Subset of Domain is Subset of Preimage of Image

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

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

where:


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

Proof
As a mapping is by definition also a relation.

Therefore Preimage of Image under Relation is Superset applies:
 * $A \subseteq S \implies A \subseteq \left({\mathcal R^{-1} \circ \mathcal R}\right) \left[{A}\right]$

where $\mathcal R$ is a relation.

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