# Categorical properties of soft sets.

1. Introduction

It is well known that many traditional mathematical tools such as fuzzy set theory, probability theory, rough set theory, and interval mathematic theory have their own limitations in dealing with some uncertain problems caused by the incompatibility of various parameter tools. To overcome the difficulties mentioned above, Molodtsov [1] initiated soft set theory by introducing enough compatible parameters. In the context of soft set, researchers can choose freely the form of parameters to simplify the decision-making process, which often makes the process more efficient under the absence of partial information. Consequently, Ali et al. [2] further introduced some new operations in soft set theory. Recently, soft set theory has opened up keen insights and has a rich potential for application in many different fields such as ontology [3], data analysis [4, 5], forecasting [6], simulation [7], decision making [8-11], medical science [12], rule mining [13], algebraic systems [14-20], optimization [21], and textures classification [22]. However, being originated from relatively simple information models, classical soft set theory may not be suitable for those complex information models. In order to solve practical problems better by employing soft set theory, it is important to allure capable pure mathematicians to participate in the study of soft set theory. On the other hand, category theory is not only a basic tool for characterizing all kinds of mathematical structures, but also a tie which can connect easily the fields of mathematics and theoretical computer science (see [23-28]). Many researchers (see [29]) even argue that it is category theory, rather than set theory, that provides the proper setting for the study of pure mathematics. Based on the above analysis, a natural question is whether we can research by combining soft set theory with category theory. The fact is that there exists some categorical concepts, such as product, in soft set theory. Moreover, category theory has been successfully applied to fuzzy set theory [30, 31] and rough set theory [32, 33]. In 2007, Aktas and Cagman [15] showed that both a fuzzy set and a rough set can be regarded as a soft set, which makes it possible to investigate soft set theory and category theory in a common setting. Inspired by this, recently, Zahiri [34] introduced a category whose objects are soft sets. Sardar and Gupta [35] defined another soft category which is a parameterized family of subcategories of a category. Varol et al. [36] defined a new category of soft sets and soft mapping. These studies have presented a preliminary, but potentially interesting, research direction. However, some basic problems still need further investigation. Based on these analyses, we further study the categorical framework of soft set theory in the present paper.

The main contributions of the paper have 3-fold. First, we show that the category SFun of soft sets and soft functions is Cartesian closed. On the one hand, because of the consistency of expression function between Cartesian closed category and [lambda]-calculation with types, many researchers have been devoted to establishing all kinds of Cartesian closed categories in the universe theory for denotational semantics of computer programming language. On the other hand, soft set theory has been widely applied to many fields. Based on this, we further study the category SFun of soft sets and soft functions and prove that it is Cartesian closed. Second, we give a new characterization on soft set relations by employing category theory. There is no doubt that soft set relations play a significant role in the study of soft set theory and they can not only characterize the theoretical relations of two soft sets but also enrich the soft set theory. Presently, researches on soft set relations have received widespread attention and have made great progress (see [37-42]). Meanwhile, it is worth noting that category of binary relations has been widely applied to mathematics and computer science [43,44]. Inspired by this, we make a further discussion on the category SRel of soft sets and Z-soft set relations. Third, we construct a concrete adjoint situation between the category SFun and SRel and characterize its basic relationships.

The remaining parts of the paper are arranged as follows. Section 2 shows some preliminaries. We present in Section 3 the concept of soft functions and discuss the fundamental properties of the category SFun. In Section 4, the characterizations of the category SRel are investigated. Section 5 focuses on studying the intrinsic connections between SFun and SRel.

2. Preliminaries

In this section, we recall some elementary notions and facts related to soft set theory [1], category theory (see [23-27]) which will be often used in this paper. In what follows, we denote by U an initial universe of objects and by E the set of parameters that relate to objects in U. P(U) presents the power set of U. A, B, C, and J are the subsets of E.

Definition 1 (see [1]). A pair (F, A) is called a soft set over U, where F is a function given by F : A [right arrow] P(U).

In other words, a soft set over U is a parameterized family of subsets of U. For any parameter x [member of] A, F(x) may be considered as the set of v-approximate elements of the soft set (F, A).

Proposition 2 (see [24]). If a category C has finite products and equalizers, then C has pullbacks.

Definition 3 (see [26]). Let Z be an object in a category C. One calls Z initial if for each object A there is exactly one morphism from Z to A; one calls Z terminal if for each object A there is exactly one morphism from A to Z; and one calls Z a zero object if it is both initial and terminal.

For objects A,B in a category with zero object Z, we use [0.sub.A,B] for the unique morphism A [right arrow] Z [right arrow] B.

Definition 4 (see [26]). A category C with zero has biproducts if for each family [{[A.sub.i]}.sub.i[member of]I] of objects there is an object [[direct sum].sub.I][A.sub.i], together with families of morphisms [[mu].sub.i]: [A.sub.i] [right arrow] [[direct sum].sub.I][A.sub.j] and [[pi].sub.i] : [[direct sum].sub.I][A.sub.j] [right arrow] [A.sub.i] such that

(i) the morphisms [[mu].sub.i]: [A.sub.i] [right arrow] [[direct sum].sub.I][A.sub.j] are a coproduct of the family [{[A.sub.i]}.sub.i[member of]I];

(ii) the morphisms [[pi].sub.i] : [[direct sum].sub.I][A.sub.j] [right arrow] [A.sub.i] are a product of the family [{[A.sub.i]}.sub.i[member of]I];

(iii) [[pi].sub.i] o [[mu].sub.j] = [[delta].sub.ij] for each i,j [member of] I; here [[delta].sub.ij] is the identity map [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] if i = j and the zero map [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Example 5 (see [23]). In category Rel of sets and the relations between them, for a family of sets [{[X.sub.i]}.sub.i[member of]I], let X be their disjoint union [[??].sub.I] [X.sub.i] = [(x,i) | x [member of] [X.sub.i] for some i [member of] I} and define relations [[mu].sub.i] from [X.sub.i] to X and [[pi].sub.i] from X to [X.sub.i] by setting [[mu].sub.i] = [(x, (x, i)) | x [member of] [X.sub.i] and [[pi].sub.i] = [((x, i), x) | x [member of] [X.sub.i]}. Then the disjoint union X with morphisms [[mu].sub.i] and [[pi].sub.i] is a biproduct of the family [{[X.sub.i]}.sub.i[member of]I].

Definition 6 (see [26]). A semiadditive category is a category C where each homset C(B,C) is equipped with the structure of a commutative monoid with operation + such that, for any f: A [right arrow] B,g,h:B [right arrow] C, and k : C [right arrow] D,

(g + h) x f = (g x f) + (h x f),

k x (g + h) = (k x g) + (k x h). (1)

Definition 7 (see [26]). An involution on a category C is a contravariant functor from C to itself of period two.

Definition 8 (see [26]). Let Z be a C-object. Then Z is injective if, for every monic f : X [right arrow] Y and each g:X [right arrow] Z, there is an h : Y [right arrow] Z with g = h x f:

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The map e : X [right arrow] Z is called an injective hull of X if e is monic, Z is injective, and for any k : Z [right arrow] V we have k x e being monic which implies that k is monic.

Definition 9 (see [27]). For categories C and D and functors F : C [right arrow] D and G : D [right arrow] C, one says (F, G) is an adjoint situation if F is left adjoint to G and G is right adjoint to F. This implies that, for objects X [member of] C and Y [member of] D, there is a natural isomorphism between the homsets C(X,G(Y)) [approximately equal to] D(F(X),Y).

Definition 10. One calls that a category C has exponential properties if it has finite products and for each of the C-objects A,B, there exists a C-object [B.sup.A] and a C-morphism ev : [B.sup.A] x A [right arrow] B such that, for each C-object D and C-morphism F:DxA [right arrow] B, there exists a unique C-morphism [bar.F] : D [right arrow] [B.sup.A] with ev x ([bar.F] x [id.sub.A]) = F. That is, the diagram is commutative.

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Definition 11 (see [26]). A category C is called Cartesian closed if it has equalizers, finite products, terminal objects, and exponential properties.

For the other standard terminology of category theory, see [24, 26].

3. The Category SFun of Soft Sets and Soft Functions

The properties of the category SFun will be investigated in this section. Particularly, we will prove that SFun is a topological construct and Cartesian closed.

Definition 12. Let (F, A) and (G, B) be two soft sets over U. Then one says that the mapping f: A [right arrow] B is a soft function from (F, A) to (G, B) if it satisfies F(a) [subset or equal to] (G x f)(a) for each a [member of] A.

Example 13. Let U = {[u.sub.1], [u.sub.2], [u.sub.3], [u.sub.4], [u.sub.5], [u.sub.6]} be the set of candidate dresses and E = {[e.sub.1], [e.sub.2], [e.sub.3], [e.sub.4]} the set of parameters, where e; (i = 1,2,3,4) stands for expensive, beautiful, elegant, and classical, respectively. Let A = {[e.sub.1], [e.sub.2]}, B = {[e.sub.1], [e.sub.4]}, F([e.sub.1]) = {[u.sub.3}], F([[e.sub.2]]) = {[u.sub.1], [u.sub.2], [u.sub.6]}, G([e.sub.1] ) = {[u.sub.3], [u.sub.5]}, and G([e.sub.4]) = {[u.sub.1], [u.sub.2], [u.sub.5], [u.sup.6]}. It is easy to check that (F,A) and (G,B) are two soft sets over U. Define a function f : A [right arrow] B by f([e.sub.1]) = e1,f([e.sub.2]) = [e.sub.4]. By routine calculations, we can prove that F([e.sub.1]) [subset or equal to] (G x f)([e.sub.1]) and F([e.sub.2]) [subset or equal to] (G x f)([e.sub.2]). By Definition 12, f is a soft function from (F,A) to (G,B).

Remark 14. The concept of soft functions is different from soft set functions defined in [37].

Let SFun denote the category of all soft sets over U and soft functions. We next discuss the properties of the category SFun.

Lemma 15. SFun has equalizers

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Proof. Suppose that (F,A) and (G,B) are two SFun-objects over U; f and g are two SFun-morphisms from (F,A) to (G,B). Define C = {a [member of] A : f(a) = g(a)}, e : C [right arrow] A, an embedding, and H = F x e. From the assumption, we can easily know that (H, C) is a SFun-object, f x e = g x e, and H(c) = (F x e)(c) for each c [member of] C. Thus e is a SFun-morphism. We next show that ((H, C), e) is the equalizer of f and g. Assume that (H', C') is a SFun-object and e is a SFun-morphism from (H', C') to (F, A) satisfying foe' = g x e . Define a mapping [bar.e] : C' [right arrow] C and e = e' .In what follows we focus on showing that e is a SFun-morphism from (H', C') to (H, C) and e' = e x [bar.e]. Firstly, by f x e' = g x e', we can infer that f(e'(c')) = g(e'(c')) for each c' [member of] C', which means that e'(c') [member of] C. Hence [bar.e] = e' is well defined. Secondly, according to H = F x e, [bar.e] = e', and e' being a SFun-morphism, we have

H' (C') [subset or equal to] F([bar.e]' F))-F(e(c' )) = F(e([bar.e](c')))

= (Fx e) ([bar.e](c')) = H([bar.e](c')), (5)

where c' [member of] C'. Therefore, [bar.e] is a SFun-morphism. At last, from the assumption, we know that e' = e x [bar.e] and [bar.e] is unique. In conclusion, ((H, C), e) is the equalizer of f and g.

Lemma 16. SFun has finite products

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Proof. Firstly, let (F,A) and (G,B) be two SFun-objects. Define three mappings

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (7)

whence, for each (a, b) [member of] A x B,

H (a, b) = P (a) [intersection] G (b) [subset or equal to] P (a) = F ([P.sub.1] (a, b)). (8)

It follows that p1 is a SFun-morphism. By the same argument, [p.sub.2] is also a SFun-morphism. Secondly, for each SFun-object (I, D), suppose that f and g are SFun-morphisms from (I, D) to (F,A) and (G,B), respectively. Then 1(d) [subset or equal to] F(f(d)) and 1(d) [subset or equal to] G(g(d)) for every d [member of] D. Further, define a mapping

h:D [right arrow] A x B

d [right arrow] (f (d),g(d)). (9)

Then we can infer that, for each d [member of] D,

1(d) [subset or equal to] (f (d)) [intersection] G (g (d)) = H(f(d),g (d))

= H(h(d)) = (H x h) (d), (10)

which yields that h is a SFun-morphism. At last, for every d [member of] D, one obtains

([p.sub.1] x h) (d) = [p.sub.1] (h (d)) = [p.sub.1] (f (d), g (d)) = f(d). (11)

Therefore, [p.sub.1] x h = f. Analogously, [p.sub.2] x h = g. Apparently, h is unique. In conclusion, {(H, C), [p.sub.1], [p.sub.2]} is a finite product of (F, A) and (G, B).

Theorem 17. SFun has pullbacks.

Proof. By Proposition 2, it is a direct consequence of Lemmas 15 and 16.

Lemma 18. SFun has terminal objects.

Proof. Define a mapping

[T.sub.{[empty set]}] : {[empty set]} [right arrow] P (U)

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[empty set] [right arrow] U.

Trivially, [T.sub.{[empty set]}], {[empty set]}) is a SFun-object. For every SFun-object ([T.sub.M], M), define a mapping

f: M [right arrow] {[empty set]} by m [right arrow] [empty set]. (13)

Then for each m [member of] M, it holds that

[T.sub.M] (m) [subset or equal to] [T.sub.{[empty set]}] (f(m)) = [T.sub.{[empty set]}] ([empty set]) = U, (14)

which implies that f is a SFun-morphism from ([T.sub.M], M) to ([T.sub.{[empty set]}], {[empty set]}). It is easy to know that f is unique. By Definition 3, ([T.sub.{[empty set]}], {[empty set]}) is a terminal object of SFun.

Proposition 19. SFun has initial objects.

Proof. The proof runs parallel to that of Lemma 18.

According to Definition 3, we can easily obtain the following proposition.

Proposition 20. SFun has zero objects.

Lemma 21. SFun has exponential properties.

Proof. Assume that (F,A) and (G,B) are two SFun-objects over U; [B.sub.A] = {f | f : A [right arrow] B is a mapping}. For all f [member of] [B.sup.A], define

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (15)

Then C [not equal to] [empty set]. In fact, choose b [member of] B and define a mapping f' : A [right arrow] B by a [right arrow] b such that t = G(b) [not equal to] [empty set]. whence t [member of] [[alpha]'.sub.f]:, which meas that f' [member of] C. That is, C = [empty set].

From the definitions, we can easily know that (H, C) is a SFun-object. Define the evaluation mapping as follows:

ev : (H, C) x (F, A) (G, B) (16)

(f a) [right arrow] f (a). (17)

Then t [intersection] F(a) [subset or equal to] G(f(a)) for all t [member of] [[alpha].sub.f]. It is immediate that

G (f (a)) [contains or equal to] ([union] {t | t [member of] [[alpha].sub.f]}). [intersection] F (a) = H(f) [intersection] F (a), (18)

which implies that ev is a SFun-morphism. Furthermore, we show that ev has the couniversal property. Assume that (Z, ]) is a SFun-object such that Z(j) [not equal to] [empty set] for every j [member of] J and g : (Z, J) x (F, A) [right arrow] (G, B) is a SFun-morphism. It remains to prove that there exists a unique SFun-morphism [bar.g] : (Z,J) [right arrow] (H, C) such that ev x ([bar.g] x [id.sub.A]) = g. Firstly, for every j [member of] J, define

[bar.g](j) : A [right arrow] B

a [right arrow] [bar.g](j)(a) = g(j,a) (19)

Since g is a SFun-morphism, one has

Z (j) [intersection] P (a) [subset or equal to] G(g (j, a)) for each j [member of] J, a [member of] A. (20)

Consequently,

Z (j) [intersection] P (a) [subset or equal to] G([bar.g] (j) (a)) for each j [member of] J, a [member of] A, (21)

which implies that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Further, according to the assumption, H([bar.g](j)) [contains or equal to] Z(j) [not equal o] [empty set]. Hence [bar.g](j) [member of] C and [bar.g] is a SFun-morphism from (Z, ]) to (H, C). Secondly, for each (j, a) [member of] J x A, it holds that

(ev x ([bar.g] x [id.sub.A])) (j, a) = ev (([bar.g] x [id.sub.A]) [right arrow] a))

= ev ([bar.g](j), a) = [bar.g] (j) (a) = g (j, a), (22)

whence ev x ([bar.g] x [id.sub.A]) = g. Namely, the diagram

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is commutative. Furthermore, suppose that g' : (Z, J) [right arrow] (H,C) is a SFun-morphism satisfying ev z (g' x [id.sub.A]) = g. Then for every j [member of] J and a [member of] A,

g (j, a) = ev x (g' x [id.sub.A]) (j, a) = ev (g (j),a) = g (j) (a). (24)

On the other hand, we have

g (j, a) = evo ([bar.g] x [id.sub.A]) (j, a) = ev ([bar.g] (j), a) = [abr.g] (j) (a).

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Thus [bar.g](j)(a) = g'(j)(a). Since j and a are arbitrary, [bar.g] = g . This completes the proof.

Now we are ready to present two of our main results as follows.

Theorem 22. SFun is Cartesian closed.

Proof. Lemmas 15,16,18, and 21 prove the claim. ?

Theorem 23. SFun is a topological construct.

Proof. Let {([F.sub.i], [[A.sub.i])}.sub.i[member of]I] be a family of SFun-objects indexed by a class I and {[f.sub.i] | A [right arrow] A i}isI a family of mappings. Define a soft set over U as follows:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (26)

Then (F,A) e Ob(SFun). It suffices to show that {f : (F,A) [right arrow] ([F.sub.i], [A.sub.i])}.sub.i[member of]I is the unique SFun initial lift of {[f.sub.i] : A [right arrow] [[A.sub.i]}i[member of]I]. Next, we complete the proof by the following two steps.

Step 1. We show that {[f.sub.i] : (F,A) [right arrow] ([([F.sub.i], [[A.sub.i])}.sub.i[member of]I] is a SFun initial lift of [{[f.sub.i] : A [right arrow] [A.sub.i]}.sub.i[member of]I]. Firstly, we claim that [f.sub.i] : (F,A) [right arrow] ([F.sub.i], [[A.sub.i]) is a family of SFun-morphisms for every i [member of] I. By the assumption, for each a [member of] A and i [member of] I, one yields

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (27)

id

whence [{[f.sub.i]}.sub.i[member of]I] is a family of SFun-morphisms. Furthermore, suppose that (G,B) [meber of] Ob(SFun), g : B [right arrow] A is a mapping such that [g.sub.i] = [f.sub.i] x g for every i [member of] I, and [g.sub.i] : (G,B) [right arrow] (Ft, A;) is a family of SFun-morphisms. Then, we can infer that G(b) [subset or equal to] [F.sub.i]([g.sub.i](b)) for all i [member of] I and b [member of] B. It follows that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (28)

Therefore, g is a SFun-morphism from (G, B) to (F, A). By definition, we can know that [{[f.sub.i] : (F, A) [right arrow] ([F.sub.i], [A.sub.i])}.sub.i[member of]I] is a SFun initial lift of {[f.sub.i] : A [right arrow] [A.sub.i])}.sub.i[member of]I].

Step 2. We show the uniqueness of the initial lift. If [{[f.sub.i] : ([bar.F], A) [right arrow] ([F.sub.i], [A.sub.i])}.sub.i[member of]I] is also a SFun initial lift of [{[f.sub.i] : A [right arrow] [A.sub.i]}.sub.i[member of]I] which is different from [{[f.sub.i] :(F,A) [right arrow] ([F.sub.i], [A.sub.i])} .sub.i[member of]I], then [{[f.sub.i] : ([bar.F], A) [right arrow] ([F.sub.i], [A.sub.i])}.sub.i[member of]I] is a family of SFun-morphisms. It is immediate that [bar.F](a) [subset or equal to] [F.sub.i] ([f.sub.i](a)) for each i [member of] I and a [member of] A. Consequently, [bar.F](a) [subset or equal to] [[intersection].sub.i[member of]I] [F.sub.i] ([f.sub.i](a)) = F(a). That is, [bar.F] [subset or equal to] F. On the other hand, for the SFun-object (F, A) and identity mapping [Id.sub.A] : A [right arrow] A, since [{[f.sub.i] : ([bar.F], A) [right arrow] ([F.sub.i], [A.sub.i])}.sub.i[member of]I] is a SFun initial lift of [{[f.sub.i] : A [right arrow] [A.sub.i]}.sub.i[member of]I], we have [f.sub.i] x [id.sub.A] = [f.sub.i], [f.sub.i] is a family of SFun-morphisms for all i [member of] I, and [id.sub.A] : (F, A) [right arrow] ([bar.F], A) is also a SFun-morphism. Therefore, F(a) [subset or equal to] [bar.F]([id.sub.A](a)) = [bar.F](a) for each a [member of] A, which means that F [subset or equal to] [bar.F]. To sum up, F = [bar.F]. Based on Steps 1 and 2, SFun is a topological construct.

4. The Category SRel of Soft Sets and Z-Soft Set Relations

The main aim of this section is to investigate the properties of the category SRel. We will begin with the analysis of the existence of the zero object, biproduct, and additive identity of SRel. Then the injective object, projective object, injective hull, and projective cover of SRel will be studied.

Definition 24 (see [37]). Let (F, A) and (G, B) be two soft sets over U. Then the Cartesian product of (F, A) and (G, B) is defined as (F, A) x (G, B) = (H, A x B), where H : AxB [right arrow] P(U x U) and H(a, b) = F(a) x G(b) for all (a, b) e Ax B; that is, H(a,b) = {([h.sub.i], [h.sub.j]) | [h.sub.i] [member of] F(a) and [h.sub.j] [member of] G(b)}.

Definition 25 (see [37]). Let (F, A) and (G, B) be two soft sets over U. Then a relation from (F, A) to (G, B) is a soft subset of (F, A) x (G, B).

In other words, a relation from (F, A) to (G, B) is of the form ([H.sub.1], S), where S c A x B and [H.sub.1] (a,b) = H(a,b), for every (a, b) [member of] S; here (H, A x B) = (F, A) x (G, B) has been defined in Definition 24. Any subset of (F, A) x (F, A) is called a relation on (F, A).

From Definition 25, we can see that the condition for soft set relation between two soft sets is very weak but just the weak conditions of the soft set relation make those many elements satisfying the above definition in fact unrelated in actual problems. It can be illustrated clearly by the following example.

Example 26. Consider the soft set (F, A) which describes "the cost of the mobile phones" and the soft set (G, B) which describes the "attractiveness of mobile phones." Assume that U = {[m.sub.1], [m.sub.2], [m.sub.3], [m.sub.4], [m.sub.5], [m.sub.6]} is the universe consisting of six mobile phones, and the parameter sets is given by A = {[e.sub.1], [e.sub.2], [e.sub.3]} and B = {[e.sub.1], [e.sub.4], [e.sub.5]}, respectively, where e; (i = 1,2,3,4,5) stands for "very cheap," "costly," "very costly," "beautiful," and "accessible," respectively. Let F([e.sub.1]) = {[m.sub.1], [m.sub.3]}, F([e.sub.2]) = {[m.sub.1], [m.sub.4]}, F([e.sub.3]) = {[m.sub.1], [m.sub.5]}, G([e.sub.1]) = {[m.sub.1], [m.sub.3], [m.sub.5]}, G([e.sub.4]) = {[m.sub.1], [m.sub.2], [m.sub.3], [m.sub.4], [m.sub.5], [m.sub.6]}, and G([e.sub.5]) = {[m.sub.5]}. Let (F,A) x (G,B) = (H, A x B). The relation R from (F,A) to (G,B) is given by ([H.sub.1], S), where S = {([e.sub.1], [e.sub.1]), ([e.sub.1], [e.sub.5])} c AxB, [H.sub.1] (a, b) = H(a, b) for every (a, b) [member of] S. By Definition 25, R = {F([e.sub.1]) x G([e.sub.1]), F([e.sub.1]) x G([e.sub.5])}.

In the above example, F([e.sub.1]) = {[m.sub.2], [m.sub.3]}, G([e.sub.5] ) = {[m.sub.5]}. It is obvious that there is no relation between them. However, F([e.sub.1]) x G([e.sub.5]) [member of] R, whence Definition 25 cannot describe precisely the relation between soft sets. To overcome this limitation, we strengthen the concept of soft set relations by defining a new soft set relation.

Definition 27. Let (F, A) and (G, B) be two soft sets over U. Then a Z-soft set relation R from (F, A) to (G, B) is a subset of A x B defined as

(a, b) [member of] R [??] F (a) [subset or equal to] G (b), where a [member of] A, b [member of] B. (29)

The definition can be illustrated by diagrams of the form

(30)

Example 28. Let (F,A) and (G,B) be the soft sets defined in Example 26. By Definition 27, we have R = {([e.sub.1], [e.sub.1]), ([e.sub.1], [e.sub.4]), ([e.sub.2], [e.sub.4]), ([e.sub.3], [e.sub.1])}.

Definition 29. Let R be a Z-soft set relation from (F, A) to (G, B) and S a Z-soft set relation from (G, B) to (H, C). Then the composition of R and S, denoted by S x R, is defined as follows:

S x R = {(a,c) | [for all]a [member of] A, [for all]c [member of] C, (31)

there is a b in B with aRb and bSc}.

Definition 30. Let (F, A) be a soft set over U. The identity Z-soft set relation [I.sub.A] on (F,A) is defined as [I.sub.A] = {(a, a) | a [member of] A}.

Proposition 31. Let R be a Z-soft set relation from (F, A) to (G, B), S a Z-soft set relation from (G, B) to (F, A), and IA an identity Z-soft set relation on (F, A). Then R x [I.sub.A] = R and [I.sub.A] x S = S.

Remark 32. From the aforementioned definitions and propositions, we can construct a category, denoted by SRel, whose objects are all soft sets and morphisms are all Z-soft set relations.

Proposition 33. The category SFun of soft sets and soft functions is the subcategory of SRel.

Proposition 34. The empty set (with the empty function into P(U))is a zero object in SRel.

Proof. For each SRel-object (F,A), there exist unique morphisms

(32)

The inequalities are satisfied by default, whence the sets SRel((F, A), ([empty set], [empty set])) and SRel(([empty set], [empty set]), (F, A)) each contain exactly one morphism.

Proposition 35. SRel has biproducts.

Proof. Let [{([F.sub.i], [A.sub.i])}.sub.i[member of]I] be a family of SRel-objects and A = [[??].sub.I] [A.sub.i] = {([a.sub.i]) | [member of] [A.sub.i] for some i [member of] 1} the disjoint union of [A.sub.i]. Define a mapping F : A [right arrow] P(U) as F([a.sub.i], i) = [F.sub.i] ([a.sub.i]). Further, define relations [q.sub.i] from A t to A and [p.sub.i] from A to [A.sub.i] as follows:

[q.sub.i] = {([a.sub.i], {[a.sub.i], i)) | [a.sub.i] [member of] [A.sub.i]}, [p.sub.i] = {{{[a.sub.i], i), [a.sub.i]) | [member of] [A.sub.i]}. (33)

We next show that (F,A) with morphisms [p.sub.i] and [q.sub.i] is a biproduct of the family [{([F.sub.i], [A.sub.i])}.sub.i[member of]I] by the following four steps.

Step 1. We first prove that [p.sub.i] and [q.sub.i] are SRel-morphisms. In fact, take an element ([a.sub.i], ([a.sub.i], i)) in [q.sub.i]; since [F.sub.i] ([a.sub.i]) = F([a.sub.i], i) for each i [member of] I and [a.sub.i] [member of] [A.sub.i], we have [F.sub.i]([a.sub.i]) [subset or equal to] F([a.sub.i], i), which means that [q.sub.i] is a Z-soft set relation from (Ft, A;) to (F, A). That is, [q.sub.i] is a SRel -morphism from ([F.sub.i], [A.sub.i]) to (F, A) for each i [member of] I. Analogously, we can prove that [p.sub.i] is a SRel-morphism from (F, A) to ([F.sub.i], [A.sub.i]) for every i [member of] I.

Step 2. We show that [q.sub.i] are the morphisms for a coproduct. Suppose that (G,B) is a SRel-object and [R.sub.i] : ([F.sub.i], [A.sub.i]) [right arrow] (G,B) is a family of SRel-morphisms. Define a relation R from A to B by ([a.sub.i], i)Rb if and only if [a.sub.i] [R.sub.i] b for each i [member of] I. Firstly, we claim that R is a SRel-morphism from (F, A) to (G,B). If [a.sub.i] [R.sub.i] b for each [a.sub.i] [member of] [A.sub.i] and b [member of] B, since [R.sub.i] : ([F.sub.i], [A.sub.i]) [right arrow] (G,B) is a family of SRel-morphisms for each i [member of] I, we have [F.sub.i] ([a.sub.i]) [subset or equal to] G(b) for every [a.sub.i] e [A.sub.i] and b [member of] B. On the other hand, [F.sub.i] ([a.sub.i]) = F([a.sub.i], i), which implies that F([a.sub.i], i) [subset or equal to] G(b). By Definition 27, we have ([a.sub.i], i)Rb, whence R is a SRel -morphism from (F, A) to (G, B). Secondly, we prove that R x [q.sub.i] = [R.sub.i] for all i [member of] I. Let [a.sub.i] [member of] [A.sub.i] and b [member of] B; then by Definition 29, [a.sub.i] (R x [q.sub.i])b is equivalent to [a.sub.i] [q.sub.i] ([a.sub.i], i) and ([a.sub.i], i)Rb for some ([a.sub.i], i) [member of] A. According to assumption, ([a.sub.i], i)Rb if and only if [a.sub.i] [R.sub.i]b, whence R x [q.sub.i] = [R.sub.i]. At last, the uniqueness is obvious. In conclusion, [q.sub.i] are the morphisms for a coproduct.

Step 3. We further show that pt are morphisms for a product. Assume that (G, B) is SRel-object and [S.sub.i] : (G, B) [right arrow] ([F.sub.i], [A.sub.i]) is a family of SRel-morphisms. Define a relation S from B to A by setting bS([a.sub.i], i) if and only if b [S.sub.i] [a.sub.i]. Similar to Step 2, we can infer that S is a unique morphism from (G, B) to (F, A) in SRel with [p.sub.i] x S = [S.sub.i]. Thus [p.sub.i] are morphisms for a product.

Step 4. Finally, a calculation shows that [p.sub.i] x [q.sub.i] is the identical relation on [A.sub.i] if i = j and the empty relation from [A.sub.j] to [A.sub.i] if i [not equal to] j. Therefore, [p.sub.i] x [q.sub.j] = [[delta].sub.ij].

From the above discussion, we know that SRel has biproducts by Definition 4.

Any category with biproducts carries a unique semiadditive structure that can be defined via biproducts [26]. Next we briefly describe some properties of SRel.

Proposition 36. Let R and S be two SRel-morphisms from (F, A) to (G, B). Then the semiadditive structure on homesets in SRel is given by taking R + S to be the union of Z-soft set relations RuS. In this case, the empty Z-soft: set relation serves as the additive identity.

Proof. Let R and S be two SRel-morphisms from (F, A) to (G, B). Firstly, we show that Ru S is a SRel-morphism from (F, A) to (G, B). In fact, let a [member of] A, b [member of] B, and a(R [union] S)b; then aRb or aSb. In the first case, R is a SRel-morphism given by F(a) [subset or equal to] G(b). And in the second case, S is a SRel-morphism given by F(a) [subset or equal to] G(b), whence Ru S is a morphism in SRel. Secondly, we can easily verify that u gives a commutative monoid structure on SRel((F, A), (G,B)) with the empty Z-soft set relation as identity, and composition distributes over union.

Proposition 37. For each SRel-object (F,A) and SRel-morphism R : (F, A) [right arrow] (G, B), there is an involution ' on SRel defined as follows:

(i) (F,A)' = ([F.sup.r], A), where [F.sup.r] (a) = U-F(a) for all a [member of] A;

(ii) R' : (G, B)' [right arrow] (F, A)' is the converse M-soft set

relation [R.sup.-1], where [R.sup.-1] = {(b,a) | aRb, [for all]a [member of] A, [for all]b [member of] B}.

Proof. Let a e A, b [member of] B, and bR1 a. Since R' is the converse Z-soft set relation of R, we have aRb. In addition, R is a SRel-morphism, so F(a) [subset or equal to] G(b) for every a [member of] A and b [member of] B, which means that U - G(b) [subset or equal to] U - F(a). It is immediate that R! is a SRel -morphism from (G, B)' to (F, A)'. Furthermore, assume that S is a Z-soft set relation from (G, B) to (H, C); then by Definition 29, we can easily obtain that (R x S)-1 = [S.sup.-1] x [R.sup.-1], whence ' is compatible with composition. At last, obviously, ' takes the identity map on (F, A) to the identity map on (F,A) . Hence, ' is a contravariant functor that is obviously period two. By Definition 7, ' is an involution on SRel.

Remark 38. It should be noted that the notion of involution ' gives a bijective mapping from homset SRel((F, A), (G, B)) to SRel(([G.sup.r], F), ([F.sup.r], A)).

Proposition 39. Let R be a SRel-morphism from (F, A) to (G,B). Then the following statements are equivalent:

(i) R is monic;

(ii) if C [subset or equal to] A, then the map R[x] : P(A) [right arrow] P(B), defined by F[C] = {b [member of] B | cRb for some c [member of] C}, is one-one;

(iii) for every a [member of] A, there exists b [member of] B such that a is the only element related to b.

Proof. (i)[??](ii) Suppose that C, D [subset or equal to] A and R[C] = R[D]. Take a singleton {*} and assume that the map 0 : {*} [right arrow] P(U) sends * to [empty set]. Define two relations S, T from {*} to A by setting S = {(*, c) | c [member of] C} and T = {(*,d) | d [member of] D}. As [empty set] is the subset of any sets, we have S, T : ([empty set], {*}) [right arrow] (F, A) being SRel -morphisms and *Sa, *Ta for each a [member of] A. Since R[C] = R[D], by the definition, for each b [member of] B, there exist c [member of] C and d [member of] D such that cRb if and only if dRb. It is immediate that there exist ceCc A, deDcA such that *Sc and cRb if and only if *Td and dRb. By Definition 29, R x S = R x T. Since R is monic, we have S = T. Consequently, C = D, which means that R[x] is one-one.

(ii) [??](iii) Assume that R[x] : P(A) [right arrow] P(B) is one-one; then F[A- {a}] [not equal] R[A]. It follows from the definition that for each a [member of] A there exists b [member of] B such that a is the only element related to b.

(iii) [??](i) Let S, T : (H, C) [right arrow] (F, A) and S [not equal to] T; then we claim that there exist c [member of] C and a [member of] A such that (c, a) [member of] S, but (c, a) [not member of] T. By (iii), take b [member of] B with aRb [??] a' = a, and then it follows from Definition 29 that c(R x S)b, but c(R x T)b does not hold, which means that R x S = R x T. Hence R is monic.

Proposition 40. Let R be a SRel-morphism from (F, A) to (G,B). Then the following statements are equivalent:

(i) R is epic;

(ii) the mapping [R.sup.-1] : P(B) [right arrow] P(A), defined by [R.sup.-1] [D] = {a [meber of] A | aRd for some d [member of] D}, is one-one;

(iii) for every b [member of] B, there exists a [member of] A with b being the only element related to a.

Proof. The proof is similar to that of Proposition 39.

Lemma 41. Let U : C [right arrow] P(U) be a map defined by U(c) = U for each c [member of] C. Then (U, C) is injective in SRel.

Proof. For each SRel-objects (F, A), (G, B), let R : (F, A) [right arrow] (G,B) be monic and S : (F,A) [right arrow] (U,C). Assume that [B.sub.1] = {b [member of] B | there exists exactly one a such that aRb}. Since R is monic, it follows from Proposition 39 that for every a [member of] A there exists b [member of] B such that a is the only element related to b. Define T : (G,B) [right arrow] (U,C) by T = {(b,c) | b [member of] [B.sub.1] and aSc for some aRb}. Apparently, T is a SRel-morphism. According to the definition of T, we can infer that T xR = S. That is, the following diagram

(34)

is commutative. By Definition 8, (U, C) is injective.

Theorem 42. Let (F,A) be a SRel-object. Then the identity embedding I: (F, A) [right arrow] (U, A) is an injective hull.

Proof. By Proposition 39, I is monic. In addition, according to Lemma 41, (U, A) is injective. Assume that R : (U, A) [right arrow] (G, B) such that R x I is monic. Because of a relation rather than a morphism, R x I = R, whence Proposition 39 gives that R is monic. It follows from Definition 8 that I is an injective hull.

Remark 43. It is well known that projective objects of a category are dual to that of injective objects and projective covers are dual to that of injective hulls (see 26]). So we can easily obtain the following proposition.

Theorem 44. Let [empty set] : C [right arrow] P(U) be a mapping defined by [empty set] (c) = [empty set] for every c [empty set] C. Then ([empty set], C) is the projective object of SRel. Further, for each SRel-object (F, A), the mapping I : ([empty set], A) [right arrow] (F, A) is a projective cover of (F, A).

We in this section mostly consider the relations among the categories SFun, Set, SRel, and Rel. In particular, we investigate the essential connections of SFun and SRel by means of adjoint situations.

Definition 45. Let [F.sub.1] : SFun [right arrow] Set be a forgetful functor which sends an object (F, A) to A and sends a morphism R : (F, A) [right arrow] (G,B) to R: A [right arrow] B.

Analogously, we can define another forgetful functor [G.sub.1] : SRel [right arrow] Rel.

Definition 46. Define [F.sub.2], [G.sub.2] : Set [right arrow] SFun and [F.sub.3], [G.sub.3] : Rel [right arrow] SRel for an object A and morphism R : A [right arrow] B by setting

(i) [F.sub.2] (A) and F3(A) to be the object ([empty set], A), where [empty set] : A [right arrow] P(U) defined by [empty set] (a) = [empty set] for all a [member of] A;

(ii) [G.sub.2] (A) and [G.sub.3] (A) to be the object (U, A), where U : A [right arrow] P(U) defined by U(a) = U for all a [member of] A;

(iii) [F.sub.2] (R), [F.sub.3] (R), [G.sub.2] (R), and [G.sub.3] (R) to be R, where R is considered with the appropriate domain and codomain.

Theorem 47. (i) The pair ([F.sub.3], [G.sub.1]) is an adjoint situation.

(ii) The pair ([G.sub.1], [G.sub.3]) is an adjoint situation.

(iii) The pair ([F.sub.2], [F.sub.1]) is an adjoint situation.

(iv) The pair ([F.sub.1], [G.sub.2]) is an adjoint situation.

Proof. We just prove (i) and (ii) because the proofs of (iii) and (iv) are similar.

(i) Let A,B be Rel-objects and (G,B) SRel-object. By Definition 46(i), [empty set] (a) = [empty set] [subset or equal to] G(b) for all a [member of] A and b [member of] B; we can infer that a morphism R : A [right arrow] B in Rel will lift to a morphism R : ([empty set], A) [right arrow] (G, B) in SRel, whence SRel(([empty set], A), (G, B)) [approximately equal to] Rel(A, B). According to Definitions 45 and 46(i), SRel([F.sub.3](A),(G,B)) [approximately equal to] Rel(A, [G.sub.1] (G, B)). It follows from Definition 9 that ([F.sub.3], [G.sub.1]) is an adjoint situation.

(ii) Assume that A, B are Rel-objects and (G, B) is a SRel object. By Definition 46 (ii), F(a) [subset or equal to] U = U(b) for every a [member of] A and b [member of] B, which means that a Rel-morphism R : A [right arrow] B can lift to a SRel-morphism from (F, A) to (U, B). Thus Rel((A, B)) [approximately equal to] SRel((F, A),(U, B)). By Definitions 45 and 46 (ii), Rel([G.sub.1] (F, A), B) [approximately equal to] SRel((F, A), [G.sub.3] (B)). Therefore, ([G.sub.1], [G.sub.3]) is an adjoint situation according to Definition 9.

Theorem 48. Assume that [F.sub.4] : SFun [right arrow] SRel is the inclusion functor and define a functor

[G.sub.4] : SRel [right arrow] SFun

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (35)

where [??]F(C) = [intersection]{F(c) | c [member of] C}. Then ([F.sub.4], [G.sub.4]) is an adjoint situation.

Proof. Firstly, we show that [G.sub.4] is well defined. Let R : (F, A) [right arrow] (G, B) be a SRel-morphism. For C [subset or equal to] A, b [member of] R[C], where R[C] is defined like Proposition 39, there exists c [member of] C such that cRb. Since F(c) [subset or equal to] G(b) for every c [member of] A,b [member of] B, we have [??]F(C) [subset or equal to] [??]G(F[C]), which implies that.R[x] : ([??]F,P(A)) [right arrow] ([??]G, P(B)) is a SFun-morphism.

That is, the definition of [G.sub.4] is reasonable. Secondly, we claim that ([F.sub.4], [G.sub.4]) is an adjoint situation. It remains to show that SRel((F, A), (G, B)) [approximately equal to] SFun((F, A), ([??]G, P(B))). In fact, if R : (F,A) [right arrow] (G,B) is a SRel-morphism, then F(a) [subset or equal to] G(b) for all a [member of] A and b [member of] B, which implies that F(a) [subset or equal to] [??]G(F[|a}]). That is, R[{x}] : (F,A) [right arrow] ([??]G, P(B)) is a SFun-morphism. Consequently, assume that f : (F,A) [right arrow] ([??]G, P(B)) is a SFun-morphism. Define a relation R : A [right arrow] B by aRb if b [member of] f(a). Then aRb implies F(a) [subset or equal to] [??]G(f(a)), whence F(a) [subset or equal to] G(b) and R : (F, A) [right arrow] (G, B) is a SRel-morphism.

Definition 49. Assume that P, I : SRel [right arrow] SRel are defined by P = [F.sub.3] x [G.sub.3] and I = [G.sub.3] x [G.sub.1]. Then P and I are called the projective cover and injective hull functor, respectively.

Proposition 50. Consider

' x P = I x '. (36)

Proof. It is straightforward from Theorems 42 and 44. ?

6. Conclusions

Soft set theory, a new powerful mathematical tool for dealing with uncertain problems, has recently received wide attention in both the real-life applications and the theory studies. In recent years, the combination of soft set theory and category theory has resulted in many interesting research topics. In this paper, we mostly focus on offering theoretical results by combining soft set theory and category theory. In other words, we have provided a categorical viewpoint for soft set theory and the results given in this paper can further enrich soft set theories. Particularly, we have proved that the category SFun is Cartesian closed, which will provide an important theoretical background for theoretical computer sciences. Naturally, applying our results to other fields such as information sciences and logic is also a valuable work and we will present it in the future work.

http://dx.doi.org/10.1155/2014/783056

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work was supported by the National Natural Science Foundation of China (Grant no. 11071151) and the Special Fund of Shaanxi Provincial Education Department (Grant no. 2013JK0568).

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Min Zhou, (1) Shenggang Li, (1) and Muhammad Akram (2)

(1) College of Mathematics and Information Sciences, Shaanxi Normal University, Xi'an 710119, China

(2) Department of Mathematics, University of the Punjab, New Campus, Lahore, Pakistan

Correspondence should be addressed to Shenggang Li; shengganglinew@126.com

Received 11 June 2014; Accepted 14 July 2014; Published 18 August 2014