Inverse Image Mapping of Codomain is Preimage Set of Mapping

Theorem

Let $S$ and $T$ be sets.

Let $\powerset S$ and $\powerset T$ be their power sets.

Let $f \subseteq S \times T$ be a mapping from $S$ to $T$.

Let $f^\gets: \powerset T \to \powerset S$ be the inverse image mapping of $f$:

$\forall Y \in \powerset T: \map {f^\gets} Y = \begin {cases} \set {s \in S: \exists t \in Y: \map f s = t} & : \Img f \cap Y \ne \O \\ \O & : \Img f \cap Y = \O \end {cases}$

Then:

$\map {f^\gets} T = \Preimg f$

where $\Preimg f$ is the preimage set of $f$.

Proof

\(\ds x\)

\(\in\)

\(\ds \map {f^\gets} T\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \exists y \in T: \, \)

\(\ds \map f x\)

\(=\)

\(\ds y\)

Definition of Inverse Image Mapping of Mapping

\(\ds \leadstoandfrom \ \ \)

\(\ds x\)

\(\in\)

\(\ds \Preimg f\)

Definition of Preimage Set of Mapping

$\blacksquare$