Abstract
Finding necessary and sufficient conditions for isomorphism between two semigroups of order-preserving transformations over an infinite domain with restricted range was an open problem. In this paper, we show a proof strategy to answer that question.
1. Introduction
For a nonempty set , let be the full transformation semigroup under composition of all maps from to . When is a partially ordered set (poset), a mapping in is called order-preserving if implies for all , and is regressive if for all . We denote by and the subsemigroups of of all order-preserving maps and all regressive maps on , respectively. The semigroups of order-preserving maps were first introduced by Howie in [1].
For , let denote the range of . In 1975, Symons [2] introduced and studied the subsemigroup where of consisting of with . Subsemigroups of transformations (with restricted range) of of this type have been studied extensively, including our work which we will mention later on. Regarding the semigroups of regressive type, in 1996, Umar proved that for any chains and , if and only if and are order-isomorphic (see in [3]). Later in [4], Saitô et al. generalized this result to partially ordered sets. They introduced the adjusted partially ordered set of a poset and proved that the order-isomorphism between and is a necessary and sufficient condition for the two semigroups to be isomorphic.
In this paper, we are also interested in studying the isomorphisms of subsemigroups of transformations with restricted range. Now, let us introduce the subsemigroups which will be of particular interest to us in this paper.
For a partially ordered set and a subset of , we let Then both of these are subsemigroups of .
In 2012, Udomkavanich and Jitjankarn proved in [5] that if and only if two adjusted chains and are order-structural isomorphic. This result leads us to study the isomorphism theorems for the semigroups of order-preserving type. It is known (e.g., [6], pages 222-223) that, for posets and , if and only if and are either order-isomorphic or order-anti-isomorphic. These necessary and sufficient conditions also hold for the isomorphisms on the semigroups of partial order-preserving transformations (see in [7]). In 2014, Fernandes et al. [8] show that these conditions apply for and to be isomorphic when and are finite as well. In this paper, we study the case when and are infinite chains. Since is trivial when , we omit this case.
Throughout the paper, we assume that and are chains, , and . The following statement is known.
If there is an order-(anti)-isomorphism such that then .
It is natural to ask whether the converse of the above result holds. Nevertheless, our work shows that it may not be the case if . To be precise, we derive that the converse of the statement (2) holds when .
To prove the statements, we apply in a similar fashion to [5] the idea of using adjusted chains. To do so, we will first introduce some notation and definitions that will be useful in Section 2. In Section 3, some homomorphism properties which are preserved under isomorphism will be given. Lastly, the isomorphism theorems for the semigroups of the type when is an infinite chain are determined in Section 4.
2. Basic Notations and Results
Let be a subchain of a chain . Let denote the set of all equivalence classes of such that each class contains all elements in with no elements in lying between them. Then we consider as a chain under the partial order induced by the chain in the natural way. This chain is an adjusted chain, denoted by .
For each with , the intervals , , , in are defined naturally and we define the following intervals:
For a nonempty subset of a chain , is said to be convex if for such that , implies ; is called an upper(lower)-convex subset of if () for all and .
For a convex subset of , we define
For convenience, if , let be the element of whose range is .
Given , . We will define some order-preserving maps of as follows.(i)For a convex subset of and such that (or ), we write where and if .(ii)When , for a lower-convex subset of and such that , we write where and .(iii) When , for an upper-convex subset of and such that , we write where and .
For , we denote .
For , we define the partial graph of transformation , denoted by , in the following way: is the set of upper vertices, is the set of lower vertices such that all vertices are placed in order, and is the set of (directed) edges which each element is in the form , where for . Notice that the number of components in each partial graph is equal to the number of elements in its range. Furthermore, the components, considered from left to right, are placed in the same order as their related elements in the range.
Example 1. For the transformation defined by
the set of upper vertices is , the set of lower vertices is , and . Then the graph has the form shown in Figure 1.
The partial graph has four components placed in order from left to right.
Theorem 2. If , then and are either order-isomorphic or order-anti-isomorphic.
Proof. Let be an isomorphism. For each , there is an element such that by idempotent and right zero properties of and . The map becomes a bijective map from onto . It remains to show that this map is either order-preserving or order-anti-preserving. Let be such that and . Since is a chain and the map is one-to-one, it must be that or and or . Now, we have such that
Then
Consequently,
Since , it follows that implies and implies . This proves that and are either order-isomorphic or order-anti-isomorphic.
From now on, let denote an isomorphism from and . The order-(anti)-isomorphism from onto , defined in the proof of Theorem 2, is denoted by . It is easy to see that the order-(anti)-isomorphism from onto , induced by the isomorphism , is the inverse function of . That is, Notice that, by considering and instead of and , respectively, all results that hold for also hold for .
3. Some Homomorphism Properties
In this section, we study some properties of transformations which will be preserved under a homomorphism. First, we will study the structure of and through when . Then we derive that two graphs of and are isomorphic. Moreover, the order of components (in the sense of partial graph) is also preserved.
Without loss of generality, we assume that is order-preserving from now on. The other case that is order-anti-preserving can be done by the same process.
Lemma 3. For each , the following statements hold. (i).(ii)For such that , In particular, if is an idempotent, then .
Proof. (i) Let . Then . Since , it follows that . Similarly, if , then ; that is, . Then .
(ii) For such that , let . Then , by (i), . That is, . Since , it follows that . Then . Similarly, by considering instead of , . Thus the equality is obtained.
Lemma 4. For each , if and , then .
Proof. Let and . Assume that is neither maximum nor minimum in . Choose such that and let . Then is an idempotent with . By Lemma 3, . Suppose in the contrary that . Then we have . Since is finite, this guarantees the existence of an idempotent in with . Then is an idempotent in , by Lemma 3, . However, which is a contradiction. If is either maximum or minimum, it can be proved in the same way by defining as before and choosing if is minimum, and if is maximum.
By Lemmas 3 and 4, the following proposition is directly obtained.
Proposition 5. For each , we have the following.(i).(ii)For any , .
This proposition leads us to define an interesting equivalence relation on the semigroup of full transformations with restricted range.
Given a transformation , the -structure is the partial graph which its components will be placed in the same order as their related elements in the range. Here we define an equivalence relation on by Indeed, it is equivalent to and . The -class containing is denoted by . It is very clear that when , is -trivial. By Proposition 5, we have that and have the same structure for all .
Next, we will construct an extension of to be an order-isomorphism on the adjusted chains.
Lemma 6. Suppose that two classes and are the minimum and the maximum of , respectively. Let be such that , and and as a lower-convex subset and an upper-convex subset of and , respectively. Then for some lower-convex and upper-convex of the minimum and the maximum of , respectively.
Proposition 7. For each , there is a corresponding such that the extended map of from onto is an order-isomorphism. Moreover, .
Proof. Let be such that for some . We choose as an idempotent in whose range is . Since two partial graphs of transformations and have the same structure, by Proposition 5, it follows that . Due to the structure of , the cardinality of is depending only on . Indeed, . This implies the existence of with and .
Suppose is maximum (or minimum) in . For any such that , we consider (or ). By Lemma 6 and using the same argument, our proof is finished.
From Proposition 7, the union of all these extensions form an order-isomorphism, denoted by (with respect to ), from onto such that for and . We notice that is an order-structural isomorphism (as defined in [5]). This conclusion results in the isomorphism theorems between the two semigroups for an infinite discrete chain.
Theorem 8. Let and be infinite discrete chains. Then if and only if there is an order-(anti)-isomorphism such that .
Nevertheless, the property that is not sufficient to determine the isomorphism for an uncountable chain. As a result, we study more of homomorphism properties associated with a class of .
Lemma 9. Let be such that (or ) for some . Then for each convex subset of , for some convex subset of with (or ).
Proof. By Proposition 5, it follows that . Since , by Lemma 3, we have that . As is order-preserving such that is in its range, there exists the unique class in , namely , containing a convex subset .
Proposition 10. Let be such that (or ) for some . Then for each , for some .
Proof. Let and stand for two idempotents in such that with and . Let be a convex subset of such that . By Lemma 9, we obtain that for some convex subset of . Suppose in the contrary that . We choose and as two convex subsets of which form a partition of , and is a lower bound of . Since , it follows that Then is an upper-convex subset of . Since , we have Then is a lower-convex subset of . It can be seen that . Then which contradicts to
Proposition 11. For with , the following statements hold. (i)If , then, for , (ii)If , then, for ,
Proof. (i) Suppose . Let . Suppose that . We let and . Clearly, . Then . By applying the same process as in the proof of Proposition 10, we obtain that . It is clear that where is a lower-convex subset of . These imply that for some .
(ii) can be proved similarly to (i).
4. Isomorphism Theorems
In the last section, we take care of the case . For convenience, we here denote by where , and are three classes in . We observe that there are only 5 classes in whose partial graph of transformations is one of the forms shown in Figure 2.
The following results are directly derived.
Lemma 12. For , , we have that(i) and are trivial,(ii),(iii) and .
Proof. Since there are only two constant maps, (i) is proved. To show (ii), it is easy to see that each element in determines the consequent map in and vice versa. Hence the bijection between the two sets is constructed. The same idea can also be applied to show and .
Theorem 13. if and only if for all .
Proof. Suppose that, for , and are two representations of order-preserving maps having the same partial graph in and , respectively. By Lemma 12, we let be a bijection from onto for . To show that is an isomorphism, we let . It is easy to see that the pairwise composition of five graph structures can be one of the following maps: for , either , or :
Suppose . One of the following statements hold: (i),(ii),(iii) and . It is clear that .
For the rest, it can be proved directly.
Example 14. Let , , , . Theorem 13 tells us that , yet it is clear that and are not order or order-anti-isomorphic.
Next, we will prove that when , the converse of (2) holds.
Theorem 15. Suppose that . Then if and only if there is an order-(anti)-isomorphism from onto such that .
Proof. It remains to show that, for each , and are order-isomorphic. Let be a class in . We will consider in two cases.
Case 1. Consider or .
Without loss of generality we can assume that . We choose with . For any with . Consider and . By Proposition 11, we have and . Let . Then
It follows that
Since and , these imply that .
Case 2. is neither nor .
Then there are such that or .
By using Proposition 10 and following the same proof as in Case 1, we derive the result.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
This research was partially supported by SWU endowment fund year 2014 and also by WU. The authors thank the (anonymous) referees for helpful comments.