Preimage of Mapping equals Domain

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.

$\blacksquare$