# Definition:Preimage/Mapping

## Definition

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

Let $f^{-1} \subseteq T \times S$ be the inverse of $f$, considered as a relation:

$f^{-1} = \left\{{\left({t, s}\right): f \left({s}\right) = t}\right\}$

### Preimage of Element

Every $s \in S$ such that $\map f s = t$ is called a preimage of $t$.

The preimage of an element $t \in T$ is defined as:

$\map {f^{-1} } t := \set {s \in S: \map f s = t}$

This can also be expressed as:

$\map {f^{-1} } t := \set {s \in \Img {f^{-1} }: \tuple {t, s} \in f^{-1} }$

That is, the preimage of $t$ under $f$ is the image of $t$ under $f^{-1}$.

### Preimage of Subset

Let $Y \subseteq T$.

The preimage of $Y$ under $f$ is defined as:

$f^{-1} \sqbrk Y := \set {s \in S: \exists y \in Y: \map f s = y}$

That is, the preimage of $Y$ under $f$ is the image of $Y$ under $f^{-1}$, where $f^{-1}$ can be considered as a relation.

If no element of $Y$ has a preimage, then $f^{-1} \sqbrk Y = \O$.

### Preimage of Mapping

The preimage of $f$ is defined as:

$\Preimg f := \set {s \in S: \exists t \in T: f \paren s = t}$

That is:

$\Preimg f := f^{-1} \sqbrk T$

where $f^{-1} \sqbrk T$ is the image of $T$ under $f^{-1}$.

In this context, $f^{-1} \subseteq T \times S$ is the the inverse of $f$.

It is a relation but not necessarily itself a mapping.

## Also known as

A preimage is also known as an inverse image.

## Also see

• Results about preimages under mappings can be found here.

## Technical Note

The $\LaTeX$ code for $\Preimg {f}$ is \Preimg {f} .

When the argument is a single character, it is usual to omit the braces:

\Preimg f