Definition:Preimage
Contents |
Definition
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 of $t \in T$ 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 $\mathcal R^{-1}$.
Preimage of a Subset
Let $Y \subseteq T$.
The preimage 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 $\mathcal R \subseteq S \times T$ is:
- $\operatorname{Im}^{-1} \left ({\mathcal R}\right) := \mathcal R^{-1} \left ({T}\right) = \left\{{s \in S: \exists t \in T: \left({s, t}\right) \in \mathcal R}\right\}$
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 an Element
Every $s \in S$ such that $f \left({s}\right) = t$ is called a preimage of $t$.
The preimage 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 $f^{-1}$.
Preimage of a Subset
Let $Y \subseteq T$.
The preimage of $Y$ under $f$ 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 $f^{-1}$, where $f^{-1}$ can be considered as a relation.
If no element of $Y$ has a preimage, then $f^{-1} \left ({Y}\right) = \varnothing$.
Preimage of a Mapping
The preimage of $f$ is defined as:
- $\operatorname{Im}^{-1} \left ({f}\right) := \left\{{s \in S: \exists t \in T: f \left({s}\right) = t}\right\}$
That is:
- $\operatorname{Im}^{-1} \left ({f}\right) := f^{-1} \left ({T}\right)$
where $f^{-1} \left ({T}\right)$ is the image of $T$ under $f$.
Also known as
A preimage is also known as an inverse image.
