Subset of Domain is Subset of Preimage of Image
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Theorem
Let $f: S \to T$ be a mapping.
Then:
- $A \subseteq S \implies A \subseteq \paren {f^{-1} \circ f} \sqbrk A$
where:
- $f \sqbrk A$ denotes the image of $A$ under $f$
- $f^{-1} \sqbrk A$ denotes the preimage of $A$ under $f$
- $f^{-1} \circ f$ denotes composition of $f^{-1}$ and $f$.
This can be expressed in the language and notation of direct image mappings and inverse image mappings as:
- $\forall A \in \powerset S: A \subseteq \map {\paren {f^\gets \circ f^\to} } A$
Equality does Not Necessarily Hold
It is not necessarily the case that:
- $A \subseteq S \implies A = \paren {f^{-1} \circ f} \sqbrk A$
Proof
As a mapping is by definition a left-total relation.
Therefore Preimage of Image under Left-Total Relation is Superset applies:
- $A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$
where $\RR$ is a relation.
Hence:
- $A \subseteq S \implies A \subseteq \paren {f^{-1} \circ f} \sqbrk A$
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 10$: Inverses and Composites
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.13 \ \text{(c)}$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 12 \alpha$
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.7 \ \text{(b)}$
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.8$
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $1 \ \text{(a})$
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 2$: Functions