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:

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

The result follows from the definition of set equality.