Element in Preimage of Image under Mapping
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Theorem
Let $f: S \to T$ be a mapping.
Let $f^{-1} \sqbrk {\map f x}$ denote the preimage of $\map f x$ under $f$.
Then:
- $\forall x \in S: x \in f^{-1} \sqbrk {\map f x}$
Proof
A mapping is by definition a left-total relation.
Therefore Preimage of Image under Left-Total Relation is Superset applies:
- $A \subseteq S \implies A \subseteq f^{-1} \sqbrk {f \sqbrk A}$
Thus:
- $\set x \subseteq S \implies \set x \subseteq f^{-1} \sqbrk {f \sqbrk A}$
Hence the result.
$\blacksquare$
Sources
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.7 \ \text{(a)}$