Image of Preimage under Mapping/Corollary

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

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

Proof
From Image of Subset under Relation is Subset of Image/Corollary 3 we have:
 * $B \subseteq \Img S \implies f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {\Img S}$

and from Intersection with Subset is Subset we have:


 * $f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {\Img S} \implies f^{-1} \sqbrk B \cap f^{-1} \sqbrk {\Img S} = f^{-1} \sqbrk B$

Hence the result.