Image of Preimage under Mapping/Corollary

Corollary to Image of Preimage under Mapping
Let $f: S \to T$ be a mapping.

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

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

Proof
From Preimage of Subset is Subset of Preimage:
 * $B \subseteq \Img f \implies f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {\Img f}$

and from Intersection with Subset is Subset:
 * $f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {\Img f} \implies f^{-1} \sqbrk B \cap f^{-1} \sqbrk {\Img f} = f^{-1} \sqbrk B$

Hence the result.