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:
 * $X \subseteq \mathcal R^{-1} \left({Y}\right) \iff \mathcal R \left({X}\right) \subseteq Y$

In the language of induced mappings, that would be written:
 * $X \subseteq f_{\mathcal R^{-1}} \left({Y}\right) \iff f_{\mathcal R} \left({X}\right) \subseteq Y$

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


 * $X \subseteq f^{-1} \left({Y}\right) \iff f \left({X}\right) \subseteq Y$

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

Let $\mathcal R \left({X}\right) \subseteq Y$.

Thus:

So:
 * $\mathcal R \left({X}\right) \subseteq Y \implies X \subseteq \mathcal R^{-1} \left({Y}\right)$

Now let $X \subseteq \mathcal R^{-1} \left({Y}\right)$.

The same argument applies:

So:
 * $X \subseteq \mathcal R^{-1} \left({Y}\right) \implies \mathcal R \left({X}\right) \subseteq Y$

Thus we have:
 * $X \subseteq \mathcal R^{-1} \left({Y}\right) \implies \mathcal R \left({X}\right) \subseteq Y$
 * $\mathcal R \left({X}\right) \subseteq Y \implies X \subseteq \mathcal R^{-1} \left({Y}\right)$

Hence the result.

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

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


 * $X \subseteq f^{-1} \left({Y}\right) \iff f \left({X}\right) \subseteq Y$

holds on the strength of the main result.