Question about homology groups of $\mathbb{S}^2$ with antipodes identified on the equator

Question

I'm doing Hatcher's exercise 2.2.10 that asks to calculate the homology groups for $\mathbb{S}^2$ with antipodes identified on the equator $\mathbb{S}^1$. By using Mayer-Vietoris on the sets $A=q(\mathbb{S}^2\setminus N)$, $B=q(\mathbb{S}^2\setminus S)$, where $N$ and $S$ are the north and south poles, respectively, and $q:\mathbb{S}^2\rightarrow X$ is the quotient map. Now, $A$ and $B$ deformation retracts to $q(\mathbb{S}^2_-)\cong\mathbb{RP}^2\cong q(\mathbb{S}^2_+)$ (whre $\mathbb{S}^2_\pm $ are the north and south pole, with the equator) respectively, and $A\cap B$ to $q(\mathbb{S}^1)\cong\mathbb{S}^1$. Thus Mayer-Vietories on the reduced homologies gives us the exact sequence:

$$ 0\rightarrow H_2(X)\rightarrow H_1(\mathbb{S}^1)\rightarrow H_1(\mathbb{RP}^2)\oplus H_1(\mathbb{RP}^2) \rightarrow H_1(X)\rightarrow 0 $$

If $\phi_* : H_1(\mathbb{S^1})\rightarrow H_1(\mathbb{RP}^2)\oplus H_1(\mathbb{RP}^2)$, $\phi_* :[\sigma]=([\sigma],[-\sigma])$,is the middle map, we should have that $H_1(X)\cong (H_1(\mathbb{RP}^2)\oplus H_1(\mathbb{RP}^2))/\text{im}\phi_* $. Now, we can pick generator $H_1(\mathbb{S^1})=<[\sigma]>$ such that in $[\sigma]=\pm 2[\tau]=0 $ in each of the $H_1(\mathbb{RP}^2)=<\pm[\tau] \|2[\tau]=0 > $. Hence we would have that $\text{im}\phi_*=0$ therefore $H_1(X)\cong\mathbb{Z}_2\oplus\mathbb{Z}_2 $. However, when I went to check my answer on mathexchange, most answers were saying that $H_1(X)\cong\mathbb{Z}_2$.

I'm trying to verify if I did something wrong, but I'm not finding the mistake. Most of the answers don't use the reduced homology, but the "normal" homology, which we would have an extra $\mathbb{Z}\rightarrow\mathbb{Z}$ at the end of the above chain. But this shouldn't change the result. If anyone could tell if and where I'm making the mistake, I would be greateful.

Answer 1

I found the error. When I took the generator $H_1(\mathbb{S}^1/~)=<[\sigma]>$ I was actually looking the generator of $\mathbb{S}^2$ without the quotient. When we quotient by the antipode, actually we have that the generator of $H_1(\mathbb{S}^1/~)$ is $\pm$ the generator of each $H_1(\mathbb{RP^2})$ - the reason is that a loop on $\mathbb{S}^1$ actually goes two times around $\mathbb{S}^1/~$. So actually we have: $\phi_*[\sigma]=([\sigma]_+,-[\sigma]_-)$, where the $\pm$ sign is just to distinguish on which $H_1(\mathbb{RP}^2)$ is it from. Hence we are killing a generator from the direct sum, arriving at $\mathbb{Z}_2$