A further study on ordered regular equivalence relations in ordered semihypergroups

Example 2.1

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

We give the covering relation “≺” and the figure of S as follows:

Then (S, ∘, ≤) is an ordered semihypergroup (see [26]). Let ρ₁, ρ₂be equivalence relations on S defined as follows:

Then

1. ρ₁is a strongly regular relation on S.

2. ρ₂is a regular relation on S, but it is not a strongly regular relation on S. In fact, since aρc, while a ∘ aρ̿c ∘ a 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 = ⋃c∈a∘b(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)_ρfor all ∈ a ∘ b. 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., ρ ∘ ρ ⊆ ρ and (3) aρb implies a ∘ cρ̿b ∘ c and c ∘ aρ̿c ∘ b, for all c ∈ S.

Example 2.3

Consider the ordered semihypergroup (S, ∘, ≤) given in Example 2.1, and define a relation ρ on S as follows:

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

Let (S, ∘, ≤), (T, ⋄, ⪯) be two ordered semihypergroups and f: S → T a mapping from S to T. f is called isotone if x ≤ y implies f(x)⪯ f(y), for all x, y ∈ S. f is called reverse isotone if x, y ∈ S, f(x) ⪯ f(y) implies x ≤ y. f is called homomorphism [25] if it is isotone and satisfies f(x) ⋄ f(y) = ⋃z∈x∘yf(z), for all x, y ∈ S. f is called isomorphism if it is homomorphism, onto and reverse isotone. The ordered semihypergroups S and T are called isomorphic, in symbol S ≅ T, 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 S ≅ Im(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 : S → S/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.

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 ρ_I on an ordered semihypergroup S as follows:

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 ρ_I 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 ρ_I on S such that the corresponding quotient structure S/ρ_I 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 “≤”:

We give the covering relation “≺” and the figure of S as follows:

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, B ∈ P^∗(S), then we set

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 a ∘ cρ⃗ b ∘ c and c ∘ a ρ⃗ c ∘ b, for all c ∈ S;

4. aρb and bρa imply a ∘ c ρ̃ b ∘ c and c ∘ a ρ̃ c ∘ b, for all c ∈ S.

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 “≤”:

We give the covering relation “≺” and the figure of S as follows:

Then (S, ∘, ≤) is an ordered semihypergroup. Let ρ be a relation on S defined as follows:

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 a ∘ bρ̿c ∘ b 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

First, we claim that ρ^* is a regular equivalence relation on S. In fact, for any a ∈ S, clearly, (a, a) ∈≤⊆ ρ, so aρ^*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 aρ^*b and c ∈ S. Then aρb and bρa. Since ρ is a weak pseudoorder on S, by condition (4) of Definition 3.4, we have

Thus, for every x ∈ a ∘ c, there exists y ∈ b ∘ c such that xρy and yρx which imply that xρ^*y, and for every y′ ∈ b ∘ c, there exists x′ ∈ a ∘ c such that x′ρ y′ and y′ρx’ which mean that x′ρ^*y′. It thus follows that a ∘ cρ^*b ∘ c. Similarly, it can be obtained that c ∘ aρ^*c ∘ b. 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:

Now, we define a relation ⪯_ρ on S/ρ^* as follows:

Then (S/ρ^*, ⪯_ρ) is a poset. Indeed, suppose that (x)_ρ^* ∈ S/ρ^*, where x ∈ S. 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 xρ^*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 z ∈ S. Since ρ is a weak pseudoorder on S, by condition (3) of Definition 3.4, we have x ∘ z ρ⃗ y ∘ z and z ∘ x ρ⃗ z ∘ y. Thus, for any a ∈ x ∘ z there exists b ∈ y ∘ z such that aρb. This implies that (a)_ρ^* ⪯_ρ (b)_ρ^*. Hence we have

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^* = ρ ∩ ρ⁻¹.

Proof

By Theorem 3.8, (S/ρ^*, ⊗, ⪯_ρ) is an ordered semihypergroup, where the order relation ⪯_ρ is defined as follows:

Also, let x, y ∈ S and x ≤ y. 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 “≤”:

We give the covering relation “≺” and the figure of S as follows:

Then (S, ∘, ≤) is an ordered semihypergroup (see [26]). Let ρ be a weak pseudoorder on S defined as follows:

Applying Theorem 3.8, we get

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 “ ⪯_ρ” are given below:

We give the covering relation “≺_ρ” and the figure of S/ρ^*as follows:

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

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:

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) ∈ ρ ⊆ θ, which implies that ((x)_ρ^*, (y)_ρ^*) ∈ θ′. Thus, ⪯_ρ ⊆ θ′. 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 z ∈ S. Since θ is a weak pseudoorder on S, we have x ∘ zθ⃗y ∘ z and z ∘ xθ⃗z ∘ y. Thus, for every a ∈ x ∘ z there exists b ∈ y ∘ z, we have (a, b) ∈ θ. This implies that ((a)_ρ^*, (b)_ρ^*) ∈ θ′, and thus we have

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 z ∈ S. By hypothesis, we have x ∘ zθ͠y ∘ z and z ∘ xθ͠z ∘ y. 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^*) ∈ θ1′ we have (x, y) ∈ θ₁ = θ₂. It implies that ((x)_ρ^*, (y)_ρ^*) ∈ θ2′. Hence θ1′⊆θ2′ By symmetry, it can be obtained that θ2′⊆θ1′.

2. f is one to one. In fact, let θ₁, θ₂ ∈ 𝓐 and θ1′=θ2′. Assume that (x, y) ∈ θ₁. Then, ((x)_ρ^*, (y)_ρ^*) ∈ θ1′ and thus ((x)_ρ^*, (y)_ρ^*) ∈ θ2′. This implies that (x, y) ∈ θ₂. Thus, θ₁ ⊆ θ₂. Similarly, we obtain θ₂ ⊆ θ₁.

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 ρ ⊆ θ. Assume that (x, y) ∈ ρ. Then, by Theorem 3.8, ((x)_ρ^*, (y)_ρ^*) ∈⪯_ρ ⊆ δ, and thus (x, y) ∈ θ. This implies that ρ ⊆ θ. If (x, y) ∈ ≤, then (x, y) ∈ ρ ⊆ θ. 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, z ∈ S 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., ⋃a∈x∘z(a)ρ∗ δ→ ⋃b∈y∘z(b)ρ∗ and ⋃a′∈z∘x(a′)ρ∗ δ→ ⋃b′∈z∘y(b′)ρ∗ Thus, for every a ∈ x ∘ z there exists b ∈ y ∘ z such that ((a)_ρ^*, (b)_ρ^*) ∈ δ, and for every a′ ∈ z ∘ x there exists b′ ∈ z ∘ y such that ((a′)_ρ^*, (b′)_ρ^*) ∈ δ. It implies that (a, b) ∈ θ and (a′, b′) ∈ θ. Hence we conclude that x ∘ zθ⃗y ∘ z and z ∘ xθ⃗z ∘ y. Furthermore, let x, y, z ∈ S be such that (x, y) ∈ θ and (y, x) ∈ θ. Then, similarly as in the previous proof, we can deduce that x ∘ zθ͠y ∘ z and z ∘ xθ͠z ∘ y. Thus θ is a weak pseudoorder on S and ρ ⊆ θ. 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 ρ ⊆ θ. We define a relation θ/ρ on S/ρ^*as follows:

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

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 : S → T 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

Then, by Theorem 3.8, the set S/σ^* := {(a)_σ^* | a ∈ S} with the hyperoperation (a)σ∗⊗(b)σ∗=⋃c∈a∘b(c)σ∗, for all a, b ∈ S and the order

is an ordered semihypergroup. Let T = (S/σ^*, ⊗, ⪯_σ) and φ be the mapping of S onto S/σ^* defined by φ :S → S/σ^* | a ↦ (a)_σ^*. Then, by Proposition 3.9, φ is a homomorphism from S onto S/σ^* and clearly, ker→φ = σ.

(2) ⇒(1). Let (S, ∘, ≤) be an ordered semihypergroup. If there exist an ordered semihypergroup (T, ⋄, ⪯) and a homomorphism φ : S → T such that ker→φ = σ, then σ is a weak pseudoorder on S. Indeed, let (a, b) ∈ ≤. Then, by hypothesis, φ(a) ⪯ φ(b). Thus (a, b) ∈ ker→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) ∈ ker→φ = σ. Also, if (a, b) ∈ σ, then φ(a) ⪯ φ(b). Since (T, ⋄, ⪯) is an ordered semihypergroup, for any c ∈ S we have φ(a) ⋄ φ(c) ⪯ φ(b)⋄ φ(c). Since φ is a homomorphism from S to T, we have

Thus, for every x ∈ a ∘ c there exists y ∈ b ∘ c such that φ(x) ⪯ φ(y), and we have (x, y) ∈ ker→φ = σ, which implies that a ∘ cσ⃗b ∘ c. In the same way, it can be shown that c ∘ aσ⃗c ∘ b. Furthermore, let (a, b) ∈ σ, (b, a) ∈ σ and c ∈ S. 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., ⋃x∈a∘cφ(x)=⋃y∈b∘cφ(y). Hence, for any x ∈ a ∘ c, there exists y ∈ b ∘ c such that φ(x) = φ(y), and we have φ(x) ⪯φ(y) and φ(y) ⪯φ(x). It thus follows that xσy and yσ x. On the other hand, for any y′ ∈ b ∘ c, there exists x′ ∈ a ∘ c such that φ(x′) = φ(y′). Hence we have φ(x′) ⪯φ(y′) and φ(y′) ⪯φ(x′), which imply that x′σy′ and y′σx′. Thus, a ∘ cσ͠b ∘ c, exactly as required. Similarly, it can be obtained that c ∘ aσ͠c ∘ b. □

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 = σ ∩ σ⁻¹.

3. There exist an ordered semihypergroup T and a homomorphism φ: S → T 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: S → S/ρ is a homomorphism. Let σ = ker→φ. By Lemma 4.1, σ is a weak pseudoorder on S and it is easy to check that ρ = σ ∩ σ⁻¹.

(2) ⇒(3). For a weak pseudoorder σ on S, by Lemma 4.1, there exist an ordered semihypergroup T and a homomorphism φ: S → T such that σ = ker→φ. Then we have

(3) ⇒(1). By hypothesis and Lemma 4.1, ker→φ is a weak pseudoorder on S. Then, by Theorem 3.8, ρ = ker→φ ∩ (ker→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”:

We give the covering relation “≺_S” and the figure of S as follows:

Then (S, ∘, ≤_S) is an ordered semihypergroup (see [28]). Let ρ be an equivalence relation on S defined as follows:

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/ρ₂as follows:

   ≺={(m,p),(p,q)}.

   

   Then (S/ρ, ⊗, ⪯) is an ordered semihypergroup and the mapping ψ :S → S/ ρ, x ↦ (x)_ρis 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 = σ ∩ σ⁻¹.

3. Let T:= {x, y, z, u, v} with the operation “∙” and the order relation “≤_T” below:

   ∘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 “≺_T” and 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: S → T is a homomorphism and ρ = ker(φ).

Remark 4.4

(1) For an ordered regular equivalence relationρonS, since the order “⪯” such that (S/ρ, ⊗, ⪯) is an ordered semihypergroup is not unique in general, we have the weak pseudoorderσcontainingρsuch thatρ = σ ∩ σ⁻¹is not unique.

(2) Ifσis a weak pseudoorder on an ordered semihypergroupS, thenρ = σ ∩ σ⁻¹is the greatest ordered regular equivalence relation onScontained inσ. In fact, ifρ₁is an ordered regular equivalence relation onScontained inσ, thenρ₁ = ρ₁ ∩ ρ1−1 ⊆ σ ∩ σ⁻¹ = ρ.

Theorem 4.5

Let (S, ∘, ≤) and (T, ⋄, ⪯) be two ordered semihypergroups, φ : S → Ta homomorphism. Then, ifσis a weak pseudoorder onSsuch thatσ ⊆ ker→φ, then there exists the unique homomorphismf : S/ρ → T | (a)_ρ ↦ φ(a) such that the diagram

commutes, whereρ = σ ∩ σ⁻¹. Moreover, Im(φ) = Im(f). Conversely, ifσis a weak pseudoorder onSfor which there exists a homomorphismf : (S/ρ, ⊗, ⪯_σ) → (T, ⋄, ⪯) (ρ = σ ∩ σ⁻¹) such that the above diagram commutes, thenσ ⊆ ker→φ.

Proof

Let σ be a weak pseudoorder on S such that σ ⊆ ker→φ, f : S/ρ → T | (a)_ρ ↦ φ(a). Then

1. f is well defined. Indeed, if (a)_ρ = (b)_ρ, then (a, b) ∈ ρ ⊆ σ. Since σ ⊆ ker→φ, we have (φ(a), φ(b)) ∈⪯. Furthermore, since (b, a) ∈ σ ⊆ ker→φ, we have (φ(b), φ(a)) ∈ ⪯. Therefore, φ(a) = φ(b).

2. f is a homomorphism and φ = f ∘ ρ^♯. 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

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

Furthermore, for any a ∈ S, (f ∘ ρ^♯)(a) = f((a)_ρ) = φ(a), and thus φ = f ∘ ρ^♯.

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 = g ∘ ρ^♯. Then

Moreover, Im(f) = {f((a)_ρ) | a ∈ S} = {φ(a) | a ∈ S} = Im(φ).

Conversely, let σ be a weak pseudoorder on S, f : S/ρ → T is a homomorphism and φ = f ∘ ρ^♯. Then σ ⊆ ker→φ. Indeed, by hypothesis, we have

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

□

Corollary 4.6

Let (S, ∘, ≤) and (T, ⋄, ⪯) be two ordered semihypergroups andφ: S → Ta homomorphism. ThenS/kerφ ≅ Im(φ), wherekerφis the kernel ofφ.

Proof

Let σ = ker→φ and ρ = ker→φ ∩ (ker→φ)⁻¹. Then, by Theorems 4.2 and 4.5, ρ is an ordered regular equivalence relation on S and f : S/ρ → T | (a)_ρ ↦ φ(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)∈ ker→φ. Thus, by Lemma 4.1, ((a)_ρ, (b)_ρ) ∈⪯_σ, i.e., (a)_ρ ⪯_σ(b)_ρ. Clearly, ρ = kerφ. By Remark 2.4, S/kerφ ≅ Im(f). Also, by Theorem 4.5, Im(f) = Im(φ). Therefore, S/kerφ ≅ Im(φ). □

Remark 4.7

Note that ifSandTare both ordered semigroups, thenCorollary 4.6coincides with Corollary in[17].

Theorem 4.8

Let (S, ∘, ≤) be an ordered semihypergroup, ρ, θbe weak pseudoorders onSsuch thatρ ⊆θ. 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) ∈ ρ ⊆θ and (b, a) ∈ ρ ⊆θ. 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

Also, if (a)_ρ^* ⪯_ρ (b)_ρ^*, then (a, b) ∈ ρ ⊆θ. 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

It thus follows from Corollary 4.6 that (S/ρ^*)/kerφ ≅ Im(φ) = S/θ^*.

Furthermore, let ker→φ := {((a)_ρ^*, (b)_ρ^*) ∈ S/ρ^* × S/ρ^* | φ ((a)_ρ^*) ⪯_θφ ((b)_ρ^*)}. Then

Therefore, kerφ = ker→φ ∩ (ker→φ)⁻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, θbe two weak pseudoorders onSandT, respectively, and the mappingf : S → Ta homomorphism. Then, fis called a (ρ, θ)-homomorphism if (x, y) ∈ ρimplies (f(x), f(y)) ∈ θ, for allx, y ∈ S.