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