Equivalence of Definitions of Image of Mapping

Proof
Let $f: S \to T$ be a mapping.

Let:
 * $Y = \set {t \in T: \exists s \in S: \map f s = t}$

Then by definition:
 * $Y$ is the image of $f$ by definition 1.

But by definition of image of subset under mapping:
 * $Y = f \sqbrk S$

Thus $Y$ is the image of $f$ by definition 2.

Hence the result.