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