Subset of Preimage under Relation is Preimage of Subset/Corollary

Corollary to Subset of Preimage under Relation is Preimage of Subset
Let $f: S \to T$ be a mapping.

Let $X \subseteq S, Y \subseteq T$.

Then:
 * $X \subseteq f^{-1} \sqbrk Y \iff f \sqbrk X \subseteq Y$

This can be expressed in the language and notation of direct image mappings and inverse image mappings as:
 * $\forall X \in \powerset S, Y \in \powerset T: X \subseteq \map {f^\gets} Y \iff \map {f^\to} X \subseteq Y$

Proof
Let $f: S \to T$ be a mapping.

As a mapping is also a relation, it follows that $f$ is a relation and so:


 * $X \subseteq f^{-1} \sqbrk Y \iff f \sqbrk X \subseteq Y$

holds on the strength of Subset of Preimage under Relation is Preimage of Subset.