Image of Preimage under Relation is Subset

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

Then:
 * $B \subseteq T \implies \paren {\RR \circ \RR^{-1} } \sqbrk B \subseteq B$

where:
 * $\RR \sqbrk B$ denotes the image of $B$ under $\RR$
 * $\RR^{-1} \sqbrk B$ denotes the preimage of $B$ under $\RR$
 * $\RR \circ \RR^{-1}$ denotes composition of $\RR$ and $\RR^{-1}$.

Proof
Let $B \subseteq T$.

Then:

So by definition of subset:
 * $B \subseteq T \implies \paren {\RR \circ \RR^{-1} }\sqbrk B \subseteq B$

Also see

 * Subset of Domain is Subset of Preimage of Image