Complement of Direct Image Mapping of Injection equals Direct Image of Complement
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Theorem
Let $f: S \to T$ be an injection.
Let $f^\to: \powerset S \to \powerset T$ denote the direct image mapping of $f$.
Then:
- $\forall A \in \powerset S: \map {\paren {\complement_{\Img f} \circ f^\to} } A = \map {\paren {f^\to \circ \complement_S} } A$
where $\circ$ denotes composition of mappings.
Proof
As $f$ is an injection, it is a fortiori a one-to-many relation.
From Image of Set Difference under Relation: Corollary 2:
- $\forall A \in \powerset S: \map {\paren {\complement_{\Img \RR} \circ \RR^\to} } A = \map {\paren {\RR^\to \circ \complement_S} } A$
where $\RR \subseteq S \times T$ is a one-to-many relation on $S \times T$.
Substituting $f$ for $\RR$ gives the result:
- $\forall A \in \powerset S: \map {\paren {\complement_{\Img f} \circ f^\to} } A = \map {\paren {f^\to \circ \complement_S} } A$
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $3$
- 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $7 \ \text {(c)}$