On my professor's solutions for my last algebraic topology homework, he gets the following Mayer-Vietoris sequence for the figure eight space (the wedge of two circles):
$0\to H_{2}(X)\to 0\to \mathbb{Z}\oplus \mathbb {Z}\to H_{1}(X)\to \mathbb{Z}\overset{\varphi_{*}}{\to}\mathbb{Z}\oplus\mathbb{Z}\to\mathbb{Z}\to0$
I didn't have a problem getting that far, but then he appeals to the "rank theorem" for finitely generated abelian groups, which we proved on a previous homework. For a short exact sequence of finitely generated abelian groups:
$0\to A\to B\to C\to 0$
The ranks of the free parts of $A$ and $C$ sum to the rank of the free part of $B$. He appeals to this to claim that it forces $\varphi_{*}$ to be injective, but I don't see how this follows. (He doesn't have any explanation in between and my algebra is not great.) Could someone explain further how the rank theorem implies that $\varphi_{*}$ is injective?
It's because when you're given an exact sequence $$\cdots \to A_3 \stackrel{\psi}{\to} A_2 \to A_1 \to A_0 \to 0,$$ you get a short exact sequence $$0\to A_2/\psi(A_3) \to A_1 \to A_0 \to 0.$$ In your case, this gives $$0\to \mathbb{Z}/(\text{something})\to \mathbb{Z}\oplus \mathbb{Z}\to \mathbb{Z}\to 0.$$ The only way for this to happen is if the "something" is $0$.