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Given a PID $A$ and $A$-modules $M$, $M'$, and $M''$. Assume that

$$0\rightarrow M'\xrightarrow{f} M\xrightarrow{g} M''$$ is an exact sequence then prove that $$0\rightarrow TM'\xrightarrow{\bar{f}} TM\xrightarrow{\bar{g}} TM''$$ is exact.

Note: $TM$ is the torsion module of $M$. How to prove this? I proved so far: $g\circ f=0 \Rightarrow \bar{g}\circ \bar{f}=0$. So $Im(\bar{f})\subset ker(\bar{g})$. Let $a\in Ker(\bar{g})$. How to prove that $a\in Im(\bar{f})$?

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$Ker(\bar{g}) \subset Ker(g)$, we have that $\bar{a} \in Ker(g)$. So $\bar{a} \in Im(f)$ and so $\bar{a} = f(a')$ for some $a' \in M'$. Since $\bar{a} \in TM$, there exists some $c \in A$ such that $c \bar{a} = 0$. Hence $0=cf(a')=f(c a')$. But $f$ is injective and so $c a' =0$. This is in turn shows that $a' \in TM'$ and so $\bar{a} \in Im(\bar{f})$.

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