Definition:Preimage/Mapping/Element

Definition
Let $S$ and $T$ be sets

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

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


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

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 $f^{-1}$.

Also known as
The preimage of an element is also known as its inverse image.

In other contexts, this is called the fiber of $t$ (under $f$).

The UK English spelling of fiber is fibre.

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

For $t \in \operatorname{Im} \left({f}\right)$:
 * $\operatorname{Im}^{-1}_f \left ({t}\right) := f^{-1} \left ({t}\right)$

but this notation is clumsy and generally not preferred.

The term argument is popular in certain branches of mathematics.

If $\left({x, y}\right) \in f$, then $x$ is the argument (of $f$) which holds the value $y$.

In the context of computability theory, the following terms are frequently found:

If $\left({x, y}\right) \in f$, then $x$ is often called the input of $f$ which produces the output $y$.

Also see

 * Definition:Domain of Mapping
 * Definition:Codomain of Mapping
 * Definition:Range


 * Definition:Image of Element under Mapping


 * Definition:Preimage of Relation


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


 * From the definition of an injection, if $f^{-1} \left({t}\right)$ is not empty, then it is guaranteed to be a singleton $f$ is an injection.

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