Preimage of Image under Left-Total Relation is Superset

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

Then:
 * $A \subseteq S \implies A \subseteq \paren {\mathcal R^{-1} \circ \mathcal R} \sqbrk A$

where:


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

This can be expressed in the language and notation of direct image mappings and inverse image mappings as:
 * $\forall A \in \powerset S: A \subseteq \map {\paren {\mathcal R^\gets \circ \mathcal R^\to} } A$

Proof
Suppose $A \subseteq S$.

We have:

So by definition of subset:
 * $A \subseteq S \implies A \subseteq \paren {\mathcal R^{-1} \circ \mathcal R} \sqbrk A$

Also see

 * Subset of Domain is Subset of Preimage of Image