Subset equals Preimage of Image iff Mapping is Injection
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Theorem
Let $f: S \to T$ be a mapping.
Let $f^{-1}$ denote the inverse of $f$.
Then:
- $\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$ if and only if $f$ is an injection
where:
- $f \sqbrk A$ denotes the image of $A$ under $f$
- $f^{-1} \circ f$ denotes the composition of $f^{-1}$ and $f$.
This can be expressed in the language and notation of direct image mappings and inverse image mappings as:
- $\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$ if and only if $f$ is an injection
Proof
Sufficient Condition
Let $g$ be such that:
- $\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$
Then by Subset equals Preimage of Image implies Injection, $f$ is an injection.
$\Box$
Necessary Condition
Let $f$ be an injection.
Then by Preimage of Image of Subset under Injection equals Subset:
- $\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.8$
- 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): Appendix $\text{A}.5$: Identity, One-one, and Onto Functions: Proposition $\text{A}.5.1: 1 \ \text{(c)}$