Image of Preimage under Mapping/Corollary

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

Then:
 * $B \subseteq \operatorname{Im} \left({S}\right) \implies \left({f \circ f^{-1}}\right) \left[{B}\right] = B$

Proof
From Image of Subset under Relation is Subset of Image/Corollary 3 we have:
 * $B \subseteq \operatorname{Im} \left({S}\right) \implies f^{-1} \left[{B}\right] \subseteq f^{-1} \left[{\operatorname{Im} \left({S}\right)}\right]$

and from Intersection with Subset is Subset we have:


 * $f^{-1} \left[{B}\right] \subseteq f^{-1} \left[{\operatorname{Im} \left({S}\right)}\right] \implies f^{-1} \left[{B}\right] \cap f^{-1} \left[{\operatorname{Im} \left({S}\right)}\right] = f^{-1} \left[{B}\right]$

Hence the result.