In some lecture notes on commutative algebra I'm reading, one defines for a multiplicative subset $S\subseteq A$ the canonical map $f:A\to S^{-1}A$, $a\mapsto \tfrac{a}{1}$, then for an ideal $I$ of $A$, the extended ideal $I^e$ is the ideal of $S^{-1}A$ generated by $f(I)$, and for an ideal $J$ of $S^{-1}A$, the contracted ideal $J^c$ is $f^{-1}(J)$.
I can see that for any ideal $J$ of $S^{-1}A$ we have $(J^c)^e=J^{ce}= J$, so every ideal of $S^{-1}A$ is an extended ideal, and that for an ideal $I$ of $A$, $I^e=S^{-1}I$.
Now they claim that $I$ is a contracted ideal, i.e. there is some ideal $J$ of $S^{-1}A$ such that $I=J^c$, if and only if $(I^e)^c=I^{ec}\subseteq I$.
I can prove one implication. If $I=J^c$, then $I^e=J^{ce}=J$, so $I^{ec}=J^c=I$. But if $I^{ec}\subseteq I$, I can't see how to show that $I=J^c$ for some $J$.
Note that in general $I^{ec}\supseteq I$, so the condition $I^{ec}\subseteq I$ is equivalent to $I^{ec}=I$.
Then you can just take $J=I^{e}$.