Inverse Image Mapping of Mapping is Mapping

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

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

Let $\mathcal R^\gets$ 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.

Proof
$f^{-1}$, is a relation.

So Inverse Image Mapping of Relation is Mapping applies directly.

Also see

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