Preimage of Subset is Subset of Preimage

Corollary of Image of Subset 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:
 * $C \subseteq D \implies f^{-1} \left({C}\right) \subseteq f^{-1} \left({D}\right)$

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 is Subset of Image.