Element in Image of Preimage under Mapping

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

Then:
 * $\forall y \in T: \in f \sqbrk {f^{-1} \sqbrk y} = \set y$

Proof
A mapping is by definition a relation.

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

Thus:
 * $\set y \subseteq T \implies f^{-1} \sqbrk {f \sqbrk {\set y} } = \set y$

Hence the result.