Equivalence of Definitions of Preimage of Subset under Mapping

Theorem
Let $f: S \to T$ be a mapping from a set $S$ to a set $T$.

Let $Y \subseteq T$ be a subset of $T$.

Proof
The difference in definitions is no more than a difference in notations.

Let $X$ be a preimage by definition $2$.

Then by definition:
 * $X := \map {f^\gets} Y$

By definition of inverse image mapping of mapping:


 * $\forall Y \in \powerset T: \map {f^\gets} Y = \set {s \in S: \exists t \in Y: \map f s = t}$

Thus $X$ is a preimage by definition $1$.