Editing Semigroup action (section)

==Transformation semigroups==

{{main|Transformation semigroup}}

A correspondence between transformation semigroups and semigroup actions is described below. If we restrict it to [[Faithful action|faithful]] semigroup actions, it has nice properties.

Any transformation semigroup can be turned into a semigroup action by the following construction. For any transformation semigroup <math>S</math> of <math>X</math>, define a semigroup action <math>T</math> of <math>S</math> on <math>X</math> as <math>T(s, x) = s(x)</math> for <math> s\in S, x\in X</math>. This action is faithful, which is equivalent to <math>curry(T)</math> being [[injective]].

Conversely, for any semigroup action <math>T</math> of <math>S</math> on <math>X</math>, define a transformation semigroup <math>S' = \{T_s \mid s \in S\}</math>. In this construction we "forget" the set <math>S</math>. <math>S'</math> is equal to the [[Image (mathematics)|image]] of <math>curry(T)</math>. Let us denote <math>curry(T)</math> as <math>f</math> for brevity. If <math>f</math> is [[injective]], then it is a semigroup [[isomorphism]] from <math>S</math> to <math>S'</math>. In other words, if <math>T</math> is faithful, then we forget nothing important. This claim is made precise by the following observation: if we turn <math>S'</math> back into a semigroup action <math>T'</math> of <math>S'</math> on <math>X</math>, then <math>T'(f(s), x) = T(s, x)</math> for all <math>s \in S, x \in X</math>. <math>T</math> and <math>T'</math> are "isomorphic" via <math>f</math>, i.e., we essentially recovered <math>T</math>. Thus, some authors<ref>{{cite book
| editor1-first = Michael A.
| editor1-last = Arbib
| year = 1968
| title = Algebraic Theory of Machines, Languages, and Semigroups
| publisher = Academic Press
| location = New York and London
| page = 83
}}</ref> see no distinction between faithful semigroup actions and transformation semigroups.