Inverse Image Mapping of Mapping is Mapping

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Theorem

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.


Proof

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

So Inverse Image Mapping of Relation is Mapping applies directly.

$\blacksquare$


Also see


Sources