Injectivity in the zero homology I'm struggling with following step in an excercises about Mayer-Vietoris sequences:
In one step the solution says this map is injective since $A \cap B$ is path-connected:
$$ H_0(A \cap B) \stackrel{i_*^{A} \oplus i_*^{B}} \longrightarrow H_0(A) \oplus H_0(B)$$
What I've been thinking is that as $A \cap B$ is path-connected we get $H_0(A \cap B) \simeq \mathbb Z[\alpha]$ with $\alpha$ our prefered generator. My idea then was that this map is injective since we map just multiples of the generator to each summand in the target and so the kernel must be trivial. But I'm not quite sure about this argument.
I'd appreciate if someone could elaborate on my idea and tell me wether it works or not and if it does, how one may formulate it better in order for me to believe my own words.
 A: I strongly disagree with you saying that it is a proof, in fact it is not! The map $H_0(A\cap B)\rightarrow H_0(A)\oplus H_0(B)$ takes $n[\alpha]\mapsto(n[i_A\circ\alpha],n[i_B\circ\alpha])$. Since every $H_0$ is free, in particular torsion free, it suffices to show that $[i_A\circ\alpha]$ is nonzero, the argument being similar for $[i_B\circ\alpha]$. Write $\alpha=\sum_{x\in A\cap B} n_x x$; then $i_A\circ\alpha=\sum_{x\in A\cap B} n_x x$. Assume by contradiction that this is $\partial\beta=\sum_{i\in I} m_i(\sigma_i(1)-\sigma_i(0))$ for a $1$-simplex $\beta=\sum_{i\in I} m_i \sigma_i$ in $A$. Then we still have $\alpha=\partial\tilde{\beta}$, where $\tilde{\beta}$ is the $1$-simplex obtained from $\beta$ by dropping all $\sigma_i$ that have $\sigma_i(1)\notin A\cap B$ or $\sigma_i(0)\notin A\cap B$ and all $\sigma_i$ that are loops. But the we may as well take the $\sigma_i$ to be in $A\cap B$, as $A\cap B$ is path-connected. This shows that $\alpha$ is a $1$-boundary, contradicting its definition.
