Definition:Inverse Image Mapping

From ProofWiki
Jump to: navigation, search

Definition

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 $S \times T$.


Definition 1

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


Definition 2

The inverse image mapping of $\mathcal R$ is the direct image mapping of the inverse $\mathcal R^{-1}$ of $\mathcal R$:

$\mathcal R^\gets= \paren {\mathcal R^{-1} }^\to: \powerset T \to \powerset S$:
$\forall X \in \powerset T: \mathcal R^\gets \paren X = \set {s \in S: \exists s \in X: \tuple {t, s} \in \mathcal R^{-1} }$


Mapping

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


Definition 1

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


Definition 2

The inverse image mapping of $f$ is the direct image mapping of the inverse $f^{-1}$ of $f$:

$f^\gets= \paren {f^{-1} }^\to: \powerset T \to \powerset S$:
$\forall X \in \powerset T: f^\gets \paren X = \set {s \in S: \exists s \in X: \tuple {t, s} \in f^{-1} }$


Also see

  • Results about inverse image mappings can be found here.