Subset of Preimage under Relation is Preimage of Subset

Theorem
Let $\mathcal R \subseteq S \times T$ be a relation.

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

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

In the language of direct image mappings, this can be written:
 * $X \subseteq \map {\mathcal R^\gets} Y \iff \map {\mathcal R^\to} X \subseteq Y$

Corollary
If $f: S \to T$ is a mapping, the same result holds:

Proof
As $\mathcal R$ is a relation, then so is its inverse $\mathcal R^{-1}$.

Let $\mathcal R \sqbrk X \subseteq Y$.

Thus:

So:
 * $\mathcal R \sqbrk X \subseteq Y \implies X \subseteq \mathcal R^{-1} \sqbrk Y$

Now let $X \subseteq \mathcal R^{-1} \sqbrk Y$.

The same argument applies:

So:
 * $X \subseteq \mathcal R^{-1} \sqbrk Y \implies \mathcal R \sqbrk X \subseteq Y$

Thus we have:


 * $X \subseteq \mathcal R^{-1} \sqbrk Y \implies \mathcal R \sqbrk X \subseteq Y$


 * $\mathcal R \sqbrk X \subseteq Y \implies X \subseteq \mathcal R^{-1} \sqbrk Y$

Hence the result.