Preimage of Image under Left-Total Relation is Superset

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

Then:


 * $A \subseteq S \implies A \subseteq \left({\mathcal R^{-1} \circ \mathcal R}\right) \left[{A}\right]$
 * $B \subseteq T \implies \left({\mathcal R \circ \mathcal R^{-1}}\right) \left[{B}\right] \subseteq B$

Proof
Suppose $A \subseteq S$.

We have:

So by definition of composition of relations:
 * $A \subseteq S \implies A \subseteq \left({\mathcal R^{-1} \circ \mathcal R}\right) \left[{A}\right]$

Let $B \subseteq T$.

Then:

So:
 * $B \subseteq T \implies \left({\mathcal R \circ \mathcal R^{-1}}\right) \left[{B}\right] \subseteq B$

Also see

 * Subset of Domain is Subset of Preimage of Image