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Let $X$ be a set. The full transformation semigroup $T_X$ is the set of all maps from $X$ into $X$ with composition of maps as binary operation. A map $\alpha \in T_X$, is a subset of the cartesian product $X \times X$. A partial map $\beta$ is a map from a subset $Y \subset X$ into $X$ and it is also a subset of $X \times X$. We'll call $P_X$, the set of all partial maps of subsets $Y \subset X$ into $X$. Let $\beta, \gamma \in P_X$ be two partial maps. Then regular definition of maps composition doesn't work with $\beta$ and $\gamma$ because in general $\operatorname{dom} \beta \not = \operatorname{im} \gamma$ and $\operatorname{dom} \gamma \not = \operatorname{im} \beta$. However if one uses composition of binary relations as defined in J. M. Howie, Fundamentals of Semigroup Theory, 1995 (eq. 1.4.2 p.16)

$$\rho \circ \sigma = \{ (x,y) \in X \times X : (\exists z \in X) (x,z) \in \rho \text{ and } (z,y) \in \sigma \}$$

then $P_X$ can be equipped with this operation to form a semigroup. Lets just call this semigroup $P_X$ and take a partial map $\beta \in P_X$. We have $\beta^1 = \beta$, $\beta^2 = \beta \circ \beta$ and more generally $\beta^n = \beta \circ \beta^{n-1}$. For $\beta^n = \beta^m$, $n \not = m$, $\beta^n$ is either the empty set $\emptyset$, an idempotent element $\not = \emptyset$ or an element generating a subgroup of $P_X$ with order strictly greater than 1.

My question concerns the case where $\beta^n = \emptyset$. I would be tempted to call $\beta$ a nilpotent partial map. Is it the usual name given to it in the litterature or does a nilpotent partial map refers usually to something different? References on this topic would also be welcomed.

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  • $\begingroup$ There are a few things I do not understand in your definitions. First, a subset of $X \times X$ is a relation, but in general it is not a function. Thus I assume you just want to say that $\alpha$ is a subset of $X \times X$, like $\{(1,2), (2,1)\}$ for the transformation $\alpha$ defined by $\alpha(1) = 2$ and $\alpha(2) = 1$. But then I do not see why $P(\alpha)$ is a semigroup. Actually, here you have $\alpha \in P(\alpha)$, but $\alpha^2 = \{(1,1), (2,2)\} \notin P(\alpha)$. $\endgroup$ – J.-E. Pin Feb 15 '16 at 20:43
  • $\begingroup$ Not all relations are functions, but the converse is true. But you are right with the last point. There is something wrong with my definition. I'll look it up more carefully and edit my question. $\endgroup$ – Nicolas Feb 15 '16 at 22:17
  • $\begingroup$ I have edited my question. I hope it makes more sense now. $\endgroup$ – Nicolas Feb 15 '16 at 22:45
  • $\begingroup$ The property on $\beta$ requires $X$ to be finite. $\endgroup$ – J.-E. Pin Feb 15 '16 at 23:21
  • $\begingroup$ Right, I overlooked the case where $\beta$ generates an infinite set. $\endgroup$ – Nicolas Feb 16 '16 at 0:01
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A preliminary remark. The set $F_n$ of all partial maps on $\{1, \dots, n\}$ is a monoid under the composition of partial maps. It can be embedded into $T_{n+1}$ by completing a partial function $f \in F_n$ to a transformation $\hat f$ on $\{0, 1, \dots, n\}$ by setting $\hat f(0) = 0$ and, for $x \in \{1, \dots, n\}$, $\hat f(x) = 0$ if $x$ is not in the domain of $f$.

Anyway, $F_n$ is a semigroup with zero. An element $s$ of a semigroup with zero is nilpotent if some power of $s$ is equal to $0$. Thus the term nilpotent partial map is perfectly legitimate in my opinion.

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