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 Image of Preimage under Mapping: Corollary 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.

$\blacksquare$

Sources

• 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.7 \ \text{(c)}$