# Definition:Inverse Mapping

*Not to be confused with Definition:Inverse of Mapping.*

## Contents

## Definition

Let $S$ and $T$ be sets.

### Definition 1

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

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

- $f^{-1} := \set {\tuple {t, s}: \map f s = t}$

Let $f^{-1}$ itself be a mapping:

- $\forall y \in T: \tuple {y, x_1} \in f^{-1} \land \tuple {y, x_2} \in f^{-1} \implies x_1 = x_2$

and

- $\forall y \in T: \exists x \in S: \tuple {y, x} \in f$

Then $f^{-1}$ is called the **inverse mapping of $f$**.

### Definition 2

Let $f: S \to T$ and $g: T \to S$ be mappings.

Let:

- $g \circ f = I_S$
- $f \circ g = I_T$

where:

- $g \circ f$ and $f \circ g$ denotes the composition of $f$ with $g$ in either order
- $I_S$ and $I_T$ denote the identity mappings on $S$ and $T$ respectively.

That is, $f$ and $g$ are both left inverse mappings and right inverse mappings of each other.

Then:

- $g$ is
**the inverse (mapping) of $f$** - $f$ is
**the inverse (mapping) of $g$**.

## Also known as

If $f$ has an **inverse mapping**, then $f$ is an **invertible mapping**.

When $f^{-1}$ is a mapping, we say that **$f$ has an inverse mapping**.

Some sources, in distinguishing this from a left inverse and a right inverse, refer to this as the **two-sided inverse**.

Some sources use the term **converse mapping** for **inverse mapping**.

## Also defined as

Some authors gloss over the fact that $f$ needs to be a surjection for the inverse of $f$ to be a mapping:

Let $S$ and $T$ be sets.

Let $f: S \to T$ be an injection.

Then its **inverse mapping** is the mapping $g$ such that:

Thus $f$ is seen to be a surjection by tacit use of Restriction of Mapping to Image is Surjection.

Such is the approach of 1999: András Hajnal and Peter Hamburger: *Set Theory*.

## Examples

### Bijective Restrictions of $f \paren x = x^2 - 4 x + 5$

Let $f: \R \to \R$ be the real function defined as:

- $\forall x \in \R: f \paren x = x^2 - 4 x + 5$

Consider the following bijective restrictions of $f$:

\(\displaystyle f_1: \hointl \gets 2\) | \(\to\) | \(\displaystyle \hointr 1 \to\) | |||||||||||

\(\displaystyle f_2: \hointr 2 \to\) | \(\to\) | \(\displaystyle \hointr 1 \to\) |

The inverse of $f_1$ is:

- $\forall y \in \hointr 1 \to: f_1^{-1} \paren y = 2 - \sqrt {y - 1}$

The inverse of $f_2$ is:

- $\forall y \in \hointr 1 \to: f_2^{-1} \paren y = 2 + \sqrt {y - 1}$

### Arbitrary Finite Set with Itself

Let $X = Y = \set {a, b}$.

Consider the mappings from $X$ to $Y$:

\((1):\quad\) | \(\displaystyle \map {f_1} a\) | \(=\) | \(\displaystyle a\) | ||||||||||

\(\displaystyle \map {f_1} b\) | \(=\) | \(\displaystyle b\) |

\((2):\quad\) | \(\displaystyle \map {f_2} a\) | \(=\) | \(\displaystyle a\) | ||||||||||

\(\displaystyle \map {f_2} b\) | \(=\) | \(\displaystyle a\) |

\((3):\quad\) | \(\displaystyle \map {f_3} a\) | \(=\) | \(\displaystyle b\) | ||||||||||

\(\displaystyle \map {f_3} b\) | \(=\) | \(\displaystyle b\) |

\((4):\quad\) | \(\displaystyle \map {f_4} a\) | \(=\) | \(\displaystyle b\) | ||||||||||

\(\displaystyle \map {f_4} b\) | \(=\) | \(\displaystyle a\) |

We have that:

- $f_1$ is the inverse mapping of itself
- $f_4$ is the inverse mapping of itself

## Also see

- Equivalence of Definitions of Inverse Mapping (use Composite of Bijection with Inverse is Identity Mapping)

- Bijection iff Left and Right Inverse, which demonstrates that if $f$ and $f^{-1}$ are inverse mappings, they are both bijections.

- Bijection iff Inverse is Bijection, where is shown that $f^{-1}$ is a mapping if and only if $f$ is a bijection, and that $f^{-1}$ is itself a bijection.

- Results about
**inverse mappings**can be found here.