Antiisomorphism

In Category theory, a branch of formal mathematics, an antiisomorphism (or anti-isomorphism) between structured sets A and B is an isomorphism from A to the opposite of B (or equivalently from the opposite of A to B). If there exists an antiisomorphism between two structures, they are said to be antiisomorphic.

Intuitively, to say that two mathematical structures are antiisomorphic is to say that they are basically opposites of one another.

The concept is particularly useful in an algebraic setting, as, for instance, when applied to rings.

Simple example

Let A be the binary relation (or directed graph) consisting of elements {1,2,3} and binary relation $\rightarrow$ defined as follows:

• $1 \rightarrow 2;$
• $1 \rightarrow 3;$
• $2 \rightarrow 1.$

Let B be the binary relation set consisting of elements {a,b,c} and binary relation $\Rightarrow$ defined as follows:

• $b \Rightarrow a;$
• $c \Rightarrow a;$
• $a \Rightarrow b.$

Note that the opposite of B (denoted B^op) is the same set of elements with the opposite binary relation $\Leftarrow$ (that is, reverse all the arcs of the directed graph):

• $b \Leftarrow a;$
• $c \Leftarrow a;$
• $a \Leftarrow b.$

If we replace a, b, and c with 1, 2, and 3 respectively, we will see that each rule in B^op is the same as some rule in A. That is, we can define an isomorphism $\phi$ from A to B^op by

$\phi(n) = \begin{cases}a&\mbox{if }n=1;\\ b&\mbox{if }n=2;\\ c&\mbox{if }n=3.\end{cases}$

This $\phi$ is an antiisomorphism between A and B.

Ring anti-isomorphisms

Specializing the general language of category theory to the algebraic topic of rings, we have: Let R and S be rings and f: R → S a bijection between them, then if

$f(x +_R y) = f(x) +_S f(y) \text{ and } f(x \cdot_R y) = f(y) \cdot_S f(x) \text{ for all } x,y \in R$

f will be called a ring anti-isomorphism. If R = S then f will be called a ring anti-automorphism.

An example of a ring anti-automorphism is given by the conjugate mapping of quaternions:

$x_0 + x_1 \mathbf{i} + x_2 \mathbf{j} + x_3 \mathbf{k}\quad \mapsto \quad x_0 - x_1 \mathbf{i} - x_2 \mathbf{j} - x_3 \mathbf{k}.$