Definition:Preimage

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Definition

Relation

Let $\RR \subseteq S \times T$ be a relation.

Let $\RR^{-1} \subseteq T \times S$ be the inverse relation to $\RR$, defined as:

$\RR^{-1} = \set {\tuple {t, s}: \tuple {s, t} \in \RR}$


Preimage of Element

Every $s \in S$ such that $\tuple {s, t} \in \RR$ is called a preimage of $t$.


In some contexts, it is not individual elements that are important, but all elements of $S$ which are of interest.

Thus the preimage of $t \in T$ is defined as:

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


This can also be written:

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


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


Preimage of Subset

Let $Y \subseteq T$.

The preimage of $Y$ under $\RR$ is defined as:

$\RR^{-1} \sqbrk Y := \set {s \in S: \exists t \in Y: \tuple {s, t} \in \RR}$


That is, the preimage of $Y$ under $\RR$ is the image of $Y$ under $\RR^{-1}$:

$\RR^{-1} \sqbrk Y := \set {s \in S: \exists t \in Y: \tuple {t, s} \in \RR^{-1} }$


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


Preimage of Relation

The preimage of $\mathcal R \subseteq S \times T$ is:

$\Preimg {\mathcal R} := \mathcal R^{-1} \sqbrk T = \set {s \in S: \exists t \in T: \tuple {s, t} \in \mathcal R}$


Mapping

$\mathcal R$ can also be (and usually is in this context) a mapping.

Exactly the same notation and terminology concerning the concept of the preimage applies to the inverse of a mapping.


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:

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


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.


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

Some sources spell preimage with a hyphen and write pre-image.

A preimage is also known as an inverse image.


Also see