\begin{definition}[Definition:Preimage of Subset under Mapping/Definition 2]
Let $f: S \to T$ be a mapping.
Let $f^{-1} \subseteq T \times S$ be the inverse of $f$, defined as:
:$f^{-1} = \set {\tuple {t, s}: \map f s = t}$
Let $Y \subseteq T$.
The '''preimage of $Y$ under $f$''' can be seen to be an element of the codomain of the inverse image mapping $f^\gets: \powerset T \to \powerset S$ of $f$:
:$\forall Y \in \powerset T: \map {f^\gets} Y := \set {s \in S: \exists t \in Y: \map f s = t}$
Thus:
:$\forall Y \subseteq T: f^{-1} \sqbrk Y = \map {f^\gets} Y$
If no element of $Y$ has a '''preimage''', then $f^{-1} \sqbrk Y = \O$.
\end{definition}