From ProofWiki
Jump to navigation Jump to search


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

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

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

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:

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

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.

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