Suppose we have an $R$-module $M$. Let $T(M)$ denote the torsion of $M$, i.e. the set of all nonzero $x \in M$ such $ax =0$ for some nonzero $a \in R$. I have proved that if $0 \to M' \to M \to M''$ is exact, then $0 \to T(M') \to T(M) \to T(M'')$ is exact.
Similarly, I have proved that for any multiplicatively closed subset $S$ of $R$, $0 \to S^{-1}M'\to S^{-1}M\to S^{-1}M''$ is exact if $0 \to M' \to M \to M''$ is exact.
We therefore have that if $0 \to M' \to M \to M''$ is an exact sequence, then $$0 \to T(S^{-1}M') \to T(S^{-1}M) \to T(S^{-1}M'')$$ is exact, and so is $$0 \to S^{-1}T(M') \to S^{-1}T(M) \to S^{-1}T(M'')$$
Is there a way to conclude that $T(S^{-1}M) \cong S^{-1}T(M)$?
While this question did arise from Question 3.13 of Atiyah-Macdonald, I'm not looking for a solution to this problem. This is simply a question relating to flat functors and exact sequences.