Preimage of Mapping equals Domain

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$