Inverse Image Mapping is Mapping

Theorem

Inverse Image Mapping of Relation is Mapping

Let $S$ and $T$ be sets.

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

Let $\RR^\gets$ be the inverse image mapping of $\RR$:

$\RR^\gets: \powerset T \to \powerset S: \map {\RR^\gets} Y = \RR^{-1} \sqbrk Y$

Then $\RR^\gets$ is indeed a mapping.

Inverse Image Mapping of Mapping is Mapping

Let $S$ and $T$ be sets.

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

Let $f^\gets$ be the inverse image mapping of $f$:

$f^\gets: \powerset T \to \powerset S: \map {f^\gets} Y = f^{-1} \sqbrk Y$

Then $f^\gets$ is indeed a mapping.