Subset of Codomain is Superset of Image of Preimage

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

Then:
 * $B \subseteq T \implies \left({f \circ f^{-1}}\right) \left({B}\right) \subseteq B$

Proof
From Image of Preimage of Mapping:
 * $B \subseteq T \implies \left({f \circ f^{-1}}\right) \left({B}\right) = B \cap f \left({S}\right)$

The result follows from Intersection Subset.