Preimage of Image under Left-Total Relation is Superset

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

Then:
 * $A \subseteq S \implies A \subseteq \left({\mathcal R^\gets \circ \mathcal R^\to}\right) \left({A}\right)$

where:


 * $\mathcal R^\to$ denotes the mapping induced on the power set $\mathcal P \left({S}\right)$ of $S$ by $\mathcal R$
 * $\mathcal R^\gets$ denotes the mapping induced on the power set $\mathcal P \left({T}\right)$ of $T$ by the inverse $\mathcal R^{-1}$
 * $\mathcal R^\gets \circ \mathcal R^\to$ denotes composition of $\mathcal R^\gets$ and $\mathcal R^\to$.

Proof
Suppose $A \subseteq S$.

We have:

So by definition of subset:
 * $A \subseteq S \implies A \subseteq \left({\mathcal R^\gets \circ \mathcal R^\to}\right) \left({A}\right)$

Also see

 * Subset of Domain is Subset of Preimage of Image