Subset of Codomain is Superset of Image of Preimage

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

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

where:


 * $f \sqbrk B$ denotes the image of $B$ under $f$
 * $f^{-1}$ denotes the inverse of $f$
 * $f \circ f^{-1}$ denotes composition of $f$ and $f^{-1}$.