Inverse Image Mapping of Relation is Mapping

Theorem

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.

Proof

$\RR^{-1}$, being a relation, obeys the same laws as $\RR$.

So Direct Image Mapping of Relation is Mapping applies directly.

$\blacksquare$

Also see

• Inverse of Direct Image Mapping does not necessarily equal Inverse Image Mapping