S^0)\rangle$;

*$\ker(\partial_3)=0$;

*$\operatorname{im}(\partial_3)=\langle (e_\alpha^1+e_\beta^1)\times f_\alpha^1+e_\alpha^2\times(f_N^0-f_S^0),(e_\alpha^1+e_\beta^1)\times f_\alpha^1+e_\alpha^2\times(f_S^0-f_N^0)\rangle$ (compute the image of $e_\alpha^2\times f_\alpha^1$ and $e_\alpha^2\times f_\beta^1$)

If we compute the quotient groups carefully, then we actually get the same answer as using method 1.
