EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 13, No. 3, 2020, 483-497 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Hyper Homomorphism and Hyper Product of Hyper UP-algebras Rohaima M. Amairanto1,∗, Rowena T. Isla2 1 Department of Mathematics, Mindanao State University-University Training Center, 9700 Marawi City, Philippines 2 Department of Mathematics and Statistics, College of Science and Mathematics, Mindanao State University-Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. In this paper, we investigate the concept of regular congruence relation on hyper UP- algebras and establish some homomorphism theorems on such algebras. We also examine the notion of hyper product of hyper UP-algebras. 2020 Mathematics Subject Classifications: 08A30, 08A99 Key Words and Phrases: Hyper UP-algebra, Regular Congruence Relation, Hyper Homomor- phisms of Hyper UP-algebras, Hyper Product of Hyper UP-algebras 1. Introduction In 1934, F. Marty [7] first introduced the concept of hyperstructure theory at the 8th Congress of Scandinavian Mathematics. This led to the formulation of hyper BCK- algebra by Y. Jun et al. [11], hyper BCI-algebra by X. Long [6], and many other classes of algebras. R. Borzooei and H. Harizavi [1] defined the regular congruence relation on a hyper BCK-algebra, constructed a quotient hyper BCK-algebra, established some homo- morphism theorems, and got some related results involving the hyper product of hyper BCK-algebras. G. Flores and G. Petalcorin [2] introduced regular congruence relation on a hyper BCI-algebra and presented some isomorphism theorems on hyper BCI-algebras. In 2017, A. Iampan [4] defined a new algebraic structure called a UP-algebra and showed that the notion of UP-algebras is a generalization of KU-algebras that was intro- duced by C. Prabpayak and U. Leerawat [8]. Recently, D. Gomisong [3] applied hyperstuc- tures to UP-algebras in her graduate thesis following the structure of hyper KU-algebras by S. Mostafa et al. [5]. D. Romano gave an equivalent definition of hyper UP-algebra in [10] and proved that every hyper KU-algebra is a hyper UP-algebra. He also introduced the quotient of a hyper UP-algebra in [9]. In this paper, we investigate the concept of regular ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v13i3.3704 Email addresses: rohaima87@yahoo.com (R. Amairanto), rowena.isla@g.msuiit.edu.ph (R. Isla) https://www.ejpam.com 483 c© 2020 EJPAM All rights reserved. R. Amairanto, R. Isla / Eur. J. Pure Appl. Math, 13 (3) (2020), 483-497 484 congruence relation on a hyper UP-algebra and present some homomorphism theorems on hyper UP-algebras. We also examine the concept of hyper product of hyper UP-algebras and extend it to the hyper product of an arbitrary family of hyper UP-algebras. 2. Preliminaries Let H be a nonempty set and P∗(H) be the set of all nonempty subsets of H. A hyperoperation on H is a mapping from H ×H into P∗(H). Definition 1. [3] A hyper UP-algebra is a set H with constant 0 and hyperoperation ~ satisfying the following axioms: for all x, y, z ∈ H, (HUP1) [(x~ y) ~ (x~ z)]� y ~ z, (HUP2) 0 ~ x = {x}, (HUP3) x~ 0 = {0}, (HUP4) x� y and y � x imply x = y, where x� y is defined by 0 ∈ y~ x and for every A,B ⊆ H, A� B is defined by: for all a ∈ A, there exists b ∈ B such that a� b. In such case, we call “�” the hyperorder in H. A hyper UP-algebra H with constant 0 and hyperoperation ~ is denoted by (H;~, 0). By (HUP2) or (HUP3), x~ y 6= ∅ for all x, y ∈ H. Note that in [10], x� y is defined by Romano as 0 ∈ x~ y. Thus, (HUP1) in [3] and [10] are equivalent; that is, 0 ∈ (y~ z)~ [(x~ y)~ (x~ z)]. Moreover, (HUP2) to (HUP4) are identical, with “ ◦ ” denoted by “ ~ ”. Example 1. [3] Let H = {0, a, b, c} be a set. Define the hyperoperation ~ by the following Cayley table : ~ 0 a b c 0 {0} {a} {b} {c} a {0} {0,a} {0,b} {c} b {0} {a} {0,b} {c} c {0} {0,a} {0,b} {0,a,c} Then, (H;~, 0) is a hyper UP-algebra. Proposition 1. [3, 10] Let H be a hyper UP-algebra. Then the following hold for all x, y, z ∈ H and for every nonempty subsets A,B,C ⊆ H: (i) 0 ~ 0 = {0} (ii) 0 ~A = A (iii) z � z (iv) A ⊆ B implies A� B R. Amairanto, R. Isla / Eur. J. Pure Appl. Math, 13 (3) (2020), 483-497 485 (v) x~ z � z (vi) A~ 0 = {0} (vii) A� {0} implies A = {0} (viii) A~ (B ~ C) = B ~ (A~ C) (ix) 0� x (x) x ∈ (0 ~ y) implies x� y Definition 2. [3] Let (H;~, 0) and (H ′;~′, 0′) be hyper UP-algebras. A mapping f : H → K is called a hyper homomorphism if (HH1) f(0) = 0′, (HH2) f(x~ y) = f(x) ~′ f(y) for all x, y ∈ H. The following definitions are analogous to the ones given by Borzooei and Harizavi [1] for regular congruence realtions on hyper BCK-algebras. Definition 3. Let θ be an equivalence relation on a hyper UP-algebra H and A,B ⊆ H. Then (i) AθB if there exists a ∈ A and b ∈ B such that aθb; (ii) Aθ̄B if for all a ∈ A, there exists b ∈ B such that aθb and for all b ∈ B, there exists a ∈ A such that aθb; (iii) θ is called a congruence relation on H if whenever xθy and x′θy′, then (x~x′)θ̄(y~y′), for all x, y, x′, y′ ∈ H; (iv) θ is called a regular congruence relation on H if θ is a congruence relation on H and whenever (x~ y)θ{0} and (y ~ x)θ{0}, then xθy for all x, y ∈ H. The set [x]θ = {y ∈ H : yθx} is called the congruence class determined by x. 3. Regular Congruence Relations and Hyper Homomorphisms on Hyper UP-algebras All throughout, H,H ′, H ′′ are hyper UP-algebras. Proposition 2. If f : H −→ H ′ is a hyper homomorphism, then for all nonempty subsets A,B ⊆ H we have f(A~B) = f(A) ~′ f(B). Proof. Let f : H −→ H ′ be a hyper homomorphism and ∅ 6= A,B ⊆ H. Let x ∈ f(A~B) = f  ⋃ a∈A,b∈B a~ b . Then there exist a ∈ A and b ∈ B such that x ∈ f(a~ b). Since f is a hyper homomorphism, x ∈ f(a) ~′ f(b) ⊆ ⋃ f(a) ∈ f(A), f(b) ∈ f(B) f(a) ~′ f(b) = f(A) ~′ f(B). R. Amairanto, R. Isla / Eur. J. Pure Appl. Math, 13 (3) (2020), 483-497 486 Thus, f(A~B) ⊆ f(A) ~′ f(B). Now, let y ∈ f(A) ~′ f(B) = ⋃ f(a) ∈ f(A), f(b) ∈ f(B) f(a) ~′ f(b). Then there exist f(a) ∈ f(A) and f(b) ∈ f(B) such that y ∈ f(a) ~′ f(b). Since f is a hyper homomorphism, y ∈ f(a~ b) ∈ f  ⋃ a∈A,b∈B a~ b  = f(A~B). Thus, f(A) ~′ f(B) ⊆ f(A~B). Therefore, f(A~B) = f(A) ~′ f(B). Definition 4. Let f : H −→ H ′ be a hyper homomorphism. We say that f is a hyper monomorphism if f is one-to-one, and f is a hyper epimorphism if f is onto; f is a hyper isomorphism, denoted by ∼=H, if f is both one-to-one and onto. Lemma 1. Suppose f : H −→ H ′ and g : H ′ −→ H ′′ are both hyper homomorphisms (epimorphisms) of hyper UP-algebras. Then g◦f is a hyper homomorphism (epimorphism) of hyper UP-algebras. The following result establishes the transitivity of the relation θ̄ on H. Lemma 2. Let θ be an equivalence relation on H and A,B ⊆ H. If Aθ̄B and Bθ̄C, then Aθ̄C. Proof. Suppose that Aθ̄B and Bθ̄C. Since Aθ̄B, by Definition 3(ii), for each a ∈ A (respectively b ∈ B), there exists b ∈ B (respectively a ∈ A) such that aθb. Similarly, since Bθ̄C, for all b ∈ B (respectively c ∈ C), there exists c ∈ C(respectively b ∈ B) such that bθc. Since by assumption θ is an equivalence relation for each a ∈ A (respectively c ∈ C), there exists c ∈ C(respectively a ∈ A) such that aθc. Therefore, Aθ̄C. Lemma 3. Let θ be an equivalence relation on H. Then the following are equivalent: (i) θ is a congruence relation on H; (ii) if xθy, then (x~ a)θ̄(y ~ a) and (a~ x)θ̄(a~ y) for all a, x, y ∈ H. Proof. (i) =⇒ (ii) Let θ be a congruence relation on H and a, x, y ∈ H. Suppose xθy. Since θ is a congruence relation on H and aθa, (x ~ a)θ̄(y ~ a) and (a ~ x)θ̄(a ~ y), by Definition 3(iii). (ii) =⇒ (i) Assume xθy. Let x, y, x′, y′ ∈ H. Suppose that xθy and x′θy′. By (ii), (x ~ x′)θ̄(y ~ x′) and (y ~ x′)θ̄(y ~ y′), so that by Lemma 2, (x ~ x′)θ̄(y ~ y′). By Definition 3(iii), θ is a congruence relation on H. R. Amairanto, R. Isla / Eur. J. Pure Appl. Math, 13 (3) (2020), 483-497 487 Theorem 1. Suppose that θ and θ′ are regular congruence relations on H with [0]θ = [0]θ′ . Then θ = θ′. Proof. Let θ and θ′ be regular conguence relations on H with [0]θ = [0]θ′ . Since θ and θ′ are both equivalence relations on H, it suffices to show that xθy if and only if xθ′y for all x, y ∈ H. Let xθy. Since θ is a congruence relation on H, by Lemma 3, (x~x)θ̄(x~y). Note that 0 ∈ x~x by Proposition 1(iii). Thus by Definition 3(ii), there exists an element s ∈ x~ y such that 0θs. It follows that s ∈ [0]θ = [0]θ′ . Hence, (x~ y)θ′{0}. In a similar manner, since xθy, (y ~ x)θ̄(y ~ y). Also, 0 ∈ y ~ y implies that there exists t ∈ y ~ x such that 0θt. Hence, t ∈ [0]θ = [0]θ′ . Thus, (y ~ x)θ′{0}. Now, since (x ~ y)θ′{0}, (y ~ x)θ′{0}, and θ′ is a regular congruence relation, we have xθ′y by Definition 3(iv). Similarly, let xθ′y. Then (x~ x)θ̄′(x~ y). Also, 0 ∈ x~ x implies that there exists an element s ∈ x~ y such that 0θ′s. Furthermore, s ∈ [0]θ′ = [0]θ. So, (x~ y)θ{0}. By similar argument, we will obtain (y ~ x)θ̄′(y ~ y). Since 0 ∈ y ~ y, there exists v ∈ y ~ x such that 0θ′v. So, v ∈ [0]θ′ = [0]θ. Hence, (y ~ x)θ{0}. Since θ is a regular congruence relation, we have xθy. We now reformulate the quotient structure of a hyper UP-algebra presented in [9] via regular congruence relation on a hyper UP-algebra H. Theorem 2. [9] Let θ be a regular congruence relation on H, I = I0 = [0]θ and H/I = {Ix : x ∈ H}, where Ix = [x]θ for all x ∈ H. Then H/I with the hyperoperation ~ and hyperorder � which are defined as follows Ix ~ Iy = {Iz : z ∈ x~ y} and Ix � Iy if and only if I ∈ Iy ~ Ix is a hyper UP-algebra which is called the quotient hyper UP-algebra. Example 2. Let H = {0, 1, 2, 3} be a set. Define the hyperoperation ~ by the following Cayley table: ~ 0 1 2 3 0 {0} {1} {2} {3} 1 {0} {0,1} {0,2} {1,3} 2 {0} {1} {0,2} {3} 3 {0} {0,1} {0,2} {0,1,3} By routine calculations, (H;~, 0) is a hyper UP-algebra. Define a relation θ on H by θ = {(0, 0), (1, 1), (0, 2), (2, 0), (2, 2), (3, 3)}. By Lemma 3, it can be verified that θ is a congruence relation on H. Moreover, by routine calculations, θ is a regular congruence relation. Consider I0 = I = [0]θ = {0, 2}, I1 = {1}, and I3 = {3}. Then H/I = {I, I1, I3}. Thus, our Cayley table is as follows: ~ I I1 I3 I {I} {I1} {I3} I1 {I} {I, I1} {I1, I3} I3 {I} {I, I1} {I, I1, I3} R. Amairanto, R. Isla / Eur. J. Pure Appl. Math, 13 (3) (2020), 483-497 488 By routine calculations, H/I is a hyper UP-algebra. To establish the First Hyper Isomorphism Theorem on hyper UP-algebras, we first reformulate some results on hyper homomorphisms of hyper UP- algebras. Lemma 4. [9] Let θ be a regular congruence relation on H and I = [0]θ. Then the mapping π : H −→ H/I which is defined by π(x) = Ix, for all x ∈ H, is a hyper epimorphism which is called the canonical epimorphism. Theorem 3. [9] (Hyper Homomorphism Theorem) Let θ be a regular congruence on H and I = [0]θ. If f : H −→ H ′ is a hyper homomorphism of hyper UP-algebras such that I is contained in the kernel of f, then f̄ : H/I −→ H ′, which is defined by f̄(Ix) = f(x), for all x ∈ H, is a unique hyper homomorphism such that f̄ ◦ π = f, where π denotes the canonical epimorphism and ◦ is the composition map. Theorem 4. (First Hyper Isomorphism Theorem) Let θ be a regular congruence relation on H and I = [0]θ. If f : H −→ H ′ is a hyper homomorphism of hyper UP- algebras such that ker f = I, then H/ker f ∼=H Imf. Proof. Define f̄ : H/I −→ H ′ by f̄(Ix) = f(x) for all x ∈ H. Let x, y ∈ H. Then Ix, Iy ∈ H/I. From Theorem 3, f̄ is a hyper homomorphism. Thus, f̄(Ix ~ Iy) = f̄(Ix) ~′ f̄(Iy) and f̄(I) = 0′. Suppose that f̄(Ix) = f̄(Iy) with x, y ∈ H. Then f(x) = f(y). Since f is a hyper homomorphism, 0′ = f(0) ∈ f(x~ x) = f(x) ~′ f(x) = f(x) ~′ f(y) = f(x~ y). So, there exists an element u ∈ x~ y such that f(u) = 0′, that is, u ∈ kerf = I = [0]θ. Thus, uθ0 and (x ~ y)θ{0}. Also, 0′ = f(0) ∈ f(x ~ x) = f(x) ~′ f(x) = f(y) ~′ f(x) = f(y ~ x). Thus, there exists an element v ∈ y~x such that f(v) = 0′. Moreover, v ∈ kerf = I = [0]θ and vθ0. Thus, (y ~ x)θ{0}. Since θ is a regular congruence relation, it follows that xθy. Thus, Ix = Iy. Hence, f̄ is one-to-one, thus ker f̄ = (ker f)/I ⊆ H/I is trivial, which occurs if and only if ker f = I. Clearly, Im f̄ = Im f and f̄ : H/I −→ Im f is onto. Therefore, H/ker f ∼=H Imf. Lemma 5. Let f : H −→ H ′ be a hyper homomorphism on hyper UP-algebras with I = [0]θ and J = [0′]θ′ where θ and θ′ are regular congruence relations on H and H ′, respectively. Suppose that I ⊆ ker f . Then for all x, y ∈ H,xθy implies that f(x)θ′f(y). Proof. Let f : H −→ H ′ be a hyper homomorphism with I = [0]θ ⊆ ker f and J = [0′]θ′ where θ and θ′ are regular congruence relations on H and H ′, respectively. Let x, y ∈ H such that xθy. Since θ is a regular congruence relation, we have xθx and (x~ x)θ̄(x~ y) by Definition 3(iii). Since 0 ∈ x~ x by Proposition 1(iii), there exists an element u ∈ x~ y such that 0θu. Thus, u ∈ I ⊆ ker f , that is, f(u) = 0′. It follows that f(u) ∈ H ′ and f(u)θ′0′. Since f is a hyper homomorphism, f(u) ∈ f(x~y) = f(x)~′f(y), thus (f(x) ~′ f(y))θ̄′{0′}. Using similar argument, with yθy, we have (f(y) ~′ f(x))θ̄′{0′}. Since θ′ is a regular congruence relation, by Definition 3(iv) we have f(x)θ′f(y). R. Amairanto, R. Isla / Eur. J. Pure Appl. Math, 13 (3) (2020), 483-497 489 Theorem 5. Let θ and θ′ be regular congruence relations on hyper UP-algebras H and H ′, respectively, such that I = [0]θ and J = [0′]θ′ . If f : H −→ H ′ is a hyper homomorphism of hyper UP-algebras such that xθy if and only if f(x)θ′f(y), for all x, y ∈ H, then there exists a unique hyper homomorphism f∗ : H/I −→ H ′/J such that π′ ◦ f = f∗ ◦ π where π and π′ are the canonical epimorphisms and ◦ is the composition map. H f−→ H ′yπ yπ′ H/I f∗−→ H ′/J Proof. Consider the mapping f∗ : H/I −→ H ′/J defined by f∗(Ix) = Jf(x), for all x ∈ H. Let x, y ∈ H such that Ix = Iy. Then xθy and so f(x)θ′f(y) by assumption. Hence, f∗(Ix) = Jf(x) = Jf(y) = f∗(Iy) and f∗ is well-defined. Let Ix, Iy ∈ H/I and Jt ∈ f∗(Ix ~ Iy). Then there exists an element t′ ∈ x ~ y such that Jf(t′) = f∗(It′) = Jt. Now, t′ ∈ x ~ y implies f(t′) ∈ f(x ~ y) = f(x) ~′ f(y). So, Jt = Jf(t′) ∈ Jf(x) ~ ′ Jf(y) = f∗(Ix) ~′ f∗(Iy). Hence, f∗(Ix ~ Iy) ⊆ f∗(Ix) ~′ f∗(Iy). Next, let Js ∈ f∗(Ix) ~′ f∗(Iy) = Jf(x) ~ ′ Jf(y). Then s ∈ f(x) ~′ f(y) = f(x ~ y). Now, s ∈ f(x~ y) implies there exists w ∈ x~ y such that f(w) = s, that is, Iw ∈ Ix ~ Iy and Js = Jf(w) = f∗(Iw) ∈ f∗(Ix ~ Iy). Therefore, f∗(Ix) ~′ f∗(Iy) ⊆ f∗(Ix ~ Iy) and so f∗(Ix ~ Iy) = f∗(Ix) ~′ f∗(Iy). Moreover, f∗(I) = Jf(0) = J0′ = J. Also, dom(π′ ◦ f) = H = dom(f∗ ◦ π). Let x ∈ H. Then (π′ ◦ f)(x) = π′(f(x)) = Jf(x) = f∗(Ix) = f∗(π(x)) = (f∗ ◦ π)(x). Thus, π′ ◦ f = f∗ ◦ π. Next, we let φ : H/I −→ H ′/J be a homomorphism such that π′ ◦ f = φ ◦ π. Note that dom(π′ ◦ f) = H = dom(φ ◦ π). Then φ = f∗ since for all x ∈ H, we have φ(Ix) = φ(π(x)) = Jπ(x) = π′(f(x)) = (π′ ◦ f)(x) = (f∗ ◦ π)(x) = f∗(Ix). Theorem 6. Suppose f : H −→ H ′ is a hyper epimorphism of hyper UP-algebras, θ′ is a regular conruence relation on H ′ and J = [0′]θ′. Then there exists a regular congruence relation θ on H such that H/I ∼=H H ′/J, where I = [0]θ. Proof. Define θ on H by xθy if and only if f(x)θ′f(y), for all x, y ∈ H. Let x ∈ H. Then f(x) ∈ H ′ and so, by reflexivity of θ′ on H ′, we have f(x)θ′f(x). It follows that xθx and θ is a reflexive relation on H. Assume that xθy, where x, y ∈ H. So, f(x), f(y) ∈ H ′ and f(x)θ′f(y). Hence, f(y)θ′f(x) which will imply that yθx. Thus, θ is a symmetric relation on H. Suppose xθy and yθz, where x, y, z ∈ H. Then f(x)θ′f(y) and f(y)θ′f(z), for all x, y, z ∈ H. Note that f(x), f(y), f(z) ∈ H ′ and by transitivity of θ′ on H ′, we have f(x)θ′f(z). Thus, xθz on H and θ is a transitive relation on H. Therefore, θ is an equivalence relation on H. Next, we will show that θ is a congruence relation. Let a, x, y ∈ H such that xθy. Then f(x)θ′f(y). Since f(a), f(x), f(y) ∈ H ′ and θ′ is a congruence relation on H ′, from R. Amairanto, R. Isla / Eur. J. Pure Appl. Math, 13 (3) (2020), 483-497 490 Lemma 3 it follows that (f(x) ~ f(a))θ̄′(f(y) ~ f(a)) and (f(a) ~ f(x))θ̄′(f(a) ~ f(y)). Thus, (x ~ a)θ̄(y ~ a) and (a ~ x)θ̄(a ~ y). Therefore, by Lemma 3, θ is a congruence relation on H. Let x, y ∈ H such that (x~ y)θ{0} and (y ~ x)θ{0}. Then f(x), f(y) ∈ H ′ and there exist a ∈ (x~y) and b ∈ (y~x) such that aθ0 and bθ0. Since f is a hyper homomorphism and f(0) = 0′, f(a) ∈ f(x ~ y) = f(x) ~′ f(y) and f(b) ∈ f(y ~ x) = f(y) ~′ f(x) such that f(a)θ′0′ and f(b)θ′0′. Thus, (f(x) ~′ f(y))θ′{0′} and (f(y) ~′ f(x))θ′{0′}. Since θ′ is a regular congruence relation on H ′, f(x)θ′f(y), implying that xθy. Therefore, θ is a regular congruence relation on H. Next, let x ∈ I = [0]θ. Since xθ0 and f(0) = 0′, f(x)θ′0′. It follows that f(x) ∈ [0′]θ′ = J, so x ∈ f−1(J). Thus, I ⊆ f−1(J). On the other hand, let y ∈ f−1(J). Then f(y) ∈ J = [0′]θ′ and f(y)θ′0′. Hence, yθ0 and y ∈ [0]θ = I, implying that f−1(J) ⊆ I. Thus, I = f−1(J). Now, let π : H ′ −→ H ′/J be the canonical hyper epimorphism and define f̄ : H −→ H ′/J by f̄ = π ◦ f . Since π and f are both hyper epimorphisms of hyper UP-algebras, by Lemma 1, f̄ is a hyper epimorphism. Observe that ker f̄ = {x ∈ H : f̄(x) = J} = {x ∈ H : π(f(x)) = J} = {x ∈ H : Jf(x) = J} = {x ∈ H : f(x) ∈ J} = {x ∈ H : x ∈ f−1(J)} = {x ∈ H : x ∈ I} = I. Therefore, by the First Hyper Isomorphism Theorem, H/I ∼=H H ′/J . Theorem 7. Let f : H −→ H ′ be a hyper epimorphism on hyper UP-algebras and let Θ and Ω be relations on H and H ′, respectively, defined by xΘy ⇐⇒ f(x)Ωf(y) for all x, y ∈ H. Then Θ is a regular congruence relation on H if and only if Ω is a regular congruence relation on H ′. Proof. Utilizing the proof of Theorem 6, we only need to show that Θ is a regular congruence relation on H implies that Ω is a regular congruence relation on H ′. Suppose Θ is a regular congruence relation on H. Let u, v, w ∈ H ′. Then there exist x, y, z ∈ H such that f(x) = u, f(y) = v, and f(z) = w. Since Θ is an equivalence relation on H, xΘx, thus u = f(x)Ωf(x) = u and Ω is a reflexive relation on H ′. Suppose uΩv. Then xΘy and since Θ is a symmetric relation on H, yΘx, so vΩu and Ω is a symmetric relation on H ′. Suppose uΩv and vΩw. Then xΘy and yΘz. Since Θ is a transitive relation on H, xΘz, that is, uΩw. Thus, Ω is an equivalence relation on H ′. Let b, u, v ∈ H ′ and uΩv. Then there exist a, x, y ∈ H such that b = f(a), u = f(x), v = f(y), and xΘy. Since Θ is a congruence relation on H and a ∈ H, (a ~ x)Θ̄(a ~ y) by R. Amairanto, R. Isla / Eur. J. Pure Appl. Math, 13 (3) (2020), 483-497 491 Lemma 3. Hence, f(a)~′ f(x) = f(a~ x)Ω̄f(a~ y) = f(a)~′ f(y), that is, (b~′ u)Ω̄(b~′ v). Similarly, since Θ is a congruence relation on H and a ∈ H, (x ~ a)Θ̄(y ~ a). So, f(x) ~′ f(a) = f(x~ a)Ω̄f(y ~ a) = f(y) ~′ f(a), that is, (u~′ b)Ω̄(v ~′ b). Hence, Ω is a congruence relation on H ′. Now, let u, v ∈ H ′ such that (u ~′ v)Ω{0′} and (v ~′ u)Ω{0′}. Since (u ~′ v)Ω{0′} and f is a hyper epimorphism, it follows that there exist s, t ∈ H such that f(s) = u, f(t) = v, f(s ~ t) = f(s) ~′ f(t) = (u ~′ v)Ω{0′}. Similarly, (v ~′ u)Ω{0′} implies f(t~ s) = f(t) ~′ f(s) = (v ~′ u)Ω{0′}. Hence, (s~ t)Θ{0} and (t~ s)Θ{0}. Since Θ is a regular congruence relation on H, it follows that sΘt and uΩv. Therefore, Ω is a regular congruence relation on H ′. Remark 1. Let f : H −→ H ′ be a hyper epimorphism on hyper UP-algebras and let Θ and Ω be the relations on H and H ′, respectively, as defined in Theorem 7. Then (i) Ω is called the regular congruence relation induced by f and Θ, and (ii) Θ is called the regular congruence relation induced by f and Ω. Theorem 8. Let f : H −→ H ′ be a hyper epimorphism on hyper UP-algebras. Then there is a one-to-one correspondence between the regular congruence relations on H ′ and the regular congruence relations on H such that ker f is contained in the regular congruence class containing 0. Proof. Let f : H −→ H ′ be a hyper epimorphism of hyper UP-algebras and A = {Θ : Θ is a regular congruence relation on H with ker f ⊆ [0]Θ} B = {Ω : Ω is a regular congruence relation on H ′}. Define γ : A −→ B by γ(Θ) = Ω, where Ω is the regular congruence relation on H ′ induced by f and Θ. Then Ω ∈ B. Let Θ1,Θ2 ∈ A such that Ω1 = γ(Θ1) = γ(Θ2) = Ω2. Then for all x, y ∈ H,xΘ1y ⇔ f(x)Ω1f(y) ⇔ f(x)Ω2f(y) ⇔ xΘ2y. Hence, Θ1 = Θ2 and γ is well-defined and one-to-one. Now, let Ω ∈ B and consider the induced regular congruence relation Θ on H. If x ∈ ker f, then f(x) = f(0.) So, f(x)Ωf(0) implies xΘ0. Thus, ker f ⊆ [0]Θ and so, Θ ∈ A. Lastly, we show that γ is onto, that is, γ(Θ) = Ω. Suppose γ(Θ) = Ω′ for some Ω′ ∈ B. Then by the definitions of Ω and Θ, for each t ∈ H ′, tΩ′0′ ⇔ t = f(x) and xΘ0 for some x ∈ H ⇔ f(x)Ωf(0)⇔ tΩ0′. Thus, [0′]Ω = [0′]Ω′ and by Theorem 1, Ω = Ω′. Hence, γ(Θ) = Ω. Therefore, γ is a bijection. R. Amairanto, R. Isla / Eur. J. Pure Appl. Math, 13 (3) (2020), 483-497 492 4. Hyper Product of Hyper UP-algebras Throughout this section, H and K shall mean the hyper UP-algebras (H,~H , 0H) and (K,~K , 0k) with �H and �K as their hyper orders, respectively. The following introduction of the hyper product of two hyper UP-algebras is influenced by the construction of the hyper product of two hyper BCK-algebras by Borzooei et al. [12], as cited in [1]. Suppose H and K are hyper UP-algebras. Then H ×K = {(a, b)|a ∈ H and b ∈ K}. Define the hyperoperation “ ~ ” on H ×K by (a, b) ~ (c, d) = (a~H c, b~K d) and hyperorder “� ” by (a, b)� (c, d)⇐⇒ a�H c and b�K d for all (a, b), (c, d) ∈ H× K. For every (A,B), (C,D) ⊆ H×K, (A,B)� (C,D) if and only if for all (a, b) ∈ (A,B), there exists (c, d) ∈ (C,D) such that (a, b) � (c, d). Then (H ×K;~, (0H , 0K)) is called the hyper product of H and K. Theorem 9. [9] Let H and K be hyper UP-algebras. Then H ×K is a hyper UP-algebra. Theorem 10. Let α1 : H1 −→ K1 and α2 : H2 −→ K2 be hyper homomorphisms of hyper UP-algebras. Define α : H1 ×H2 −→ K1 ×K2 by α((a, b)) = (α1(a), α2(b)) for all (a, b) ∈ H1 ×H2. Then (i) α is a hyper homomorphism; (ii) ker α = ker α1 × ker α2; (iii) Im α = Im α1 × Im α2; and (iv) α is a hyper monomorphism (respectively, hyper epimorphism) if and only if αi is a hyper monomorphism (respectively, hyper epimorphism) for each i = 1, 2. Proof. Define α : H1 × H2 −→ K1 × K2 by α((a, b)) = (α1(a), α2(b)) for all (a, b) ∈ H1 ×H2. (i) Let (a, b), (c, d) ∈ H1 × H2 such that (a, b) = (c, d). Then a = c and b = d. Now, since α1 and α2 are well-defined maps, it follows that α((a, b)) = (α1(a), α2(b)) = (α1(c), α2(d)) = α((c, d)). So, α is a well-defined map. Observe that (0H1 , 0H2) ∈ H1 × H2. Since α1 and α2 are hyper homomorphisms, by (HH1) we have α((0H1 , 0H2)) = (α1(0H1), α2(0H2)) = (0K1 , 0K2) R. Amairanto, R. Isla / Eur. J. Pure Appl. Math, 13 (3) (2020), 483-497 493 and by (HH2), α((a, b) ~ (c, d)) = α((a~ c, b~ d)) = {α((u, v))|u ∈ a~ c, v ∈ b~ d} = {(α1(u), α2(v))|u ∈ a~ c, v ∈ b~ d} = (α1(a~ c), α2(b~ d)) = (α1(a) ~ α1(c), α2(b) ~ α2(d)) = α(a, b) ~ α(c, d). Hence, α is a hyper homomorphism. (ii) By definition, ker α = {(a, b) ∈ H1 ×H2|α((a, b)) = (0K1 , 0K2)} = {(a, b) ∈ H1 ×H2|(α1(a), α2(b)) = (0K1 , 0K2)} = {(a, b) ∈ H1 ×H2|α1(a) = 0K1 and α2(b) = 0K2} = {(a, b) ∈ H1 ×H2|a ∈ ker α1, b ∈ ker α2} = ker α1 × ker α2. (iii) By definition, Im α = {α((a, b))|(a, b) ∈ H1 ×H2} = {(α1(a), α2(b))|(a, b) ∈ H1 ×H2} = {(α1(a), α2(b))|α1(a) ∈ Im α1, α2(b) ∈ Im α2} = Im α1 × Im α2. (iv) Suppose that α is one-to-one. Let a, c ∈ H1 and b, d ∈ H2 such that α1(a) = α1(c) and α2(b) = α2(d). Then α((a, b)) = (α1(a), α2(b)) = (α1(c), α2(d)) = α((c, d)). Since α is one-to-one, (a, b) = (c, d), that is, a = c and b = d. Thus, α1 and α2 are one-to-one maps. Conversely, assume that α1 and α2 are one-to-one maps. Suppose (a, b), (c, d) ∈ H1×H2 such that α((a, b)) = α((c, d)). Then (α1(a), α2(b)) = α((a, b)) = α((c, d)) = (α1(c), α2(d)). This means that α1(a) = α1(c) and α2(b) = α2(d) and since α1 and α2 are both one-to-one, it follows that a = c and b = d. Hence, (a, b) = (c, d). Therefore, α is one-to-one. Suppose α is onto. Let x ∈ K1 and y ∈ K2. It follows that (x, y) ∈ K1×K2. Since α is onto, there exists (a, b) ∈ H1 ×H2 such that (α1(a), α2(b)) = α((a, b)) = (x, y), that is, α1(a) = x and α2(b) = y for some a ∈ H1 and b ∈ H2. So, α1 and α2 are onto R. Amairanto, R. Isla / Eur. J. Pure Appl. Math, 13 (3) (2020), 483-497 494 maps. Next, suppose α1 and α2 are onto maps. Let (x, y) ∈ K1 ×K2. Then x ∈ K1 and y ∈ K2. Since α1 and α2 are onto maps, we can pick some elements a ∈ H1 and b ∈ H2 such that α1(a) = x and α2(b) = y, that is, α((a, b)) = (α1(a), α2(b)) = (x, y) for some (a, b) ∈ H1 ×H2. Therefore, α is onto and (iv) holds. Recall that if {Ak : k ∈ I} is a family of sets, the Cartesian product ∏ k∈I Ak is the set of all functions p : I −→ ⋃ k∈I Ak such that p(k) ∈ Ak, for all k ∈ I. If p ∈ ∏ k∈I Ak such that p(i) = ai ∈ Ai for all i ∈ I, then we will denote p as {ai}. We now extend the hyper product H × K of H and K to the hyper product of an arbitrary family of hyper UP-algebras. Let {Hk : k ∈ I} be a family of hyper UP-algebras. For each k ∈ I, let ~k, 0k, and �k be the hyperoperation, the zero element, and the hyperorder of Hk, respectively. Let G = ∏ k∈I Hk and define the hyperoperation ~ as follows: for {xk}, {yk} ∈ G, {xk} ~ {yk} = ∏ k∈I (xk ~ yk). Since xk ~ yk 6= ∅ for each k ∈ I, the Axiom of Choice ensures us that ∏ k∈I (xk ~ yk) 6= ∅, and so ~ is indeed a hyperoperation. The zero element of G is {0k}, and under the hyperoperation ~, the hyperorder � is established as follows: for {xk}, {yk} ∈ G, {xk} � {yk} ⇐⇒ {0k} ∈ {yk}~ {xk} ⇐⇒ {0k} ∈ ∏ k∈I (yk ~ xk) ⇐⇒ for all k ∈ I, 0k ∈ yk ~ xk ⇐⇒ for all k ∈ I, xk �k yk, and for all ∏ k∈I Ak, ∏ k∈I Bk ⊆ ∏ k∈I Hk, ∏ k∈I Ak � ∏ k∈I Bk ⇐⇒ ∀{ak} ∈ ∏ k∈I Ak, ∃{bk} ∈ ∏ k∈I Bk such that {ak} � {bk} ⇐⇒ ∀k ∈ I, ∀ak ∈ Ak,∃bk ∈ Bk such that ak �k bk ⇐⇒ ∀k ∈ I, Ak �k Bk. Then (G,~, {0K}) is called the hyper product of {Hk : k ∈ I}. Lemma 6. Let {Hk : k ∈ I} be a nonempty family of hyper UP-algebras. Suppose that Ak, Bk ⊆ Hk, for all k ∈ I. Then for each k ∈ I, R. Amairanto, R. Isla / Eur. J. Pure Appl. Math, 13 (3) (2020), 483-497 495∏ k∈I Ak ~ ∏ k∈I Bk = ∏ k∈I (Ak ~Bk). Theorem 11. Suppose that {Hk : k ∈ I} is a nonempty family of hyper UP-algebras. Then (∏ k∈I Hk,~, {0k} ) is a hyper UP-algebra. Proof. Suppose {Hk : k ∈ I} is a nonempty family of hyper UP-algebras. Let {ak}, {bk}, {ck}, {dk} ∈ ∏ k∈I Hk. Then ak, bk, ck, dk ∈ Hk for all k ∈ I. We will show first that ~ is a well-defined hyperoperation on ∏ k∈I Hk. Assume that {ak} = {bk} and {ck} = {dk}, for all k ∈ I. Then ak = bk and ck = dk for all k ∈ I. So, {ak}~ {ck} = ∏ k∈I (ak ~k ck) = ∏ k∈I (bk ~k dk) = {bk}~ {dk} for all k ∈ I. Thus, ~ is a well-defined hyperoperation on ∏ k∈I Hk. Let {xk}, {yk}, {zk} ∈∏ k∈I Hk. Then xk, yk, zk ∈ Hk for all k ∈ I. Now, for each k ∈ I, we have ({xk}~ {yk}) ~ ({xk}~ {zk}) = (∏ k∈I (xk ~k yk) ) ~ (∏ k∈I (xk ~k zk) ) = (∏ k∈I (xk ~k yk) ~ (xk ~k zk) ) . Since for each k ∈ I, (xk ~k yk) ~ (xk ~k zk)�k yk ~k zk, it follows that∏ k∈I (xk ~k yk) ~ (xk ~k zk)� ∏ k∈I (yk ~k zk), that is, ({xk ~ yk}) ~ ({xk ~ zk})� {yk}~ {zk}. This means that (HUP1) holds on ∏ k∈I Hk. Since for each k ∈ I, 0k ~k xk = {xk}, it follows that {0k}~ {xk} = ∏ k∈I (0k ~k xk) = ∏ k∈I {xk}. REFERENCES 496 Thus, (HUP2) holds on ∏ k∈I Hk. Moreover, since for each k ∈ I, xk ~k 0k = {0k}, it follows that {xk}~ {0k} = ∏ k∈I (xk ~k 0k) = ∏ k∈I {0k}. Hence, (HUP3) holds on ∏ k∈I Hk. Furthermore, suppose {xk} � {yk} and {yk} � {xk} for all k ∈ I. Then xk �k yk and yk �k xk for all k ∈ I. Hence, xk = yk for all k ∈ I and so {xk} = {yk}. This means that (HUP4) holds on ∏ k∈I Hk. Therefore, (∏ k∈I Hk,~, {0k} ) is a hyper UP-algebra. Acknowledgements This research is funded by the Philippine Department of Science and Technology- Accelerated Science and Technology Human Resource Development Program (DOST- ASTHRDP) and the Mindanao State University-Iligan Institute of Technology. The au- thors wish to express their sincere thanks to the reviewers for their valuable suggestions for the improvement of this paper. References [1] R. Borzooei and H. Harizavi. Regular congruence relations on hyper BCK-algebras. Scientiae Mathematicae Japonicae Online, pages 217–231, 2004. [2] G. Flores and G. Petalcorin. Some hyper isomorphism theorems on hyper BCI- algebra. Journal of Algebra and Applied Mathematics, 13(1):15–31, 2015. [3] D. Gomisong. On Fully UP-semigroups and Hyper UP-algebras. Mindanao State University-Iligan Institute of Technology, Philippines, 2019. [4] A. Iampan. A new branch of the logical algebra: UP-algebras. Journal of Algebra and Related Topics, 5(1):35–54, 2017. [5] S. Mostafa F. Kareem and B. Davvaz. Hyper structure theory applied to KU-algebra. Journal of Hyperstructures, 6(2):82–95, 2017. [6] X. Long. Hyper BCI-algebra. Discuss Math. Soc., 26:5–19, 2006. [7] F. Marty. Sur une generalisation de la notion de groupe. In 8th Congress des Math- ematician Scandinaves, pages 45–49, Stockholm, 1934. [8] C. Prabpayak and U. Leerawat. On ideals and congruence in KU-algebras. Scientia Magna Journal, 5(1):54–57, 2009. REFERENCES 497 [9] D. Romano. Quotient of hyper UP-algebras. Preprint. [10] D. Romano. Hyper UP-algebras. Journal of Hyperstructures online, 8(2):1–11, 2019. [11] Y. Jun M. Zahedi X. Xin and R. Borzooei. On hyper BCK-algebra. Italian Journal of Pure and Applied Mathematics, 10:127–136, 2000. [12] R. Borzooei A. Hasankhani M. Zahedi and Y. Jun. On hyper K-algebras. Mathemat- icae Japonicae, 52(1):113–121, 2000.