Preimage of Subset is Subset of Preimage

Corollary of Image of Subset under Relation is Subset of Image
Let $S$ and $T$ be sets.

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

Let $C, D \subseteq T$.

Then:
 * $C \subseteq D \implies f^{-1} \sqbrk C \subseteq f^{-1} \sqbrk D$

This can be expressed in the language and notation of inverse image mappings as:
 * $\forall C, D \in \powerset T: C \subseteq D \implies \map {f^\gets} C \subseteq \map {f^\gets} D$

Proof
As $f: S \to T$ is a mapping, it is also a relation, and thus so is its inverse:
 * $f^{-1} \subseteq T \times S$

The result follows directly from Image of Subset under Relation is Subset of Image.