Ideal is Contained in Contraction of Extension

Theorem
Let $A$ and $B$ be commutative rings with unity.

Let $f : A \to B$ be a ring homomorphism.

Let $\mathfrak a \subseteq A$ be an ideal.

Then $\mathfrak a$ is contained in the contraction of its extension by $f$:
 * $\mathfrak a \subseteq \mathfrak a^{ec}$

Proof
By definition of extension and generated ideal:
 * $f \sqbrk {\mathfrak a} \subseteq \mathfrak a^e$

By Subset of Domain is Subset of Preimage of Image:
 * $\mathfrak a \subseteq f^{-1} \sqbrk {f \sqbrk {\mathfrak a}}$

By Preimage Preserves Inclusion of Subsets:
 * $\mathfrak a \subseteq f^{-1} \sqbrk {f \sqbrk {\mathfrak a }} \subseteq f^{-1} \sqbrk {\mathfrak a^e}$