# Equivalence of Definitions of Image of Mapping

## Theorem

The following definitions of the concept of image of a mapping are equivalent:

### Definition 1

The image of a mapping $f: S \to T$ is the set:

$\Img f = \set {t \in T: \exists s \in S: \map f s = t}$

That is, it is the set of values taken by $f$.

### Definition 2

The image of a mapping $f: S \to T$ is the set:

$\Img f = f \sqbrk S$

where $f \sqbrk S$ is the image of $S$ under $f$.

## Proof

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

Let:

$Y = \set {t \in T: \exists s \in S: f \left({s}\right) = t}$

Then by definition:

$Y$ is the image of $f$ by definition 1.

But by definition of image of subset under mapping:

$Y = f \left[{S}\right]$

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

Hence the result.

$\blacksquare$