# 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:

$X \subseteq f^{-1} \sqbrk Y \iff f \sqbrk X \subseteq Y$

This can be expressed in the language and notation of direct image mappings and inverse image mappings as:

$\forall X \in \powerset S, Y \in \powerset T: X \subseteq \map {f^\gets} Y \iff \map {f^\to} X \subseteq Y$

## Proof

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

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

Thus:

 $\displaystyle \mathcal R \sqbrk X$ $\subseteq$ $\displaystyle Y$ $\displaystyle \leadsto \ \$ $\displaystyle \mathcal R^{-1} \sqbrk {\mathcal R \sqbrk X}$ $\subseteq$ $\displaystyle \mathcal R^{-1} \sqbrk Y$ Image of Subset under Relation is Subset of Image: Corollary 1 $\displaystyle \leadsto \ \$ $\displaystyle X$ $\subseteq$ $\displaystyle \mathcal R^{-1} \sqbrk Y$ Image of Preimage under Relation is Subset, Subset Relation is Transitive

So:

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

$\Box$

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

The same argument applies:

 $\displaystyle X$ $\subseteq$ $\displaystyle \mathcal R^{-1} \sqbrk Y$ $\displaystyle \leadsto \ \$ $\displaystyle \mathcal R \sqbrk X$ $\subseteq$ $\displaystyle \mathcal R \sqbrk {\mathcal R^{-1} \sqbrk Y}$ Image of Subset under Relation is Subset of Image $\displaystyle \leadsto \ \$ $\displaystyle \mathcal R \sqbrk X$ $\subseteq$ $\displaystyle Y$ Image of Preimage under Relation is Subset, Subset Relation is Transitive

So:

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

$\Box$

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.

$\blacksquare$