Definition:Preimage

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Relation


Let $\mathcal R \subseteq S \times T$ be a relation.

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

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


Preimage of an Element

Every $s \in S$ such that $\left({s, t}\right) \in \mathcal R$ 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 (or inverse image) of an element $t \in T$ is defined as:

$\mathcal R^{-1} \left ({t}\right) := \left\{{s \in S: \left({s, t}\right) \in \mathcal R}\right\}$


This can also be written:

$\mathcal R^{-1} \left ({t}\right) := \left\{{s \in \operatorname{Im} \left({\mathcal R^{-1}}\right): \left({t, s}\right) \in \mathcal R^{-1}}\right\}$


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


The preimage of $t \in T$ is also known as the fiber of $t$.


Note that:

  • $t \in T$ may have more than one preimage.
  • It is possible for $t \in T$ to have no preimages at all, in which case $\mathcal R^{-1} \left ({t}\right) = \varnothing$.


Preimage of a Subset

Let $Y \subseteq T$.

The preimage (or inverse image) of $Y$ under $\mathcal R$ is defined as:

$\mathcal R^{-1} \left ({Y}\right) := \left\{{s \in S: \exists y \in Y: \left({s, y}\right) \in \mathcal R}\right\}$


That is, the preimage of $Y$ under $\mathcal R$ is the image of $Y$ under $\mathcal R^{-1}$.


Clearly:

$\displaystyle \mathcal R^{-1} \left ({Y}\right) = \bigcup_{y \in Y} \mathcal R^{-1} \left({y}\right)$

... the union of the preimages of each of the elements of $Y$.


If no element of $Y$ has a preimage, then $\mathcal R^{-1} \left ({Y}\right) = \varnothing$.


Preimage of a Relation

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

$\operatorname{Im}^{-1} \left ({\mathcal R}\right) := \mathcal R^{-1} \left ({T}\right) = \left\{{s \in S: \exists t \in T: \left({s, t}\right) \in \mathcal R}\right\}$


Some sources, for example T.S. Blyth: Set Theory and Abstract Algebra (1975), call this the domain of $\mathcal R$.


Mapping

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


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

Exactly the same notation and terminology concerning the concept of the preimage applies to its inverse $f^{-1}$.


Thus:

  • Every $s \in S$ such that $f \left({s}\right) = t$ is called a preimage of $t$.


  • The preimage (or inverse image) of an element $t \in T$ is defined as:
$f^{-1} \left ({t}\right) := \left\{{s \in S: f \left({s}\right) = t}\right\}$


This can also be written:

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


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


  • The preimage of $Y \subseteq \operatorname{Im} \left({f}\right)$ is defined as:
$f^{-1} \left ({Y}\right) := \left\{{s \in S: \exists y \in Y: f \left({s}\right) = y}\right\}$


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


Clearly:

$\displaystyle f^{-1} \left ({Y}\right) = \bigcup_{y \in Y} f^{-1} \left({y}\right)$

... the union of the preimages of each of the elements of $Y$.


If no element of $Y$ has a preimage, then $\mathcal R^{-1} \left ({Y}\right) = \varnothing$.


The preimage of a mapping $f: S \to T$ is:

$\operatorname{Im}^{-1} \left ({f}\right) = f^{-1} \left ({T}\right) = \left\{{s \in S: \exists t \in T: f \left({s}\right) = t}\right\}$


Note that:

Thus, while $f^{-1}$ is always a relation, it is not actually a mapping unless $f$ is a bijection.


Alternative Notation

As well as using the notation $\operatorname{Im}^{-1} \left ({\mathcal R}\right)$ to denote the preimage of an entire relation, the symbol $\operatorname{Im}^{-1}$ can also be used as:

  • For $t \in \operatorname{Im} \left({\mathcal R}\right)$, we have: $\operatorname{Im}^{-1}_\mathcal R \left ({t}\right) = \mathcal R^{-1} \left ({t}\right)$
  • For $Y \subseteq \operatorname{Im} \left({\mathcal R}\right)$, we have: $\operatorname{Im}^{-1}_\mathcal R \left ({Y}\right) = \mathcal R^{-1} \left ({Y}\right)$

but this notation is clumsy and generally not preferred.


Also see


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