Inverse Image Mapping of Relation is Mapping

Theorem
Let $S$ and $T$ be sets.

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

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


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

Then $\mathcal R^\gets$ is indeed a mapping.

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

So Direct Image Mapping of Relation is Mapping applies directly.

Also see

 * Inverse of Induced Mapping does not necessarily equal Mapping Induced by Inverse