Preimage of Image under Left-Total Relation is Superset

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

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

where:


 * $\mathcal R \left[{A}\right]$ denotes the image of $A$ under $\mathcal R$
 * $\mathcal R^{-1} \left[{A}\right]$ denotes the preimage of $A$ under $\mathcal R$
 * $\mathcal R^{-1} \circ \mathcal R$ denotes composition of $\mathcal R^{-1}$ and $\mathcal R$.

Proof
Suppose $A \subseteq S$.

We have:

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

Also see

 * Subset of Domain is Subset of Preimage of Image