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.