Let $A$ be a ring and $S\subset A$ be a multiplicatively closed subset. Let $M$ be an $A$-module and let $\iota:M\to S^{-1}M$ be the natural map given by $m\mapsto m/1$. let $N'$ be a $S^{-1}A$-submodule of $S^{-1}M$, and let $N:=\iota^{-1}N'$ be the preimage of $N$ under $\iota$. Show that $S^{-1}N=N'$.
I want to show that $\forall n/s\in S^{-1}N$, we also have $n/s\in N'$. Then for some $n'/s'\equiv n/s$, $n'/1\in N'$. But I don't know how to show that $n/s\in N'$.
