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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

# A further study on ordered regular equivalence relations in ordered semihypergroups

Jian Tang
/ Xinyang Feng
/ Bijan Davvaz
/ Xiang-Yun Xie
• Corresponding author
• School of Mathematics and Computational Science, Wuyi University, Jiangmen, Guangdong, 529020, China
• Email
• Other articles by this author:
Published Online: 2018-03-20 | DOI: https://doi.org/10.1515/math-2018-0016

## Abstract

In this paper, we study the ordered regular equivalence relations on ordered semihypergroups in detail. To begin with, we introduce the concept of weak pseudoorders on an ordered semihypergroup, and investigate several related properties. In particular, we construct an ordered regular equivalence relation on an ordered semihypergroup by a weak pseudoorder. As an application of the above result, we completely solve the open problem on ordered semihypergroups introduced in [B. Davvaz, P. Corsini and T. Changphas, Relationship between ordered semihypergroups and ordered semigroups by using pseuoorders, European J. Combinatorics 44 (2015), 208–217]. Furthermore, we establish the relationships between ordered regular equivalence relations and weak pseudoorders on an ordered semihypergroup, and give some homomorphism theorems of ordered semihypergroups, which are generalizations of similar results in ordered semigroups.

MSC 2010: 20N20; 06F05; 20M12

## 1 Introduction

The theory of algebraic hyperstructures was first introduced by Marty [1], which is a generalization of the theory of ordinary algebraic structures. Later on, people have observed that hyperstructures have many applications to several branches of both pure and applied sciences (see [2, 3, 4, 5]). In particular, a semihypergroup is the simplest algebraic hyperstructure which possess the properties of closure and associativity. Nowadays, semihypergroups have been found useful for dealing with problems in different areas of algebraic hyperstructures. Many authors have studied different aspects of semihypergroups, for instance, see [6, 7, 8, 9, 10, 11, 12, 13]. Especially, regular and strong regular relations on semihypergroups have been introduced and investigated in [14].

On the other hand, the theory of ordered algebraic structure was first studied by Fuchs [15] in 1960’s. In particular, an ordered semigroup is a semigroup together with a partial order that is compatible with the semigroup operation. Ordered semigroups have several applications in the theory of sequential machines, formal languages, computer arithmetics and error-correcting codes. There are several results which have been added to the theory of ordered semigroups by many researchers, for example, N. Kehayopulu, M. Tsingelis, X. Y. Xie, M. Kuil and others. For partial results, the reader is referred to [16, 17, 18, 19, 20, 21, 22].

A theory of hyperstructures on ordered semigroups has been recently developed. In [23], Heidari and Davvaz applied the theory of hyperstructures to ordered semigroups and introduced the concept of ordered semihypergroups, which is a generalization of the concept of ordered semigroups. Also see [2]. The work on ordered semihypergroup theory can be found in [24, 25, 26, 27]. It is worth pointing out that Davvaz et al. [25] introduced the concept of a pseudoorder on an ordered semihypergroup, and extended some results in [16] on ordered semigroups to ordered semihypergroups. In particular, they posed an open problem about ordered semihypergroups: Is there a regular relation ρ on an ordered semihypergroup (S, ∘, ≤) for which S/ρ is an ordered semihypergroup? To answer the above open problem, Gu and Tang attempted in [28] to introduce the concept of ordered regular equivalence relations on ordered semihypergroups, and construct an ordered regular equivalence relation ρI on an ordered semihypergroup S by a proper hyperideal I of S such that the corresponding quotient structure is also an ordered semihypergroup. However, they only provide a partial solution to the above problem. In fact, for an ordered semihypergroup S, S doesn’t necessarily exist a proper hyperideal (see Example 3.3). As a further study, in the present paper we study the ordered regular equivalence relations on ordered semihypergroups in detail and completely solve the open problem given by Davvaz et al. in [25].

The rest of this paper is organized as follows. In Section 2 we recall some basic notions and results from the hyperstructure theory which will be used throughout this paper. In Section 3, we introduce the concept of weak pseudoorders on an ordered semihypergroup S, and illustrate this notion is a generalization of the concept of pseudoorders on S by some examples. Furthermore, the properties of weak pseudoorders on an ordered semihypergroup are investigated. In particular, we construct an ordered regular equivalence relation on an ordered semihypergroup by a weak pseudoorder. As an application of the above result, we give a complete answer to the open problem given by Davvaz et al. in [25]. In Section 4, the relationships between ordered regular equivalence relations and weak pseudoorders on an ordered semihypergroup is established, and some homomorphism theorems of ordered semihypergroups by weak pseudoorders are given. Some conclusions are given in the last Section.

## 2 Preliminaries and some notations

Recall that a hypergroupoid (S, ∘) is a nonempty set S together with a hyperoperation, that is a map ∘ : S × SP(S), where P(S) denotes the set of all the nonempty subsets of S. The image of the pair (x, y) is denoted by xy. If xS and A, BP(S), then AB is defined by $\begin{array}{}A\circ B=\bigcup _{a\in A,b\in B}^{}a\circ b.\end{array}$ Also Ax is used for A ∘ {x} and xA for {x} ∘ A. Generally, the singleton {x} is identified by its element x.

We say that a hypergroupoid (S, ∘) is a semihypergroup if the hyperoperation “∘” is associative, that is, (xy) ∘ z = x ∘ (yz) for all x, y, zS (see [14]).

We now recall the notion of ordered semihypergroups from [23]. An algebraic hyperstructure (S, ∘, ≤) is called an ordered semihypergroup (also called po-semihypergroup in [23]) if (S, ∘) is a semihypergroup and (S, ≤) is a partially ordered set such that: for any x, y, aS, xy implies axay and xaya. Here, if A, BP(S), then we say that AB if for every aA there exists bB such that ab. Clearly, every ordered semigroup can be regarded as an ordered semihypergroup, see [26]. By a subsemihypergroup of an ordered semihypergroup S we mean a nonempty subset A of S such that AAA. A nonempty subset A of a semihypergroup (S, ∘) is called a left (resp. right) hyperideal of S if (1) SAA (resp. ASA) and (2) If aA and Sba, then bA. If A is both a left and a right hyperideal of S, then it is called a hyperideal of S (see [26]).

Let ρ be a relation on a semihypergroup (S, ∘) or an ordered semihypergroup (S, ∘, ≤). If A and B are nonempty subsets of S, then we set $Aρ→B ⇔ (∀a∈A)(∃b∈B) aρb,Aρ←B ⇔ (∀b′∈B)(∃a′∈A) a′ρb′,Aρ¯B ⇔ Aρ→B and Aρ←B,Aρ¯¯B ⇔ (∀a∈A,∀b∈B) aρb.$

An equivalence relation ρ on S is called regular [14, 25] if $(∀x,y,a∈S) xρy⇒a∘x ρ¯ a∘y and x∘a ρ¯ y∘a;$

ρ is called strongly regular [14, 25] if $(∀x,y,a∈S) xρy⇒a∘x ρ¯¯ a∘y and x∘a ρ¯¯ y∘a.$

#### Example 2.1

We consider a set S:= {a, b, c, d} with the following hyperoperation “∘” and the order “≤”: $∘abcda {a,d} {a,d} {a,d} {a} b {a,d} {b} {a,d} {a,d} c {a,d} {a,d} {c} {a,d} d {a} {a,d} {a,d} {d}≤:={(a,a),(a,b),(a,c),(b,b),(c,c),(d,b),(d,c),(d,d)}.$

We give the covering relation “≺” and the figure of S as follows: $≺={((a,b),(a,c),(d,b),(d,c)}.$

Then (S, ∘, ≤) is an ordered semihypergroup (see [26]). Let ρ1, ρ2 be equivalence relations on S defined as follows: $ρ1:={(a,a),(a,d),(b,b),(c,c),(d,a),(d,d)},ρ2:={(a,a),(a,c),(b,b),(c,a),(c,c),(d,d)}.$

Then

1. ρ1 is a strongly regular relation on S.

2. ρ2 is a regular relation on S, but it is not a strongly regular relation on S. In fact, since aρc, while aa ρ̿ ca doesn’t hold.

#### Lemma 2.2

([14]). Let (S, ∘) be a semihypergroup and ρ an equivalence relation on S. Then

1. If ρ is regular, then (S/ρ, ⊗) is a semihypergroup with respect to the following hyperoperation: (a)ρ ⊗ (b)ρ = $\begin{array}{}\bigcup _{c\in a\circ b}\end{array}$(c)ρ. In this case, we call (S/ρ, ⊗) a quotient semihypergroup.

2. If ρ is strongly regular, then (S/ρ, ⊗) is a semigroup with respect to the following operation: (a)ρ ⊗ (b)ρ = (c)ρ for allab. In this case, we call (S/ρ, ⊗) a quotient semigroup.

A relation ρ on an ordered semihypergroup (S, ∘, ≤) is called pseudoorder [25] if it satisfies the following conditions: (1) ≤⊆ ρ, (2) aρb and bρc imply aρc, i.e., ρρρ and (3) aρb implies ac ρ̿ bc and ca ρ̿ cb, for all cS.

#### Example 2.3

Consider the ordered semihypergroup (S, ∘, ≤) given in Example 2.1, and define a relation ρ on S as follows: $ρ:={(a,a),(a,b),(a,c),(a,d),(b,b),(c,c),(d,a),(d,b),(d,c),(d,d)}.$

It is not difficult to verify that ρ is a pseudoorder on S.

Let (S, ∘, ≤), (T, ⋄, ⪯) be two ordered semihypergroups and f: ST a mapping from S to T. f is called isotone if xy implies f(x)⪯ f(y), for all x, yS. f is called reverse isotone if x, yS, f(x) ⪯ f(y) implies xy. f is called homomorphism [25] if it is isotone and satisfies f(x) ⋄ f(y) = $\begin{array}{}\bigcup _{z\in x\circ y}\end{array}$f(z), for all x, yS. f is called isomorphism if it is homomorphism, onto and reverse isotone. The ordered semihypergroups S and T are called isomorphic, in symbol ST, if there exists an isomorphism between them.

#### Remark 2.4

Let S and T be two ordered semihypergroups. If f is a homomorphism and reverse isotone mapping from S to T, then SIm(f).

In order to investigate the structure of quotient ordered semihypergroups, Gu and Tang [28] introduced the concept of (strongly) ordered regular equivalence relations on an ordered semihypergroup. A regular (resp. strongly regular) equivalence relation ρ on an ordered semihypergroup S is called ordered regular (resp. strongly ordered regular) if there exists an order relation “⪯” on (S/ ρ, ⊗) such that:

1. (S/ρ, ⊗, ⪯) is an ordered semihypergroup (resp. ordered semigroup), where the hyperoperation “ ⊗” is defined as one in Lemma 2.2.

2. The mapping φ : SS/ρ, x ↦ (x)ρ is isotone, that is, φ is a homomorphism from S onto S/ρ.

The reader is referred to [2, 19] for notation and terminology not defined in this paper.

## 3 Weak pseudoorders on ordered semihypergroups

In [25], Davvaz et al. obtained an ordered semigroup from an ordered semihypergroup by means of pseudoorders. In the same paper they posed the following open problem about ordered semihypergroups.

#### Problem 3.1

Is there a regular equivalence relation ρ on an ordered semihypergroup (S, ∘, ≤) for which S/ρ is an ordered semihypergroup?

To answer the above open problem, Gu and Tang [28] defined an equivalence relation ρI on an ordered semihypergroup S as follows: $ρI:={(x,y)∈S∖I×S∖I | x=y}∪(I×I),$

where I is a proper hyperideal of S. Furthermore, they provided the following theorem:

#### Theorem 3.2

([28]). Let (S, ∘, ≤) be an ordered semihypergroup and I a hyperideal of S. Then ρI is an ordered regular equivalence relation on S.

As we have seen in Theorem 3.2, for an ordered semihypergroup S, there exists a regular equivalence relation ρI on S such that the corresponding quotient structure S/ρI is also an ordered semihypergroup, where I is a proper hyperideal I of S. However, in general, there need not exist proper hyperideals in S. We can illustrate it by the following example.

#### Example 3.3

We consider a set S:= {a, b, c, d} with the following hyperoperation “∘” and the order “≤”: $∘abcda {a} {a,b} {a,c} Sb {b} {b} {b,d} {b,d}c {c} {c,d} {c} {c,d}d {d} {d} {d} {d}≤:={(a,a),(a,b),(b,b),(c,c),(c,d),(d,d)}.$

We give the covering relation “≺” and the figure of S as follows: $≺={(a,b),(c,d)}.$

Then (S, ∘, ≤) is an ordered semihypergroup. Moreover, it is a routine matter to verify that there do not exist proper hyperideals in S.

By the above example, it can be seen that Theorem 3.2 only provides a partial solution to Problem 3.1, and not completely. In order to fully solve the open problem, we need define and study the weak pseudoorders on an ordered semihypergroup.

If A, BP(S), then we set $A ρ~ B ⇔ (∀a∈A)(∃b∈B) aρb and bρa & (∀b′∈B)(∃a′∈A) a′ρb′ and b′ρa′.$

#### Definition 3.4

Let (S, ∘, ≤) be an ordered semihypergroup and ρ a relation on S. ρ is called weak pseudoorder if it satisfies the following conditions:

1. ≤⊆ ρ;

2. aρb and bρc imply aρc;

3. aρb implies ac ρ⃗ bc and ca ρ⃗ cb, for all cS;

4. aρb and bρa imply ac ρ̃ bc and ca ρ̃ cb, for all cS.

#### Remark 3.5

Note that if (S, ∘, ≤) is an ordered semigroup, then Definition 3.4 coincides with Definition 1 in [16].

#### Lemma 3.6

Let (S, ∘, ≤) be an ordered semihypergroup. Then there exists a weak pseudoorder relation on S.

#### Proof

With a small amount of effort one can verify that the order relation “≤” on S is a weak pseudoorder relation on S. □

One can easily observe that every pseudoorder relation on an ordered semihypergroup S is a weak pseudoorder on S. However, the converse is not true, in general, as shown in the following example.

#### Example 3.7

We consider a set S:= {a, b, c, d} with the following hyperoperation “∘” and the order “≤”: $∘abcda {a,d} {a,d} {a,d} {a}b {a,d} {b} {a,d} {a,d}c {a,d} {a,d} {c} {a,d}d {a} {a,d} {a,d} {d}≤:={(a,a),(a,c),(b,b),(c,c),(d,c),(d,d)}.$

We give the covering relation “≺” and the figure of S as follows: $≺={(a,c),(d,c)}.$

Then (S, ∘, ≤) is an ordered semihypergroup. Let ρ be a relation on S defined as follows: $ρ:={(a,a),(b,b),(c,c),(d,d),(a,c),(b,a),(b,c),(d,a),(d,c)}.$

We can easily verify that ρ is a weak pseudoorder on S, but it is not a pseudoorder on S. In fact, since aρc, while ab ρ̿ cb doesn’t hold.

In the following, we shall construct a regular equivalence relation on an ordered semihypergroup by a weak pseudoorder, from which we can answer Problem 3.1 completely.

#### Theorem 3.8

Let (S, ∘, ≤) be an ordered semihypergroup and ρ a weak pseudoorder on S. Then, there exists a regular equivalence relation ρ* on S such that S/ρ* is an ordered semihypergroup.

#### Proof

We denote by ρ* the relation on S defined by $ρ∗:={(a,b)∈S×S | aρb and bρa} (=ρ∩ρ−1).$

First, we claim that ρ* is a regular equivalence relation on S. In fact, for any aS, clearly, (a, a) ∈≤⊆ ρ, so * a. If (a, b) ∈ ρ*, then aρb and bρa. Thus (b, a) ∈ ρ*. Let (a, b) ∈ ρ* and (b, c) ∈ ρ*. Then aρb, bρa, bρc and cρb. Hence aρc and cρa. It implies that (a, c) ∈ ρ*. Thus ρ* is an equivalence relation on S. Now, let * b and cS. Then aρb and bρa. Since ρ is a weak pseudoorder on S, by condition (4) of Definition 3.4, we have $a∘c ρ~ b∘c and c∘a ρ~ c∘b.$

Thus, for every xac, there exists ybc such that xρy and yρx which imply that *y, and for every y′ ∈ bc, there exists x′ ∈ ac such that xρ y′ and yρ x’ which mean that xρ*y′. It thus follows that ac ρ* bc. Similarly, it can be obtained that ca ρ* cb. Hence ρ* is indeed a regular equivalence relation on S. Thus, by Lemma 2.2, (S/ρ*, ⊗) is a semihypergroup with respect to the following hyperoperation: $(∀(a)ρ∗,(b)ρ∗∈S/ρ∗) (a)ρ∗⊗(b)ρ∗=⋃c∈a∘b(c)ρ∗.$

Now, we define a relation ⪯ρ on S/ρ* as follows: $⪯ρ:={((x)ρ∗,(y)ρ∗)∈S/ρ∗×S/ρ∗ | (x,y)∈ρ}.$

Then (S/ρ*, ⪯ρ) is a poset. Indeed, suppose that (x)ρ*S/ρ*, where xS. Then (x, x) ∈≤⊆ ρ. Hence, (x)ρ*ρ (x)ρ*. Let (x)ρ*ρ (y)ρ* and (y)ρ*ρ (x)ρ*. Then xρy and yρx. Thus * y, and we have (x)ρ* = (y)ρ*. Now, if (x)ρ*ρ (y)ρ* and (y)ρ*ρ (z)ρ*, then xρy and yρz. Hence, by hypothesis, xρz, and we conclude that (x)ρ*ρ (z)ρ*.

Furthermore, let (x)ρ*, (y)ρ*, (z)ρ*S/ρ* be such that (x)ρ*ρ (y)ρ*. Then xρy and zS. Since ρ is a weak pseudoorder on S, by condition (3) of Definition 3.4, we have xz ρ⃗ yz and zx ρ⃗ zy. Thus, for any axz there exists byz such that aρb. This implies that (a)ρ*ρ (b)ρ*. Hence we have $(x)ρ∗⊗(z)ρ∗=⋃a∈x∘z(a)ρ∗⪯ρ⋃b∈y∘z(b)ρ∗=(y)ρ∗⊗(z)ρ∗.$

In a similar way, it can be obtained that (z)ρ* ⊗ (x)ρ*ρ (z)ρ* ⊗ (y)ρ*. Therefore, (S/ρ*, ⊗, ⪯ρ) is an ordered semihypergroup. □

Furthermore, we have the following proposition:

#### Proposition 3.9

Let (S, ∘, ≤) be an ordered semihypergroup and ρ a weak pseudoorder on S. Then ρ* is an ordered regular equivalence relation on S, where ρ* = ρρ−1.

#### Proof

By Theorem 3.8, (S/ρ*, ⊗, ⪯ρ) is an ordered semihypergroup, where the order relation ⪯ρ is defined as follows: $⪯ρ:={((x)ρ∗,(y)ρ∗)∈S/ρ∗×S/ρ∗ | (x,y)∈ρ}.$

Also, let x, yS and xy. Then, since ρ is a weak pseudoorder on S, (x, y) ∈≤⊆ ρ, and thus ((x)ρ*, (y)ρ*) ∈ ⪯ρ, i.e., (x)ρ*ρ (y)ρ*. Therefore, ρ* is an ordered regular equivalence relation on S. □

#### Example 3.10

We consider a set S:= {a, b, c, d, e} with the following hyperoperation “∘” and the order “≤”: $∘abcdea {a,b} {a,b} {c} {c} {c}b {a,b} {a,b} {c} {c} {c}c {a,b} {a,b} {c} {c} {c}d {a,b} {a,b} {c} {d,e} {d}e {a,b} {a,b} {c} {d} {e}≤:={(a,a),(a,c),(a,d),(a,e),(b,b),(b,c),(b,d),(b,e),(c,c),(c,d),(c,e),(d,d),(e,e)}.$

We give the covering relation “≺” and the figure of S as follows: $≺={(a,c),(b,c),(c,d),(c,e)}.$

Then (S, ∘, ≤) is an ordered semihypergroup (see [26]). Let ρ be a weak pseudoorder on S defined as follows: $ρ={(a,a),(a,c),(a,d),(a,e),(b,b),(b,c),(b,d),(b,e),(c,c),(c,d),(c,e),(d,c),(d,d),(d,e),(e,c),(e,d),(e,e)}.$

Applying Theorem 3.8, we get $ρ∗={(a,a),(b,b),(c,c),(c,d),(c,e),(d,c),(d,d),(d,e),(e,c),(e,d),(e,e)}.$

Then S/ρ* = {w1, w2, w3, where w1 = {a}, w2 = {b} and w3 = {c, d, e}. Moreover, (S/ρ*, ⊗, ⪯ρ) is an ordered semihypergroup with the hyperoperation “ ⊗” and the order “ ⪯ρare given below: $∘w1w2w3w1 {w1,w2} {w1,w2} {w3}w2 {w1,w2} {w1,w2} {w3}w3 {w1,w2} {w1,w2} {w3}≤:={(w1,w1),(w1,w3),(w2,w2),(w2,w3),(w3,w3)}.$

We give the covering relation “≺ρand the figure of S/ρ* as follows: $≺ρ={(w1,w3),(w2,w3)}.$

Similar to Theorem 4.3 in [25], we have the following theorem.

#### Theorem 3.11

Let (S, ∘, ≤) be an ordered semihypergroup and ρ a weak pseudoorder on S. Let $A:={θ | θ isaweakpseudoorderon S suchthat ρ⊆θ}.$

Let 𝓑 be the set of all weak pseudoorders on S/ρ*. Then, card(𝓐) = card(𝓑).

#### Proof

For θ ∈ 𝓐, we define a relation θ′ on S/ρ* as follows: $θ′:={((x)ρ∗,(y)ρ∗)∈S/ρ∗×S/ρ∗ | (x,y)∈θ}.$

To begin with, we claim that θ′ is a weak pseudoorder on S/ρ*. To prove our claim, let ((x)ρ*, (y)ρ*) ∈⪯ρ. Then, by Theorem 3.8, (x, y) ∈ ρθ, which implies that ((x)ρ*, (y)ρ*) ∈ θ′. Thus, ⪯ρθ′. Now, assume that ((x)ρ*, (y)ρ*) ∈ θ′ and ((y)ρ*, (z)ρ*) ∈ θ′. Then, (x, y) ∈ θ and (y, z) ∈ θ. It implies that (x, z) ∈ θ. Hence, ((x)ρ*, (z)ρ*) ∈ θ′. Also, let ((x)ρ*, (y)ρ*) ∈ θ′ and (z)ρ*S/ρ*. Then, (x, y) ∈ θ and zS. Since θ is a weak pseudoorder on S, we have xz θ⃗ yz and zx θ⃗ zy. Thus, for every axz there exists byz, we have (a, b) ∈ θ. This implies that ((a)ρ*, (b)ρ*) ∈ θ′, and thus we have $(x)ρ∗⊗(z)ρ∗=⋃a∈x∘z(a)ρ∗ θ′→ ⋃b∈y∘z(b)ρ∗=(y)ρ∗⊗(z)ρ∗.$

Similarly, we obtain that (z)ρ* ⊗ (x)ρ* θ′⃗ (z)ρ* ⊗ (y)ρ*. Furthermore, let ((x)ρ*, (y)ρ*), (z)ρ*S/ρ* be such that ((x)ρ*, (y)ρ*) ∈ θ′ and ((y)ρ*, (x)ρ*) ∈ θ′. Then (x, y) ∈ θ, (y, x) ∈ θ and zS. By hypothesis, we have xz θ͠ yz and zx θ͠ zy. Then, similarly as in the above proof, it can be obtained that (x)ρ* ⊗ (z)ρ* θ′͠ (y)ρ* ⊗ (z)ρ* and (z)ρ* ⊗ (x)ρ* θ′͠ (z)ρ* ⊗ (y)ρ*. Therefore, θ′ is indeed a weak pseudoorder on S/ρ*.

Now, we define the mapping f: 𝓐 → 𝓑 by f(θ) = θ′, ∀ θ ∈ 𝓐. Then, f is a bijection from 𝓐 onto 𝓑. In fact,

1. f is well defined. Indeed, let θ1, θ2 ∈ 𝓐 and θ1 = θ2. Then, for any ((x)ρ*, (y)ρ*) ∈ $\begin{array}{}{\theta }_{1}^{\prime }\end{array}$ we have (x, y) ∈ θ1 = θ2. It implies that ((x)ρ*, (y)ρ*) ∈ $\begin{array}{}{\theta }_{2}^{\prime }\end{array}$. Hence $\begin{array}{}{\theta }_{1}^{\prime }\subseteq {\theta }_{2}^{\prime }\end{array}$ By symmetry, it can be obtained that $\begin{array}{}{\theta }_{2}^{\prime }\subseteq {\theta }_{1}^{\prime }\end{array}$.

2. f is one to one. In fact, let θ1, θ2 ∈ 𝓐 and $\begin{array}{}{\theta }_{1}^{\prime }={\theta }_{2}^{\prime }.\end{array}$ Assume that (x, y) ∈ θ1. Then, ((x)ρ*, (y)ρ*) ∈ $\begin{array}{}{\theta }_{1}^{\prime }\end{array}$ and thus ((x)ρ*, (y)ρ*) ∈ $\begin{array}{}{\theta }_{2}^{\prime }\end{array}$. This implies that (x, y) ∈ θ2. Thus, θ1θ2. Similarly, we obtain θ2θ1.

3. f is onto. In fact, let δ ∈ 𝓑. We define a relation θ on S as follows: $θ:={(x,y)∈S×S | ((x)ρ∗,(y)ρ∗)∈δ}.$

We show that θ is a weak pseudoorder on S and ρθ. Assume that (x, y) ∈ ρ. Then, by Theorem 3.8, ((x)ρ*, (y)ρ*) ∈⪯ρδ, and thus (x, y) ∈ θ. This implies that ρθ. If (x, y) ∈ ≤, then (x, y) ∈ ρθ. Hence, ≤⊆ θ. Let now (x, y) ∈ θ and (y, z) ∈ θ. Then ((x)ρ*, (y)ρ*) ∈ δ and ((y)ρ*, (z)ρ*) ∈ δ. Hence ((x)ρ*, (z)ρ*) ∈ δ, which implies that (x, z) ∈ θ. Also, let x, y, zS be such that (x, y) ∈ θ. Then ((x)ρ*, (y)ρ*) ∈ δ and (z)ρ*S/ρ*. Since δ is a weak pseudoorder on S/ρ*, we have (x)ρ* ⊗ (z)ρ* δ⃗ (y)ρ* ⊗ (z)ρ* and (z)ρ* ⊗ (x)ρ* δ⃗ (z)ρ* ⊗ (y)ρ*, i.e., $\begin{array}{}\bigcup _{a\in x\circ z}\left(a{\right)}_{{\rho }^{\ast }}\text{\hspace{0.17em}}\stackrel{\to }{\delta }\text{\hspace{0.17em}}\bigcup _{b\in y\circ z}\left(b{\right)}_{{\rho }^{\ast }}\text{\hspace{0.17em}and\hspace{0.17em}}\bigcup _{{a}^{\prime }\in z\circ x}\left({a}^{\prime }{\right)}_{{\rho }^{\ast }}\text{\hspace{0.17em}}\stackrel{\to }{\delta }\text{\hspace{0.17em}}\bigcup _{{b}^{\prime }\in z\circ y}\left({b}^{\prime }{\right)}_{{\rho }^{\ast }}\end{array}$ Thus, for every axz there exists byz such that ((a)ρ*, (b)ρ*) ∈ δ, and for every a′ ∈ zx there exists b′ ∈ zy such that ((a′)ρ*, (b′)ρ*) ∈ δ. It implies that (a, b) ∈ θ and (a′, b′) ∈ θ. Hence we conclude that xz θ⃗ yz and zx θ⃗ zy. Furthermore, let x, y, zS be such that (x, y) ∈ θ and (y, x) ∈ θ. Then, similarly as in the previous proof, we can deduce that xz θ͠ yz and zx θ͠ zy. Thus θ is a weak pseudoorder on S and ρθ. In other words, θ ∈ 𝓐. Moreover, clearly, θ′ = δ. It thus follows that f(θ) = δ.

Therefore, f is a bijection from 𝓐 onto 𝓑, and card(𝓐) = card(𝓑). □

By the proof of Theorem 3.11, we immediately obtain the following corollary:

#### Corollary 3.12

Let (S, ∘, ≤) be an ordered semihypergroup and ρ, θ be weak pseudoorders on S such that ρθ. We define a relation θ/ρ on S/ρ* as follows: $θ/ρ:={((x)ρ∗,(y)ρ∗)∈S/ρ∗×S/ρ∗ | (x,y)∈θ}.$

Then θ/ρ is a weak pseudoorder on S/ρ*.

## 4 Homomorphism theorems of ordered semihypergroups

In the current section, we shall establish the relationships between ordered regular equivalence relations and weak pseudoorders on an ordered semihypergroup, and discuss homomorphism theorems of ordered semihypergroups by weak pseudoorders.

In order to establish the relationships between ordered regular equivalence relations and weak pseudoorders on an ordered semihypergroup, the following lemma is essential.

#### Lemma 4.1

Let (S, ∘, ≤) be an ordered semihypergroup and σ a relation on S. Then the following statements are equivalent:

1. σ is a weak pseudoorder on S.

2. There exist an ordered semihypergroup (T, ⋄, ⪯) and a homomorphism φ : ST such that $ker→φ:={(a,b)∈S×S | φ(a)⪯φ(b)}=σ.$

#### Proof

(1) ⇒(2). Let σ be a weak pseudoorder on S. We denote by σ* the regular equivalence relation on S defined by $σ∗:={(a,b)∈S×S | (a,b)∈σ,(b,a)∈σ}(=σ∩σ−1).$

Then, by Theorem 3.8, the set S/σ* := {(a)σ* | aS} with the hyperoperation $\begin{array}{}\left(a{\right)}_{{\sigma }^{\ast }}\otimes \left(b{\right)}_{{\sigma }^{\ast }}=\bigcup _{c\in a\circ b}\left(c{\right)}_{{\sigma }^{\ast }},\end{array}$ for all a, bS and the order $⪯σ:={((x)σ∗,(y)σ∗)∈S/σ∗×S/σ∗ | (x,y)∈σ}$

is an ordered semihypergroup. Let T = (S/σ*, ⊗, ⪯σ) and φ be the mapping of S onto S/σ* defined by φ :SS/σ* | a ↦ (a)σ*. Then, by Proposition 3.9, φ is a homomorphism from S onto S/σ* and clearly, $\begin{array}{}\stackrel{\to }{ker}\end{array}$φ = σ.

(2) ⇒(1). Let (S, ∘, ≤) be an ordered semihypergroup. If there exist an ordered semihypergroup (T, ⋄, ⪯) and a homomorphism φ : ST such that $\begin{array}{}\stackrel{\to }{ker}\end{array}$φ = σ, then σ is a weak pseudoorder on S. Indeed, let (a, b) ∈ ≤. Then, by hypothesis, φ(a) ⪯ φ(b). Thus (a, b) ∈ $\begin{array}{}\stackrel{\to }{ker}\end{array}$φ = σ, and we have ≤⊆ σ. Moreover, let (a, b) ∈ σ and (b, c) ∈ σ. Then φ(a) ⪯ φ(b) ⪯ φ(c). Hence φ(a) ⪯ φ(c), i.e., (a, c) ∈ $\begin{array}{}\stackrel{\to }{ker}\end{array}$φ = σ. Also, if (a, b) ∈ σ, then φ(a) ⪯ φ(b). Since (T, ⋄, ⪯) is an ordered semihypergroup, for any cS we have φ(a) ⋄ φ(c) ⪯ φ(b)⋄ φ(c). Since φ is a homomorphism from S to T, we have $⋃x∈a∘cφ(x)=φ(a)⋄φ(c)⪯φ(b)⋄φ(c)=⋃y∈b∘cφ(y).$

Thus, for every xac there exists ybc such that φ(x) ⪯ φ(y), and we have (x, y) ∈ $\begin{array}{}\stackrel{\to }{ker}\end{array}$φ = σ, which implies that ac σ⃗ bc. In the same way, it can be shown that ca σ⃗ cb. Furthermore, let (a, b) ∈ σ, (b, a) ∈ σ and cS. Then φ(a) ⪯ φ(b) and φ(b) ⪯ φ(a). Thus φ(a) = φ(b), which implies that φ(a) ⋄ φ(c) = φ(b)⋄ φ(c), i.e., $\begin{array}{}\bigcup _{x\in a\circ c}\phi \left(x\right)=\bigcup _{y\in b\circ c}\phi \left(y\right).\end{array}$ Hence, for any xac, there exists ybc such that φ(x) = φ(y), and we have φ(x) ⪯φ(y) and φ(y) ⪯φ(x). It thus follows that xσ y and yσ x. On the other hand, for any y′ ∈ bc, there exists x′ ∈ ac such that φ(x′) = φ(y′). Hence we have φ(x′) ⪯φ(y′) and φ(y′) ⪯φ(x′), which imply that xσ y′ and yσ x′. Thus, ac σ͠ bc, exactly as required. Similarly, it can be obtained that ca σ͠ cb. □

In the following, we give a characterization of ordered regular equivalence relations in terms of weak pseudoorders.

#### Theorem 4.2

Let (S, ∘, ≤) be an ordered semihypergroup and ρ an equivalence relation on S. Then the following statements are equivalent:

1. ρ is an ordered regular equivalence relation on S.

2. There exists a weak pseudoorder σ on S such that ρ = σσ−1.

3. There exist an ordered semihypergroup T and a homomorphism φ: ST such that ρ = ker(φ), where kerφ = {(a, b) ∈ S × S | φ (a) = φ (b)} is the kernel of φ.

#### Proof

(1) ⇒(2). Let ρ be an ordered regular equivalence relation on S. Then there exists an order relation “ ⪯” on the quotient semihypergroup (S/ρ, ⊗) such that (S/ρ, ⊗, ⪯) is an ordered semihypergroup, and φ: SS/ρ is a homomorphism. Let σ = $\begin{array}{}\stackrel{\to }{ker}\end{array}$φ. By Lemma 4.1, σ is a weak pseudoorder on S and it is easy to check that ρ = σσ−1.

(2) ⇒(3). For a weak pseudoorder σ on S, by Lemma 4.1, there exist an ordered semihypergroup T and a homomorphism φ: ST such that σ = $\begin{array}{}\stackrel{\to }{ker}\end{array}$φ. Then we have $kerφ=ker→φ∩(ker→φ)−1=σ∩σ−1=ρ.$

(3) ⇒(1). By hypothesis and Lemma 4.1, $\begin{array}{}\stackrel{\to }{ker}\end{array}$φ is a weak pseudoorder on S. Then, by Theorem 3.8, ρ = $\begin{array}{}\stackrel{\to }{ker}\end{array}$φ ∩ ($\begin{array}{}\stackrel{\to }{ker}\end{array}$φ)−1 is a regular equivalence relation on S. Thus, by the proof of Lemma 4.1, ρ is an ordered regular equivalence relation on S. □

#### Example 4.3

We consider a set S:= {a, b, c, d, e} with the following hyperoperation “∘” and the order “≤S”: $∘abcdea {a} {e} {c} {d} {e}b {a} {b,e} {c} {d} {e}c {a} {e} {c} {d} {e}d {a} {e} {c} {d} {e}e {a} {e} {c} {d} {e}≤S:={(a,a),(a,d),(a,e),(b,b),(c,c),(c,e),(d,d),(e,e)}.$

We give the covering relation “≺Sand the figure of S as follows: $≺S={(a,d),(a,e),(c,e)}.$

Then (S, ∘, ≤S) is an ordered semihypergroup (see [28]). Let ρ be an equivalence relation on S defined as follows: $ρ={(a,a),(b,b),(c,c),(c,d),(d,c),(d,d),(e,e)}.$

Then S/ρ = {{a}, {b}, {c, d}, {e}}, and ρ is regular. Moreover, we have

1. ρ is an ordered regular equivalence relation on S. In fact, let S/ρ = {m, n, p, q}, where m = {a}, n = {b}, p = {c, d}, q = {e}. The hyperoperation “ ⊗” and the order “ ⪯” on S/ρ are the follows: $⊗mnpqm {m} {q} {p} {q}n {m} {n,q} {p} {q}p {m} {q} {p} {q}q {m} {q} {p} {q}⪯:={(m,m),(n,n),(p,p),(q,q),(m,p),(m,q),(p,q)}.$

We give the covering relation “≺” and the figure of S/ρ2 as follows: $≺={(m,p),(p,q)}.$

Then (S/ρ, ⊗, ⪯) is an ordered semihypergroup and the mapping ψ :SS/ ρ, x ↦ (x)ρ is isotone. Hence ρ is an ordered regular equivalence relation on S.

2. Let σ be a relation on S defined as follows: $σ={(a,a),(a,c),(a,d),(a,e),(b,b),(c,c),(c,d),(c,e),(d,c),(d,d),(e,e)}.$

With a small amount of effort one can verify that σ is a weak pseudoorder on S, and ρ = σσ−1.

3. Let T:= {x, y, z, u, v} with the operation “∙” and the order relation “≤Tbelow: $∘xyzuvx {x} {y} {z} {u} {u}y {x} {y} {z} {u} {u}z {x} {y} {z} {u} {u}u {x} {y} {z} {u} {u}v {x} {y} {z} {u} {u,v}≤T:={(x,x),(x,z),(x,u),(x,v),(y,y),(y,z),(y,u),(y,v),(z,z),(z,u),(z,v),(u,u),(v,v)}.$

We give the covering relation “≺Tand the figure of S as follows: $≺T={(x,z),(y,z),(z,u),(z,v)}.$

It is easy to check that (T, ∙, ≤T) is also an ordered semihypergroup. We define a mapping φ from S to T such that φ(a) = x, φ(b) = v, φ(c) = φ(d) = z, φ(e) = u. It is a routine matter to verify that φ: ST is a homomorphism and ρ = ker(φ).

#### Remark 4.4

(1) For an ordered regular equivalence relation ρ on S, since the order “⪯” such that (S/ρ, ⊗, ⪯) is an ordered semihypergroup is not unique in general, we have the weak pseudoorder σ containing ρ such that ρ = σσ−1 is not unique.

(2) If σ is a weak pseudoorder on an ordered semihypergroup S, then ρ = σσ−1 is the greatest ordered regular equivalence relation on S contained in σ. In fact, if ρ1 is an ordered regular equivalence relation on S contained in σ, then ρ1 = ρ1$\begin{array}{}{\rho }_{1}^{-1}\end{array}$σσ−1 = ρ.

Let σ be a weak pseudoorder on an ordered semihypergroup (S, ∘, ≤). Then, by Theorem 4.2, ρ = σσ−1 is an ordered regular equivalence relation on S. We denote by ρ the mapping from S onto S/ρ, i.e., ρ : SS/ρ | x ↦ (x)ρ, which is a homomorphism. In the following, we give out a homomorphism theorem of ordered semihypergroups by weak pseudoorders, which is a generalization of Theorem 1 in [17].

#### Theorem 4.5

Let (S, ∘, ≤) and (T, ⋄, ⪯) be two ordered semihypergroups, φ : ST a homomorphism. Then, if σ is a weak pseudoorder on S such that σ$\begin{array}{}\stackrel{\to }{ker}\phi \end{array}$, then there exists the unique homomorphism f : S/ρT | (a)ρφ(a) such that the diagram

commutes, where ρ = σσ−1. Moreover, Im(φ) = Im(f). Conversely, if σ is a weak pseudoorder on S for which there exists a homomorphism f : (S/ρ, ⊗, ⪯σ) → (T, ⋄, ⪯) (ρ = σσ−1) such that the above diagram commutes, then σ$\begin{array}{}\stackrel{\to }{ker}\phi \end{array}$.

#### Proof

Let σ be a weak pseudoorder on S such that σ$\begin{array}{}\stackrel{\to }{ker}\phi \end{array}$, f : S/ρT | (a)ρφ(a). Then

1. f is well defined. Indeed, if (a)ρ = (b)ρ, then (a, b) ∈ ρσ. Since σ$\begin{array}{}\stackrel{\to }{ker}\phi \end{array}$, we have (φ(a), φ(b)) ∈⪯. Furthermore, since (b, a) ∈ σ$\begin{array}{}\stackrel{\to }{ker}\phi \end{array}$, we have (φ(b), φ(a)) ∈ ⪯. Therefore, φ(a) = φ(b).

2. f is a homomorphism and φ = fρ. In fact, by Lemma 4.1, there exists an order relation “⪯σ” on the quotient semihypergroup (S/ρ, ⊗) such that (S/ρ, ⊗, ⪯σ) is an ordered semihypergroup and the mapping ρ is a homomorphism. Moreover, we have

$(a)ρ⪯σ(b)ρ⇒(a,b)∈σ⊆ker→φ⇒φ(a)⪯φ(b)⇒f((a)ρ)⪯f((b)ρ).$

Also, let (a)ρ, (b)ρS/ρ. Since φ is a homomorphism from S to T, we have

$f((a)ρ)⋄f((b)ρ)=φ(a)⋄φ(b)=⋃c∈a∘bφ(c)=⋃(c)ρ∈(a)ρ⊗(b)ρf((c)ρ).$

Furthermore, for any aS, (fρ)(a) = f((a)ρ) = φ(a), and thus φ = fρ.

We claim that f is the unique homomorphism from S/ρ to T. To prove our claim, let g is a homomorphism from S/ρ to T such that φ = gρ. Then

$f((a)ρ)=φ(a)=(g∘ρ♯)(a)=g((a)ρ).$

Moreover, Im(f) = {f((a)ρ) | aS} = {φ(a) | aS} = Im(φ).

Conversely, let σ be a weak pseudoorder on S, f : S/ρT is a homomorphism and φ = fρ. Then σ$\begin{array}{}\stackrel{\to }{ker}\phi \end{array}$. Indeed, by hypothesis, we have

$(a,b)∈σ⇔(a)ρ⪯σ(b)ρ⇒f((a)ρ)⪯f((b)ρ)⇒(f∘ρ♯)(a)⪯(f∘ρ♯)(b)⇒φ(a)⪯φ(b)⇒(a,b)∈ker→φ,$

where the order ⪯σ on S/ρ is defined as in the proof of Lemma 4.1, that is

$⪯σ:={((x)ρ,(y)ρ)∈S/ρ×S/ρ | (x,y)∈σ}.$

#### Corollary 4.6

Let (S, ∘, ≤) and (T, ⋄, ⪯) be two ordered semihypergroups and φ: ST a homomorphism. Then S/kerφIm(φ), where kerφ is the kernel of φ.

#### Proof

Let σ = $\begin{array}{}\stackrel{\to }{ker}\phi \end{array}$ and ρ = $\begin{array}{}\stackrel{\to }{ker}\phi \end{array}$ ∩ ($\begin{array}{}\stackrel{\to }{ker}\phi \end{array}$)−1. Then, by Theorems 4.2 and 4.5, ρ is an ordered regular equivalence relation on S and f : S/ρT | (a)ρφ(a) is a homomorphism. Moreover, f is inverse isotone. In fact, let (a)ρ, (b)ρ be two elements of S/ρ such that f((a)ρ) ⪯ f((b)ρ). Then φ(a) ⪯ φ(b), and we have (a, b)∈ $\begin{array}{}\stackrel{\to }{ker}\phi \end{array}$. Thus, by Lemma 4.1, ((a)ρ, (b)ρ) ∈⪯σ, i.e., (a)ρσ(b)_ρ. Clearly, ρ = kerφ. By Remark 2.4, S/kerφIm(f). Also, by Theorem 4.5, Im(f) = Im(φ). Therefore, S/kerφIm(φ). □

#### Remark 4.7

Note that if S and T are both ordered semigroups, then Corollary 4.6 coincides with Corollary in [17].

Let (S, ∘, ≤) be an ordered semihypergroup, ρ, θ be weak pseudoorders on S such that ρθ. We define a relation θ/ρ on (S/ρ*, ⊗, ⪯ρ) as follows:

$θ/ρ:={((a)ρ∗,(b)ρ∗)∈S/ρ∗×S/ρ∗ | (a,b)∈θ},$

where ⪯ρ := {((a)ρ*, (b)ρ*) | (a, b) ∈ ρ}, ρ* = ρρ−1. By Corollary 3.12, θ/ρ is a weak pseudoorder on S/ρ*. Moreover, we have the following theorem.

#### Theorem 4.8

Let (S, ∘, ≤) be an ordered semihypergroup, ρ, θ be weak pseudoorders on S such that ρθ. Then (S/ρ*)/(θ/ρ)*S/θ*.

#### Proof

We claim that the mapping φ : S/ρ*S/θ* | (a)ρ* ↦ (a)θ* is a homomorphism. In fact:

1. φ is well-defined. Indeed, let (a)ρ* = (b)ρ*. Then (a, b) ∈ ρ*. Thus, by the definition of ρ*, (a, b) ∈ ρθ and (b, a) ∈ ρθ. This implies that (a, b) ∈ θ*, and thus (a)θ* = (b)θ*.

2. φ is a homomorphism. In fact, let (a)ρ*, (b)ρ*S/ρ*. Then, since ρ*,θ* are both ordered regular equivalence relations on S, we have

$(a)ρ∗⊗ρ(b)ρ∗=⋃x∈a∘b(x)ρ∗,(a)θ∗⊗θ(b)θ∗=⋃x∈a∘b(x)θ∗,$

where “ ⊗ρ ” and “ ⊗θ ” are the hyperactions on S/ρ* and S/θ*, respectively. Thus

$φ((a)ρ∗)⊗θφ((b)ρ∗)=(a)θ∗⊗θ(a)θ∗=⋃x∈a∘b(x)θ∗=⋃(x)ρ∗∈(a)ρ∗⊗ρ(b)ρ∗φ((x)ρ∗).$

Also, if (a)ρ*ρ (b)ρ*, then (a, b) ∈ ρθ. It implies that (a)θ* ⪯_θ (b)θ*, and thus φ is isotone. On the other hand, it is easy to see that φ is onto, since

$Im(φ)={φ((a)ρ∗)|a∈S}={(a)θ∗|a∈S}=S/θ∗.$

It thus follows from Corollary 4.6 that (S/ρ*)/kerφIm(φ) = S/θ*.

Furthermore, let $\begin{array}{}\stackrel{\to }{ker}\phi \end{array}$ := {((a)ρ*, (b)ρ*) ∈ S/ρ* × S/ρ* | φ ((a)ρ*) ⪯θ φ ((b)ρ*)}. Then

$((a)ρ∗,(b)ρ∗)∈ker→φ⟺(a)θ∗⪯θ(b)θ∗⟺(a,b)∈θ⟺((a)ρ∗,(b)ρ∗)∈θ/ρ.$

Therefore, kerφ = $\begin{array}{}\stackrel{\to }{ker}\phi \end{array}$ ∩ ($\begin{array}{}\stackrel{\to }{ker}\phi \end{array}$)−1 = (θ/ρ) ∩ (θ/ρ)−1 = (θ/ρ)*. We have thus shown that (S/ρ*)/(θ/ρ)*S/θ*. □

#### Definition 4.9

Let (S, ∘, ≤) and (T, ⋄, ⪯) be two ordered semihypergroups, ρ, θ be two weak pseudoorders on S and T, respectively, and the mapping f : ST a homomorphism. Then, f is called a (ρ, θ)-homomorphism if (x, y) ∈ ρ implies (f(x), f(y)) ∈ θ, for all x, yS.

#### Example 4.10

Consider the ordered semihypergroups (S, ∘, ≤S) and (T, ∙, ≤T) given in Example 4.3, and define the relations ρ, θ on S and T, respectively, as follows:

$ρ:={(a,a),(a,c),(a,d),(a,e),(b,b),(c,c),(c,d),(c,e),(d,c),(d,d),(e,e)},θ:={(x,x),(x,z),(x,u),(x,v),(y,y),(y,z),(y,u),(y,v),(z,z),(z,u),(z,v),(u,u),(v,v)}.$

It is not difficult to verify that ρ, θ is pseudoorders on S and T, respectively. Furthermore, we define a mapping f from S to T such that f(a) = x, f(b) = v, f(c) = f(d) = z, f(e) = u. By Example 4.3, f : ST is a homomorphism. Hence f is a (ρ, θ)-homomorphism.

#### Lemma 4.11

Let (S, ∘, ≤) and (T, ⋄, ⪯) be two ordered semihypergroups, ρ, θ be two weak pseudoorders on S and T, respectively, and the mapping f : ST a (ρ, θ)-homomorphism. Then, the mapping f : (S/ρ*, ⊗ρ, ⪯ρ) → (T/θ*, ⊗θ, ⪯θ) defined by

$(∀x∈S)f¯((x)ρ∗):=(f(x))θ∗$

is a homomorphism of ordered semihypergroups, where the ordersρ, ⪯θ on S/ρ* and T/θ*, respectively, are both defined as in the proof of Lemma 4.1.

#### Proof

Let f : ST be a (ρ, θ)-homomorphism and f: S/ρ*T/θ* | (x)ρ* ↦ (f(x))θ*. Then

1. f is well defined. In fact, let (x)ρ*, (y)ρ*S/ρ* be such that (x)ρ* = (y)ρ*. Then (x, y) ∈ ρ*ρ. Since f is a (ρ, θ)-homomorphism, we have (f(x),f(y)) ∈ θ. It implies that ((f(x))θ*,(f(y))θ*) ∈⪯θ. Similarly, since (y, x) ∈ ρ, we have ((f(y))θ*,(f(x))θ*) ∈⪯θ. Therefore, ((f(x))θ* = (f(y))θ*, i.e., f((x)ρ*)=f((y)ρ*).

2. f is a homomorphism. Indeed, let (x)ρ*,(y)ρ*S/ρ*. By Theorem 4.2, ρ*,θ* are ordered regular equivalence relations on S and T, respectively. Since f is a homomorphism, by Lemma 2.2 we have

$f¯((x)ρ∗)⊗θf¯((y)ρ∗)=(f(x))θ∗⊗θ(f(y))θ∗=⋃a∈x∘y(f(a))θ∗=⋃(a)ρ∗∈(x)ρ∗⊗ρ(y)ρ∗f¯((a)ρ∗).$

Also, since f is a (ρ, θ)-homomorphism, we have

$(x)ρ∗⪯ρ(y)ρ∗⇒(x,y)∈ρ⇒(f(x),f(y))∈θ⇒(f(x))θ∗⪯θ(f(y))θ∗⇒f¯((x)ρ∗)⪯θf¯((y)ρ∗).$

Hence f is isotone. We have thus shown that f is a homomorphism, as required. □

#### Lemma 4.12

Let (S, ∘, ≤) and (T, ⋄, ⪯) be two ordered semihypergroups, ρ, θ be two weak pseudoorders on S and T, respectively, and the mapping f : ST a (ρ, θ)-homomorphism. Then, We define a relation on S/ρ* denoted by ρf as follows:

$ρf:={((x)ρ∗,(y)ρ∗)∈S/ρ∗×S/ρ∗|(f(x))θ∗⪯θ(f(y))θ∗}.$

Then ρf is a weak pseudoorder on S/ρ*.

#### Proof

Assume that ((x)ρ*,(y)ρ*) ∈⪯ρ. By Lemma 4.11, f is a homomorphism. Then f((x)ρ*) ⪯θ f((y)ρ*), i.e., (f(x))θ*θ (f(y))θ*. It implies that ((x)ρ*,(y)ρ*)∈ ρf, and thus ⪯ρρf. Now, let ((x)ρ*,(y)ρ*) ∈ ρf and ((y)ρ*,(z)ρ*) ∈ ρf. Then (f(x))θ*θ (f(y))θ* and (f(y))θ*θ (f(z))θ*. Thus, by the transitivity of ⪯θ, (f(x))θ*θ (f(z))θ*. This implies that ((x)ρ*,(z)ρ*)∈ ρf. Also, let ((x)ρ*,(y)ρ*) ∈ ρf and (z)ρ*S/ρ*. Then (f(x))θ*θ (f(y))θ*. Since (T/θ*, ⊗θ, ⪯θ) is an ordered semihypergroup, it can be obtained that (f(x))θ*θ (f(z))θ*θ (f(y))θ*θ (f(z))θ*, that is, f((x)ρ*)⊗θ f((z)ρ*) ⪯θ f((y)ρ*)⊗θ f((z)ρ*). Then, since f is a homomorphism, we have

$⋃(a)ρ∗∈(x)ρ∗⊗ρ(z)ρ∗f¯((a)ρ∗)⪯θ⋃(b)ρ∗∈(y)ρ∗⊗ρ(z)ρ∗f¯((b)ρ∗),$

which means that $\begin{array}{}\bigcup _{a\in x\circ z}\end{array}$ (f(a))θ*θ $\begin{array}{}\bigcup _{b\in y\circ z}\end{array}$(f(b))θ*. Thus, for any axz, there exists byz such that (f(a))θ*θ (f(b))θ*. Hence ((a)ρ*, (b)ρ*) ∈ ρf. It thus implies that (x)ρ*ρ (z)ρ* ρ⃗f (y)ρ*ρ (z)ρ*. Similarly, it can be shown that (z)ρ*ρ (x)ρ* ρ⃗f (z)ρ*ρ (y)ρ*. Furthermore, let (x)ρ*,(y)ρ*,(z)ρ*S/ρ* be such that ((x)ρ*,(y)ρ*) ∈ ρf and ((y)ρ*,(x)ρ*) ∈ ρf. Then, similarly as in the above proof, it can be verified that (x)ρ*ρ (z)ρ* ρ̃f (y)ρ*ρ (z)ρ* and (z)ρ*ρ (x)ρ* ρ̃f (z)ρ*ρ (y)ρ*. Therefore, ρf is a weak pseudoorder on S/ρ*. □

By Lemmas 4.11 and 4.12, we immediately obtain the following corollary:

#### Corollary 4.13

Let (S, ∘, ≤) and (T, ⋄, ⪯) be two ordered semihypergroups, ρ, θ be two weak pseudoorders on S and T, respectively, and the mapping f : ST be a (ρ, θ)-homomorphism. Then, the following diagram

commutates.

#### Theorem 4.14

Let (S, ∘, ≤) and (T, ⋄, ⪯) be two ordered semihypergroups, ρ, θ be two weak pseudoorders on S and T, respectively, and the mapping f : ST a (ρ, θ)-homomorphism. If σ is a weak pseudoorder on S/ρ* such that σρf, then there exists the unique homomorphism φ : (S/ρ*)/σ*T/θ* | ((a)ρ*)σ*f((a)ρ*) such that the diagram

commutes.

Conversely, if σ is a weak pseudoorder on S/ρ* for which there exists a homomorphism φ : (S/ρ*)/σ*T/θ* such that the above diagram commutes, then σρf.

#### Proof

The proof is straightforward by Lemma 4.11, Lemma 4.12 and Theorem 4.5, and we omit the details. □

Let (S, ∘, ≤S) and (T, ⋄, ≤T) be two ordered semihypergroups. In [25], it is shown that (S × T, ⋆, ≤S×T) is also an ordered semihypergroup with the hyperoperation “ ⋆ ” and the order relation “ ≤S×T ” below: for any (s1, t1), (s2, t2) ∈ S × T,

$(s1,t1)⋆(s2,t2)=(s1∘s2)×(t1⋄t2);(s1,t1)≤S×T(s2,t2)⇔s1≤Ss2andt1≤Tt2.$

#### Lemma 4.15

Let (S, ∘, ≤S) and (T, ⋄, ≤T) be ordered semihypergroups, ρ1, ρ2 be two weak pseudoorders on S and T, respectively. We define a relation ρ on S × T as follows:

$(∀(s1,t1),(s2,t2)∈S×T)(s1,t1)ρ(s2,t2)⇔s1ρ1s2andt1ρ2t2.$

Then ρ is a weak pseudoorder on S × T.

#### Proof

The verifications of the conditions (1), (2) and (3) of Definition 3.4 are straightforward. We only need show that the condition (4) of Definition 3.4 is satisfied.

Let (s1, t1), (s2, t2), (s, t) ∈ S × T be such that (s1, t1)ρ (s2, t2) and (s2, t2)ρ (s1, t1). Then s1ρ1 s2, t1ρ2 t2, s2ρ1 s1, t2ρ2 t1 and sS, tT. Since ρ1, ρ2 are weak pseudoorders on S and T, respectively, we have

$s1∘sρ~1s2∘sandt1∘tρ~1t2∘t.$

Thus

$(∀x∈s1∘s)(∃u∈s2∘s)xρ1uanduρ1x&(∀u′∈s2∘s)(∃x′∈s1∘s)x′ρ1u′andu′ρ1x′,(∀y∈t1⋄t)(∃v∈t2⋄t)yρ2vandvρ2y&(∀v′∈t2⋄t)(∃y′∈t1⋄t)y′ρ2v′andv′ρ2y′.$

Hence we have

$(∀(x,y)∈(s1∘s)×(t1⋄t))(∃(u,v)∈(s2∘s)×(t2⋄t))(x,y)ρ(u,v)and(u,v)ρ(x,y),(∀(u′,v′)∈(s2∘s)×(t2⋄t))(∃(x′,y′)∈(s1∘s)×(t1⋄t))(x′,y′)ρ(u′,v′)and(u′,v′)ρ(x′,y′).$

It thus follows that (s1s) × (t1t) ρ̃ (s2s) × (t2t), i.e., (s1, t1) ⋆ (s, t) ρ̃ (s2, t2) ⋆ (s, t). Similarly, it can be verified that (s, t) ⋆ (s1, t1) ρ̃ (s, t)⋆ (s2, t2). □

By Theorem 3.8 and Lemma 4.15, ((S × T)/ρ*, ⊗ρ, ⪯ρ) and (S/$\begin{array}{}{\rho }_{1}^{\ast }\end{array}$ × T/$\begin{array}{}{\rho }_{2}^{\ast }\end{array}$, ⋆′, ⪯S×T) are both ordered semihypergroups, where the hyperoperation ⋆′ and the order relation ⪯S×T on S/$\begin{array}{}{\rho }_{1}^{\ast }\end{array}$ × S/$\begin{array}{}{\rho }_{2}^{\ast }\end{array}$ are similar to the hyperoperation ⋆ and the order relation ≤S×T on S × T, respectively. Furthermore, we have the following theorem:

#### Theorem 4.16

Let (S, ∘, ≤S) and (T, ⋄, ≤T) be ordered semihypergroups, ρ1, ρ2 be two weak pseudoorders on S and T, respectively. Then (S × T)/ρ*S/$\begin{array}{}{\rho }_{1}^{\ast }\end{array}$ × T/$\begin{array}{}{\rho }_{2}^{\ast }\end{array}$.

#### Proof

We claim that the mapping φ : (S × T)/ρ*S/$\begin{array}{}{\rho }_{1}^{\ast }\end{array}$ × T/$\begin{array}{}{\rho }_{2}^{\ast }\end{array}$ | ((s, t))ρ* ↦ ((s)$\begin{array}{}{\rho }_{1}^{\ast }\end{array}$, (t)$\begin{array}{}{\rho }_{2}^{\ast }\end{array}$) is an isomorphism. In fact:

1. φ is well-defined. Indeed, let ((s1, t1))ρ* = ((s2, t2))ρ*. Then (s1, t1)ρ*(s2, t2). Hence, by the definition of ρ*, (s1, t1)ρ(s2, t2) and (s2, t2)ρ(s1, t1). It thus follows that s1ρ1 s2, t1ρ2 t2, s2ρ1 s1 and t2ρ2 t1, which imply s1$\begin{array}{}{\rho }_{1}^{\ast }\end{array}$ s2 and t1 $\begin{array}{}{\rho }_{2}^{\ast }\end{array}$ t2. Thus, ((s1)$\begin{array}{}{\rho }_{1}^{\ast }\end{array}$,(t1)$\begin{array}{}{\rho }_{2}^{\ast }\end{array}$) = ((s2)$\begin{array}{}{\rho }_{1}^{\ast }\end{array}$, (t2)$\begin{array}{}{\rho }_{2}^{\ast }\end{array}$).

2. φ is an isomorphism. In fact, for any ((s1, t1))ρ*,((s2, t2))ρ*∈ (S × T)/ρ*, we have

$φ(((s1,t1))ρ∗)⋆′φ(((s2,t2))ρ∗)=((s1)ρ1∗,(t1)ρ2∗)⋆′((s2)ρ1∗,(t2)ρ2∗)=((s1)ρ1∗⊗ρ1(s2)ρ1∗,(t1)ρ2∗⊗ρ2(t2)ρ2∗)=(⋃s∈s1∘s2(s)ρ1∗,⋃t∈t1⋄t2(t)ρ2∗)=⋃s∈s1∘s2t∈t1⋄t2((s)ρ1∗,(t)ρ2∗)=⋃(s,t)∈(s1∘s2)×(t1∘t2)φ(((s,t))ρ∗)=⋃(s,t)∈(s1,t1)⋆(s2,t2)φ(((s,t))ρ∗)=⋃((s,t))ρ∗∈((s1,t1))ρ∗⊗ρ((s2,t2))ρ∗φ(((s,t))ρ∗).$

Moreover, it can be easily seen that φ is onto. Also, φ is isotone and reverse isotone. Indeed, let ((s1, t1))ρ*,((s2, t2))ρ* ∈ (S × T)/ρ*. Then we have

$((s1,t1))ρ∗⪯ρ((s2,t2))ρ∗⟺(s1,t1)ρ(s2,t2)⟺s1ρ1s2andt1ρ2t2⟺(s1)ρ1∗⪯ρ1(s2)ρ1∗and(t1)ρ2∗⪯ρ2(t2)ρ2∗⟺((s1)ρ1∗,(t1)ρ2∗)⪯S×T((s2)ρ1∗,(t2)ρ2∗)⟺φ(((s1,t1))ρ∗)⪯S×Tφ(((s2,t2))ρ∗).$

Therefore, φ is indeed an isomorphism, which means that (S × T)/ρ*S/$\begin{array}{}{\rho }_{1}^{\ast }\end{array}$ × T/$\begin{array}{}{\rho }_{2}^{\ast }\end{array}$. □

## 5 Conclusions

As we know, (strongly) regular equivalence relations on ordered semihypergroups play an important role in studying the structure of ordered semihypergroups. In this paper, we further studied the ordered regular equivalence relations on ordered semihypergroups. We introduced the concept of weak pseudoorders on an ordered semihypergroup, and established the relationships between ordered regular equivalence relations and weak pseudoorders on an ordered semihypergroup. Especially, we constructed an ordered regular equivalence relation on an ordered semihypergroup by a weak pseudoorder, and gave a complete answer to the open problem given by Davvaz et al. in [25].

In [25], for a pseudoorder ρ on an ordered semihypergroup S, Davvaz et al. obtained an ordered semigroup S/ρ*, where ρ*(= ρρ−1) is a strongly regular equivalence relation on S. Moreover, they provided some homomorphism theorems of ordered semigroups by pseudoorders, for example see Theorems 4.5, 4.11 and 4.15 in [25]. However, they have not established the homomorphisms of ordered semihypergroups. As a further study, in the present paper we also discussed the homomorphic theory of ordered semihypergroups by weak pseudoorders, and generalized some similar results in ordered semigroups.

## Acknowledgement

This research was partially supported by the National Natural Science Foundation of China (No. 11361027), the University Natural Science Project of Anhui Province – “The algebraic structure of (ordered) semihypergroups and its applications” in 2018, the Key Project of Department of Education of Guangdong Province (No. 2014KZDXM055) and the Natural Science Foundation of Guangdong Province (No. 2014A030313625).

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Accepted: 2017-11-15

Published Online: 2018-03-20

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 168–184, ISSN (Online) 2391-5455,

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