Let $f : A \longrightarrow B$ be a ring homomorphism. Let $I \subseteq A$ be an ideal of $A$ and $J \subseteq B$ be an ideal of $B$. By definition, the extension of $I$ is: $$ I^{e} = \{ \sum_{i = 1}^{n} a_{i}f(x_{i}) : a_{i} \in B,\; x_{i} \in I,\; \forall n \in \mathbb{N} \} $$ and the contraction of $J$ is $f^{-1}(J) = J^{c}$.
The following property holds: $(J^{c})^{e} \subseteq J$.
My Proof: Suppose $J \subsetneq (J^{c})^{e} $. Let $x \in J$. Then $f^{-1}(x) \in f^{-1}(J) = J^{c}$. Now, \begin{align*} (f^{-1}(J))^{e} = (J^{c})^{e} &= \{ \sum_{i = 1}^{n} a_{i}f(f^{-1}(x_{i})) : a_{i} \in B,\; f^{-1}(x_{i}) \in J^{c},\; \forall n \in \mathbb{N}\}\\ &= \{\sum_{i = 1}^{n} a_{i}x_{i} : a_{i} \in B,\; x_{i} \in J,\; \forall n \in \mathbb{N}\}\\ &= J, \end{align*} contradiction to our assumption that $J \subsetneq (J^{c})^{e} $. Hence, $J \supseteq (J^{c})^{e}$.
Can somebody confirm either or not the proof is correct?