Definition:Inverse Mapping

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Not to be confused with Definition:Inverse of Mapping.

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:

its domain $\Dom g$ equals the image $\Img f$ of $f$
$\forall y \in \Img f: \map f {\map g y} = y$


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
the inverse of neither $f_2$ nor $f_3$ are mappings.


Also see

  • Results about inverse mappings can be found here.