Let $A$ be a ring, $S \subset A$ a multiplicative set and $M$ an $A$-module.
Tha natural map $M \rightarrow S^{-1}M$, $m \mapsto \frac{m}{1}$, is bijective iff for all $s \in S$ the map $M \rightarrow M$, $m \mapsto s \cdot m$ is bijective.
Any help is highly appreciated!
Suppose the condition holds, then $ m/1=0 \Rightarrow$ there exists $s\in S $ such that $sm=0 \Rightarrow m=0$ now for any $m/s \in S^{-1}M$, $m/s=st/s=t/1$ for some $t\in M$.
Suppose the natural map $i$ is an isomorphism. For any $s\in S$, $sm=0 \Rightarrow sm/s=0=m/1 $ in $S^{-1}M$ so, $m=0$. For $n\in M$ we have $sn/s=n/1= i(n) $.
Now $n/s=i(m) $ for some $m\in M$ , so $i(sm) =si(m) =i(n) \Rightarrow sm=n$.