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

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.

• I think I found my error. It is not that $\sigma$ is $2\tau$, but $\sigma$ is $\pm\tau$ May 2 at 19:39

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 $$\mathbb{Z}_2$$