Homology of $\mathbb{R}P^2$ using cellular homology - verification. As in the title, I want to calculate the homology of $\mathbb{R}P^2$ using cellular approach. $\mathbb{R}P^2$ has a cell structure with one $0$-cell $x_0$, one $1$-cell $a$ and one $2$-cell $e$, The cellular chain complex is then of the form: 
\begin{equation}
0 \rightarrow \mathbb{Z} \xrightarrow[{}]{d_2}\mathbb{Z} \xrightarrow[{}]{d_1}\mathbb{Z} \rightarrow 0.
\end{equation}
The boundary map $d_1$ is zero: we have the composition
\begin{equation}
S^0 \xrightarrow[{}]{\gamma} \{x_0\} \rightarrow \{x_0\}/\emptyset \xrightarrow[{}]{=} S^{0},
\end{equation}
and the degree of this composition is $0$. Then $d_1$ is trivial, as claimed. For $d_2$, we consider:
\begin{equation}
S^1 \xrightarrow[{}]{\gamma} S^1/\sim=\mathbb{RP^1} \xrightarrow[{}]{q} \mathbb{RP^1}/\mathbb{RP^0} \cong S^{1}.
\end{equation}
The attaching map $\gamma$ is given by $a^2$ and so the composition above has degree $2$. We then end up with:
\begin{equation}
0 \rightarrow \mathbb{Z} \xrightarrow[{}]{2}\mathbb{Z} \xrightarrow[{}]{0}\mathbb{Z} \rightarrow 0.
\end{equation}
Now, $H_2 \cong 0$ since $\times 2$ map is injective in $\mathbb{Z}$ and so its kernel is zero. Putting all the above together, we have that $H_0 \cong \mathbb{Z}, H_1 \cong \mathbb{Z}/2\mathbb{Z}$ and $H_i \cong 0$ for $i > 1$. Is that a correct argument?
 A: To finish off the comments...
Everything is good up through the cellular chain complex, given below, which is only nonzero in degrees zero through two: 
$\cdots \to 0 \to \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \to 0$
Then you had the right answer, and a comment I made caused a bit of confusion which led me to point out something important: the cellular chain complex is (in general) not exact, and computing its homology gives you the homology of your space. So everything you added in your edit starting with "The exact sequence of homology groups is..." is not how you'll want to finish it off (my comment had meant something more like, "finish it off by adding a sentence for how you can tell $H_2 \cong 0$ when you're reading off the homology of your chain complex").
Edit: The OP got edited so this above paragraph is more or less irrelevant.
Anyways, now let's compute the homology of $\mathbb{R}P^2$, which amounts to the homology of the above chain complex. Label the maps $\delta_1$ and $\delta_2$. Then $H_0(\mathbb{R}P^2) \cong \mathbb{Z}/\text{im }\delta_1 \cong \mathbb{Z}/0 \cong \mathbb{Z}$.
$H_1(\mathbb{R}P^2) \cong \ker\delta_1/\text{im }\delta_2$. Since $\delta_1$ is zero its kernel is all of $\mathbb{Z}$, and since $\delta_2$ is multiplication by $2$ its image is $2\mathbb{Z}$, so $H_1(\mathbb{R}P^2) \cong \mathbb{Z}/2\mathbb{Z}$.
$H_2(\mathbb{R}P^2) \cong \ker\delta_2/\text{im }0 \cong \ker \delta_2$. Since $\delta_2$ is multiplication by $2$, it is injective ($\mathbb{Z}$ is an integral domain), so $\ker \delta_2 = 0$. Thus $H_2(\mathbb{R}P^2) \cong 0$.
Since the cellular chain complex consists of zeros above degree $2$, $H_i(\mathbb{R}P^2) \cong 0$ for $i > 2$ as well.
