I am trying to compute the first homology group of the Klein bottle $K$ using its descomposition in two Mobius bands $M_1,M_2$ by using the reduced Mayer-Vietoris sequence: $$\dots \longrightarrow \tilde H_1(M_1\cap M_2) \stackrel{\Phi}{\longrightarrow}\tilde H_1(M_1)\oplus \tilde H_1(M_2)\longrightarrow\tilde H_1(K)\longrightarrow 0 \longrightarrow \dots$$ The sequence is exact, so the first homology group of the Klein bottle must be given by: $$\tilde H_1(K)\cong \frac{\tilde H_1(M_1)\oplus \tilde H_1(M_2)}{\textrm{im}\,\Phi}.$$ Lets compute the image of $\Phi$. Denote by $b,a_1,a_2$ the generators of $\tilde H_1(M_1\cap M_2)$, $\tilde H_1(M_1)$, $\tilde H_1(M_2)$ respectively and recall that $\Phi$ is given by the induced maps $i_1^\star,i_2^\star$ of the inclusions $i_1:M_1\cap M_2\hookrightarrow M_1$ and $i_2:M_1\cap M_2\hookrightarrow M_2$ as follows: $$\Phi(b)=\big(i_1^\star(b),i_2^\star(b)\big).$$ It is clear that 1-degree loops in $M_1\cap M_2$ are mapped to 2-degree loops in the Mobius band by the inclusions. This implies: $$\Phi(b)=(a_1^2,a_2^2),$$ Thus $\textrm{im}\, (\Phi)\cong \langle a_1^2\rangle \oplus \langle a_2^2\rangle$. From here, I dont know how to compute the quotient: $$\tilde H_1(K)\cong \frac{\tilde H_1(M_1)\oplus \tilde H_1(M_2)}{\textrm{im}\,\Phi}\cong \frac{\langle a_1\rangle \oplus \langle a_2\rangle}{ \langle a_1^2\rangle \oplus \langle a_2^2\rangle}$$ This is probabily a very basic fact of group theory.
Any help would be appreciated!