# Definition:Image (Set Theory)/Mapping/Element

(Redirected from Definition:Functional Value)

## Definition

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

Let $s \in S$.

The image of $s$ (under $f$) is defined as:

$\Img s = \map f s = \displaystyle \bigcup \set {t \in T: \tuple {s, t} \in f}$

That is, $\map f s$ is the element of the codomain of $f$ related to $s$ by $f$.

By the nature of a mapping, $\map f s$ is guaranteed to exist and to be unique for any given $s$ in the domain of $f$.

## Also known as

The image of an element $s$ under a mapping $f$ is also called the functional value, or value, of $f$ at $s$.

The terminology:

$f$ maps $s$ to $\map f s$
$f$ assigns the value $\map f s$ to $s$
$f$ carries $s$ into $\map f s$

can be found.

The modifier by $f$ can also be used for under $f$.

Thus, for example, the image of $s$ by $f$ means the same as the image of $s$ under $f$.

In the context of computability theory, the following terms are frequently found:

If $\tuple {x, y} \in f$, then $y$ is often called the output of $f$ for input $x$, or simply, the output of $f$ at $x$.

## Also denoted as

The notation $\Img f$ to denote the image of a mapping $f$ is specific to $\mathsf{Pr} \infty \mathsf{fWiki}$.

The usual notation is $\image f$ or a variant, but this is too easily confused with $\map \Im z$, the imaginary part of a complex number.

Hence the non-standard usage $\Img f$.

Some sources use $f \sqbrk S$, where $S$ is the domain of $f$.

Others just use $\map f S$, but that notation is deprecated on $\mathsf{Pr} \infty \mathsf{fWiki}$ so as not to confuse it with the notation for the image of an element.

The notation $\Img f$ is specific to $\mathsf{Pr} \infty \mathsf{fWiki}$.

## Examples

### Image of $2$ under $x \mapsto x^4 - 1$

Let $f: \R \to \R$ be the mapping defined as:

$\forall x \in \R: \map f x = x^4 - 1$

The image of $2$ is:

$\map f 2 = 15$

### Images of Various Numbers under $x \mapsto x^2 + 2 x + 1$ in $\closedint 0 1$

Let $f: \closedint 0 1 \to \R$ be the mapping defined as:

$\forall x \in \closedint 0 1: \map f x = x^2 + 2 x + 1$

where $\closedint 0 1$ denotes the closed real interval from $0$ to $1$.

The images of various real numbers under $f$ are:

 $\displaystyle \map f 0$ $=$ $\displaystyle 0^2 + 2 \times 0 + 1$ $\displaystyle = 1$ $\displaystyle \map f 1$ $=$ $\displaystyle 1^2 + 2 \times 1 + 1$ $\displaystyle = 4$ $\displaystyle \map f {\dfrac 1 2}$ $=$ $\displaystyle \paren {\dfrac 1 2}^2 + 2 \times \dfrac 1 2 + 1$ $\displaystyle = 2 \tfrac 1 4$ $\displaystyle \map f 2$  $\displaystyle \text {is undefined}$ $\displaystyle \text {as 2 is not in the domain of f}$ $\displaystyle \map f {-1}$  $\displaystyle \text {is undefined}$ $\displaystyle \text {as -1 is not in the domain of f}$