Subset of Domain is Subset of Preimage of Image

Theorem
Let $f: S \to T$ be a mapping.

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

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

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

Proof
As a mapping is by definition a left-total relation.

Therefore Preimage of Image under Left-Total Relation is Superset applies:
 * $A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$

where $\RR$ is a relation.

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