Compute homology groups I'm not confident with this kind of problem, so I post my solution here to ask for checking it. I want to compute the homology groups of the space obtained from two copies of $\mathbb{R} P^2$ by
    gluing them along standard copies of $\mathbb{R} P^1$.
First I give it a cell structure: we know that $\mathbb{R}P^2$ has the cell structure of one $0$-cell $x$, one $1$-cell $a$ and one $2$-cell $A$ that glues to $2a$ (i.e. go around $a$ 2 times). So I believe the cell struture of our space is one $0$-cell $x$, one $1$-cell $a$ and two $2$-cells $A,B$ that both glue to $2a$. Hence we have the chain complex
$$0\to \mathbb{Z}^2 \xrightarrow{d_2} \mathbb{Z} \xrightarrow{d_1}\mathbb{Z} \to 0$$
where $d_1=0$ and $d_2(A)=d_2(B)=2a$. Hence $$H_0=\mathbb{Z}/0=\mathbb{Z}$$
$$H_1=\langle a\rangle /2\langle a \rangle = \mathbb{Z}_2$$
$$H_2=\langle A-B\rangle / 0 = \mathbb{Z}$$
Is this correct?
 A: What you wrote down looks right.
You can alternatively compute the homology groups by coming up with a homotopy equivalence to a simpler space. We'll use the following lemma (see for example this question for a weaker version):

Let $A, B, C$ be CW complexes, and let $f\colon A \to B$ and $g\colon A \to C$ be cellular maps. Then if $f'$ is homotopic to $f$ there is a homotopy equivalence $B\cup_{f, g} C\simeq B\cup_{f',g}C$.

Now define $D$ as the quotient $\mathbb{RP}^2 \cup_{\iota, \iota} \mathbb{RP}^2$ where $\iota\colon \mathbb{RP}^1\to \mathbb{RP}^2$ is the standard inclusion. This has a cell structure as you've described. We can think about it as starting with a copy of $\mathbb{RP}^2$ and attaching a $2$-cell via a map $\varphi\colon\partial D^2 \cong S^1\to \mathbb{RP}^1\cong S^1$ of degree $2$, i.e. $D$ is homeomorphic to $\mathbb{RP}^2\cup_{\varphi} D^2$. 
However, since $\pi_1(\mathbb{RP}^2)\cong \mathbb{Z}/2$ this attaching map is null-homotopic so in fact $D$ is homotopy-equivalent to $\mathbb{RP^2}\vee S^2$ by the above lemma (cf this related question), whose homology groups are easily computable via $\tilde{H}_k(A\vee B)\cong \tilde{H}_k(A) \oplus \tilde{H}_k(B)$.
