Let $A$, $B$ and $C$ be abelian groups and let $$0\to A\overset{\varphi}{\longrightarrow} B\overset{\psi}{\longrightarrow} C\to 0$$ be a short exact sequence. Denoting by $T(X)$ the torsion subgroup of the abelian group $X$, I've already shown that $\varphi(T(A))\subseteq T(B)$ and $\psi(T(B))\subseteq T(C)$. Thus, we have a induced sequence $$0\to A/T(A)\overset{\bar{\varphi}}{\longrightarrow} B/T(B)\overset{\bar{\psi}}{\longrightarrow} C/T(C)\to 0,$$ where, of course, $\bar{\varphi}(a+T(A))=\varphi(a)+T(B)$ and $\bar{\psi}(b+T(B))=\psi(b)+T(C)$.
I'm wondering about exactness of this last sequence. I already know that $\bar{\varphi}$ is injective, $\bar{\psi}$ is surjective and $\bar{\psi}\circ\bar{\varphi}=0$.
Question: is it also true that $\ker\bar{\psi} \subseteq {\rm ran\,}\bar{\varphi}$ ? What if $A$, $B$ and $C$ are finitely generated ?
Thanks in advance!
I don't think so for your first question. Take $$0 \to \mathbb Z \to_{\times 2} \mathbb Z \to \mathbb Z/2 \to 0.$$
Then the induced sequence is $$0 \to \mathbb Z \to \mathbb Z \to 0 \to 0$$
but the first map was $\times 2$ which is not surjective, so this sequence cannot be exact.