# Is this sequence exact?

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!

## 1 Answer

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.