Direct Image Mapping of Domain is Image 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 $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$:

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

Then:

$\map {f^\to} S = \Img f$

where $\Img f$ is the image set of $f$.

$\ds y$

$\in$

$\ds \map {f^\to} S$

$\ds \leadstoandfrom \ \ $

$\ds \exists x \in S: \, $

$\ds \map f x$

$=$

$\ds y$

$\ds \leadstoandfrom \ \ $

$\ds y$

$\in$

$\ds \Img f$

$\blacksquare$