Definition:Inverse Mapping

Not to be confused with Definition:Inverse of Mapping.

Definition

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.

Hence, when the inverse (relation) of $f^{-1}$ 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 $f: S \to T$ be an injection.

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

$(1): \quad$ its domain $\Dom g$ equals the image $\Img f$ of $f$

$(2): \quad \forall y \in \Img f: \map f {\map g y} = y$

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

$x^3$ Function on Real Numbers

Let $f: \R \to \R$ be the mapping defined on the set of real numbers as:

$\forall x \in \R: \map f x = x^3$

The inverse of $f$ is:

$\forall y \in \R: \map {f^{-1} } y = \sqrt [3] y$

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 $f$:

\(\ds f_1: \hointl \gets 2\)

\(\to\)

\(\ds \hointr 1 \to\)

\(\ds f_2: \hointr 2 \to\)

\(\to\)

\(\ds \hointr 1 \to\)

The inverse of $f_1$ is:

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

The inverse of $f_2$ is:

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

Arbitrary Finite Set with Itself

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

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

\(\text {(1)}: \quad\)

\(\ds \map {f_1} a\)

\(=\)

\(\ds a\)

\(\ds \map {f_1} b\)

\(=\)

\(\ds b\)

\(\text {(2)}: \quad\)

\(\ds \map {f_2} a\)

\(=\)

\(\ds a\)

\(\ds \map {f_2} b\)

\(=\)

\(\ds a\)

\(\text {(3)}: \quad\)

\(\ds \map {f_3} a\)

\(=\)

\(\ds b\)

\(\ds \map {f_3} b\)

\(=\)

\(\ds b\)

\(\text {(4)}: \quad\)

\(\ds \map {f_4} a\)

\(=\)

\(\ds b\)

\(\ds \map {f_4} b\)

\(=\)

\(\ds a\)

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

• 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.

• Left and Right Inverses of Mapping are Inverse Mapping

• Inverse Mapping is Unique

• Results about inverse mappings can be found here.

Sources

• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): inverse mapping