Preimage of Image under Left-Total Relation is Superset

Theorem
Let $\RR \subseteq S \times T$ be a left-total relation.

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

where:


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

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

Proof
Suppose $A \subseteq S$.

We have:

Then we have that $\RR$ is a left-total relation.

Thus:
 * $\exists t \in T: \tuple {x, t} \in \RR$

Hence by definition of inverse relation:


 * $\exists t \in T: \tuple {t, x} \in \RR^{-1}$

Hence:

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

Also see

 * Subset of Domain is Subset of Preimage of Image