Consider the Mayer-Vietoris sequence for $\mathbb{RP}^2$, where the two open sets are $U:= \{ [x;y;z] \in \mathbb{RP}^2 | z \neq 0 \}$ and $V = \mathbb{RP}^2 \setminus [0;0;1]$. I've proved that $U \sim *$, $V$ and $U \cap V$ are homotopically equivalent to $S^1$.
I want some indications about how formalising this passage, because I need it to be as formal as possible:
$$ H_2(S^1) \rightarrow H_2(*)\oplus H_2(S^1) \rightarrow H_2(\mathbb{RP}^2) \rightarrow H_1(S^1) \rightarrow H_1(*)\oplus H_1(S^1) \rightarrow $$ which boils down to $$ 0 \rightarrow 0\oplus 0 \rightarrow H_2(\mathbb{RP}^2) \overset{\delta}{\rightarrow} \mathbb{Z} \overset{i_{1*},-i_{2*}}{\longrightarrow} 0\oplus \mathbb{Z} \rightarrow $$ where $i_{1*}$ is the map induced by the inclusion $U \cap V \in U$ and $i_{1*}$ by the one in $V$. Thanks to the exactness $H_2(\mathbb{RP}^2) \cong Im \delta \cong \ker(i_{1*},-i_{2*})$. The first $i_{1*}$ is the trivial map, so I need to check the second one. My professor made this sketch to explain why it is injective
where $c$ is the generator of $H_1(S^1) \cong H_1 (U \cap V)$. the letter $a$ represent the identification on the border of the disk. He said that using the simplices $\sigma_1,\sigma_2$ it's immediate that $[2a]=[c]$ and so the map is injective.
I want to write down a formal explanation of this fact, without the necessity of a picture, and I want to be sure why the map is injective. Is it because the image of the generator is a generator of the second? How can I prove this?
Thanks in advance!
You are asking to compute the map $i_2 : \mathbb{Z} = H_1(U \cap V) \to H_1(V) = \mathbb{Z}$ induced by the inclusion $U \cap V \subset V$. As you say, both of the spaces $U \cap V$ and $V$ are homotopy equivalent to circles. You can use this to good effect if you can prove a little bit more, namely, there exist homotopy equivalences $f : U \cap V \to S^1$ and $g : V \to S^1$ such that the composition $$U \cap V \subset V \xrightarrow{g} S^1 $$ and the composition $$U \cap V \xrightarrow{f} S^1 \xrightarrow{z \to z^2} S^1 $$ are homotopic; in other words, you are getting a homotopy commutative square.
I'm sure you can formally compute the map on homology $H_1(S^1) \mapsto H_1(S^1)$ that is induced by the map $z \mapsto z^2$, and that will give you the map on homology that is induced by the inclusion map $U \cap V \subset V$.