# Equivalence of Definitions of Image of Mapping

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## 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:

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

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.

$\blacksquare$