Compute the homology groups $H_* \mathbb RP^2$ using Mayer-Vietoris sequence.
I know that $\mathbb R P^2$ is obtained from the Möbius strip by attaching a $2$-cell along the boundary circle of the Möbius strip i.e. $\mathbb R P^2 = M \cup_{\partial} D^2$ and $M \cap D^2 \approx S^1.$ In order to use Mayer-Vietoris sequence I first need to cover $\mathbb R P^2$ by the interior of two of it's subspaces. I think one of the subspaces is $M \simeq S^1$ and the other subspaces is a slightly larger disk which is obtained from $D^2$ by attaching a small annular region along the boundary. Let $\mathscr D$ be the disk obtained in this way. Then clearly the interior of $M$ and $\mathscr D$ covers $\mathbb R P^2$ and $M \cap \mathscr D \simeq S^1.$ By using Mayer-Vietoris sequence it is easy to see that $H_n (\mathbb R P^2) = 0,$ for $n \geq 3.$ For $n \leq 2$ Mayer-Vietoris sequence yields
$$H_{2}(\mathscr D) \oplus H_{2}(S^{1}) \longrightarrow H_{2}(\mathbb{R}P^{2})\longrightarrow H_{1}(S^{1}) \longrightarrow H_{1}(\mathscr D) \oplus H_{1}(S^{1}) \longrightarrow H_{1}(\mathbb{R}P^{2}) \longrightarrow H_{0}(S^{1}) \longrightarrow H_{0}(\mathscr D) \oplus H_{0}(S^{1}) \longrightarrow H_{0}(\mathbb{R}P^{2})\longrightarrow 0.$$
Which further simplifies to
$$0 \longrightarrow H_{2}(\mathbb{R}P^{2})\longrightarrow \mathbb{Z} \xrightarrow {\varphi} \mathbb{Z} \longrightarrow H_{1}(\mathbb{R}P^{2}) \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z} \oplus \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow 0.$$
After some observation I realized that in order to understand what happens to $H_2 (\mathbb R P^2)$ we need to analyze the map $\varphi.$ To understand this we need to know how do we include the annular region to $M.$ But I find it difficult to figure that out. Could anyone please help me in this regard?
Thanks!