Homology of quotient of 3-sphere by identifying antipodal points on equator

Question

I'm trying to solve exercise 2.2.10 in Hatcher's Algebraic Topology:

Let $X$ be the quotient space of $S^{3}$ under the identifications $x\sim-x$ for $x$ in equator $S^2$. Compute the homology groups $H_i(X)$.

$H_i(X) = 0$ for $i > 3$. $H_0(Z)\cong\mathbb{Z}$. I think I found $H_3(X),H_2(X)$. I need help to see if my work is correct and to compute $H_1(X)$.

I know that $\mathbb{R}\textbf{P}^2$ has homology groups $\mathbb{Z}$ for $i=0$, $\mathbb{Z}_2$ for $i=1$ and $0$ otherwise. $\mathbb{R}\textbf{P}^3$ has homology groups $\mathbb{Z}$ for $i=0,3$, $\mathbb{Z}_2$ for $i=1$ and $0$ otherwise.

Let $A,B$ be the upper and lower hemispheres of $X$. $A=B=\mathbb{R}\textbf{P}^3$ and $A\cap B = \mathbb{R}\textbf{P}^2$. Add a small neighborhood of the intersection that deformation retracts to the intersection to each of $A,B$. Apply Mayer-Vietoris.

$$H_3(A\cap B)\to H_3(A)\oplus H_3(B)\to H_3(X)\to H_2(A\cap B)$$ gives $H_3(X)\cong\mathbb{Z}\oplus\mathbb{Z}$

$$H_2(A)\oplus H_2(B)\to H_2(X)\to H_1(A\cap B)$$ gives $H_2(X)\cong\mathbb{Z}_2$

For $H_1(X)$ I get this which doesn't help $$\mathbb{Z}_2\to\mathbb{Z}_2\oplus\mathbb{Z_2}\to H_1(X)\to\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}$$

This my work correct? How to compute $H_1(X)$?

Note: A simpler version of the exercise was asked on this site. I was able to solve that via cellular chains with no problem. Homology of some quotient of $S^2$

EDIT: As soon as I post this I get an idea for $H_1(X)$. Since $\mathbb{Z}_2\oplus\mathbb{Z_2}$ is finite and all nontrivial subgroups of $\mathbb{Z}$ are infinite, $H_1(X)$ has to be infinite cyclic. Is this correct?

Answer 1

Your $H_3$ calculation is correct because $H_2(A\cap B)=0$. The $H_2$ calculation is a little fishy. Here is the relevant fragment of the MV sequence: $$0\to H_2(X)\to H_1(A\cap B) \to H_1(A)\oplus H_1(B),$$ which turns into $$0\to H_2(X) \to \mathbb Z_2\to \mathbb Z_2\oplus\mathbb Z_2.$$ There is no reason to conclude that $H_2(X)=\mathbb Z_2$. Indeed, based on the topology, the map $\mathbb Z_2\to \mathbb Z_2\oplus\mathbb Z_2$ should be given by $x\mapsto (x,x)$, which is injective. Hence $H_2(X)=0$.

But now you have enough to figure out $H_1(X)$. This is easiest if you use reduced homology. Then, similarly to what you've written you get $$0\to \mathbb Z_2\to \mathbb Z_2\oplus\mathbb Z_2\to H_1(X)\to 0.$$ Thus $H_1(X)\cong (\mathbb Z_2\oplus\mathbb Z_2)/\langle (x,x)\rangle\cong \mathbb Z_2$.

An alternative approach is to use cellular homology. This is basically $\mathbb{RP}^2$ with two $3$-cells attached. The cellular chain complex is $$0\to \mathbb Z^2\overset{0}{\to} \mathbb Z\overset{\times 2}{\to}\mathbb Z\overset{0}{\to}\mathbb Z\to 0.$$ You can deduce these maps since you know what they are for $\mathbb{RP}^3$, and this is essentially $\mathbb{RP}^3$ with an additional $3$ cell attached in the same way. Taking the homology gives the same answer as Mayer-Vietoris.

Answer 2

Alternatively, view $S^3$ as the suspension of $S^2$; a quotient $q':S^2\times I\twoheadrightarrow S^3$. We also have a quotient $q'':S^3\twoheadrightarrow X$ where $X$ is our space with the antipodal points of our equatorial $S^2$ identified.

Overall we have a quotient map $q=q''\circ q':S^2\times I\twoheadrightarrow X$ which is a homeomorphism on $S^2\times(I\setminus\{0,1/2,1\})$.

Decompose along the lines of, say, $A:=q(S^2\times([0,1/3]\cup[2/3,1]))$ and $B:=q(S^2\times[1/4,3/4])$. Evidently $A\simeq S^0$, $B\simeq\Bbb RP^2$ and $A\cap B\simeq S^2\sqcup S^2$.

Then the reduced Mayer-Vietoris sequence is just this: $$\cdots\to0\to H_3(X)\to\Bbb Z^2\to0\to H_2(X)\to0\to\Bbb Z_2\to H_1(X)\to\Bbb Z\overset{1}{\to}\Bbb Z\to\widetilde{H_0}(X)\to0$$Making the calculation of $H_\ast(X)$ trivial. This method generalises to the similar spaces where one takes $S^n$ and identifies antipodal points on the equator $S^{n-1}$, for $n\ge1$.