Pascal's simplex

In modern algebra, 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 antiisomorphic.

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

An example may clarify the idea. Let A be the binary relation (or graph) consisting of elements {1,2,3} and binary relation $\to$ defined as follows:

• $1\to 2;$
• $1\to 3;$
• $2\to 1.$

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 (called B^op) is the same set of elements with the opposite binary relation $\Leftarrow$:

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

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 $\varphi$ from A to B^op by

$\varphi (n)=\{\begin{array}{ll}a& \text{if }n=1;\\ b& \text{if }n=2;\\ c& \text{if }n=3.\end{array}$

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

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