# 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.

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.

• Thank you for the detailed answer. After some thought I think I have another solution. If you notice that $S^1=[0,1], 0 \sim 1$, then $X$ is the mapping torus of $S^2$ with $f:S^2 \rightarrow S^2, f(x)=-x$ as the monodromy map. Now, by the Wang long exact sequence, after computing $f_*-I([\gamma])=-2[\gamma]$, we get $coker(f_*-I) \cong \mathbb{Z}_2 \cong H_2(X)$ and $ker(f_*-I) = 0 = H_3(X)$. Also, $H_1(X) \cong H_0(X) \cong \mathbb{Z}$ following arrows. May 19, 2022 at 23:28