Definition:Preimage

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

Let:
 * $$\operatorname{Dom} \left({\mathcal{R}}\right)$$ denote the domain of $$\mathcal{R}$$;
 * $$\operatorname{Rng} \left({\mathcal{R}}\right)$$ denote the range of $$\mathcal{R}$$.

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 $$s \in \operatorname{Dom} \left({\mathcal{R}}\right)$$ 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 $$\operatorname{Dom} \left({\mathcal{R}}\right)$$ which are of interest.

Thus the preimage (or inverse image) of an element $$t \in \operatorname{Rng} \left({\mathcal{R}}\right)$$ is defined as:


 * $$\mathcal{R}^{-1} \left ({t}\right) \ \stackrel {\mathbf {def}} {=\!=} \ \left\{{s \in \operatorname {Dom} \left({\mathcal{R}}\right): \left({s, t}\right) \in \mathcal{R}}\right\}$$

This can also be written:
 * $$\mathcal{R}^{-1} \left ({t}\right) \ \stackrel {\mathbf {def}} {=\!=} \ \left\{{s \in \operatorname {Rng} \left({\mathcal{R}^{-1}}\right): \left({t, s}\right) \in \mathcal{R}^{-1}}\right\}$$

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

The preimage of $$t \in \operatorname{Rng} \left({\mathcal{R}}\right)$$ is also known as the fiber of $$t$$.

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

Preimage of a Subset
Let $$Y \subset \operatorname{Rng} \left({\mathcal{R}}\right)$$.

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


 * $$\mathcal{R}^{-1} \left ({Y}\right) \ \stackrel {\mathbf {def}} {=\!=} \ \left\{{s \in \operatorname{Dom} \left({\mathcal{R}}\right): \exists y \in Y: \left({s, y}\right) \in \mathcal{R}}\right\}$$

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

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

... 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 $$\mathcal{R}$$ is:


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

Some sources, for example, call this the domain of $$\mathcal{R}$$.

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

Let $$f$$ be a mapping.

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

Thus:


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


 * The preimage (or inverse image) of an element $$t \in \operatorname{Rng} \left({f}\right)$$ is defined as:


 * $$f^{-1} \left ({t}\right) \ \stackrel {\mathbf {def}} {=\!=} \ \left\{{s \in \operatorname {Dom} \left({f}\right): f \left({s}\right) = t}\right\}$$

This can also be written:
 * $$f^{-1} \left ({t}\right) \ \stackrel {\mathbf {def}} {=\!=} \ \left\{{s \in \operatorname {Rng} \left({f^{-1}}\right): \left({t, s}\right) \in f^{-1}}\right\}$$

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


 * The preimage of $$Y \subset \operatorname{Rng} \left({f}\right)$$ is defined as:


 * $$f^{-1} \left ({Y}\right) \ \stackrel {\mathbf {def}} {=\!=} \ \left\{{s \in \operatorname{Dom} \left({f}\right): \exists y \in Y: f \left({s}\right) = y}\right\}$$

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

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

... 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 $$f$$ is:


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

Note that:


 * From Preimages All Exist iff Surjection $$f^{-1} \left({t}\right)$$ is guaranteed not to be empty iff $$f$$ is a surjection.


 * From Preimages All Unique iff Injection, if $$f^{-1} \left({t}\right)$$ is not empty, then it is guaranteed to be a singleton iff $$f$$ is an injection;

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

 * Domain
 * Range (or Codomain)
 * Image