EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 10, No. 4, 2017, 850-857 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Monomorphism and Epimorphism Properties of Soft Categories Simge Öztunç1,∗, Ali Mutlu1, Aysun Erdoğan Sert1 1 Manisa Celal Bayar University, Faculty of Science and Arts, Department of Mathematics, Turkey Abstract. In this paper, firstly we recall some definitions and basic properties of soft set theory, category theory and soft category theory. We study on soft monomorphism, soft epimorphism, equalizer and coequalizer for soft categories. We gave some properties of these soft morphisms. We proved that the soft Category SFun has coequalizers, equalizer of a morphism pair in SFun category is soft monic and coequalizers of a morphism pair in SFun category is soft epic. 2010 Mathematics Subject Classifications: 46A80, 47H10, 54H25 Key Words and Phrases: Soft Set, Soft Category, Monomorphism, Epimorphism 1. Introduction The concept of soft sets was introduced by D. Molodtsov [6] in 1999 and Soft set theory became an alternative and useful tool for computer science, modeling problems in engineering, economics, medical and social science. Theorical properties of soft set theory has also been studied some mathematicians. Maji et all [5] defined some operations on soft sets. On the other hand Aras, Sönmez, Çakallı [1] and Zorlutuna, Çakır [11] worked on continuity of soft mappings. Also Probabilistic Soft Set Theory has been studied by Aras and Poşul in [2] and soft topological spaces have been studied by Shabir and Naz in [9]. The soft category theory studied by Sardar and Gupta in [8] and Zhou, Li and Akram in [10] and Öztunç [7]. They introduced the basic notions of the theory of soft categories and gave some introductory results of the soft category theory. The purpose of this paper is to study some new properties of soft category theory. Zhou, Li and Akram [10] defined the SFun Category and gave some results such as SFun has equalizers in [10]. Then we present monomorphism and epimorphism properties of SFun category and also prove SFun has coequalizers. We used some fundamental books from Category theory such as [3] and [4]. ∗Corresponding author. Email addresses: simge.oztunc@cbu.edu.tr (S. Öztunç), abgamutlu@gmail.com (A. Mutlu), aysn−erdogn@hotmail.com (A. Erdoğan Sert) http://www.ejpam.com 850 c© 2017 EJPAM All rights reserved. S. Öztunç, A. Mutlu, A. Erdoğan Sert / Eur. J. Pure Appl. Math, 10 (4) (2017), 850-857 851 2. Preliminaries We express a series of definitions of some fundamental notions related to soft set theory and category theory. Definition 1. [6] A pair (F,A) is said to be a soft set over the universe X, where F is a mapping given by F : A→ P (X) and A ⊆ E. Any soft set (F,A) can be extended to a soft set of type (F,E), where F (e) 6= ∅ for all e ∈ A and F (e) = ∅ for all e ∈ E\A. S(X,E) indicates the family of all soft sets over X. Definition 2. [10] Let (F,A) and (G,B) be two soft sets over the set X. Then one says that the mapping f : (F,A)→ (G,B) is a soft function from (F,A) to (G,B) if it satisfies F (a) ⊆ (G ◦ f)(a) for each a ∈ A. Definition 3. [3] A category C consists of the data which is given below: • Objects: A,B,C, ... • Arrows: f, g, h, ... • For each arrow f , there are given objects dom(f), cod(f) which is called the domain and codomain of f . It is written f : A→ B to indicate that A = dom(f) and B = cod(f) • Given arrows f : A→ B and g : B → C, that is, with cod(f) = dom(g) there is an arrow given by g ◦ f : A→ C called the composite of f and g. • For each object A, there is given an arrow 1A : A→ A called the identity arrow of A. This property must satisfy the following laws: Associativity : h ◦ (g ◦ f) = (h ◦ g) ◦ f for all f : A→ B,g : B → C and h : C → D. Unit : f ◦ 1A = f = 1B ◦ f for all f : A→ B. Let SFun denote the category of all soft sets over X and soft functions. [10] S. Öztunç, A. Mutlu, A. Erdoğan Sert / Eur. J. Pure Appl. Math, 10 (4) (2017), 850-857 852 3. Monomorphism and Epimorphism of the Category SFun Definition 4. Let SFun be a soft category and (F,A) and (G,B) be two SFun−objects. If a SFun−morphism fs : (F,A) → (G,B) in SFun Category is left cancellable, then fs is said to be a soft monomorphism. Theorem 1. Let (F,A), (G,B) and (H,C) be SFun−objects over X. Suppose that fs : (F,A) → (G,B) and gs : (G,B) → (H,C) be two soft functions. If fs and gs are soft monic, then gs ◦ fs is soft monic. Proof. If fs : (F,A) → (G,B) and gs : (G,B) → (H,C) are SFun−morphisms, then there is a y ∈ B such that fs(x) = y for every x ∈ A and there is a z ∈ C such that gs(y) = z for every y ∈ B. We have F (x) ⊆ (G ◦ fs)(x) for all x ∈ A, since fs is a soft function and G(y) ⊆ (H ◦ gs)(y) for all y ∈ B, since gs is a soft function. We must show that F (x) ⊆ (H ◦ gs ◦ fs)(x) and gs ◦ fs SFun−morphism is left cancellable in order to prove that gs ◦ fs : (F,A)→ (H,C) is monic. Thus we have the following: F (x) ⊆ (G ◦ fs)(x) = G(fs(x)) = G(y) (1) G(y) ⊆ (H ◦ gs)(y) = H(gs(y)) = H(z) (2) If F (x) ⊆ G(y) and G(y) ⊆ H(z), then F (x) ⊆ H(z). We obtain that F (x) ⊆ H(z) = H(gs(y)) = H(gs(f(x))) = (H ◦ gs ◦ fs)(x) by 1 and 2. Let now show that the left cancellable property. Suppose that (gs◦fs)◦h = (gs◦fs)◦ks for any hs, ks : (K,D) → (F,A) SFun−morphisms. We get gs ◦ (fs ◦ hs) = gs ◦ (fs ◦ ks) because of associativity of morphisms and obtain that fs ◦ hs = fs ◦ ks since gs SFun−morphism is left cancellable. fs is left cancellable since it is a SFun−monomorphism. Thus we conclude that hs = ks. � Theorem 2. Let (F,A), (G,B) and (H,C) be SFun−objects over X and let fs : (F,A) → (G,B) and gs : (G,B) → (H,C) be two SFun−morhisms. If gs ◦ fs is soft monic, then fs is soft monic. S. Öztunç, A. Mutlu, A. Erdoğan Sert / Eur. J. Pure Appl. Math, 10 (4) (2017), 850-857 853 Proof. If fs : (F,A) → (G,B) and gs : (G,B) → (H,C) are SFun−morphisms, then there is a b ∈ B such that fs(a) = b for all a ∈ A and there is a c ∈ C such that gs(b) = c for all b ∈ B. For every a ∈ A we have F (a) ⊆ (G ◦ fs)(a), since fs is a soft function and for every b ∈ B, we have G(b) ⊆ (H ◦ gs)(b), since gs is a soft function. At first we must show that gs ◦ fs : (F,A) → (H,C) is a SFun− morphism and then fs is left cancellable. Thus we have the following inclusions: F (a) ⊆ (G ◦ fs)(a) = G(fs(a)) = G(b) (3) G(b) ⊆ (H ◦ gs)(b) = H(gs(b)) = H(c) (4) By 3 and 4 we obtain that F (a) ⊆ H(c) and F (a) ⊆ H(c) = H(gs(b)) = H(gs(fs(a))) = (H ◦ gs ◦ fs)(a). Therefore gs ◦ fs is on SFun− morphism. Let now show that the left cancellable property. Assume that fs ◦ hs = fs ◦ ks for any hs, ks : (K,D)→ (F,A) SFun− morphisms. Applying the SFun− morphism gs gs ◦ fs ◦ hs = gs ◦ fs ◦ ks. gs ◦ fs is left cancellable since gs ◦ fs is a SFun− monomorphism. Therefore we obtain that hs = ks. � Definition 5. Let SFun be a soft category and (F,A) and (G,B) be two SFun− objects. If fs : (F,A)→ (G,B) SFun− morphism is right cancellable, then fs is said to be a soft epimorphism. Theorem 3. Let (F,A), (G,B) and (H,C) be SFun− objects over X. Suppose that fs : (F,A)→ (G,B) and gs : (G,B)→ (H,C) be two soft functions. If fs and gs are soft epic, then gs ◦ fs is soft epic. Proof. If fs : (F,A) → (G,B) and gs : (G,B) → (H,C) are SFun− morphisms, then there is a b ∈ B such that fs(a) = b for all a ∈ A and there is a c ∈ C such that gs(b) = c for all b ∈ B. Since fs is a soft function, F (a) ⊆ (G ◦ fs)(a) for every a ∈ A and since gs is a soft function G(b) ⊆ (H ◦ gs)(b) for every b ∈ B. At first we must show that F (a) ⊆ (H ◦ gs ◦ fs)(a) and SFun− morphism gs ◦ fs is right cancellable in order to show that gs ◦ fs : (F,A) → (H,C) is epic. Thus we have the following: F (a) ⊆ (G ◦ fs)(a) = G(fs(a)) = G(b) (5) S. Öztunç, A. Mutlu, A. Erdoğan Sert / Eur. J. Pure Appl. Math, 10 (4) (2017), 850-857 854 G(b) ⊆ (H ◦ gs)(b) = H(gs(b)) = H(c) (6) By (5) and (6) it is obtained that F (a) ⊆ H(c) and then, F (a) ⊆ H(c) = H(gs(b)) = H(gs(fs(a))) = (H ◦ gs ◦ fs)(a). Let now show the right cancellable property. Let hs ◦ (gs ◦ fs) = ks ◦ (gs ◦ fs) for any hs, ks : (K,D)→ (F,A) SFun− morphisms. We have (hs ◦ gs) ◦ fs = (ks ◦ gs) ◦ fs, since morphisms are associative and fs is right cancellable since it is a SFun− epimorphism. Similarly gs is right cancellable since it is a SFun− epimorphism. Therefore we obtained that hs = ks. � Theorem 4. Let (F,A), (G,B) and (H,C) be SFun− objects over X. Suppose that fs : (F,A) → (G,B) and gs : (G,B) → (H,C) be two soft functions. If gs ◦ fs is epic, then gs is soft epic. Proof. If fs : (F,A) → (G,B) and gs : (G,B) → (H,C) are SFun− morphisms, then there is a b ∈ B such that fs(a) = b for all a ∈ A and there is a c ∈ C such that gs(b) = c for all b ∈ B. Also we have F (a) ⊆ (G ◦ fs)(a) for every a ∈ A since fs is a soft function and we have G(b) ⊆ (H ◦ gs)(b) for every b ∈ B since gs is a soft function. Now we must show the map gs ◦ fs : (F,A) → (H,C) is a SFun− morphism. Thus we have the following inclusions: F (a) ⊆ (G ◦ fs)(a) = G(fs(a)) = G(b) (7) G(b) ⊆ (H ◦ gs)(b) = H(gs(b)) = H(c) (8) By (8) and (9) we obtained that F (a) ⊆ H(c) and F (a) ⊆ H(c) = H(g(b)) = H(gs(fs(a))) = (H ◦ gs ◦ fs)(a). Hence gs ◦ fs is a SFun− morphism. Next show that gs is right cancellable. Let hs ◦ gs = ks ◦ gs for any hs, ks : (K,D)→ (F,A) soft morphisms. Applying SFun−morphism fs hs ◦ gs ◦ fs = ks ◦ gs ◦ fs . Thus we get hs = ks, since gs ◦ fs is a soft epimorphism. � Theorem 5 (11). SFun has equalizers. (H ′, C ′) ē �� e′ ## (H,C) e // (F,A) gs // fs // (G,B) S. Öztunç, A. Mutlu, A. Erdoğan Sert / Eur. J. Pure Appl. Math, 10 (4) (2017), 850-857 855 Theorem 6. In category SFun , if the equalizer of a morphism pair is ((H,C), e), then ((H,C), e) is monic. Proof. (H ′, C ′) ¯̄e �� ē �� e′ ## (H,C) e // (F,A) gs // fs // (G,B) Suppose that ((H,C), e), is equalizer of fs and gs. Let ē and ¯̄e be two soft morphisms as illustrated above diagram. We have H ′(c′) ⊆ (F ◦ e′)(c′) since e′ : (H ′, C ′) → (F,A) is soft morphism, H ′(c′) ⊆ (H ◦ ē)(c′) since ¯̄e : (H ′, C ′) → (H,C) is soft morphism and H ′(c′) ⊆ (H ◦ ¯̄e)(c′) since ¯̄e : (H ′, C ′) → (H,C) is soft morphism. Thus we have the following inclusion: H ′(c′) ⊆ (F ◦ e′)(c′) = F (e′)(c′) = F (e ◦ ē)(c′) = F (e ◦ ¯̄e)(c′). Assume that eē = e¯̄e. Then we want to show that ē = ¯̄e. Put e′ = eē = e¯̄e. Then fse ′ = fseē = fse¯̄e fse ′ = fseē = gse¯̄e = gse ′ . By using Universal Mapping Property , there is unique u : (H ′, C ′)→ (H,C) such that eu = e′. Hence we obtain that ē = u = ¯̄e, because we have eē = e′ and e¯̄e = e′. Since eē = e¯̄e implies that ē = ¯̄e, e is left cancellable. Therefore e is monic. � Theorem 7. SFun has coequalizers. (F,A) fs // gs // (G,B) e // e′ ## (H,C) ē �� (H ′, C ′) Proof. Define the set C = {b ∈ B : fs(b) = gs(b)}, the embedding map e : B → C and G = H ◦ e. From the diagram e ◦ fs = e ◦ gs and G(b) = (H ◦ e)(b) for every b ∈ B. Thus e is a SFun−morphism. Now show that ((G,B), e) is coequalizer of fs and gs. Let (H,C) is a SFun−object and be a morphism of (G,B) into (H,C) satisfying e′ ◦ fs = e′ ◦ gs. Define the map ē : (H,C) → (H ′, C ′) such that ē = e′. We obtain that e′ = ē ◦ e from the above diagram. S. Öztunç, A. Mutlu, A. Erdoğan Sert / Eur. J. Pure Appl. Math, 10 (4) (2017), 850-857 856 Since e′ ◦ fs = e′ ◦ gs, we have e′(fs(c)) = e′(gs(c)) for every c ∈ C. Hence gs(c) ∈ C and ē ◦ e′ is well defined. Since G = H ◦ e, ē = e′ and e′ is a SFun− morphism we have the following inclusion: G(b) ⊆ (H ◦ ē)(b) = (H ◦ e′)(b) = H(ē(e(b))) = (H ◦ ē)(e(b)) = (H ◦ e)(b) = G(b) . Hence ē is a SFun− morphism. Since e′ = ē ◦ e and ē is unique, we conclude that ((G,B), e) is a coequalizer of fs and gs. � Theorem 8. If ((G,B), e) is a coequalizer of a morphism pair in SFun category, then ((G,B), e) is epic. Proof. (F,A) fs // gs // (G,B) e // e′ ## (H,C) ¯̄e �� ē �� (H ′, C ′) Suppose that ((G,B), e) is coequalizer of fs and gs. Let ē and ¯̄e be two SFun− morphisms as illustrated above diagram. We have G(b) ⊆ (H ′ ◦ e′)(b) since e′ : (G,B)→ (H ′, C ′) is a SFun− morphism, H(c) ⊆ (H ′ ◦ ē)(c) since ē : (H,C) → (H ′, C ′) is a SFun−morphism and H(c) ⊆ (H ′ ◦ ¯̄e)(c) since ¯̄e : (H,C)→ (H ′, C ′) is a SFun− morphism. From the above diagram G(b) ⊆ (H ′ ◦ e′)(b) = H ′(e′)(b) = H ′(ē ◦ e)(b) = H ′(¯̄e ◦ e)(b). Next suppose that ēe = ¯̄ee. We want to show that ē = e¯̄e. Put e′ = ēe = ¯̄ee. Then e′fs = ēefs = ¯̄eefs e′fs = ēefs = ēegs = e′gs. By using universal mapping property, there is unique u : (H,C) → (H ′, C ′) such that ue = e′. Hence we obtain ē = u = ¯̄e from ēe = e′ and ¯̄ee = e′. 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