Ratio Mathematica Vol.34, 2018, pp. 85–93 ISSN : 1592-7415 eISSN : 2282-8214 Some Kinds of Homomorphisms on Hypervector Spaces Elham Zangiabadi∗, Zohreh Nazari† Received: 24-05-2018 . Accepted: 24-06-2018. Published: 30-06-2018 doi: 10.23755/rm.v34i0.416 Abstract In this paper, we introduce the concepts of homomorphism of type 1 , 2 and 3 and good homomorphism . Then we investigate some properties of them. Keywords : Hypervector space , Homomorphism , Homomorphism of type 1 , 2 and 3 , good homomorphism . 2010 AMS subject classifications : 20N20 , 22A30 . 1 Introduction and Preliminaries The concept of hyperstructure was first introduced by Marty [13] in 1934 . He defined hypergroups and began to analysis their properties and applied them to groups and rational algebraic functions . Tallini introduced the notion of hyper- vector spaces [14] , [15] and studied basic properties of them . Homomorphisms of hypergroups are studied by several authers ([2] - [12]) . Since some kinds of homomorphisms on hypergroup were defined , we encourage to define them on hypervector spaces . In this paper , we introduce the concept of homomorphism of type 1 , 2 and 3. And give an example of a homomorphism that is not a ho- momorphism of type 1, 2 and 3. We show that if f be a homomorphism of type 1, 2 and 3, then f is a homomorphism and every homomorphism of type 2 or 3 ∗Department of Mathematics, Vali-e-asr University, Rafsangan, Iran. e.zangiabadi@vru.ac.ir †Department of Mathematics, Vali-e-asr University, Rafsangan, Iran. z.nazari@vru.ac.ir 85 Elham Zangiabadi and Zohreh Nazari is a homomorphism of type 1. Also, we define a good homomorphism and obtain that every homomorphism of type 2 is a good homomorphism and every good homomorphism is a homomorphism. Finally, w e prove that every onto strong homomorphism is a good homomorphism. Let us recall some definitions which are useful in our results . Definition 1.1. A hypervector space over a field K is a quadruplet (V, +,◦, K) such that (V, +) is an abelian group and ◦ : K ×V → P∗(V ) is a mapping of K × V into the power set of V (deprived of the empty set) , such that (a + b)◦x ⊆ (a◦x) + (b◦x), ∀a, b ∈ K, ∀x ∈ V, (1) a◦ (x + y) ⊆ (a◦x) + (a◦y), ∀a ∈ K, ∀x, y ∈ V, (2) a◦ (b◦x) = (ab)◦x, ∀a, b ∈ K, ∀ x ∈ V, (3) x ∈ 1◦x, ∀x ∈ V, (4) a◦ (−x) = −a◦x, ∀a ∈ K, ∀x ∈ V. (5) Definition 1.2. Let (V, +,◦, K) be a hypervector space . Then H ⊆ V is a subspace of V , if 1) the zero vector , 0 , is in H , 2) U, V ∈ H, then U + V ∈ H , 3) U ∈ H, r ∈ K, then r ◦U ⊆ H . Definition 1.3. Let (V, +,◦, K) and (W,⊕,∗, K) be two hypervector spaces . A mapping f : V → W is called 1) a homomorphism , if ∀r ∈ K, ∀x, y ∈ V : f(x + y) = f(x)⊕f(y), (6) f(r ◦x) ⊆ r ∗f(x). (7) 2) a strong homomorphism, if ∀r ∈ K, ∀x, y ∈ V : f(x + y) = f(x)⊕f(y), (8) f(r ◦x) = r ∗f(x). (9) 86 Some Kinds of Homomorphisms on Hypervector Spaces 2 The main results In this paper, the ground field of a hypervector space V is presented with K, This field is usually considered by R or C. Let (V, +,◦) and (W,⊕,∗) be two hypervector spaces and f : V → W be a mapping. We employ for simplicity of notation xf = f−1(f(x)) and for a subset A of V , Af = f−1(f(A)) = ⋃ {xf : x ∈ A}. Lemma 2.1. Let r ∈ K and x ∈ V . Then the following statements are valid: i) r ◦x ⊆ (r ◦x)f, ii) r ◦x ⊆ r ◦xf, iii) (r ◦x)f ⊆ (r ◦xf)f, iv) r ◦xf ⊆ (r ◦xf)f. Definition 2.1. Let (V, +,◦, K) and (W,⊕,∗, K) be two hypervector spaces and f : V → W be a map such that f(x + y) = f(x) ⊕ f(y), for all a, b ∈ V . Then, for any r ∈ K and x, y ∈ V, f is called a homomorphism of i) type 1, if f−1(r ∗f(x)) = (r ◦xf)f, ii) type 2, if f−1(r ∗f(x)) = (r ◦x)f, iii) type 3, if f−1(r ∗f(x)) = (r ◦xf). Theorem 2.1. Let (V, +,◦, K) and (W,⊕,∗, K) be two hypervector spaces, A be a non-empty subset of V and f : V → W be a map such that f(a + b) = f(a)⊕f(b), for all a, b ∈ V . Then, f is a homomorphism of i) type 1 implies f−1(r ∗f(A)) = (r ◦Af)f, ii) type 2 implies f−1(r ∗f(A)) = (r ◦A)f, iii) type 3 implies f−1(r ∗f(A)) = (r ◦Af). Proof. Each part is established by a straightforward set theoretic argument. Example 2.1. Let (W, +, ·, K) be a classical vector space, P be a proper sub- space of W , W1 = (W, +, ·, K) and W2 = (W,⊕,◦, K) that r◦a = r ·a + P for r ∈ K and a ∈ W. Then W1 and W2 are hypervector spaces. Let f : W1 → W2 be the function defined by f(x) = k ·x, where k ∈ K. We show 87 Elham Zangiabadi and Zohreh Nazari that f is a homomorphism, but not a homomorphism of type 1, 2 and 3. For every r ∈ K and x ∈ W1 we have f(r ·x) = rk ·x rk ·x + P = r ◦f(x). Thus f is a homomorphism. Since f is one to one, we obtain xf = x, for x ∈ W. It followes that (r ·xf)f = (r ·x)f = (r ·xf) = (r ·x). On the other hand, f−1(r ◦f(x)) = f−1(kr ·x + P) = {t ∈ W1 : f(t) ∈ kr ·x + P} = {t ∈ W1 : k · t ∈ kr ·x + P} = {t ∈ W1 : k · t−kr ·x ∈ P}. Hence, f−1(r ◦f(x)) 6= r ·x. Therefore, f is not a homomorphism of type 1, 2 and 3. Theorem 2.2. Let (V, +,◦, K) and (W,⊕,∗, K) be two hypervector spaces and f : V → W be a homomorphism of type n, for n=1,2,3. Then f is a homomor- phism map. Proof. If f be a homomorphism of type 1. Then by using Lemma 2.1, we have f(r ◦x) ⊆ f(r ◦xf) ⊆ f((r ◦xf)f) = f(f−1(r ∗f(x)) ⊆ r ∗f(x). Suppose f is a homomorphism of type 2. Then f(r ◦x) ⊆ f((r ◦x)f) = f(f−1(r ∗f(x)) ⊆ r ∗f(x). Similarly, if f is a homomorphism of type 3, then f(r ◦x) ⊆ f(r ◦xf) = f(f−1(r ∗f(x))) ⊆ r ∗f(x). Lemma 2.2. Let f be a homomorphism. Then (r ◦xf)f ⊆ f−1(r ∗f(x)). 88 Some Kinds of Homomorphisms on Hypervector Spaces Proof. Since f is a homomorphism, for all r ∈ K and x ∈ V, we have f(r ◦xf) ⊆ r ∗f(xf). Since r ∗ f(xf) = r ∗ f(f−1(f(x)) ⊆ r ∗ f(x), hence, f(r ◦ xf) ⊆ r ∗ f(x). Therefore, (r ◦xf)f ⊆ f−1(r ∗f(x)). Proposition 2.1. Let (V, +,◦, K) and (W,⊕,∗, K) be two hypervector spaces and f : V → W be a homomorphism of type 2 or 3. Then f is a homomorphism of type 1. Proof. Suppose that r ∈ K, x ∈ V and f : V → W be a homomorphism of type 2, then by Lemma 2.2 we have (r ◦x)f ⊆ (r ◦xf)f ⊆ f−1(r ∗f(x)) = (r ◦x)f. Similarly, if f is a homomorphism of type 3, then r ◦xf ⊆ (r ◦xf)f ⊆ f−1(r ∗f(x)) = r ◦xf. Proposition 2.2. Let (V, +,◦, K) and (W, +⊕,∗, K) be two hypervector spaces and f : V → W be an onto mapping. Then, given r ∈ K and x ∈ V , f is a homomorphism of i) type 1 if and only if f(r ◦xf) = r ∗f(x), ii) type 2 if and only if f(r ◦x) = r ∗f(x). Proof. Since f is onto, we obtain ff−1(r ∗f(x)) = r ∗f(x). Thus, (i) and (ii) are trivial. Corolary 2.1. Let (V, +,◦, K) and (W,⊕,∗, K) be two hypervector spaces, A and B be non-empty subsets of V and f : V → W be an onto mapping.Then, f is homomorphism of i) type 1 implies f(r ◦Af) = r ∗f(A), ii) type 2 implies f(r ◦A) = r ∗f(A). 89 Elham Zangiabadi and Zohreh Nazari Remark 2.1. On onto homomorphisms between hypervector spaces, a homomor- phism of type 2 is equivalent with a strong homomorphism. Theorem 2.3. Let (V1, +1,◦1, K), (V2, +2,◦2, K) and (V3, +3,◦3, K) be hyper- vector spaces. For n = 1, 2, 3, let f be a homomorphism of type n of V1 onto V2 and g be a homomorphism of type n of V2 onto V3. Then, gf is a homomorphism of type n of V1 onto V3. Proof. Let x, y ∈ V . We have gf(x+1 y) = g(f(x)+2 f(y)) = gf(x)+3 gf(y)). One can easily seen that xgf = f−1(f(x)g). Let n = 1. By above relation, we obtain gf(r ◦xgf) = gf(r ◦f−1(f(x)g)). Since f is onto, there exists a subset A of V such that f(A) = f−1(f(x)g). By Corollary 2.1, we obtain gf(r ◦1 f−1(f(x)g)) = g(r ◦2 f(x)g). Then, by Proposition 2.2, we have g(r ◦2 f(x)g) = r ◦3 gf(x). Let n = 2. Similar to the previous case, but simpler. Let n = 3. Since g is of type 3, (gf)−1(r ◦3 (gf)(x)) = f−1g−1r ◦3 (gf)(x)) = f−1(r ∗f(x)g). Since f is onto, the item (iii) of Theorem 2.1 implies f−1(r ◦2 f(x)g) = r ◦1 f−1(f(x)g) = r ◦1 xgf. Definition 2.2. Let a ∈ V and r ∈ K. We define a/r = {x ∈ V : a ∈ r ◦x}. Proposition 2.3. Let (V1, +,◦, K) and (V2,⊕,∗, K) be two hypervector spaces. If f : V1 → V2 be an onto mapping. Then we have 1) f(a/r) = f(a)/r, if f is a homomorphism of type 2. 2) f(a)/r ⊆ f(af)/r, if f is a homomorphism of type 3. 90 Some Kinds of Homomorphisms on Hypervector Spaces Proof. 1) We know that an onto homomorphism of type 2 is a strong homomor- phism. Suppose that y ∈ f(a/r). Then, there exists t ∈ a/r such that f(t) = y, so a ∈ r ◦ t and f(a) ∈ r ∗ f(t). It implies that y = f(t) ∈ f(a)/r. Therefore, f(a/r) ⊂ f(a)/r. Note that the inverse inclusion is always true. 2) If y ∈ f(a)/r, there is t ∈ V1 such that f(t) = y. Since f is homomorphism of type 3, we have af ∈ r ◦ tf , which means that tf ∈ af/r, therefore y ∈ f(af)/r. Definition 2.3. Let (V, +,◦, K) and (W,∗,⊕, K) be two hypervector spaces and f : V → W be a map such that f(a + b) = f(a)⊕f(b). Then f is called a good homomorphism if f(a/r) = f(a)/r, for any a, b ∈ V and r ∈ K. Remark 2.2. According to Proposition 2.3, if f is a homomorphism of type 2, then f is a good homomorphism. Theorem 2.4. Let (V, +,◦, K) and (W,⊕,∗, K) be two hypervector spaces. If f : V → W be a good homomorphism then, f is a homomorphism. Proof. Let r ∈ K and a ∈ V1. If y ∈ f(r◦a), then, there exists t ∈ r◦a such that y = f(t). Hence, f(a) ∈ f(t/r) = f(t)/r. Abviously, y = f(t) ∈ r ∗f(a). Theorem 2.5. Let (V1, +1,◦1, K), (V2, +2,◦2, K), and (V3, +3,◦3, K) be hyper- vector spaces. Let f be a good homomorphism of V1 to V2 and g be a good homomorphism of V2 to V3. Then, gf is a good homomorphism of V1 to V3. Proof. For every r ∈ K and a ∈ V1, we have gf(a/r) = g(f(a)/r) = gf(a)/r. Proposition 2.4. Let V and W be two hypervector spaces over K and f : V → W be a good homomorphism. Then f(A/K) = f(A)/K, where A ⊆ V and A/K = ⋃ {a/r : a ∈ A, r ∈ K}. Proof. Let y ∈ f(A/K). There exist r ∈ K and a ∈ A such that y ∈ f(a/r) = f(a)/r ⊆ f(A)/K. Conversely, let y ∈ f(A)/k. Then, there exist r ∈ k and a ∈ V such that y ∈ f(a)/r = f(a/r) and so y ∈ f(A/K). Theorem 2.6. Let (V, +,◦, K) and (W,⊕,∗, K) be two hypervector spaces, f be onto strong homomorphism from V to W . Then f is a good homomorphism. 91 Elham Zangiabadi and Zohreh Nazari Proof. Let f(t) ∈ f(x/r). So x ∈ r◦t. It followes that f(t) ∈ f(x)/r. Therefore f(x/r) ⊆ f(x)/r. On the other hand, let y ∈ f(x)/r. Since f is an onto mapping, there exists a t ∈ V such that y = f(t). Hence, f(x) ∈ r ∗ f(t) = f(r ◦ t). Thus x ∈ r ◦ t and then we have t ∈ x/r and y = f(t) ∈ f(x/r). Therefore f(x)/r ⊆ f(x/r). This implies that f(x/r) = f(x)/r. References [1] R. Ameri, O. R. Dehghan, On dimension of hypervector spaces, Eur. J. Pure Appl. Math, 1 (2008), 32-50. [2] P. Corsini, Recent results in the theory of hypergroups, Boll. Unione Mat. Ital. (9), 2 (1983), 133-138. [3] P. Corsini, V. Leoreanu, Applications of hyperstructure theory, Kluwer Aca- demic Publishers, Advances in Mathematics, 2003. [4] B. Davvaz, Isomorphism theorems of polygroups, Bull. Malays. Math. Sci. Soc. (2), 33 (2010), 385-392. [5] B. Davvaz, Groups in polygroups, Iran. J. Math. Sci. Inform., 1 (2006), 25- 31. [6] J. E. Eaton, Theory of cogroups,Duke Math. J., 6 (1940), 101-107. [7] D. Freni, Strongly transitive geometric spaces : Applications to hypergroups and semigroups theory, Comm. Algebra, 32 (2004), 969-988. [8] D. Freni, A new characterization of the derived hypergroup via strongly reg- ular equivalences, Comm. Algebra, 30 (2002), 3977-3989. [9] T. W. Hungerford, Algebra, graduate texts in mathematics, 73. Springer- Verlag, New York-Berlin, 1980. [10] J. Jantosciak, Homomorphisms, equivalences and reductions in hypergroups, Riv. Mat., 9 (1991), 23-47. [11] M. Koskas, Groupoides, demi-hypergroupes et hypergroupes. (French), J. Math. Pures Appl. (9), 49 (1970), 155-192. [12] M. Krasner, A class of hyperrings and hyperfiels, internat. J. Math. Math. Sci., 6 (1983), 307-311. 92 Some Kinds of Homomorphisms on Hypervector Spaces [13] F. Marty, Sur nue generalization de la notion do group, 8nd congress of the Scandinavic Mathematics, Stockholm, (1934), 45-49. [14] M. S. Tallini, Weak hypervector space and norms in such spaces, Algebraic Hyper Structurs and Applications, Jast, Rumania, Hadronic Press, (1994), 199-206. [15] M. S. Tallini, Hypervector spaces, Proceedings of the fourth international congress of algebraic hyperstructures and applications, Xanthi, Greece, ( 1990), 167-174. 93