Image of Preimage under Relation is Subset

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

Then:
 * $B \subseteq T \implies \left({\mathcal R^\to \circ \mathcal R^\gets}\right) \left({B}\right) \subseteq B$

where:
 * $\mathcal R^\to$ denotes the mapping induced on the power set $\mathcal P \left({S}\right)$ of $S$ by $\mathcal R$
 * $\mathcal R^\gets$ denotes the mapping induced on the power set $\mathcal P \left({T}\right)$ of $T$ by the inverse $\mathcal R^{-1}$
 * $\mathcal R^\to \circ \mathcal R^\gets$ denotes composition of $\mathcal R^\to$ and $\mathcal R^\gets$.

Proof
Let $B \subseteq T$.

Then:

So by definition of subset:
 * $B \subseteq T \implies \left({\mathcal R^\to \circ \mathcal R^\gets}\right) \left({B}\right) \subseteq B$

Also see

 * Subset of Domain is Subset of Preimage of Image