Preimage of Mapping equals Domain

Theorem

The preimage of a mapping is the same set as its domain:

$\operatorname{Im}^{-1} \left({f}\right) = \operatorname{Dom} \left({f}\right)$

Proof

Let $f \subseteq S \times T$ be a mapping. Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle \forall x \in S: \exists y \in T: \left({x, y}\right) \in f\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of mapping
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle \forall x \in S: x \in \operatorname{Im}^{-1} \left({f}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of preimage
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle S \subseteq \operatorname{Im}^{-1} \left({f}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of subset

From Preimage of Relation is Subset of Domain, we have that $\operatorname{Im}^{-1} \left({f}\right) \subseteq S$.

The result follows from the definition of set equality.

$\blacksquare$