Subset of Domain is Subset of Preimage of Image

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

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

where:


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

Proof
As a mapping is by definition also a relation.

Therefore Preimage of Image is Superset applies:
 * $A \subseteq S \implies A \subseteq \left({\mathcal R^\gets \circ \mathcal R^\to}\right) \left({A}\right)$

where $\mathcal R$ is a relation.

Hence:
 * $A \subseteq S \implies A \subseteq \left({f^\gets \circ f^\to}\right) \left({A}\right)$