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
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections