Preimage of Image under Left-Total Relation is Superset

Theorem
Let $$\mathcal{R} \subseteq S \times T$$ be a relation. Then:


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

Proof
We have:

$$ $$

Suppose $$s \notin \left({\mathcal{R}^{-1} \circ \mathcal{R}}\right) \left({A}\right)$$.

$$ $$ $$ $$

The proof of the second part follows from Inverse of Inverse Relation.