����������� ������ 55 (2003), 65–82

UDK 510.8

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SOME REMARKS ON THE CATEGORY SET(L), Part II

Sergey A. Solovyov

Abstract. This paper considers some intrinsic properties of the category SET(L) of Lsubsets of sets with a fixed basis L and is a continuation of our previous work [4]. Here we study properties of some abstract functors when applied to the category SET(L) as well as some special objects related to them. In the last section we consider two standard constructions, namely, inverse and direct systems in this category.

1. Introduction

The notion of a fuzzy set introduced in [5] induced many researchers to study different mathematical structures involving fuzzy sets and their generalization Lfuzzy sets [2] or just L-sets for short. In particular, some authors considered the category SET(L) of all L-subsets of all sets with a fixed lattice L. The purpose of our work is to study systematically some intrinsic properties of this category. The article is a continuation of our previous investigation of the category SET(L) in [4]. In particular, in [4] we considered some special morphisms and objects (as, e.g., section, retraction, isomorphism) as well as some standard constructions (as, e.g., product and coproduct of objects and morphism, pullback and pushout) in this category. The aim of this paper is to develop further the study of the category SET(L).

The paper starts with an introductory section, i.e., Preliminaries, where we recall the definition of the category SET(L) and discuss some results from [4]. The next section is devoted to some applications of abstract functors to the category SET(L). We consider, for example, the properties of the so-called set-valued homfunctor and evaluation functor. Some properties of the former are also considered in the next section devoted to some special objects in the category SET(L). Among other problems, the section deals with the properties of subobjects and quotient objects. Two last sections of the article are devoted to, accordingly, special morphisms and standard constructions in the category SET(L), where we generalize some results of our previous work, i.e., the properties of equalizers and consider some more concepts, i.e., inverse and direct systems.

AMS Subject Classification: 03E72, 04A72, 18B99 Keywords and phrases: Category theory, L-fuzzy set, functor, special morphism, special object, standard construction.

We use standard terminology accepted in Category theory (see, e.g., [3]).

2. Preliminaries

In this section we will discuss some basic properties of the category SET(L). Let us start by recalling its definition (see [2]).

Suppose L is a complete lattice (L, ⩽), i.e., a partially ordered set such that for every subset A ⊂ L the join ⋁ A and the meet ⋀ A are defined. In particular, ⋁ L =: 1L and ⋀ L =: 0L. We assume that 0L ≠ 1L, i.e., L has at least two elements. Then the category SET(L) can be defined as follows.

The objects of SET(L) are all L-subsets of sets, i.e., mappings X: X̃ → L where X̃ is an arbitrary set (maybe empty). Henceforth the objects of SET(L) will be denoted by X, Y or Z and arbitrary sets by X̃, Ỹ or Z̃. By saying that an object X ∈ Obj SET(L) is given we will always mean that X is a mapping X: X̃ → L.

Given two objects X, Y ∈ Obj SET(L), the set of morphisms from X to Y Mor_{SET(L)}(X, Y) consists of all mappings f: X̃ → Ỹ such that X(x) ⩽ Y ∘ f(x) for all x ∈ X̃. Given an object X ∈ Obj SET(L), we denote its identity morphism by e_{X}.

Now we will list some properties of the category SET(L) which we will need throughout the article and whose proofs can be found in [4]. All of them are related to special morphisms and objects in the category SET(L); also notice that we use “iff” for “if and only if”.

A morphism f: X → Y is

(1) a monomorphism iff f is injective;

(2) an epimorphism iff f is surjective;

(3) a bimorphism iff f is bijective;

(4) an isomorphism iff f is bijective and X(x) = Y ∘ f(x) for all x ∈ X̃;

(5) a constant morphism iff f(X̃) = {y_{0}} ⊂ Ỹ;

(6) an equalizer iff f is injective and X(x) = Y ∘ f(x) for all x ∈ X̃. An object X is an initial object iff X̃ = ∅.

3. Special functors

In this section we will consider the properties of some abstract functors in case of one distinct category, i.e., the category SET(L).

Suppose we have a bifunctor hom_{SET(L)} : SET(L)^{op} × SET(L) → SET (notice that SET(L)^{op} denotes the dual category of the category SET(L), “×” denotes the product of two categories and SET stands for the category of sets) where hom_{SET(L)}(X, Y) = Mor_{SET(L)}(X, Y) and hom_{SET(L)}(f, g)(h) = g ∘ h ∘ f.

hom_{SET(L)}(X, Y) hom_{SET(L)}(f, g)

hom_{SET(L)}(X′, Y′)

X  w h  Yg

X′ u f w g ∘ h ∘ f = hom_{SET(L)}(f, g)(h) Y′

The functor is called the set-valued hom-functor or morphism functor. For shortness sake, later on, we will denote the category SET(L)^{op} × SET(L) by C.

Theorem 3.1. The morphism functor hom_{SET(L)} is dense, not full, not faithful and not an embedding.

Proof. Let us prove that hom_{SET(L)} is dense, that is, for each object Z̃ ∈ Obj SET there exist such two objects X, Y ∈ Obj SET(L) that hom_{SET(L)}(X, Y) is isomorphic to Z̃. Let X̃ = {x_{0}}, X(x_{0}) = 0L and Ỹ = Z̃, Y (y) = 1L for all y ∈ Ỹ. Then obviously hom_{SET(L)}(X, Y) = Mor_{SET(L)}(X, Y) is isomorphic to Z̃.

Now we will verify that hom_{SET(L)} is not full. For the proof of this property it will be enough to find such two objects (X, Y), (X′, Y′) ∈ Obj C that Mor_{C}((X, Y), (X′,Y′)) = ∅ and Mor_{C}(hom_{SET(L)}(X,Y), hom_{SET(L)}(X′,Y′)) ≠ ∅. It is easy to see that such two objects can be each four mappings X′ ≡ X ≡ Y′ ≡ 0L and Y ≡ 1L with arbitrary non-empty sets X̃, Ỹ , X̃′, Ỹ′. Indeed, there are no morphisms from (X, Y) to (X′, Y′) in the category C, but there exists at least one morphism from hom_{SET(L)}(X, Y) to hom_{SET(L)}(X′, Y′) in the category SET.

Now let us prove that hom_{SET(L)} is not faithful. We have to find such two objects (X, Y), (X′, Y′) ∈ Obj C and such two morphisms (f, g), (f′, g′): (X, Y) → (X′, Y′) that (f, g) ≠ (f′, g′) but hom_{SET(L)}(f, g) = hom_{SET(L)}(f′, g′). Let

X̃′ = X̃ = ∅ and |Ỹ | = |Ỹ′| > 1, Y′ ≡ 1L. Then hom_{SET(L)}(X, Y) = {x_{0}}, hom_{SET(L)}(X′, Y′) = {y_{0}} and there exists a unique morphism f_{0} ∈ Mor_{SET}({x_{0}}, {y_{0}}). Obviously, one can find at least two distinct morphisms (f, g_{1}), (f, g_{2}) ∈ Mor_{C}((X, Y), (X′, Y′)) such that both hom_{SET(L)}(f, g_{1}) and hom_{SET(L)}(f, g_{2}) are equal to f_{0}.

The last property follows immediately from the fact that hom_{SET(L)} is not faithful.

Now let us try to make an L-set of hom_{SET(L)}(X, Y) and thus to get a bifunctor Hom_{SET(L)} : C → SET(L).

The first way is to assume that Hom_{SET(L)}(X, Y) = X^{∗} where X^{∗} is a mapping from Mor_{SET(L)(X,Y)} to L defined in the following way:

X^{∗}(h) = \{ \begin{matrix}⋁ Y (Ỹ), & Ỹ ≠ ∅ \\ 0L, & Ỹ = ∅. \end{matrix}

Suppose we have an object (X′, Y′) ∈ Obj C and a morphism (f, g): (X, Y) → (X′, Y′). Then Hom_{SET(L)}(X′, Y′) = Y^{∗} and Hom_{SET(L)}(f, g)(h) = g∘h∘f = \overline{h}. We need to prove that Hom_{SET(L)}(f, g) ∈ Mor_{SET(L)(X^{∗},Y^{∗})}, that is, Y^{∗}(\overline{h}) ⩾ X^{∗}(h). Indeed, from the existence of the morphism g ∈ Mor_{SET(L)}(Y, Y′) we have that

⋁ Y′(Ỹ′) ⩾ Y′ ∘ g(y_{0}) ⩾ Y (y_{0}) for all y_{0} ∈ Ỹ. Then ⋁ Y′(Ỹ′) ⩾ ⋁ Y (Ỹ) and therefore Y^{∗}(\overline{h}) ⩾ X^{∗}(h).

The second way assumes that the lattice L is endowed with an operation^{c}: L → L (called involution) with the following properties:

(1) for all a, b ∈ L if a ⩽ b then b^{c} ⩽ a^{c};

(2) (a^{c})^{c} = a for all a ∈ L.

From these properties it can easily be derived that (1L)^{c} = 0L and (0L)^{c} = 1L. Then Hom_{SET(L)}(X, Y) can be defined as X^{∗}: Mor_{SET(L)}(X, Y) → L where X^{∗}(h) = (⋁ X(X̃))^{c} ∨ (⋁ Y (Ỹ)) with 0L for ⋁ ∅. Suppose again we have an object (X′, Y′) ∈ Obj C and a morphism (f, g): (X, Y) → (X′, Y′). Then Hom_{SET(L)}(X′, Y′) = Y^{∗}, Y^{∗}(h̃) = (⋁ X′(X̃′))^{c} ∨ (⋁ Y′(Ỹ′)) and then Hom_{SET(L)}(f, g)(h) = g ∘ h ∘ f = \overline{h}. We have to prove that Hom_{SET(L)}(f, g) ∈ Mor_{SET(L)(X^{∗},Y^{∗})}, that is, Y^{∗}(\overline{h}) ⩾ X^{∗}(h).

(1) If X̃′ = ∅ then Y^{∗}(\overline{h}) = 1L ⩾ X^{∗}(h).

(2) If X̃′ ≠ ∅ then X̃ ≠ ∅ and from the existence of the morphism f: X′ → X we get that ⋁ X(X̃) ⩾ ⋁ X′(X̃′) and therefore (⋁ X′(X̃′))^{c} ⩾ (⋁ X(X̃))^{c}.

(a) If Ỹ′ = ∅ then Ỹ = ∅ and then Y^{∗}(\overline{h}) = (⋁ X′(X̃′))^{c} ⩾ (⋁ X(X̃))^{c} = X^{∗}(h).

(b) If Ỹ′ ≠ ∅ then from the existence of the morphism g: Y → Y′ we get that ⋁ Y′(Ỹ′) ⩾ ⋁ Y (Ỹ) and therefore Y^{∗}(\overline{h}) ⩾ X^{∗}(h).

It is easy to see that for both ways of defining the functor Hom_{SET(L)} the following proposition holds. (Notice that for a given object X ∈ Obj SET(L) there exists the functor Hom_{SET(L)}( , X): SET(L)^{op} → SET(L) defined by Hom_{SET(L)}( , X)(Y)= Hom_{SET(L)}(Y, X), Hom_{SET(L)}( , X)(f) = Hom_{SET(L)}(f, e_{X}). By analogy one can define the functor hom_{SET(L)}( , X).)

Proposition 3.1. For each X ∈ Obj SET(L) the triangle

SET(L)^{op}  w Hom_{SET(L)}( , X) [ [ [ [ [ [ [ [ [ [ [ [[] hom_{SET(L)}( , X)

{SET(L) \atop U}

SET

where U is a forgetful functor commutes.

We will consider some more properties of the functor hom_{SET(L)} in the next section.

Now we will consider two more special functor types in the category SET(L). Let us begin with definitions (see [1]). Suppose we have a category C′ and two functors F_{1}, F_{2} : C′ → C′. Then F_{1} is said to be a subfunctor of F_{2} provided that there exists a natural transformation η: F_{1} → F_{2} whose elements η_{A}: F_{1}(A) → F_{2}(A) are monomorphisms for all A ∈ Obj C′. Dually, F_{1} is said to be a superfunctor of F_{2} provided that there exists a natural transformation η: F_{1} → F_{2} whose elements are epimorphisms. By F_{∗}: C′ → C′ we will denote a functor which is a subfunctor of each functor F: C′ → C′ and by F^{∗}: C′ → C′ such a functor that each functor F: C′ → C′ is a superfunctor of F^{∗}.

Theorem 3.2. The category SET(L) has a unique functor F_{∗} and has no functor F^{∗}.

Proof. Suppose X ∈ Obj SET(L) is an initial object in the category SET(L). Then the set X̃ is empty. Let F: SET(L) → SET(L) be such that F (Y) = X for all Y ∈ Obj SET(L) and F (f) = e_{X} for all f ∈ Mor SET(L). Obviously, F is a functor. Suppose another functor G: SET(L) → SET(L) is given. Then for every two objects Y, Z ∈ Obj SET(L) and each morphism f: Y → Z the following diagram

\begin{matrix}F (Y)  w η_{Y} F (f) \\ G(Y) G(f) \end{matrix} ; \begin{matrix}F (Z)  w η_{Z} \\ G(Z) \end{matrix}

where η_{Y} and η_{Z} are the unique morphisms from X to, accordingly, G(Y) and G(Z) commutes. Thus, η = {η_{Y} }_{Y ∈Obj SET(L)} is a natural transformation from F

to G. Since X̃ = ∅, all η_{Y} are monomorphisms. Therefore, F is a subfunctor of G and F = F_{∗}.

Suppose there exists another functor F∗^{′}. From the fact that F∗^{′} is a subfunctor of F_{∗} we derive that there exists a natural transformation η: F∗^{′} → F_{∗} and then { \atop F∗^{′}(Y)} = ∅ for all Y ∈ Obj SET(L). Thus, F∗^{′} = F_{∗}.

Suppose there exists a functor F^{∗}. Then F_{∗} is a superfunctor of F^{∗} and there exists a natural transformation η: F_{∗} → F^{∗} whose elements are epimorphisms, therefore, { \atop F^{∗}(Y)} = ∅ for all Y ∈ Obj SET(L). Suppose F (Y) = Z_{0} ∈ Obj SET(L), Z̃_{0} ≠ ∅ for all Y ∈ Obj SET(L) and F (f) = e_{Z_{0}} for all f ∈ Mor SET(L). Then F is a superfunctor of F^{∗} and there exists a natural transformation η: F → F^{∗}. Choose any two objects Y_{1}, Y_{2} ∈ Obj SET(L) and a morphism f: Y_{1} → Y_{2}. Then there exist such two morphisms η_{Y_{1}}, η_{Y_{2}} that the following diagram

\begin{matrix}F (Y_{1})  w η_{Y_{1}} F (f) \\ F^{∗}(Y_{1}) F^{∗}(f) \end{matrix} ; F (Y_{2})  w η_{Y_{2}} F^{∗}(Y_{2})

commutes. Since the sets Mor_{SET(L)}(F (Y_{1}), F^{∗}(Y_{1})), Mor_{SET(L)}(F (Y_{2}), F^{∗}(Y_{2})) are empty, we immediately get a contradiction.

The last functor we will consider here is a bifunctor E: SET(L)^{SET(L)} × SET(L) → SET(L) (notice that SET(L)^{SET(L)} denotes the (quasi)category of all functors F: SET(L) → SET(L)) defined in the following way: E(F, X) = F (X) for all F ∈ Obj SET(L)^{SET(L)}, X ∈ Obj SET(L) and E(η, f) = G(f) ∘ η_{X} = η_{Y} ∘ F (f) for all (η, f): (F, X) → (G, Y). The functor is called the evaluation functor for SET(L)^{SET(L)}. For shortness sake, later on, we will denote the category SET(L)^{SET(L)} × SET(L) by C.

Theorem 3.3. The evaluation functor E is dense, not full, not faithful and not an embedding.

Proof. For each object X ∈ Obj SET(L) E(e_{SET(L)}, X) = e_{SET(L)}(X) = X where e_{SET(L)} is the identity functor on SET(L). Thus, E is dense.

Now let us prove that E is not full. Let F (Y) = X ∈ Obj SET(L), G(Y) = Z ∈ Obj SET(L) for all Y ∈ Obj SET(L) and F (f) = e_{X}, G(f) = e_{Z} for all f ∈ Mor SET(L) where X̃ = {x_{0}}, X(x_{0}) = 0L and Z̃ = {z_{0}}, Z(z_{0}) = 1L. Suppose X̃_{1} = {x_{1}}, X_{1}(x_{1}) = 1L and X̃_{2} = {x_{2}}, X_{2}(x_{2}) = 0L. Then E(F, X_{1}) = X, E(G, X_{2}) = Z and the set Mor_{SET(L)}(X, Z) is not empty. The fact that Mor_{C((F,X_{1}),(G,X_{2}))} = ∅ implies that E is not full.

Now we will verify that E is not faithful. Let X ∈ Obj SET(L) be such that X̃ = ∅. Then let F (Y) = G(Y) = X for all Y ∈ Obj SET(L) and F (f) = G(f) = e_{X} for all f ∈ Mor SET(L). Suppose X_{1}, X_{2} ∈ Obj SET(L) are such two objects that |X̃_{1}| = |X̃_{2}| > 1 and X_{2}(x) = 1L for all x ∈ X̃_{2}. Then Mor_{SET(L)}(F (X_{1}), G(X_{2})) = {f} and there exist at least two distinct morphisms g_{1}, g_{2} : X_{1} → X_{2} such that E(η, g_{1}) = E(η, g_{2}) = f where η: F → G is the identity natural transformation.

The last property follows immediately from the fact that E is not faithful.

4. Special objects

This section is devoted to some special objects in the category SET(L). To begin with, we will consider some objects related to the functor hom_{SET(L)} discussed in the previous section. In fact, all these objects can be defined through this functor, but here we will use another “internal” characterizations of them. Let us begin by considering separator and coseparator in the category SET(L).

Theorem 4.1. An object X ∈ Obj SET(L) is a separator iff X̃ ≠ ∅ and X(x) = 0L for all x ∈ X̃.

Proof. Let us prove the necessity first and therefore assume that X is a separator. Then for every two objects Y, Z ∈ Obj SET(L), whenever f, g: Y → Z are two distinct morphisms, there exists such a morphism h: X → Y that f ∘ h ≠ g ∘ h.

Obviously, X̃ ≠ ∅. Suppose there exists a point x_{0} ∈ X̃ such that X(x_{0}) ≠ 0L. Then let Ỹ = {y_{1}}, Y (y_{1}) = 0L and Z̃ = {z_{1}, z_{2}}, z_{1} ≠ z_{2}, Z(z_{1}) = Z(z_{2}) = 1L.

In this case every two mappings f, g: Ỹ → Z̃ will be morphisms and thus, we can take f(y_{1}) = z_{1}, g(y_{1}) = z_{2}. Obviously, f ≠ g. Since Mor_{SET(L)}(X, Y) = ∅, we conclude that X is not a separator that contradicts our former assumption.

The sufficiency is obvious.

Given a separator X ∈ Obj SET(L), we can get a faithful functor from SET(L) to SET, i.e., hom_{SET(L)}(X, ).

Theorem 4.2. An object X ∈ Obj SET(L) is a coseparator iff there exist such two points x_{1}, x_{2} ∈ X̃, x_{1} ≠ x_{2} that X(x_{1}) = X(x_{2}) = 1L.

Proof. Let us prove the necessity first and therefore assume that X is a coseparator. Then for every two objects Y, Z ∈ Obj SET(L), whenever f, g: Y → Z are two distinct morphisms, there exists such a morphism h: Z → X that h ∘ f ≠ h ∘ g.

Suppose Ỹ = {y_{1}}, Y (y_{1}) = 1L and Z̃ = {z_{1}, z_{2}}, z_{1} ≠ z_{2}, Z(z_{1}) = Z(z_{2}) = 1L. In this case every two mappings f, g: Ỹ → Z̃ will be morphisms and thus we can take f(y_{1}) = z_{1}, g(y_{1}) = z_{2}. Obviously, f ≠ g. Then there exists such a morphism h: Z → X that h ∘ f ≠ h ∘ g, therefore, h ∘ f(y_{1}) = x_{1} ≠ x_{2} = h ∘ g(y_{1}) and X(x_{1}) = X(x_{2}) = 1L.

The sufficiency is obvious.

Now we will consider projective and injective objects in the category SET(L).

Theorem 4.3. An object X ∈ Obj SET(L) is a projective object iff X(x) = 0L for all x ∈ X̃.

Proof. Let us prove the necessity first and therefore assume that X is a projective object. Then for every two objects Y, Z ∈ Obj SET(L), each epimorphism f: Y → Z and each morphism g: X → Z there exists a morphism h: X → Y such that the triangle

X h h h h k h  g ; {Y  w f \atop Z}

commutes.

Suppose there exists such a point x_{0} ∈ X̃ that X(x_{0}) ≠ 0L. Then let Z = X, g = e_{X} and Ỹ = X̃, Y (x) = 0L for all x ∈ X̃, f = idX̃. Obviously, f ∈ Mor_{SET(L)}(Y, Z) and is surjective, therefore, f is an epimorphism in SET(L). Since Mor_{SET(L)}(X, Y) = ∅, we conclude that X is not a projective object that contradicts our former assumption.

Now let us prove the sufficiency. Since f is an epimorphism, f is surjective and then we can define h as follows: h(x) = y ∈ f^{−1} ∘ g(x) for all x ∈ X̃. Obviously, h ∈ Mor_{SET(L)}(X, Y) and g = f ∘ h.

The theorem implies the following result.

Theorem 4.4. If X ∈ Obj SET(L) is a separator then X is a projective object.

Given a separator X ∈ Obj SET(L), we can get a functor from SET(L) to SET which is faithful and preserves epimorphisms, i.e., hom_{SET(L)}(X, ).

Theorem 4.5. An object X ∈ Obj SET(L) is an injective object iff X̃ ≠ ∅ and X(x) = 1L for all x ∈ X̃.

Proof. Let us prove the necessity first and therefore assume that X is an injective object. Then for every two objects Y, Z ∈ Obj SET(L), each monomorphism f: Y → Z and each morphism g: Y → X there exists a morphism h: Z → X such that the triangle

\begin{matrix}Y w f g \\ Z h h h h k h \end{matrix} ; X

commutes.

Suppose X̃ = ∅. Then let Ỹ = ∅ and Z̃ ≠ ∅. Obviously, there exists a monomorphism f: Y → Z and a morphism g: Y → X. Since the set Mor_{SET(L)}(Z, X) is empty, we conclude that X is not an injective object that contradicts our former assumption.

Now suppose there exists such a point x_{0} ∈ X̃ that X(x_{0}) ≠ 1L. Then let Ỹ = {y_{0}}, Y (y_{0}) = 0L and g(y_{0}) = x_{0}. Obviously, g ∈ Mor_{SET(L)}(Y, X). Let Z̃ = {z_{0}}, Z(z_{0}) = 1L and f(y_{0}) = z_{0}. Obviously, f ∈ Mor_{SET(L)}(Y, Z). Since f is injective, f is a monomorphism in the category SET(L). If there exists such a morphism h: Z → X that h ∘ f = g then x_{0} = g(y_{0}) = h ∘ f(y_{0}) = h(z_{0}), therefore, X ∘ h(z_{0}) ⩾̸ Z(z_{0}) and h is not a morphism. Thus, X is not an injective object.

Now let us prove the sufficiency. Since f is a monomorphism, f is injective and then we can define h as follows:

h(z) = \{ \begin{matrix}g ∘ f^{−1}(z), & z ∈ f(Ỹ) \\ x ∈ X̃, & z ∉ f(Ỹ). \end{matrix}

Obviously, h ∈ Mor_{SET(L)}(Z, X) and g = h ∘ f.

Lastly, we will consider some properties of subobjects and quotient objects in the category SET(L). Let us start with the former.

Suppose we have an object Y ∈ Obj SET(L) and two arbitrary subobjects of Y (X, f), (Z, g) where f: X → Y and g: Z → Y are some monomorphisms. Then (X, f) is said to be smaller than (Z, g) – denoted by (X, f) ⩽ (Z, g) – iff there exists some morphism h: X → Z such that the triangle

{X w f h \atop Y} ; Z^{h h hh j} g

commutes.

If (X, f) ⩽ (Z, g) and (Z, g) ⩽ (X, f) then (X, f) and (Z, g) are said to be isomorphic subobjects of Y; denoted by (X, f) ≈ (Z, g).

Theorem 4.6. (X, f) ⩽ (Z, g) iff the following conditions are fulfilled:

(1) f(X̃) ⊂ g(Z̃);

(2) X(x) ⩽ Z(z) for x ∈ X̃, z ∈ Z̃, f(x) = g(z).

Proof. Let us prove the necessity first and therefore assume that (X, f) ⩽ (Z, g).

Suppose there exists such y_{0} ∈ f(X̃) that y_{0} ∉ g(Z̃). Then there exists some x_{0} ∈ X̃ such that f(x_{0}) = y_{0} and for all mappings h: X̃ → Z̃ g∘h(x_{0}) ≠ y_{0} = f(x_{0}). Thus, (X, f) ⩽̸ (Z, g).

Suppose there exist such x_{0} ∈ X̃ and z_{0} ∈ Z̃ that f(x_{0}) = g(z_{0}) but X(x_{0}) ⩽̸ Z(z_{0}). Since (X, f) ⩽ (Z, g), there exists some morphism h: X → Z such that g ∘ h = f. Thus, g(z_{0}) = f(x_{0}) = g ∘ h(x_{0}) and then z_{0} = h(x_{0}), since g is a monomorphism and therefore injective. Since h is a morphism, Z ∘ h(x_{0}) = Z(z_{0}) ⩾ X(x_{0}) that contradicts our former assumption.

Now let us prove the sufficiency and therefore assume that all conditions of the theorem are fulfilled. Let h(x) = g^{−1} ∘ f(x) for all x ∈ X̃. Then g ∘ h(x) = g ∘ g^{−1} ∘ f(x) = f(x), since g is injective. Further, the second condition of the theorem implies that Z ∘ h(x) ⩾ X(x), since g ∘ h(x) = f(x). Therefore, h ∈ Mor_{SET(L)}(X, Z).

The theorem implies the following result.

Theorem 4.7. (X, f) ≈ (Z, g) iff the following conditions are fulfilled:

(1) f(X̃) = g(Z̃);

(2) X(x) = Z(z) for x ∈ X̃, z ∈ Z̃, f(x) = g(z).

Since ≈ is an equivalence relation on the class of all subobjects of Y, the class can be partitioned into equivalence classes of isomorphic subobjects. By choosing one representative from each class we get a system of representatives called a representative class of subobjects of Y. It can be easily seen that the pairs (X, h) where X̃ ⊂ Ỹ , X(x) ⩽ Y (x) for all x ∈ X̃ and h is the inclusion map form a representative class of subobjects of Y, since (X_{1}, h_{1}) ≈ (X_{2}, h_{2}) iff X_{1} = X_{2} and for each subobject (Z, g) of Y there exists a pair (X, h) ≈ (Z, g) where X̃ = g(Z̃) and X(x) = Z(g^{−1}(x)) for all x ∈ X̃. Since the class of all pairs (X, h) is a set, the following theorem holds.

Theorem 4.8. The category SET(L) is well-powered, i.e., each object Y ∈ Obj SET(L) has a representative class of subobjects that is a set.

Now suppose we have an object Y ∈ Obj SET(L) and two arbitrary quotient objects of Y (f, X), (g, Z) where f: Y → X and g: Y → Z are some epimorphisms. Then (f, X) is said to be larger than (g, Z) – denoted by (f, X) ⩾ (g, Z) – iff there exists some morphism h: X → Z such that the triangle

\begin{matrix}Y w f_{4 4 44 6} g \\ Xh \end{matrix} ; Z

commutes.

If (f, X) ⩾ (g, Z) and (g, Z) ⩾ (f, X) then (f, X) and (g, Z) are said to be isomorphic quotient objects of Y; denoted by (f, X) ≈ (g, Z).

Theorem 4.9. (f, X) ⩾ (g, Z) iff the following conditions are fulfilled:

(1) for all y_{1}, y_{2} ∈ Ỹ f(y_{1}) = f(y_{2}) implies g(y_{1}) = g(y_{2});

(2) X(f(y)) ⩽ Z(g(y)) for all y ∈ Ỹ.

Proof. Let us prove the necessity first and therefore assume that (f, X) ⩾ (g, Z).

Suppose there exist such y_{1}, y_{2} ∈ Ỹ that f(y_{1}) = f(y_{2}) and g(y_{1}) ≠ g(y_{2}). Since (f, X) ⩾ (g, Z), there exists some morphism h: X → Z such that h ∘ f = g. Thus, g(y_{1}) = h∘f(y_{1}) = h∘f(y_{2}) = g(y_{2}) that contradicts our former assumption.

Suppose there exists such y_{0} ∈ Ỹ that X(f(y_{0})) ⩽̸ Z(g(y_{0})). Since (f, X) ⩾ (g, Z), there exists some morphism h: X → Z such that h∘f = g. Thus, Z(g(y_{0})) = Z(h ∘ f(y_{0})) ⩾ X(f(y_{0})) that contradicts out former assumption.

Now let us prove the sufficiency and therefore assume that all conditions of the theorem are fulfilled. Let h(x) = g(y), y ∈ f^{−1}(x). If x_{1}, x_{2} ∈ X̃ and x_{1} = x_{2} then h(x_{1}) = g(y_{1}), f(y_{1}) = x_{1} and h(x_{2}) = g(y_{2}), f(y_{2}) = x_{2}. Since f(y_{1}) = x_{1} = x_{2} = f(y_{2}) implies g(y_{1}) = g(y_{2}) then h(x_{1}) = h(x_{2}) and the definition of h is correct. For each y ∈ Ỹ it follows that h ∘ f(y) = g(y_{0}), y_{0} ∈ f^{−1} ∘ f(y). Since f(y_{0}) = f(y) implies g(y_{0}) = g(y) then h ∘ f(y) = g(y). Further, for an arbitrary point x_{0} ∈ X̃ Z ∘ h(x_{0}) = Z ∘ g(y_{0}), y_{0} ∈ f^{−1}(x_{0}) and Z(g(y_{0})) ⩾ X(f(y_{0})) = X(x_{0}). Therefore, h ∈ Mor_{SET(L)}(X, Z).

The theorem implies the following result.

Theorem 4.10. (f, X) ≈ (g, Z) iff the following conditions are fulfilled:

(1) for all y_{1}, y_{2} ∈ Ỹ f(y_{1}) = f(y_{2}) iff g(y_{1}) = g(y_{2});

(2) X(f(y)) = Z(g(y)) for all y ∈ Ỹ.

Since ≈ is an equivalence relation on the class of all quotient objects of Y, the class can be partitioned into equivalence classes of isomorphic quotient objects. By choosing one representative from each class we get a system of representatives called a representative class of quotient objects of Y. It can be easily seen that the pairs (h, X) where X̃ = Ỹ /Q for some equivalence relation Q on Ỹ, X([y]) ⩾ ⋁ Y ([y]) where [y] denotes the equivalence class generated by y and h is the induced quotient map form a representative class of quotient objects of Y, since (h_{1}, X_{1}) ≈ (h_{2}, X_{2}) iff X_{1} = X_{2} and for each quotient object (g, Z) of Y there exists a pair (h, X) ≈ (g, Z) where X̃ = Ỹ /Q with (y_{1}, y_{2}) ∈ Q iff g(y_{1}) = g(y_{2}) for all y_{1}, y_{2} ∈ Ỹ and X([y]) = Z(g(y)) for all y ∈ Ỹ. Since the class of all pairs (h, X) is a set, the following theorem holds.

Theorem 4.11. The category SET(L) is co-(well-powered), i.e., each object Y ∈ Obj SET(L) has a representative class of quotient objects that is a set.

5. Special morphisms

In this section we will consider the properties of some special morphisms in the category SET(L). To begin with, we will note that the category SET(L) is not balanced, that is, there exists a bimorphism f ∈ Mor SET(L) which is not an isomorphism. Consider, for example, a morphism f: X → Y where X̃ = {x_{0}}, X(x_{0}) = 0L, Ỹ = {y_{0}}, Y (y_{0}) = 1L and f(x_{0}) = y_{0}. Since f is bijective, f is a bimorphism but Y ∘ f(x_{0}) ≠ X(x_{0}), therefore, f is not an isomorphism.

We will continue by considering a coconstant morphism in the category SET(L).

Theorem 5.1. A morphism f: X → Y is a coconstant morphism iff X̃ = ∅. Proof. Let us prove the necessity first and therefore assume that f is a cocon-

stant morphism. Then for each object Z ∈ Obj SET(L) and every two morphisms g, h: Y → Z it follows that g ∘ f = h ∘ f.

X̃  w f_{4 4 44 6} X  Ỹ Y  w g w h Z̃ ; h h h h k ; Z

L

Suppose there exists a point x_{0} ∈ X̃. Then f(x_{0}) = y_{0} ∈ Ỹ. Let Z̃ = {z_{1}, z_{2}}, z_{1} ≠ z_{2} and Z(z_{1}) = Z(z_{2}) = 1L. In this case every mappings g, h: Ỹ → Z̃ will be morphisms and thus we can take g(y_{0}) = z_{1} and h(y_{0}) = z_{2}. It is easy to see that g ∘ f ≠ h ∘ f.

The sufficiency is obvious.

From this theorem, the theorem about constant morphism in the category SET(L) and the definition of zero morphism (notice that a morphism is said to be a zero morphism provided that it is both constant and coconstant) the following theorem can be derived.

Theorem 5.2. A morphism f: X → Y is a zero morphism iff X̃ = ∅.

Obviously, there exist such two objects X, Y ∈ Obj SET(L) that the set Mor_{SET(L)}(X, Y) contains no zero morphism, therefore, the category SET(L) is not pointed.

Now we will consider a coequalizer in the category SET(L). Suppose we have two morphisms f, g: X → Y. We will find a coequalizer of f and g. Following the definition we need such a pair (h, Z) where h ∈ Mor_{SET(L)}(Y, Z) that (1) h ∘ f = h ∘ g;

(2) for each object W ∈ Obj SET(L) and each morphism h′: Y → W such that h′ ∘f = h′ ∘g there exists a unique morphism m: Z → W such that the triangle

\begin{matrix}Y w h h′ \\ Z i i i i k m \end{matrix} ; W

commutes.

(Notice that the definition of coequalizer implies that (h, Z) is a quotient object of Y.)

Let Q be the smallest equivalence relation on Ỹ that contains all pairs (f(x), g(x)) for x ∈ X̃. Then let Z̃ = Ỹ /Q = {[y]|y ∈ Ỹ } where [y] denotes the equivalence class generated by y and Z([y_{0}]) = ∨{Y (y)|y ∈ [y_{0}]} for all [y_{0}] ∈ Z̃. The mapping h: Ỹ → Z̃ will be the induced quotient map, therefore, h(y) = [y] for all y ∈ Ỹ. Obviously, h ∈ Mor_{SET(L)}(Y, Z).

Proposition 5.1. (h, Z) is a coequalizer of f and g.

Proof. Obviously, h ∘ f = h ∘ g. Suppose there exists another morphism h′: Y → W such that h′ ∘ f = h′ ∘ g. From the definition of Q we derive that h′([y]) ⊂ {w_{y}} ⊂ W̃ , w_{y} ∈ W̃ for all y ∈ Ỹ, therefore, there exists a unique mapping m: Z̃ → W̃ such that m ∘ h = h′, i.e., m([y]) = h′(y) for all [y] ∈ Z̃. Let us prove that m ∈ Mor_{SET(L)}(Z, W), that is, W ∘ m([y]) ⩾ Z([y]) for all [y] ∈ Z̃. Fix an arbitrary point [y_{0}] ∈ Z̃. Since h′ ∈ Mor_{SET(L)}(Y, W), it follows that W ∘ h′(y_{0}) = W ∘ h′(y) ⩾ Y (y) for all y ∈ [y_{0}] and then W ∘ m([y_{0}]) = W ∘ h′(y_{0}) ⩾ ∨{Y (y)|y ∈ [y_{0}]} = Z([y_{0}]). Thus, m is indeed a morphism.

Since f and g were two arbitrary morphisms, the following result holds.

Theorem 5.3. The category SET(L) has coequalizers, i.e., every two morphisms f, g: X → Y have a coequalizer.

The notion of coequalizer gives rise to defining a special kind of epimorphisms called regular epimorphisms. Given an arbitrary morphism h: Y → Z, the pair (h, Z) is called a regular quotient object of Y and h is called a regular epimorphism iff there exist such two morphisms f, g: X → Y that (h, Z) is a coequalizer of f and g. The following theorem shows the necessary and sufficient conditions for a morphism to be a regular epimorphism.

Theorem 5.4. A morphism h: Y → Z is a regular epimorphism iff the following conditions are fulfilled:

(1) h is surjective;

(2) Z(z) = ∨{Y (y)|h(y) = z} for all z ∈ Z̃. Proof. Let us prove the necessity first and therefore assume that h is a regular

epimorphism. Then (h, Z) is a coequalizer of some morphisms f, g: X → Y and thus, a quotient object of Y. Therefore h is an epimorphism that implies that h is surjective.

Now let us prove that Z(z) = ∨{Y (y)|h(y) = z} for all z ∈ Z̃. Let Q be such an equivalence relation on Ỹ that for every two points y_{1}, y_{2} ∈ Ỹ (y_{1}, y_{2}) ∈ Q iff h(y_{1}) = h(y_{2}). Let W̃ = Ỹ /Q = {[y]|y ∈ Ỹ } and W ([y_{0}]) = ∨{Y (y)|y ∈ [y_{0}]} for all [y_{0}] ∈ W̃. The mapping h′ will be the induced quotient map, therefore, h′(y) = [y] for all y ∈ Ỹ. Obviously, h′ ∈ Mor_{SET(L)}(Y, W) and h′ ∘ f = h′ ∘ g. Then there exists such a morphism m: Z → W that h′ = m ∘ h. Now suppose we have an arbitray point z_{0} ∈ Z̃. Since h ∈ Mor_{SET(L)}(Y, Z), it follows that Z(z_{0}) ⩾ ∨{Y (y)|h(y) = z_{0}}. The fact that m ∈ Mor_{SET(L)}(Z, W) implies that W ∘ m(z_{0}) ⩾ Z(z_{0}). The mapping h is surjective, therefore, there exists such a point y_{0} ∈ Ỹ that h(y_{0}) = z_{0}. Then W ∘ m(z_{0}) = W ∘ m ∘ h(y_{0}) = W ∘ h′(y_{0}) = W ([y_{0}]) = ∨{Y (y)|y ∈ [y_{0}]} = ∨{Y (y)|h(y) = h(y_{0}) = z_{0}} ⩾ Z(z_{0}). Thus, Z(z_{0}) = ∨{Y (y)|h(y) = z_{0}}.

Now let us prove the sufficiency and therefore assume that all the conditions of the theorem are fulfilled. Obviously, one can find a set X̃ and such two mappings f, g: X̃ → Ỹ that Q defined in the first part of the theorem is the smallest equivalence relation that contains all pairs (f(x), g(x)) for x ∈ X̃ (for every y_{1}, y_{2} ∈ Ỹ such that h(y_{1}) = h(y_{2}) take a point x_{y_{1}y_{2}} and define f(x_{y_{1}y_{2}}) = y_{1} and g(x_{y_{1}y_{2}}) = y_{2}). Then let X(x) = 0L for all x ∈ X̃. Thus, f, g ∈ Mor_{SET(L)}(X, Y). The proof that (h, Z) is a coequalizer of f and g is the same as in the proposition 5.1.

The theorem implies that there are epimorphisms in the category SET(L) that are not regular epimorphisms.

Now let us consider a generalization of coequalizer, i.e., multiple coequalizer in the category SET(L). Suppose we have two objects X, Y ∈ Obj SET(L) and a nonempty indexed family of morphisms (f_{i})_{i∈I} contained in the set Mor_{SET(L)}(X, Y). A pair (h, Z) where h ∈ Mor_{SET(L)}(Y, Z) called a multiple coequalizer of (f_{i})_{i∈I} provided that

(1) h ∘ f_{i} = h ∘ f_{j} for all i, j ∈ I;

(2) for each object W ∈ Obj SET(L) and each morphism h′: Y → W such that h′ ∘ f_{i} = h′ ∘ f_{j} for all i, j ∈ I there exists a unique morphism m: Z → W such that m ∘ h = h′. Let Q be the smallest equivalence relation on Ỹ that contains all pairs

(f_{i}(x), f_{j}(x)) for i, j ∈ I and x ∈ X̃. Then let Z̃ = Ỹ /Q = {[y]|y ∈ Ỹ } and Z([y_{0}]) = ∨{Y (y)|y ∈ [y_{0}]} for all [y_{0}] ∈ Z̃. The mapping h: Ỹ → Z̃ will be the induced quotient map. One can easily prove that (h, Z) is a multiple coequalizer of (f_{i})_{i∈I}. Thus, the following theorem holds.

Theorem 5.5. The category SET(L) has multiple coequalizers, i.e., each nonempty indexed family of morphisms (f_{i})_{i∈I} contained in Mor_{SET(L)}(X, Y) has a multiple coequalizer.

Lastly, we will consider one special case of coequalizers, i.e., cokernels in the category SET(L). Suppose we have a morphism f: X → Y. If g_{0} : X → Y is the unique zero morphism from X to Y then (if it exists) a coequalizer of f and g_{0} is called the cokernel of f. From our previous investigations it follows that if g_{0} is a zero morphism then X̃ = ∅ and then f = g_{0}. Thus, the cokernel of f is a pair (h, Z) where h is bijective and Z ∘ h(y) = Y (y) for all y ∈ Ỹ. Therefore, h is an isomorphism. Since the category SET(L) is not pointed, there exist such morphisms in this category which have no cokernel. Further, suppose we have a morphism h: Y → Z. Then (h, Z) is called a normal quotient object of Y and h is called a normal epimorphism provided that there exists such a morphism f: X → Y that (h, Z) is the cokernel of f. The following theorem shows that in the category SET(L) normal epimorphisms and isomorphisms are equivalent.

Theorem 5.6. A morphism h: Y → Z is a normal epimorphism iff h is an isomorphism.

Proof. Let us prove the necessity first and therefore assume that h is a normal epimorpsim. Then it follows that (h, Z) is a coequalizer of some f, g_{0} ∈ Mor_{SET(L)}(X, Y) where X̃ = ∅. Since e_{Y} ∘ f = e_{Y} ∘ g_{0}, there exists such a morphism m: Z → Y that m∘h = e_{y}. Hence, h is a section. The fact that (h, Z) is a subobject of Y implies that h is an epimorphism and then h is an isomorphism.

Now let us prove the sufficiency and therefore assume that h is an isomorphism. Then h is bijective and Z ∘ h(y) = Y (y) for all y ∈ Ỹ. Obviously, h ∘ f = h ∘ g_{0} where f, g_{0} ∈ Mor_{SET(L)}(X, Y), X̃ = ∅. For each morphism h′: Y → W such that h′ ∘ f = h′ ∘ g_{0} there exists a unique mapping m: Z̃ → W̃ such that m ∘ h = h′, i.e., m(z) = h′ ∘ h^{−1}(z) for all z ∈ Z̃. Then W ∘ m(z) = W ∘ h′ ∘ h^{−1}(z) ⩾ Y ∘ h^{−1}(z) = Z ∘ h ∘ h^{−1}(z) = Z(z) and therefore m ∈ Mor_{SET(L)}(Z, W).

Now let us consider the dual of coequalizer, i.e., equalizer in the category SET(L). Suppose we have two morphisms f, g: X → Y. We will find an equalizer of f and g. Following the definition, we need such a pair (Z, h) where h ∈ Mor_{SET(L)}(Z, X) that

(1) f ∘ h = g ∘ h;

(2) for each object W ∈ Obj SET(L) and each morphism h′: W → X such that f ∘h′ = g∘h′ there exists a unique morphism m: W → Z such that the triangle

{Z  w h \atop X} ; W^{5 5 5 57} m  uh′

commutes.

(Notice that the definition of equalizer implies that (Z, h) is a subobject of X.) Let Z̃ = {x|x ∈ X̃, f(x) = g(x)} and Z(x) = X(x) for all x ∈ Z̃. The mapping

h: Z̃ → X̃ will be the inclusion map. Obviously, h ∈ Mor_{SET(L)}(Z, X).

Proposition 5.2. (Z, h) is an equalizer of f and g.

Proof. Obviously, f ∘ h = g ∘ h. Suppose there exists another morphism h′: W → X such that f ∘ h′ = g ∘ h′. Then there exists a unique mapping m: W̃ → Z̃ such that h ∘ m = h′, i.e., m(w) = h^{−1} ∘ h′(w) for all w ∈ W̃. Fix an arbitrary point w_{0} ∈ W̃. Then Z ∘ m(w_{0}) = X ∘ h ∘ m(w_{0}) = X ∘ h′(w_{0}) ⩾ W (w_{0}). Thus, m is indeed a morphism.

Since f and g were two arbitrary morphisms, the following result holds.

Theorem 5.7. The category SET(L) has equalizers, i.e., every two morphisms f, g: X → Y have an equalizer.

By analogy with coequalizers a special kind of monomorphisms, i.e., regular monomorphisms can be defined. From the theorem about the properties of equalizers in the category SET(L) we can derive the necessary and sufficient conditions for an arbitrary morphism h: Z → X to be a regular monomorphism.

Theorem 5.8. A morphism h: Z → X is a regular monomorphism iff the following conditions are fulfilled:

(1) h is injective;

(2) X ∘ h(x) = Z(x) for all x ∈ X̃.

The theorem implies that there are monomorphisms in the category SET(L) that are not regular monomorphisms.

Now let us consider a generalization of equalizer, i.e., multiple equalizer in the category SET(L) whose definition can be obtained by dualizing the notion of multiple coequalizer.

Suppose we have two objects X, Y ∈ Obj SET(L) and a non-empty indexed family of morphisms (f_{i})_{i∈I} contained in the set Mor_{SET(L)}(X, Y). Let Z̃ = {x|x ∈ X̃, f_{i}(x) = f_{j}(x) for all i, j ∈ I} and Z(x) = X(x) for all x ∈ Z̃. The mapping h: Z̃ → X̃ will be the inclusion map. Obviously, h ∈ Mor_{SET(L)}(Z, X). One can easily prove that (Z, h) is a multiple equalizer of (f_{i})_{i∈I}. Thus, the following theorem holds.

Theorem 5.9. The category SET(L) has multiple equalizers, i.e., each nonempty indexed family of morphisms (f_{i})_{i∈I} contained in Mor_{SET(L)}(X, Y) has a multiple equalizer.

Lastly, let us consider one special case of equalizers, i.e., kernels in the category SET(L). Suppose we have a morphism f: X → Y. If g_{0} : X → Y is the unique zero morphism from X to Y then (if it exists) an equalizer of f and g_{0} is called the kernel of f. If g_{0} is a zero morphism then X̃ = ∅ and then the kernel of f is a pair (Z, h) where Z̃ = ∅ and h: Z̃ → X̃ is the empty mapping. Obviously, there exist such morphisms in the category SET(L) which have no kernel. By analogy with normal epimorphisms we can obtain normal monomorphisms. It is easy to see that the only normal monomorphism in the category SET(L) is the morphism h: X → Y where X̃ = Ỹ = ∅.

6. Standard constructions

In this section we will consider two standard constructions in the category SET(L), i.e., inverse and direct systems. We will start with the former (see [1]).

Suppose C′ is a category and Ω is a partially-ordered and directed set. For each ω ∈ Ω choose an object X_{ω} ∈ Obj C′ such that X_{ω_{1}} ≠ X_{ω_{2}} for ω_{1} ≠ ω_{2} and for every ω_{1}, ω_{2} ∈ Ω such that ω_{1} ⩽ ω_{2} choose a morphism f_{ω_{1}}^{ω_{2}} : X_{ω_{2}} → X_{ω_{1}}. If the triple S = {X_{ω}, f_{ω_{1}}^{ω_{2}}, Ω} is a subcategory of C′ then S is called an inverse system. (Notice that from the properties of S it follows that for every ω_{1}, ω_{2}, ω_{3} ∈ Ω such that ω_{1} ⩽ ω_{2} ⩽ ω_{3} f_{ω_{1}}^{ω_{2}} ∘ f_{ω_{2}}^{ω_{3}} = f_{ω_{1}}^{ω_{3}} and for each ω ∈ Ω f_{ω}^{ω} = e_{X_{ω}}.) A pair {X, (h_{ω})_{ω∈Ω}} where X ∈ Obj C′ and h_{ω} ∈ MorC^{′}(X, X_{ω}) for all ω ∈ Ω is called a limit of S iff the following conditions are fulfilled:

(1) f_{ω_{1}}^{ω_{2}} ∘ h_{ω_{2}} = h_{ω_{1}} for all ω_{1}, ω_{2} ∈ Ω, ω_{1} ⩽ ω_{2};

(2) for each pair {Y, (g_{ω})_{ω∈Ω}} where Y ∈ Obj C′, g_{ω} ∈ MorC^{′}(Y, X_{ω}) for all ω ∈ Ω and f_{ω_{1}}^{ω_{2}} ∘g_{ω_{2}} = g_{ω_{1}} for all ω_{1}, ω_{2} ∈ Ω, ω_{1} ⩽ ω_{2} where exists a unique morphism f: Y → X such that h_{ω} ∘ f = g_{ω} for all ω ∈ Ω.

Theorem 6.1. In the category SET(L) each inverse system has a limit.

Proof. Suppose we have an inverse system S = {X_{ω}, f_{ω_{1}}^{ω_{2}}, Ω}. We have to find a limit of S, i.e., a pair {X, (h_{ω})_{ω∈Ω}}. We will proceed as in [FF]. Let

X̃ = {{x} = (x_{ω})_{ω∈Ω} ∈ ∏_{ω∈Ω} X̃_{ω}|f_{ω_{1}}^{ω_{2}}(x_{ω_{2}}) = x_{ω_{1}} for all ω_{1}, ω_{2} ∈ Ω, ω_{1} ⩽ ω_{2}}

and X({x}) = ⋀_{x_{ω}∈{x}} X_{ω}(x_{ω}). For each ω ∈ Ω let h_{ω} be the restriction to X̃ of

the projective mapping p_{ω}: ∏_{ω∈Ω} X̃_{ω} → X̃_{ω}: {x} ↦ x_{ω}. Since X_{ω} ∘ h_{ω}({x}) =

X_{ω}(x_{ω}) ⩾ ⋀_{x_{ω}∈{x}} X_{ω}(x_{ω}) = X({x}), the mapping h_{ω} is a morphism. For every

ω_{1}, ω_{2} ∈ Ω, ω_{1} ⩽ ω_{2} f_{ω_{1}}^{ω_{2}} ∘ h_{ω_{2}}({x}) = f_{ω_{1}}^{ω_{2}}(x_{ω_{2}}) = x_{ω_{1}} = h_{ω_{1}}({x}).

Suppose we have another pair {Y, (g_{ω})_{ω∈Ω}} and f_{ω_{1}}^{ω_{2}} ∘g_{ω_{2}} = g_{ω_{1}} for all ω_{1}, ω_{2} ∈ Ω, ω_{1} ⩽ ω_{2}. Let f be the diagonal mapping \begin{matrix}△ \\ ω∈Ω \end{matrix} g_{ω}: Ỹ → ∏_{ω∈Ω} X̃_{ω}: y ↦ (g_{ω}(y))_{ω∈Ω}.

Obviously, p_{ω} ∘ f = g_{ω} for all ω ∈ Ω. Fix an arbitrary point y_{0} ∈ Ỹ. Then f_{ω_{1}}^{ω_{2}} ∘ p_{ω_{2}} ∘ f(y_{0}) = f_{ω_{1}}^{ω_{2}} ∘ g_{ω_{2}}(y_{0}) = g_{ω_{1}}(y_{0}) = p_{ω_{1}} ∘ f(y_{0}). Thus, f(y_{0}) ∈ X̃ and f: Ỹ → X̃. It is easy to see that h_{ω} ∘ f = g_{ω} for all ω ∈ Ω. Let us prove that f ∈ Mor_{SET(L)}(Y, X), that is, X ∘ f(y) ⩾ Y (y) for all y ∈ Ỹ. Suppose y_{0} ∈ Ỹ. Then X_{ω} ∘ g_{ω}(y_{0}) ⩾ Y (y_{0}) and then ⋀_{ω∈Ω} X_{ω} ∘ g_{ω}(y_{0}) ⩾ Y (y_{0}). Therefore,

X ∘ f(y_{0}) = ⋀_{ω∈Ω} X_{ω} ∘ g_{ω}(y_{0}) ⩾ Y (y_{0}). For each morphism f′: Y → X such that

h_{ω} ∘ f′ = g_{ω} for all ω ∈ Ω it follows that f′ = f, since g_{ω} = h_{ω} ∘ f′ = p_{ω} ∘ f′. Thus, f is the only morphism with the required property.

Now let us consider the dual of inverse system, i.e., direct system in the category SET(L).

Suppose C′ is a category and Ω is a partially-ordered and directed set. For each ω ∈ Ω choose an object X_{ω} ∈ Obj C′ such that X_{ω_{1}} ≠ X_{ω_{2}} for ω_{1} ≠ ω_{2} and for every ω_{1}, ω_{2} ∈ Ω such that ω_{1} ⩽ ω_{2} choose a morphism f_{ω_{2}}^{ω_{1}} : X_{ω_{1}} → X_{ω_{2}}. If the triple S = {X_{ω}, f_{ω_{2}}^{ω_{1}}, Ω} is a subcategory of C′ then S is called a direct system. (Notice that from the properties of S it follows that for every ω_{1}, ω_{2}, ω_{3} ∈ Ω such that ω_{1} ⩽ ω_{2} ⩽ ω_{3} f_{ω_{3}}^{ω_{2}} ∘ f_{ω_{2}}^{ω_{1}} = f_{ω_{3}}^{ω_{1}} and for each ω ∈ Ω f_{ω}^{ω} = e_{X_{ω}}.) A pair {(h_{ω})_{ω∈Ω}, X} where X ∈ Obj C′ and h_{ω} ∈ MorC^{′}(X_{ω}, X) for all ω ∈ Ω is called a limit of S iff the following conditions are fulfilled:

(1) h_{ω_{2}} ∘ f_{ω_{2}}^{ω_{1}} = h_{ω_{1}} for all ω_{1}, ω_{2} ∈ Ω, ω_{1} ⩽ ω_{2};

(2) for each pair {(g_{ω})_{ω∈Ω}, Y } where Y ∈ Obj C′, g_{ω} ∈ MorC^{′}(X_{ω}, Y) for all ω ∈ Ω and g_{ω_{2}} ∘f_{ω_{2}}^{ω_{1}} = g_{ω_{1}} for all ω_{1}, ω_{2} ∈ Ω, ω_{1} ⩽ ω_{2} where exists a unique morphism f: X → Y such that f ∘ h_{ω} = g_{ω} for all ω ∈ Ω.

Theorem 6.2. In the category SET(L) each direct system has a limit.

Proof. Suppose we have an inverse system S = {X_{ω}, f_{ω_{2}}^{ω_{1}}, Ω}. We have to find a limit of S, i.e., a pair {(h_{ω})_{ω∈Ω}, X}. Let Q be such a relation on ⨁_{ω∈Ω} X̃_{ω} that for

every two points x_{ω_{1}}, y_{ω_{2}} ∈ ⨁_{ω∈Ω} X̃_{ω} (x_{ω_{1}}, y_{ω_{2}}) ∈ Q iff there exists such ω ∈ Ω that

ω_{1} ⩽ ω, ω_{2} ⩽ ω and f_{ω}^{ω_{1}}(x_{ω_{1}}) = f_{ω}^{ω_{2}}(y_{ω_{2}}). Let us verify that Q is an equivalence relation on ⨁_{ω∈Ω} X̃_{ω}.

(1) (x_{ω}, x_{ω}) ∈ Q for all x_{ω} ∈ ⨁_{ω∈Ω} X̃_{ω}, since f_{ω}^{ω}(x_{ω}) = x_{ω}.

(2) Obviously, (x_{ω_{1}}, y_{ω_{2}}) ∈ Q implies (y_{ω_{2}}, x_{ω_{1}}) ∈ Q for all x_{ω_{1}}, y_{ω_{2}} ∈ ⨁_{ω∈Ω} X̃_{ω}.

(3) Suppose (x_{ω_{1}}, y_{ω_{2}}) ∈ Q and (y_{ω_{2}}, z_{ω_{3}}) ∈ Q for some x_{ω_{1}}, y_{ω_{2}}, z_{ω_{3}} ∈ ⨁_{ω∈Ω} X̃_{ω}.

Then there exist such two points ω_{0}^{1}, w_{0}^{2} ∈ Ω that ω_{1} ⩽ ω_{0}^{1}, ω_{2} ⩽ ω_{0}^{1}, f_{ω_{0}^{1}}^{ω_{1}}(x_{ω_{1}}) = f_{ω_{0}^{1}}^{ω_{2}}(y_{ω_{2}}) and ω_{2} ⩽ ω_{0}^{2}, ω_{3} ⩽ ω_{0}^{2}, f_{ω_{0}^{2}}^{ω_{2}}(y_{ω_{2}}) = f_{ω_{0}^{2}}^{ω_{3}}(z_{ω_{3}}). Since the set Ω is directed, there exists such a point ω_{0}^{3} ∈ Ω that ω_{0}^{1} ⩽ ω_{0}^{3} and ω_{0}^{2} ⩽ ω_{0}^{3}. Then f_{ω_{0}^{3}}^{ω_{0}^{1}} ∘ f_{ω_{0}^{1}}^{ω_{1}}(x_{ω_{1}}) = f_{ω_{0}^{3}}^{ω_{0}^{1}} ∘ f_{ω_{0}^{1}}^{ω_{2}}(y_{ω_{2}}) and f_{ω_{0}^{3}}^{ω_{0}^{2}} ∘ f_{ω_{0}^{2}}^{ω_{2}}(y_{ω_{2}}) = f_{ω_{0}^{3}}^{ω_{0}^{2}} ∘ f_{ω_{0}^{2}}^{ω_{3}}(z_{ω_{3}}). From

the properties of S it follows that f_{ω_{0}^{3}}^{ω_{0}^{1}} ∘ f_{ω_{0}^{1}}^{ω_{1}}(x_{ω_{1}}) = f_{ω_{0}^{3}}^{ω_{1}}(x_{ω_{1}}), f_{ω_{0}^{3}}^{ω_{0}^{1}} ∘ f_{ω_{0}^{1}}^{ω_{2}}(y_{ω_{2}}) =

f_{ω_{0}^{3}}^{ω_{2}}(y_{ω_{2}}), f_{ω_{0}^{3}}^{ω_{0}^{2}} ∘ f_{ω_{0}^{2}}^{ω_{2}}(y_{ω_{2}}) = f_{ω_{0}^{3}}^{ω_{2}}(y_{ω_{2}}) and f_{ω_{0}^{3}}^{ω_{0}^{2}} ∘ f_{ω_{0}^{2}}^{ω_{3}}(z_{ω_{3}}) = f_{ω_{0}^{3}}^{ω_{3}}(z_{ω_{3}}). Thus, f_{ω_{0}^{3}}^{ω_{1}}(x_{ω_{1}}) = f_{ω_{0}^{3}}^{ω_{2}}(y_{ω_{2}}) = f_{ω_{0}^{3}}^{ω_{3}}(z_{ω_{3}}) and (x_{ω_{1}}, z_{ω_{3}}) ∈ Q.

Let X̃ = ⨁_{ω∈Ω} X̃_{ω}/Q = {[x_{ω}]|x_{ω} ∈ ⨁_{ω∈Ω} X̃_{ω}} and X([x_{ω}]) = ∨{X_{ω′} (y_{ω′})|y_{ω′} ∈

[x_{ω}]}. For each ω ∈ Ω the mapping h_{ω} will be the induced quotient map, i.e., h_{ω}(x_{ω}) = [x_{ω}] for all x_{ω} ∈ X̃_{ω}. Since X ∘ h_{ω}(x_{ω}) = X([x_{ω}]) ⩾ X_{ω}(x_{ω}) then h_{ω} ∈ Mor_{SET(L)}(X_{ω}, X). Suppose we have ω_{1}, ω_{2} ∈ Ω, ω_{1} ⩽ ω_{2}. Then for each x_{ω_{1}} ∈

X̃_{ω_{1}} it follows that (x_{ω_{1}}, f_{ω_{2}}^{ω_{1}}(x_{ω_{1}})) ∈ Q, since ω_{1} ⩽ ω_{2}, ω_{2} ⩽ ω_{2} and f_{ω_{2}}^{ω_{1}}(x_{ω_{1}}) = f_{ω_{2}}^{ω_{2}} ∘ f_{ω_{2}}^{ω_{1}}(x_{ω_{1}}). Therefore, h_{ω_{2}} ∘ f_{ω_{2}}^{ω_{1}}(x_{ω_{1}}) = [f_{ω_{2}}^{ω_{1}}(x_{ω_{1}})] = [x_{ω_{1}}] = h_{ω_{1}}(x_{ω_{1}}) and the required property is fulfilled.

Suppose we have another pair {(g_{ω})_{ω∈Ω}, Y } and g_{ω_{2}} ∘f_{ω_{2}}^{ω_{1}} = g_{ω_{1}} for all ω_{1}, ω_{2} ∈ Ω, ω_{1} ⩽ ω_{2}. Let f: X̃ → Ỹ be such that f([x_{ω}]) = g_{ω}(x_{ω}) for all [x_{w}] ∈ X̃. We have to verify that the definition of f is correct. Suppose [x_{ω_{1}}] = [y_{ω_{2}}] for some x_{ω_{1}}, y_{ω_{2}} ∈ X̃. Since (x_{ω_{1}}, y_{ω_{2}}) ∈ Q, there exists such ω ∈ Ω that ω_{1} ⩽ ω, ω_{2} ⩽ ω and f_{ω}^{ω_{1}}(x_{ω_{1}}) = f_{ω}^{ω_{2}}(y_{ω_{2}}). Then g_{ω} ∘ f_{ω}^{ω_{2}}(y_{ω_{2}}) = g_{ω_{2}}(y_{ω_{2}}) and g_{ω} ∘ f_{ω}^{ω_{1}}(x_{ω_{1}}) = g_{ω_{1}}(x_{ω_{1}}). Since g_{ω} ∘ f_{ω}^{ω_{2}}(y_{ω_{2}}) = g_{ω} ∘ f_{ω}^{ω_{1}}(x_{ω_{1}}) then g_{ω_{2}}(y_{ω_{2}}) = g_{ω_{1}}(x_{ω_{1}}). Thus, f([x_{ω_{1}}]) = f([y_{ω_{2}}]).

Obviously, f ∘ h_{ω} = g_{ω} for all ω ∈ Ω. Let us prove that f ∈ Mor_{SET(L)}(X, Y), that is, Y ∘ f([x_{ω}]) ⩾ X([x_{ω}]) for all [x_{ω}] ∈ X̃. Suppose we have [x_{ω_{0}}] ∈ X̃. Then for all y_{ω} ∈ [x_{ω_{0}}] g_{ω}(y_{ω}) = f ∘ h_{ω}(y_{ω}) = f([y_{ω}]) = f([x_{ω_{0}}]) = g_{ω}(x_{ω_{0}}). Thus, Y ∘ f([x_{ω_{0}}]) = Y ∘ g_{ω}(x_{ω_{0}}) = Y ∘ g_{ω}(y_{ω}) ⩾ X_{ω}(y_{ω}) for all y_{ω} ∈ [x_{ω_{0}}], since g_{ω} ∈ Mor_{SET(L)}(X_{ω}, Y). Then Y ∘ f([x_{ω_{0}}]) ⩾ ∨{X_{ω}(y_{ω})|y_{ω} ∈ [x_{ω_{0}}]} = X([x_{ω_{0}}]). Thus, f is indeed a morphism. For each morphism f′: X → Y such that f′∘h_{ω} = g_{ω} for all ω ∈ Ω it follows that f′ ∘ h_{ω}(x_{ω}) = f′([x_{ω}]) = g_{ω}(x_{ω}) and then f′ = f.

REFERENCES

[1] Fedorchuk, V. V., Filippov, V. V., General Topology. Basic Constructions, Moscow State Univ. Publ., 1988 (in Russian).

[2] Goguen, J. A., L-fuzzy sets, J. Math. Anal. Appl. 18 (1967), 145–174.

[3] Herrlich, H., Strecker, G., Category Theory, Heldermann Verlag, Berlin, 1979.

[4] Solovyov, S., Some remarks on the category SET(L), to appear in Acta Univ. Latv. (Ser. Math.).

[5] Zadeh, L. A., Fuzzy sets, Inform. and Control 8 (1965), 338-365.

(received 10.06.2002)

Department of Mathematics, University of Latvia, 19 Raina Blvd., Riga LV-1586, Latvia. E-mail: sergejs@lanet.lv