# Definition:Mapping Induced on Powerset

## Definition

The mapping induced on a powerset can be either of two concepts:

## Direct Image Mapping

Let $S$ and $T$ be sets.

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

### Relation

Let $\mathcal R \subseteq S \times T$ be a relation on $S \times T$.

The direct image mapping of $\mathcal R$ is the mapping $\mathcal R^\to: \powerset S \to \powerset T$ that sends a subset $X \subseteq T$ to its image under $\mathcal R$:

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

### Mapping

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

The direct image mapping of $f$ is the mapping $f^\to: \powerset S \to \powerset T$ that sends a subset $X \subseteq S$ to its image under $f$:

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

## Inverse Image Mapping

Let $S$ and $T$ be sets.

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

### Relation

Let $\mathcal R \subseteq S \times T$ be a relation on $S \times T$.

The inverse image mapping of $\mathcal R$ is the mapping $\mathcal R^\gets: \powerset T \to \powerset S$ that sends a subset $Y \subseteq T$ to its preimage $\map {\mathcal R^{-1} } Y$ under $\mathcal R$:

$\forall Y \in \powerset T: \map {\mathcal R^\gets} Y = \begin {cases} \set {s \in S: \exists t \in Y: \tuple {t, s} \in \mathcal R^{-1} } & : \Img {\mathcal R} \cap Y \ne \O \\ \O & : \Img {\mathcal R} \cap Y = \O \end {cases}$

### Mapping

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

The inverse image mapping of $f$ is the mapping $f^\gets: \powerset T \to \powerset S$ that sends a subset $Y \subseteq T$ to its preimage $f^{-1} \paren T$ under $f$:

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