Preimage of Mapping equals Domain
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Theorem
The preimage of a mapping is the same set as its domain:
- $\Preimg f = \Dom f$
Proof
Let $f \subseteq S \times T$ be a mapping.
Then:
\(\ds \) | \(\) | \(\ds \forall x \in S: \exists y \in T: \tuple {x, y} \in f\) | Definition of Mapping | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \forall x \in S: x \in \Preimg f\) | Definition of Preimage of Mapping | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds S \subseteq \Preimg f\) | Definition of Subset |
From Preimage of Relation is Subset of Domain, we have that $\Preimg f \subseteq S$.
The result follows from the definition of set equality.
$\blacksquare$