Apparently this question has been asked many times on here, but my question is more specific.
Let $X$ be the quotient space of $S^2$ under the identifications $x \sim -x$ for $x$ in the equator $S^1$. We are asked to compute the homology groups $H_i(X)$.
The method I chose is to use cellular homology. Give $X$ the following cell structure: one 0-cell $e^0$, one 1-cell $e^1$, and two 2-cells $e_1^2$ and $e_2^2$. The 2-cells are attached via the quotient projection of the boundary $x \sim -x$. In particular, the 0 and 1 skeletons are $\mathbb{R}P^0$ and $\mathbb{R}P^1$ respectively and the 2-cells are attached in the same way as in Hatcher's computation of the homology of $\mathbb{R}P^n$ using cellular homology.
The nontrivial part of cellular chain complex is
$0 \rightarrow \mathbb{Z}^2 \xrightarrow[]{d_2} \mathbb{Z} \xrightarrow[]{0} \mathbb{Z} \rightarrow 0$.
I don't understand why $d_2$ is not multiplication by $2$ here since they are attached in the same way as in Hatcher's Example 2.42. According to another answer $d_2(e_1^2)=2e^1$ and $d_2(e_2^2)=-2e^1$ "by the cellular boundary formula". I do not see why this is the case since $e_1^2$ and $e_2^2$ have the same attaching map.