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2 TOPOLOGY PROCEEDING Volume 29, No.2, 2005 Pages QUOTIENT OF TEXTURE AND OF DITOPOLOGICAL TEXTURE PACE LAWRENCE M. BROWN Abstract. The author considers equivalence direlations and the corresponding quotients of textures, using the interior relation on the texture as a tool in this investigation. A close relationship with surjective difunctions is shown, and the notions of quotient ditopology and quotient difunction are defined and studied. Finally, the existence of a T 0 reflection on the category dfditop is established. 1. Introduction In this section we recall some basic notions regarding textures and ditopological texture spaces, which should make the article accessible to the casual reader. Full details, motivation and background material may be obtained from [2 9, 17]. Textures Let be a set. We work within a subset of the power set P() called a texturing. A texturing is a point-separating, complete, completely distributive lattice with respect to inclusion, which contains and, and for which arbitrary meets coincide with intersections, and finite joins with unions Mathematics ubject Classification. 54B15, 54B30, 03G10, 06D10. Key words and phrases. Texture, Locally supercompact space, Core space, C-space, Interior relation, Equivalence direlation, Quotient texture, Difunction, Canonical quotient difunction, Ditopology, Bicontinuity, Quotient ditopology, Quotient difunction, T 0 reflection. 337

3 338 L. M. BROWN If is a texturing of the pair (, ) is called a texture [2]. The internal definition of textural concepts are expressed using the p sets and q sets, namely, for s, the sets P s = {A s A}, Q s = {A s / A}. Examples 1.1. (1) The discrete texture (X, P(X)) on the set X. For x X, P x = {x}, Q x = X \ {x}. (2) The texture (L, L), where L=(0, 1] and L={(0, r] 0 r 1}. Here, for r L, P r = Q r = (0, r]. (3) The unit interval texture (I, I), I=[0, 1], I={[0, r) r I} {[0, r] r I}. Here, for r I, P r = [0, r] and Q r = [0, r). (4) The product texture ( T, T) of textures (, ) and (T, T). Here the product texturing T of T consists of arbitrary intersections of sets of the form (A T ) ( B), A and B T. For (s, t) T, P (s,t) = P s P t and Q (s,t) = (Q s T ) ( Q t ). For A, the set A = { {Aj j J} {A j j J}, A = } {A j j J}, called the core of A is also of importance, although in general it need not belong to. We note the following fundamental properties of p sets and q sets. These expose a form of duality which is useful in defining pairs of dual properties. Lemma 1.2. [6, Theorem 1.2] (1) s / A = A Q s = s / A for all s, A. (2) For A, A = {s A Q s }. (3) For A, A is the smallest element of containing A. (4) For A, B, if A B then there exists s with A Q s and P s B. (5) A = {Q s P s A} for all A. (6) A = {P s A Q s } for all A.

4 QUOTIENT OF TEXTURE 339 Ditopology: ince a texturing of need not be closed under set complementation we must forego the traditional relationship between open and closed sets. Hence we define a dichotomous topology, or ditopology for short, as a pair (τ, κ) of generally unrelated subsets τ, κ of satisfying, τ,, κ, G 1, G 2 τ = G 1 G 2 τ, G i τ, i I = i G i τ. K 1, K 2 κ = K 1 K 2 κ, K i κ, i I = K i κ. The elements of τ are called open and those of κ closed. We refer to τ as the topology and to κ as the cotopology of (τ, κ). Any bitopology (u, v) on X gives rise to the ditopology (u, v c ) on the discrete texture (X, P(X)), while τ = {[0, r) 0 r 1} {I}, κ = {[0, r] 0 r 1} { } is a natural ditopology on the unit interval texture (I, I). Textures and ditopological texture spaces were first conceived as a means of representing fuzzy sets and topologies in a point-based setting, and this aspect of the theory continues to be of interest [4 8]. However, much of the theory has been developed quite independently of this context, and has concentrated on developing concepts appropriate to a complement-free setting which enable the expression of powerful results in the context of often quite minimal structures. As a case in point the unit interval texture with its natural ditopology is seen to play a role analogous to that of the unit interval in classical topology, despite the fact that we work within a very small subset of P(I). These aspects of the theory are likely to find applications in the modelling of negation-free logics, and in the development of structures for computation. In this paper we give a systematic method of constructing and studying quotients of textures and of ditopological texture spaces by using the notion of direlation [6]. The reader may refer to [5] for some early ideas concerning subtextures and quotient textures. At the suggestion of the referee we characterize the basic notions relating to direlations and difunctions in terms of the interior relation, thereby making the interior relation available as a tool in this investigation. For terms from lattice theory not defined here the reader may consult [14], while we follow [1] for terms from category theory.

5 340 L. M. BROWN 2. Textures and the Interior Relation It is clear that if is a texturing of then c = {A c A } is a T 0 topology on, that is may be regarded as the set of closed sets of the T 0 topological space (, c ). Viewed in this light, textures are just the closed set version of the locally supercompact spaces, core spaces or C-spaces studied initially by Hoffmann [15], Erné [11], Erné and Wilke [13]; and later by Erné [12], Lawson [16], and others. ince, in general, the sets of c do not belong to the texture, this link with C-spaces has not been exploited in the literature on textures so far 1. The referee has pointed out, however, that the interior relation for the topology (, c ), used by the aforementioned authors to characterize and study the C-spaces, can be used to give an algebraic characterization of textural concepts and results that may be more economic than the internal expression using p-sets and q-sets, and which at the same time could make the work on ditopological texture spaces of more immediate interest to workers in the field of C-spaces. With this in mind, therefore, we devote this section to a sketch of the relation between the internal and algebraic characterizations of the principle concepts used in the study of textures, particularly those involved in the main part of this paper on quotients. We recall that for a topological space (X, T) the interior relation ω is defined by x ω y y ( {G T x G}) o. For the topology (, c ) of the texture (, ) we therefore have s ω t t ( {A c A, s / A}) o t ( \ {A s / A}) o t / {A s / A} P t Q s, which immediately shows the relevance of the relation ω to the study of textures. In particular we note that the idempotency, ω = ω 2, of ω in this case can be read off directly from Lemma 1.2 (1) and (4). We will refer to ω as the interior relation of (, ), and denote it by ω if necessary, to avoid confusion. 1 With the exception of a subsection in [9] relating simple textures with sober topologies

6 QUOTIENT OF TEXTURE 341 For a relation ϕ from X to Y, A X and B Y, we set Aϕ = {y Y a A, aϕy} and ϕb = {x X b B, xϕb} as usual, abbreviating {a}ϕ to a ϕ and ϕ{b} to ϕ b, respectively. As above we will often write s ϕ t in preference to (s, t) ϕ. We note that: Lemma 2.1. Let (, ) be a texture with interior relation ω. (1) A A c ω = A c. (2) For A, A Q s s ω s A s Aω 1 (3) For A we have A = ω A = Aω 1. (4) For s, P s ω 1 = sω 1 and Q s = (sω ) c. In the present paper membership of a given set in a given texture will generally be known, so Lemma 2.1 (3) will be particularly useful in establishing A B from Aω 1 Bω 1. Direlations: Direlations give a suitable notion of (binary) relation for textures. For example, along with the notion of dicover, they play an important role in defining uniformities on textures [17]. Let (, ), (T, T) be textures. Generally, T is too small to support a useful notion of relation from (, ) to (T, T). We consider instead P() T. We use P (s,t), Q (s,t) to denote the p sets and q sets for ( T, P() T), and likewise P (t,s), Q (t,s) those for P(T ). An element of P() T is related to T but not to. We add conditions relating it to in two different ways, producing dual concepts of relation and corelation [6]. r P() T is called a relation from (, ) to (T, T) if it satisfies R1 r Q (s,t), P s Q s = r Q (s,t).. R2 r Q (s,t) = s such that P s Q s and r Q (s,t). R P() T is called a corelation from (, ) to (T, T) if it satisfies CR1 P (s,t) R, P s Q s = P (s,t) R. CR2 P (s,t) R = s such that P s Q s and P (s,t) R. A pair (r, R) consisting of a relation r and corelation R is now called a direlation.

7 342 L. M. BROWN Where necessary we will refer to an arbitrary relation between the base sets as a point relation to avoid possible confusion with the relations and corelations in the above sense. Example 2.2. For any texture (, ) the identity direlation (i, I) on (, ) is given by i = {P (s,s) s } and I = {Q (s,s) s }. Direlations are ordered by (r 1, R 1 ) (r 2, R 2 ) r 1 r 2 and R 2 R 1, and the direlation (r, R) on (, ) is called reflexive if (i, I) (r, R). Now we characterize direlations in terms of the interior relations ω, ω T of (, ), (T, T), respectively. First we note the following: Lemma 2.3. Let ϕ be a point relation from to T. (1) ϕ P() T ϕ c = ϕ c ω T. (2) If ϕ P() T then ϕ Q (s,t) s ϕω 1 T t. (3) If ϕ P() T then ϕ = ϕω 1 T. Proof. (1) Left to the interested reader. (2) If ϕ Q (s,t) then by Lemma 1.2 (4) applied to the texture ( T, P() T) we have t T with ϕ Q (s,t ) and P (s,t ) Q (s,t). By Lemma 1.2 (6) we have s ϕ t, while on the other hand P t Q t, so t ω 1 T t. Hence s ϕω 1 T t. The converse is straightforward, and we omit the details. (3) Immediate from (2) and Lemma 1.2 (2). If ϕ P() T then by (3), ϕωt 1 determines ϕ as the smallest set in the texture containing it, that is as its closure in the corresponding topology. ee the comments following Lemma 2.1. Proposition 2.4. Let r, R be point relations from to T. (1) r is a relation from (, ) to (T, T) if and only if R a : r c = r c ω T and R b : rω 1 T = ω 1 rω 1 T. (2) R is a corelation from (, ) to (T, T) if and only if CR a : R c = R c ω T and CR b : R c = ω R c.

8 QUOTIENT OF TEXTURE 343 Proof. (1) R a is just Lemma 2.3 (1). On the other hand it is easy to verify that r satisfies R1 if and only if ω 1 rω 1 T rωt 1, and that it satisfies R2 if and only if rωt 1 ω 1 rω 1 T. (2) Left to the interested reader. We note in passing that the identity direlation (i, I ) on (, ) is characterized by i ω 1 = ω 1 and I = ω c, as the reader may easily verify. Inverse of direlations: The inverse of (r, R) from (, ) to (T, T) is the direlation (r, R) = (R, r ) from (T, T) to (, ) given by r = {Q (t,s) r Q (s,t) }, R = {P (t,s) P (s,t) R}. The direlation (r, R) on (, ) is called symmetric if (r, R) = (r, R). Proposition 2.5. Let (r, R) be a direlation from (, ) to (T, T). Then (r, R) = (R, r ) : (T, T) (, ) is characterized by: (1) (R )ω 1 = (Rc ) 1, and (2) r = (ω T r 1 ) c. Proof. (1) In view of CR b it will be sufficient to prove (R )ω 1 = (ω R c ) 1. If t (R )ω 1 s then R Q (t,s) by Lemma 2.3 (2), so we have s with P (t,s ) Q (t,s) and P (s,t) R. This gives s ω s, s R c t, so s ω R c t and we have shown (2.1) (R )ω 1 (ω T R c ) 1. Conversely, if t(ω R c ) 1 s then s(ω R c )t and so for some s we have s ω s and s R c t. This gives P (s,t) R, whence P (t,s ) R and so t R s, while s ω 1 s and so t (R )ω 1 s. Thus (2.2) (ω R c ) (R )ω 1. The result now follows from (2.1) and (2.2). (2) Left to the interested reader.

9 344 L. M. BROWN Composition of direlations: If (r 1, R 1 ) : ( 1, 1 ) ( 2, 2 ) and (r 2, R 2 ) : ( 2, 2 ) ( 3, 3 ) are direlations, their composition (r 2, R 2 ) (r 1, R 1 ) = (r 2 r 1, R 2 R 1 ) from ( 1, 1 ) to ( 3, 3 ) is given by r 2 r 1 = {P (s,u) t 2 with r 1 Q (s,t) and r 2 Q (t,u) }, R 2 R 1 = {Q (s,u) t 2 with P (s,t) R 1 and P (t,u) R 2 }. It is known [6] that composition is associative and that (i, I) is the identity under composition. The direlation (r, R) on (, ) is transitive if (r, R) (r, R) (r, R). Proposition 2.6. With the notation above, and denoting by ω k the interior point relation for ( k, k ), 1 k 3, we have (1) (r 2 r 1 )ω3 1 = r 1 r 2 ω3 1, (2) R 2 R 1 = (R c 1 Rc 2 )c. Proof. traightforward. ections: Let (r, R) : (, ) (T, T) be a direlation, A. The A-sections of r, R are given by respectively. r A = {Q t s, r Q (s,t) = A Q s } T, R A = {P t s, P (s,t) R = P s A} T, Proposition 2.7. With the notation above: (1) (r A)ω 1 T = Arω 1 T, (2) (R A) = (A c R c ) c. Proof. (1) t (r A)ω 1 T gives r A Q t by Lemma 2.1 (2), so we have s with r Q (s,t) and A Q s. Now s (rω 1 ) t by Lemma 2.3 (2), and s Aω 1 by Lemma 2.1 (2), so t Aω 1 rω 1 T. Hence (2.3) (r A)ω 1 T Aω 1 rω 1 T. Conversely take t Aω 1 rω 1 T = Aω 1 rω 1 T ω 1 T since ω T is idempotent. Now we have s A with s (Aω 1 rω 1 T ω 1 T ) t, and hence there exist s, t T with s ω 1 s, s (rωt 1 ) t and t ωt 1 t. From Lemma 2.3 (2) we have r Q (s,t ). We claim that P t r A.

10 QUOTIENT OF TEXTURE 345 To see this assume the contrary. Then there exists t T with P t Q t for which (2.4) r Q (z,t ) = A Q z for all z. However we easily deduce r Q (s,t ), so implication (2.4) with z = s gives A Q s, which contradicts s ω 1 s since s A. This establishes P t r A, and P t Q t so r A Q (s,t) which gives t (r A)ω 1 T by Lemma 2.3 (2). Thus, (2.5) Aω 1 rω 1 T (r A)ωT 1. From (2.3), (2.5) and R b we have (r A)ω 1 T as required. = Aω 1 rω 1 T = Arω 1 T, (2) Essentially dual to (1), and is omitted. Presections: For B T, the B-presections of r, R, are the B- sections (r ) B, (R ) B of the inverses r, R, respectively. Normally these are written as r B, R B, respectively. Proposition 2.8. With the notation as above we have: (1) r B = (rω T B c ) c, (2) (R B)ω 1 = Rc B. Proof. (1) Noting that r is a corelation from (T, T) to (, ), we have r B = (r ) B = (B c (r ) c ) c by Proposition 2.7 (2). Now using Proposition 2.5 (2), r B = (B c ω T r 1 ) c = (B c (rω 1 T ) 1 ) c = (rω T B c ) c, as required. (2) In a similar way, using Propositions 2.7 (1) and 2.5 (1), (R B)ω 1 as required. = ((R ) B)ω 1 = B(R )ω 1 = B(Rc ) 1 = R c B, Difunctions: A difunction from (, ) to (T, T) is a direlation (f, F ) from (, ) to (T, T) satisfying the conditions DF1 For s, s, P s Q s = t T with f Q (s,t) and P (s,t) F. DF2 For t, t T and s, f Q (s,t) and P (s,t ) F = P t Q t.

11 346 L. M. BROWN A difunction (f, F ) is surjective if the inverse direlation (f, F ) satisfies DF1, and injective if (f, F ) satisfies DF2. A characteristic property of a difunction (f, F ) is that f B = F B for all B T [6]. Proposition 2.9. Given textures (, ), (T, T): (1) Let (r, R) : (, ) (T, T) be a direlation. (a) (r, R) satisfies DF1 if and only if ω R c ω T r 1. (b) (r, R) satisfies DF2 if and only if ω T r 1 R c ω T (2) Let (f, F ) : (, ) (T, T) be a difunction. (a) (f, F ) is surjective if and only if ω T ω T f 1 F c. (b) (f, F ) is injective if and only if F c ω T r 1 ω. Proof. (1) According to [6, Lemma 2.23 (1 ii)], (r, R) satisfies DF1 if and only if r R I. Now using Propositions 2.6 (2), 2.5 (2) and I = ω c gives r R I (R c (r ) c ) c ω c (R c ω T r 1 ) c ω c, which is equivalent to ω R c ω T r 1 on negating each side. This gives (a), and (b) follows similarly using [6, Lemma 2.23 (2 ii)]. (2) (f, F ) is surjective if and only if (F, f ) : (T, T) (, ) satisfies DF1, so using (1 i) we obtain ω T (r ) c ω (R ) 1, which by Proposition 2.5 (2) is equivalent to ω T ω T r 1 ω (R ) 1 = ω T r 1 ((R )ω 1 ) 1. Applying Proposition 2.5 (1) now gives (a), and the proof of (b) is similar and is omitted. In case (f, F ) : ( 1, 1, τ 1, κ 1 ) ( 2, 2, τ 2, κ 2 ) is a difunction we make the following definitions: Also: (1) (f, F ) is continuous if G τ 2 = F G τ 1. (2) (f, F ) is cocontinuous if K κ 2 = f K κ 1. (3) (f, F ) is bicontinuous if it is continuous and cocontinuous. (1) (f, F ) is open (co-open) if G τ 1 = f G τ 2 ( F G τ 2 ). (2) (f, F ) is closed (coclosed) if K κ 1 = f K κ 2 (F K κ 2 ).

12 QUOTIENT OF TEXTURE 347 It is shown in [7] that the category dfditop of ditopological texture spaces and bicontinuous difunctions is topological over the category dftex of textures and difunctions, thereby justifying our use of the term topology in naming the pair (τ, κ). Difunctions and Point Functions: Only in certain special cases do difunctions correspond to point functions between the base sets. We recall in particular [6, Lemma 3.4] that if (, ), (T, T) are textures and the point function ϕ : T satisfies (a) s, s, P s Q s = P ϕ(s) Q ϕ(s ), then the equalities f = f ϕ = {P (s,t) v with P s Q v and P ϕ(v) Q t }, F = F ϕ = {Q (s,t) v with P v Q s and P t Q ϕ(v) }, define a point function (f, F ) : (, ) (T, T). Moreover [6, Lemma 3.9], if ϕ also satisfies (b) s, P ϕ(s) B, B T = s with P ϕ(s ) B, then f B = ϕ 1 [B] = F B for all B T. Proposition With the notation as above: (1) ϕ satisfies (a) if and only if ω 1 ϕ ϕω 1 (2) (f, F ) = (f ϕ, F ϕ ) is characterized by fω 1 T T. = ω 1 ϕω 1 T and F = (ω ϕω T ) c. (3) ϕ satisfies (b) if and only if ω T ϕ 1 ω T ϕ 1 ω. Proof. (1) uppose (a) holds. Then if s ω 1 ϕ t then we have s with s ω 1 s and s ϕ t, that is t = ϕ(s ). Now P s Q s so P ϕ(s) Q t by (a). This gives s ϕωt 1 t, whence ω 1 ϕωt 1. The proof of the converse is similar, and is omitted. (2) s fω 1 T t f Q (s,t) t T, v with P (s,t ) Q (s,t), P s Q v and P ϕ(v) Q t s ω 1 ϕω 1 T ω 1 T t. Hence fωt 1 = ω 1 ϕω 1 T since ω T is idempotent. The proof of the formula for F is left to the reader. (3) traightforward on noting that in the definition of (b) it is sufficient to take B to be a q-set.

13 348 L. M. BROWN Finally we present a result which will be needed in the next section. Proposition Let ϕ be onto and satisfy (a) and (b). Then (f ϕ, F ϕ ) is surjective. Proof. For (f, F ) = (f ϕ, F ϕ ) we have ω T f 1 F c = (fω 1 T ) 1 F c = (ω 1 ϕ 1 ωt 1 ) 1 ω ϕω T = ω T ϕ 1 ω ϕω T by Proposition 2.10 (2). On the other hand, since ϕ is onto we have T ϕ 1 ϕ, where T is the diagonal of T, so using the above characterization of (b), ω T = ω T T ω T ω T ϕ 1 ϕω T ω T ϕ 1 ω ϕω T. This shows that ω T ω T f 1 F c, whence (f, F ) is surjective by Proposition 2.9 (2 a). The converse of this result is false, even if (f, F ) is bijective [7, Example 2.14]. 3. Equivalence Direlations and the Quotient Texture Definition 3.1. Let (, ) be a texture. A direlation (r, R) on (, ) is called an equivalence direlation if it is reflexive, symmetric and transitive. We have at once: Lemma 3.2. (r, R) is an equivalence direlation if and only if ω 1 rω 1, (rω 1 ) 2 rω 1 and R = r = (ωr 1 ) c. Corollary 3.3. Let (r, R) is a equivalence direlation. Then rω 1 is idempotent. We also note the following: Lemma 3.4. Let (r, R) be an equivalence direlation on (, ). Then for s, t the following are equivalent. (i) v, P s Q v = r Q (t,v). (ii) s ω 1 t rω 1. (iii) v, P s Q v = P (v,t) R. (iv) P s r P t.

14 QUOTIENT OF TEXTURE 349 Proof. (1) (2) Immediate from Lemma 2.3 (2). (3) (2) Clearly (3) is equivalent to s ω 1 t (R c ) 1, while (R c ) 1 = (R )ω 1 = rω 1 by Proposition 2.5 (1) and the fact that (r, R) is symmetric. (4) (2) P s r P t is equivalent to P s ω 1 (r P t )ω 1, and by Proposition 2.7 (1) and R b we have (r P t )ω 1 = P t rω 1 = P t ω 1 rω 1. The proof is now completed by applying Lemma 2.1 (4), and using R b again. The sense in which an equivalence direlation gives rise to a quotient texture is described in the following theorem. Theorem 3.5. Let (r, R) be an equivalence direlation on (, ). Then there exists a point equivalence relation ρ on, a texturing U of the quotient set U = /ρ and a surjective difunction (f, F ) on (, ) to (U, U) satisfying r = F f and R = f F. Proof. We will present the proof as a series of lemmas. Let us begin by associating an equivalence point relation ρ with (r, R). Lemma 3.6. If (r, R) is an equivalence direlation on (, ) then the point relation ρ defined by s ρ t s ω 1 t rω 1 and t ω 1 s rω 1 is an equivalence point relation on. Proof. By Lemma 3.2 we have ω 1 rω 1, whence s ω 1 s rω 1, which gives s ρ s for all s. It is immediate that ρ is symmetric, so it remains to show transitivity. For s, t, v let s ρ t and t ρ v. Then s ω 1 t rω 1 = t ω 1 rω 1 v rω 1 rω 1 v rω 1 by R b and Lemma 3.2. Likewise, v ω 1 s rω 1, so s ρ v as required. Lemma 3.7. Let (r, R) be an equivalence direlation on (, ) and ρ the equivalence point relation defined above. Then (1) For all s, t, s ρ t P s r P t and P t r P s. (2) For A, r A = A R A = A r A = R A. We denote by R the family of all A satisfying these equivalent conditions. (3) A R Arω 1 = Aω 1. In particular the elements of R are saturated with respect to ρ.

15 350 L. M. BROWN (4) r A R and R A R for all A. (5) R is closed under arbitrary joins and intersections. (6) For A, r A = {B R A B} and R A = {B R B A}. Proof. (1) This is just the equivalence of (ii) and (iv) in Lemma 3.4. (2) If r A = A then A r (r A) = r (A) = R A by [6, Proposition 2.10 (2)] and the fact that (r, R) is symmetric. On the other hand R A A since R is reflexive, so R A = A. The converse is proved likewise, so r A = A R A = A. The second equivalence is clear since R A A r A for all A by reflexivity. (3) A R f A = A (f A)ω 1 = Aω 1 Arω 1 = Aω 1 by Proposition 2.7 (1). For s A R, s ρ t, we have tω 1 srω 1 Arω 1 = Aω 1, whence t A and so A is saturated. (4) For A let B = r A. Then Brω 1 = (r A)ω 1 rω 1 = Arω 1 rω 1 = Arω 1 = Bω 1 by R b, Proposition 2.7 (1) and Corollary 3.3, so B R by (3). The proof of R A R is left to the interested reader. (5) Take A i R, i I. For B = i I A i, r B = r ( i I A i) = i I r A i = i I A i = B by [6, Corollary 2.12 (2)] and (2), so again by (2), i I A i R. In the same way B = i I A i satisfies R B = B, so we also have i I A i R. (6) Take A. If A B for B R then r A r B = B, so r A {B R A B}. The reverse inclusion is clear on noting that B = r A satisfies B R and A B. The dual proof of the second equality is omitted. We will denote by U the quotient set /ρ and by ϕ the canonical quotient mapping ϕ : U, ϕ : s s, where s denotes the equivalence class of s. Let us define U = {A U ϕ 1 [A] R}. Note that, since the sets in R are saturated with respect to ρ, we also have U = {ϕ[a] A R}. Now let us verify that U is a texturing of U, and identify the p-sets and q-sets in (U, U).

16 QUOTIENT OF TEXTURE 351 Lemma 3.8. With the notation above U = {A ϕ 1 [A] R} is a texturing of U. Moreover, for all s, (i) P s = ϕ[r P s ], and (ii) Q s = {ϕ[r Q s ] P s Q s } ϕ[r Q s ]. Proof. Note first that = r R implies U = ϕ[] U. Likewise, = R R gives = ϕ[ ] U. Now let us take A j U, j J, and put A = j J A j. Then ϕ 1 [A] = j J ϕ 1 [A j ] R by Lemma 3.7 (5), so j J A j U. Thus, U is a complete lattice of subsets of U. Now let A = j J A j. We will verify that ϕ 1 [A] = j J ϕ 1 [A j ]. It is clear that j J ϕ 1 [A j ] ϕ 1 [A]. On the other hand by Lemma 3.7 (5) we have j J ϕ 1 [A j ] R, so [ ] A j ϕ ϕ 1 [A j ] j J i I [ ϕ j J ] ϕ 1 [A j ] U. Thus A ϕ[ j J ϕ 1 [A j ]] U, which gives [ ] ϕ 1 [A] ϕ 1 [ϕ ϕ 1 [A i ] ] = ϕ 1 [A j ] j J j J since the sets of R are saturated. Thus ϕ 1 [A] = j J ϕ 1 [A j ]. We see from the above that both joins and intersections are preserved under the mapping ϕ 1 : U. Moreover, this mapping is 1-1 since ϕ is onto. Hence, since is completely distributive, U is completely distributive too. We have already seen that meet in U coincides with intersection, but we need to verify that finite joins coincide with unions. Let I be finite. Then ϕ 1 [ i I A i] = i I ϕ 1 [A i ] = i I ϕ 1 [A i ] = ϕ 1[ i I A i] since I is finite, whence i I A i = i I A i for A i U, i I and I finite, as required. In order to show U is a texturing it remains to show that it separates the points of U. Take s, t with ϕ(s) ϕ(t). Then s ρ c t, so by Lemma 3.7 (1) we have P s r P t or P t r P s. In the first case we have t ϕ[r P t ] U since t P t r P t and s / ϕ[r P t ] since r P t R is saturated. Likewise, the second case gives s ϕ[r P s ] and t / ϕ[r P s ], which completes the proof that (U, U) is a texture.

17 352 L. M. BROWN Finally, let us verify (i) and (ii). For (i), first note that s ϕ[r P s ] U, so P s ϕ[r P s ]. On the other hand, every element of U has the form ϕ[a] for A R and, since the elements of R are saturated, s ϕ[a] = P s A = r P s r A = A = ϕ[r P s ] ϕ[a], whence P s = ϕ[r P s ]. For (ii), first note that Q s = {ϕ[a] A R, P s ϕ[a]} and that P s ϕ[a] is equivalent to P s A since A R is saturated. Hence, for P s Q s we have ϕ 1 (Q s ) = ϕ 1 ( {ϕ[a] A R, P s A}) = {ϕ 1 [ϕ[a]] A R, P s A} = {A R P s A} since A R is saturated {A R A Q s } = R Q s = ϕ 1 [ϕ[r Q s ]], where we have used Lemma 3.7 (6). Hence ϕ[r Q s ] Q s, so {ϕ[r Q s ] P s Q s } Q s. For the reverse inclusion take t with P s P t, whence P s r P t. By the equivalence of (iii) and (iv) in Lemma 3.4 we have s with P s Q s and P (s,t) R. Now P t R Q s by [6, Lemma 2.6 (2)], whence r P t R Q s since R Q s R, and we deduce P t = ϕ[r P t ] ϕ[r Q s ]. Hence we have shown P s P t = P t ϕ[r Q s ] for some P s Q s, which gives us Q s {ϕ[r Q s ] P s Q s }, as required. To complete the proof of (ii) we need only note that if P s Q s then Q s Q s and so ϕ[r Q s ] ϕ[r Q s ], from which the required inclusion is clear. Corollary 3.9. (1) For (U, U) the interior point relation ω U is given by ϕ(s) ωu 1 ϕ(t) s ( rω 1 {rω 1 t t ω 1 t } ). (2) For s, t, r Q (s,t) = P ϕ(s) Q ϕ(t). (3) ϕ satisfies the conditions (a) and (b) mentioned earlier. Proof. (1) Using the facts proved above we have: P ϕ(s) Q ϕ(t) ϕ 1 [P ϕ(s) ] ϕ 1 [Q ϕ(t) ] r P s {R Q t P t Q t } u (r P s )ω 1 with u / R Q t t ω 1 t.

18 QUOTIENT OF TEXTURE 353 By Proposition 2.7 (1), u (r P s )ω 1 is equivalent to u srω 1, while for t ω 1 t we have u (R Q t ) c = (r Q t ) c = rω 1 Qc t by Proposition 2.8 (1). By Lemma 2.1 (4) we deduce that u rω 1 t, whence u {rω 1 t t ω 1 t }, as required. (2) uppose r Q (s,t). Then s rω 1 have t rω 1 t for all t ω 1 P ϕ(s) Q ϕ(t). t, and since ω 1 rω 1 we ϕ(t) by (1), that is t. Hence ϕ(s) ω 1 U (3) (a) If s ω 1 ϕ u then we have s ω 1 s and u = ϕ(s ). Now s rω 1 s and by (2) we deduce s ϕω 1 U u. Hence ω 1 ϕ ϕω 1 U, so ϕ satisfies (a) by Proposition 2.10 (1). (b) If we take s ϕωu 1 ϕ(t) then ϕ(s) ω 1 T ϕ(t), so by (1) s rω 1 u with u rω 1 u t ω 1 u. But by R b we have s ω 1 rω 1 u, whence s ω 1 ϕω 1 U ϕ(t). Thus ϕω 1 U ω 1 ϕω 1 U and (b) is satisfied by Proposition 2.10 (3). It follows from Corollary 3.9 (3) that ϕ gives rise to a difunction (f, F ) = (f ϕ, F ϕ ) : (, ) (U, U), which we will call the canonical quotient difunction. Moreover f B = ϕ 1 [B] = F B for all B U, while from Proposition 2.11 we see that the canonical quotient difunction (f, F ) is surjective since ϕ is onto. Moreover: Lemma f A = ϕ[a] = F A for all A R. Proof. Take A R. ince (f, F ) is surjective we have F A f A by [6, Corollary 2.33 (1 i)], so it will be sufficient to prove f A ϕ[a] F A. ince ϕ[a] U, the first inclusion will follow from (f A)ωU 1 AϕωU 1, which by Proposition 2.7 (1) is equivalent to Aω 1 ϕω 1 U Aϕω 1. By Proposition 2.10 (1) this follows from (a). U By Propositions 2.7 (2) and 2.10 (2) we have F A=(A c ω ϕω U ) c, so ϕ[a] F A gives the existence of u Aϕω 1 U Ac ω ϕω U. Hence we have a A, b / A with a ϕω 1 U ϕ 1 ω 1 b since ω U is idempotent. Now for some b with b ω 1 b we have ϕ(a) ω 1 U ϕ(b ), so by Corollary 3.9 (1) a rω 1 rω 1 b, that is a rω 1 b. Thus b A by Lemma 3.7 (3), which is a contradiction. Arω 1

19 354 L. M. BROWN Finally, we must show that r = F f, R = f F following by taking the inverse of both sides. Further, since r, F f P(), it will be sufficient to show rω 1 = (F f)ω 1. Now by Proposition 2.6 (1) and R b for F we have (F f)ω 1 = f(f )ω 1 = fωu 1 (F )ω 1, whence using Propositions 2.5 (1), 2.10 (2) and the idempotency of ωu 1 we obtain (F f)ω 1 ϕω 1ϕ 1 ω 1 = ω 1 If s ω 1 ϕω 1 U ϕ 1 ω 1 t we have s, t satisfying s ω 1 s, t ω 1 t and ϕ(s ) ω 1 U ϕ(t ). We deduce s rω 1 rω 1 t from Corollary 3.9 (1), whence s rω 1 t. Conversely let s rω 1 t. Then s ω 1 rω 1 ω 1 t, so we have s, t with s ω 1 s, s rω 1 t and t ω 1 t. By Corollary 3.9 (2) we have ϕ(s ) ωu 1 ϕ(t ), so s ω 1 ϕω 1 U ϕ 1 ω 1 t. This shows r = F f, and the proof is complete. U. Example For a discrete texture (X, P(X)) the interior relation is just the identity relation on X, so (r, R) is an equivalence direlation if and only if the point relation r is reflexive and transitive, and R = r = (r c ) 1. Clearly ρ = r r 1 and (f, F ) = (ϕ, ϕ c ). Finally ϕ(x) ωu 1 ϕ(y) x r y, so (U, U) will be a discrete texture if and only if r = ρ, that is if and only if r is also symmetric as a point relation. In the reverse direction we have: Theorem Let (, ), (T, T) be textures and (g, G) a difunction on (, ) to (T, T). Then (r, R) defined by r = G g and R = g G is an equivalence direlation on (, ). Moreover, if (U, U) is the quotient texture associated with (r, R) as in Theorem 3.5, and (f, F ) the canonical quotient difunction from (, ) to (U, U), then (h, H) = (g, G) (f, F ) is an injective difunction on (U, U) to (T, T) satisfying (h, H) (f, F ) = (g, G). Finally, if (g, G) is surjective then (h, H) is bijective. (, ) (g,g) (T, T) (f,f ) (h,h) (U, U)

20 QUOTIENT OF TEXTURE 355 Proof. ince r = (G g) = g (G ) = g G = R by [6, Proposition 2.17 (2) and Lemma 2.4 (2)], we see (r, R) is symmetric. ince (g, G) is a difunction we have R = g G I by [6, Lemma 2.23 (1 ii)], whence i = I R = r and the direlation (r, R), is reflexive. To show that (r, R) is transitive note first that r r = (G g) (G g) = G [(g G ) g] by [6, Proposition 2.17 (3)]. ince (g, G) is a difunction, g G i T by [6, Lemma 2.23 (2 ii)]. Hence by [6, Proposition 2.17 (4)] we have G [(g G ) g] G (i T g), and i T g = g by [6, Proposition 2.17 (1)], so r r G g = r. R R R follows by symmetry, whence (r, R) is transitive and hence an equivalence direlation. Let us now show that (g, G) (f, F ) = (g F, G f ) is a difunction. For B T we have g B = G B R by [6, Theorem 2.24] and the fact that r (g B) = (G g) (g B) = G (g (g B)) G B = g B by [6, Lemma 2.9 (1), Lemma 2.7], which proves r (g B) = g B since r is reflexive. Now, using Lemma 2.8 (2) we have (g F ) B = F (g B) = f (g B) = f (G B) = (G f ) B which gives the required result by [6, Theorem 2.24]. To show (g, G) (f, F ) is injective we must show that ((g, G) (f, F ) ) = (f G, F g ) satisfies DF2. Note that for B T we have (f G ) ((F g ) B) = f (G g) (F B)) = f (r (F B)). ince F B = ϕ 1 [B] R we have r (F B) = F B, while f (F B) B by [6, Theorem 2.24 (2 b)]. Hence (f G ) ((F g ) B) B and so (f G, F g ) satisfies DF2 by [6, Lemma 2.23 (2)]. It remains to show that if (g, G) is surjective, then so is (g, G) (f, F ). But in this case the direlation (g, G) satisfies DF1 by [6, Theorem 2.31 (1)], whence [(g, G) (f, F ) ] = (f, F ) (g, G) also satisfies DF1 since (f, F ) does. But now (g, G) (f, F ) is surjective, again by [6, Theorem 2.31 (1)], and the proof of the theorem is complete.

21 356 L. M. BROWN Example If (g, G) : (X, P(X)) (Y, P(Y )) is a difunction then g : X Y is a point function, r = gg 1 is symmetric as a point relation, and x ρ x g(x) = g(x ). Hence in this case Theorem 3.12 gives the discrete texturing on the usual quotient associated with g. Naturally, difunctions (g, G) from (X, P(X)) to non-discrete textures will generally produce quotients of (X, P(X)) which are not discrete. The following result will also prove useful: Theorem For k = 1, 2 let ( k, k ) be a texture, (r k, R k ) an equivalence direlation on ( k, k ), (U k, U k ) the corresponding quotient texture and (g, G) : ( 1, 1 ) ( 2, 2 ) a difunction which is compatible in the sense that A R 2 = g A R 1. Then there exists a difunction (ḡ, Ḡ) : (U 1, U 1 ) (U 2, U 2 ) making the following diagram commutative. ( 1, 1 ) (g,g) ( 2, 2 ) (f 1,F 1 ) (f 2,F 2 ) (U 1, U 1 ) (ḡ,ḡ) (U 2, U 2 ) Proof. Consider the mapping θ : U 2 U 1 defined by θ(a) = ϕ 1 [g ϕ 1 2 [A]] for all A U 2. We have noted above that the mapping A ϕ 1 2 [A] from U 2 to R 2, being an isomorphism, preserves arbitrary intersections and joins. By hypothesis, for B R 2 we have g B R 1, so B g B is a mapping from R 2 to R 1, and this too preserves arbitrary intersections and joins by [6, Corollary 2.26]. Finally the mapping C ϕ 1 [C] from R 1 to U 1 is an isomorphism and therefore preserves arbitrary intersections and joins. Thus, the same is true of the mapping θ, and by [7, Proposition 4.1] we deduce the existence of a difunction (ḡ, Ḡ) : (U 1, U 1 ) (U 2, U 2 ) satisfying ḡ A = Ḡ A = θ(a) for all A U 2. It remains to show the commutativity of the above diagram. We will establish ḡ f 1 = f 2 g, leaving the dual proof of Ḡ F 1 = F 2 G to the interested reader. It will be sufficient to show that (ḡ f 1 ) = (f 2 g), whence by [6, Lemma 2.7] we need only show

22 QUOTIENT OF TEXTURE 357 that (ḡ f 1 ) A = (f 2 g) A for all A U 2. However (ḡ f 1 ) A = f 1 (ḡ A) = ϕ 1 1 [θ(a)] which gives the required result. = ϕ 1 1 [ϕ 1[g ϕ 1 2 [A]] = g ϕ 1 2 [A] = g (f 2 A) = (f 2 g) A, We end this section by presenting a method for generating equivalence direlations. Proposition Let (, ) be a texture, B and define the direlation (r, R) on (, ) by r = {P (s,t) u with P s Q u and P t B or B Q u B B}, R = {Q (s,t) v with P v Q s and P v B or B Q t B B}. Then: (1) (a) s rω 1 t u sω 1, v ω 1 t satisfying u Bω 1 = v Bω 1 B B, (b) (s, t) R c v ω 1 s, u tω 1 satisfying u Bω 1 = v Bω 1. (2) (r, R) is an equivalence direlation on (, ). (3) B generates the set R associated with (r, R) in the sense that B R and every element of R can be written as an intersection of joins of elements of B. Proof. (1) If s rω 1 t then r Q (s,t), which gives P t Q t, P s Q u satisfying B Q u = P t B. Taking P t Q v, P v Q t now gives u sω 1, v ω 1 with u Bω 1 = v Bω 1, and the converse is easily established. This shows (a), and the proof of (b) is similar and is left to the reader. (2) If s ω 1 t then s ω 1 ω 1 t so we have u with s ω 1 u, u ω 1 t, and taking v = u in (1 a) gives s rω 1 t. Hence ω 1 rω 1. uppose s rω 1 rω 1 t and take w with s rω 1 w, w rω 1 t. Now by (1 a) we have u sω 1, v ω 1 w and u wω 1, v ω 1 t satisfying the appropriate implications. Hence u Bω 1 = v Bω 1 = v Bω 1, which shows that s rω 1 = u Bω 1 t by (1 a), and so rω 1 ω 1 ω 1 = Bω 1 rω 1 rω 1.

23 358 L. M. BROWN Finally from (1) we note that s rω 1 t t Rc s so rω 1 = (R c ) 1 = (R )ω 1 by Proposition 2.5 (1), whence r = R and so R = r. By Corollary 3.2 this completes the proof that (r, R) is an equivalence direlation. (3) Take B 0 B. By Lemma 3.7 (3), B 0 R if and only if B 0 rω 1 = B 0 ω 1, and since ω 1 rω 1 it will clearly suffice to show B 0 rω 1 B 0 ω 1. Take t B 0rω 1. Then for some b B we have b rω 1 t, so there exists u bω 1, v ω 1 t with u Bω 1 = v Bω 1 for all B B by (1 a). But u B 0 ω 1 so v B 0 ω 1, whence t B 0ω 1 ω 1 = B 0 ω 1, as required. This establishes B R. Finally, we take A R. uppose that P t A. Then P t r A since r A = A for A R, so we have t with P t Q t satisfying r Q (s,t ) = A Q s. Take any s with A Q s and take s with A Q s, P s Q s. By the above implication r Q (s,t ), whence P (s,t) r and we may choose Bs t B satisfying P t Bs t Q s. Let B t = {Bs t A Q s }. Then A B t. Indeed, if not we shall have s satisfying A Q s, P s B t and so the contradiction P s Bs t Bs. t Hence A {B t P t A}. If we do not have equality we may take w satisfying {B t P t A} Q w, P w A. In particular this gives B w Q w, so there exists s satisfying Bs w Q w, which contradicts P w Bs w. This shows that A = {B t P t A}, so A has the required representation. Corollary If (r, R) is the equivalence direlation generated by B as in Proposition 3.15, the elements of B are saturated for (r, R). Proof. Immediate from Lemma 3.7 (3). Example Consider the texture (L, L) of Examples 1.1 (2), and set B = {(0, 1 2 ]}. Then the corresponding equivalence direlation is easily seen to be: r = {(s, t) 0 < t 1 2 or 1 2 < s 1}, and R = {(s, t) 0 < t 1 2 or 1 2 s 1}.

24 QUOTIENT OF TEXTURE 359 r R Clearly, R = {, (0, 1 2 ], L}, while sρt 0 < s 1 2, 0 < t 1 2 or 1 2 < s 1, 1 2 < t 1. Hence there are two equivalence { classes which we denote by 1 2 and 12 0 < s 1 1. This gives us ϕ(s) = < s 1, U = { 2 1, 1} and U = {, { 1 2 }, U}. The following result will be useful when applying Theorem 3.14 to equivalence direlations obtained as in Proposition Lemma Let (r k, R k ) be equivalence direlations on ( k, k ) generated by B k k, k = 1, 2 as in Proposition Then if the difunction (g, G) : ( 1, 1 ) ( 2, 2 ) satisfies B B 2 = g B R 1 it satisfies A R 2 = g A R 1. Proof. By Proposition 3.15, A R 2 may be written as an intersection of joins of sets of B 2. On the other hand the mapping (g ) preserves arbitrary intersections and joins, so by hypothesis g A = (g ) A can be written as an intersection of joins of elements of R 1. However, R 1 is closed under arbitrary intersections and joins by Lemma 3.7 (5), whence g A R Quotient Ditopologies and Quotient Difunctions Now let (τ, κ) be a ditopology on (, ). If, as above, (r, R) is an equivalence direlation on (, ), (U, U) the corresponding quotient texture and (f, F ) : (, ) (U, U) the canonical quotient difunction then clearly τ U = {H U F H τ} is a topology and κ U = {K U f K κ} a cotopology on (U, U). Definition 4.1. With the notation above, (τ U, κ U ) is called the quotient ditopology of (τ, κ) on (U, U), and (U, U, τ U, κ U ) the quotient ditopological texture space of (,, τ, κ) modulo (r, R).

25 360 L. M. BROWN Clearly the quotient ditopology is the finest ditopology on the quotient (U, U) making the canonical quotient difunction bicontinuous. Just as for quotients of topological spaces we have: Proposition 4.2. Let (U, U, τ U, κ U ) be a quotient of (,, τ, κ) and (f, F ) the canonical quotient difunction. If (g, G) : (U, U, τ U, κ U ) (,, τ, κ ) is a difunction, then (g, G) is continuous (cocontinuous, bicontinuous) if and only if (g, G) (f, F ) is continuous (respectively, cocontinuous, bicontinuous). Proof. This follows from the equalities (g f) A = f (g A) and (G F ) A = F (G A) for all A, established in [6, Lemma 2.16 (2)]. Before proceeding it will be useful to have a notion of homeomorphism for difunctions. Definition 4.3. An isomorphism in the category dftex of textures and difunctions which, together with its inverse, is bicontinuous, will be called a difunctional homeomorphism, or dihomeomorphism for short. Dihomeomorphisms, therefore, coincide with the isomorphisms of the category dfditop of ditopological texture spaces and bicontinuous difunctions. It is proved in [6, Proposition 3.14 (5)] that a difunction (h, H) is an isomorphism if and only if it is bijective, and it is clear from the proof of that proposition that the inverse is then the inverse direlation (h, H) = (H, h ). The following gives several useful characterizations of a dihomeomorphism, and is the textural analogue of [10, Proposition ]. Proposition 4.4. Let (h, H) : ( 1, 1, τ 1, κ 1 ) ( 2, 2, τ 2, κ 2 ) be a bijective difunction. Then the following are equivalent. (i) (h, H) is a dihomeomorphism. (ii) (h, H) is bicontinuous, open and coclosed. (iii) (h, H) is bicontinuous, closed and co-open. (iv) h A = H A τ 2 A τ 1, and h A = H A κ 2 A κ 1. (v) h B = H B τ 1 B τ 2, and h B = H B κ 1 B κ 2.

26 QUOTIENT OF TEXTURE 361 Proof. (i) = (ii) The inverse (H, h ) of (h, H) is continuous, so U 1 τ 1 = h U 1 = ((h ) ) U 1 = (h ) U 1 τ 2, which shows that (h, H) is open. Likewise, the cocontinuity of (H, h ) implies the coclosedness of (h, H). (ii) = (iii) By [6, Corollary 2.33 (1, 2)] we see that since (h, H) is bijective, h A = H A for all A 1. Hence (h, H) is closed if and only if it is coclosed, and open if and only if it is co-open. (iii) = (iv) A τ 1 = H A τ 2 since (h, H) is co-open, and we have already noted that h A = H A since (h, H) is bijective. On the other hand if h A = H A τ 2 then A = H (h A) τ 1 by the continuity of (h, H), the injectivity of (h, H) and [6, Corollary 2.33 (2)]. The proof of the second result is dual to this and is omitted. (iv) = (v) Take B 2 and set A = H B. Then, since (h, H) is surjective, B = h (H B) = h A by [6, Corollary 2.33 (1)]. If B τ 2 then h A τ 2 and we deduce H B = A τ 1 by (iv), and we know that H B = h B since (h, H) is a difunction. On the other hand if we assume that h B = H B τ 1 then A τ 1 and by (iv) we have B = h A τ 2. The proof of the remaining results is dual to this, and is omitted. (v) = (i) It is clear from (v) that (h, H) is bicontinuous. To show (H, h ) is continuous take U 1 τ 1. Then U 1 = H (h U 1 ) τ 1 by [6, Corollary 2.33 (2)] since (h, H) is injective, whence h U 1 τ 2 by (v) applied to B = h U 1. But, as noted above, h U 1 = (h ) U 1, so (h, H ) is continuous. Likewise (H, h ) is cocontinuous and we have shown that (h, H) is a dihomeomorphism. Now let ( k, k, τ k, κ k ), k = 1, 2 be ditopological texture spaces and (g, G) : ( 1, 1, τ 1, κ 1 ) ( 2, 2, τ 2, κ 2 ) a bicontinuous difunction. As in Theorem 3.12, (g, G) gives rise to a quotient texture (U, U), a canonical quotient difunction (f, F ) and an injective difunction (h, H) from (U, U) to ( 2, 2 ) satisfying (h, H) (f, F ) = (g, G). We take the quotient ditopology (τ U, κ U ) on (U, U).

27 362 L. M. BROWN (g,g) ( 1, 1, τ 1, κ 1 ) ( 2, 2, τ 2, κ 2 ) (h,h) (f,f ) (U, U, τ U, κ U ) By Proposition 4.2 we see that the difunction (h, H) is bicontinuous. In case (g, G) is surjective (h, H) is a bijection, but in general it need not be a dihomeomorphism because the inverse may not be bicontinuous. Definition 4.5. (g, G) : ( 1, 1, τ 1, κ 1 ) ( 2, 2, τ 2, κ 2 ) is called a quotient difunction if it can be expressed as the composition of a canonical quotient difunction on ( 1, 1, τ 1, κ 1 ) and a dihomeomorphism onto ( 2, 2, τ 2, κ 2 ). Proposition 4.6. The following are equivalent for a surjective difunction (g, G) from ( 1, 1, τ 1, κ 1 ) to ( 2, 2, τ 2, κ 2 ) : (1) The difunction (g, G) is quotient. (2) G B τ 1 B τ 2 and g B κ 1 B κ 2. (3) With the notation as in Theorem 3.12 and the commutative diagram above, the difunction (h, H) is a dihomeomorphism. Proof. (1) = (2) Let (g, G) = (k, K) (f, F ), where (f, F ) : ( 1, 1, τ 1, κ 1 ) (U, U, τ U, κ U ) is a canonical quotient difunction and (k, K) : (U,, τ U, κ U ) ( 2, 2, τ 2, κ 2 ) a dihomeomorphism. By the definition of (τ U, κ U ) we have G B = F (K B) τ 1 K B τ U, while by Proposition 4.4 (v) we have K B τ U B τ 2. The proof of the second equivalence is dual to this and is omitted. (2) = (3) ince (h, H) is bijective because (g, G) is surjective we need only verify Proposition 4.4 (iv). Take A U. Then G (h A) = (H F ) (h A) = F (H (h A)) = F A by [6, Corollary 2.33 (2)] since (h, H) is injective. Hence A τ U F A τ 1 G (h A) τ 1 h A τ 2 by (2). In just the same way A κ U H A κ 2. (3) = (1) Immediate.

28 QUOTIENT OF TEXTURE 363 Corollary 4.7. A composition of quotient difunctions is a quotient difunction. On the other hand if (g 1, G 1 ), (g 2, G 2 ) are bicontinuous difunctions whose composition (g 2, G 2 ) (g 1, G 1 ) is a quotient difunction then (g 2, G 2 ) is a quotient difunction. Proof. The first statement is clear from Proposition 4.6 (2). If (g 2 g 1, G 2 G 1 ) is a quotient difunction it is surjective so by [6, Theorem 3.32 (1)] we have i (g 2 g 1 ) (G 2 G 1 ) = g 2 (g 1 G 1 ) G 2 g 2 G 2 since g 1 G 1 i by [6, Lemma 2.23 (2)] applied to the difunction (g 1, G 1 ). Hence (g 2, G 2 ) is surjective. ince (g 1, G 1 ), (g 2, G 2 ) are bicontinous the result now follows easily from Proposition 4.6. Corollary 4.8. An injective quotient difunction is a dihomeomorphism. Proof. In view of Proposition 4.6 (3) it will be sufficient to show that the canonical quotient difunction (f, F ) of the quotient generated by a bijective difunction (g, G) as in Theorem 3.12 is a dihomeomorphism. But since, as we have noted before, (G, g ) is now the inverse of (g, G) we have r = G g = i and R = g G = I. Lemma 3.7 (1) now gives sρt P s P t and P t P s P s = P t s = t. It follows that the quotient (U, U, τ U, κ U ) may be identified with (,, τ, κ), ϕ becomes the identity and so (f, F ) = (i, I), which is a dihomeomorphism. Corollary 4.9. urjective bicontinuous difunctions which are open or co-open, and closed or coclosed, are quotient difunctions. Proof. Clear from Proposition 4.6 (2) and [6, Corollary 2.33 (1) and Theorem 2.24 (3)]. Proposition Let (r, R) be an equivalence direlation on (,, τ, κ) and (f, F ) the canonical quotient difunction. Then (f, F ) is: (1) open if and only if A τ = r A τ. (2) co-open if and only if A τ = R A τ. (3) closed if and only if A κ = r A κ. (4) coclosed if and only if A κ = R A τ.

29 364 L. M. BROWN Proof. We establish (1), leaving the other cases to the interested reader. By Theorem 3.5 we have r = F f, so for A we have r A = (F f) A = F (f A). By the definition of the quotient topology we deduce that r A τ f A τ U, from which (1) follows at once. Corollary Let (g, G) : ( 1, 1, τ 1, κ 1 ) ( 2, 2, τ 2, κ 2 ) be a quotient difunction. Then (g, G) is: (1) open if and only if A τ 1 = G (g A) τ 1. (2) co-open if and only if A τ 1 = G (G A) τ 1. (3) closed if and only if A κ 1 = g (g A) κ 1. (4) coclosed if and only if A κ 1 = g (G A) κ 1. Proof. With the notation as in Theorem 3.12 we have (g, G) = (h, H) (f, F ), where (f, F ) is the canonical quotient difunction for the equivalence direlation (r, R) given by r = G g, R = g G, and (h, H) is a dihomeomorphism by Proposition 4.6. Hence (g, G) and (f, F ) share the properties in question, so the results follow at once from Proposition Definition An equivalence direlation (r, R) on a ditopological texture space is called open (co-open, closed, coclosed) if the canonical quotient difunction is open (co-open, closed, coclosed). We recall from [8] that (,, τ, κ) is T 1 if each set A can be written in the form A = j J K j with K j κ, j J; and that it is co-t 1 if each A can be written in the form A = j J G j with G j τ, j J. Proposition Let (,, τ, κ) be a ditopological texture space and (r, R) an equivalence direlation on (, ). (1) If (,, τ, κ) is T 1 and (r, R) is closed then every set in R can be written as a join of closed sets in R. (2) If (,, τ, κ) is co-t 1 and (r, R) is co-open then every set in R can be written as an intersection of open sets in R. Proof. (1) For K j κ, j J, we have r ( j J K j) = j J r K j by [6, Corollary 2.12 (2)], and r K j κ by Proposition 4.10 (3). ince R = {r A A } by Lemma 3.7 (2,4), the required result follows at once. (2). Dual to (1).

30 QUOTIENT OF TEXTURE The T 0 reflector We begin by showing that every ditopological texture space has a quotient that is a T 0 ditopological texture space. Let us recall the following characteristic property of T 0 ditopological spaces [8]. Definition 5.1. The ditopology (τ, κ) on (, ) is T 0 if given Q s Q t there exists B τ κ satisfying P s B Q t. Now we may give: Theorem 5.2. Let (,, τ, κ) be a ditopological texture space, (r, R) the equivalence direlation generated by B = τ κ as in Proposition 3.15, (U, U) the quotient texture of (, ) with respect to (r, R) and (τ U, κ U ) the quotient ditopology on (U, U). Then the quotient ditopological space (U, U, τ U, κ U ) is T 0. Proof. Take s, t with Q s Q t. By Lemma 3.8 (ii) we have Q s = {ϕ[r Q s ] P s Q s } and so there exists s with P s Q s satisfying ϕ[r Q s ] Q t. On the other hand Q t U so we have A R with ϕ[a] = Q t, whence ϕ[r Q s ] ϕ[a] and so R Q s A. Now we have w with P w A for which P (z,w) R = P z Q s. If we choose s with P s Q s, P s Q s, and set z = s in the above implication we obtain P (s,w) R. If now take w with P w Q w and P w A we obtain R Q (s,w ). From the definition of R, and bearing in mind that P s Q s, we see that there exists B B = τ κ satisfying P s B Q w. From P s / B, and the fact that B is saturated with respect to ρ gives s / ϕ[b], that is P s ϕ[b]. Also, B A and since A and B are both saturated with respect to ρ we obtain ϕ[b] ϕ[a] = Q t. Finally, ϕ[b] ϕ[b] = ϕ[τ κ] = τ U κ U, and we have established that (U, U, τ U, κ U ) is T 0. We obtain the subcategory dfditop 0 of dfditop by restricting to T 0 ditopological texture spaces. Theorem 5.3. The category dfditop 0 is a full reflexive subcategory of dfditop. Proof. It is clear that dfditop 0 is a full subcategory of dfditop. Take (,, τ, κ) ObdfDitop and let (f, F ) : (,, τ, κ) (U, U, τ U, κ U ) be the canonical difunction onto the T 0 quotient.

31 366 L. M. BROWN We claim (f, F ) is a reflection [1]. To show this take ( 0, 0, τ 0, κ 0 ) ObdfDitop 0 and (g, G) dfditop((,, τ, κ), ( 0, 0, τ 0, κ 0 )). We must show the existence of a dfditop 0 morphism (ḡ, Ḡ) making the following diagram commutative. (f,f ) (,, τ, κ) (U, U, τ U, κ U ) (ḡ,ḡ) (g,g) ( 0, 0, τ 0, κ 0 ) Let us now show that since ( 0, 0, τ 0, κ 0 ) is T 0 the corresponding equivalence direlation (r 0, R 0 ) is the identity (i 0, I 0 ) on ( 0, 0 ). By symmetry it will be sufficient to show that R 0 = I 0, and by reflexivity that I 0 R 0. uppose that I 0 R 0 and take s, t with I 0 Q (s,t) and P (s,t) I 0. The first result leads to Q s Q t, whence we have B 0 τ κ satisfying P s B 0 Q s. From the second result we have t with P (s,t) Q (s,t ), and then v with P v Q s for which P v B or B Q t for all B τ κ. But P t Q t gives Q t Q t, and P v Q s gives P s P v, so P s B or B Q t B τ κ. Taking B = B 0 now gives a contradiction, so (r 0, R 0 ) = (i 0, I 0 ) as required. Arguing as in the proof of Corollary 4.8 we may identify the quotient with ( 0, 0, τ 0, κ 0 ), and the canonical quotient difunction becomes (i 0, I 0 ). Also, the bicontinuity of (g, G) implies that g (τ 0 κ 0 ) τ κ = B R, so by Lemma 3.18, Theorem 3.14 gives a difunction (ḡ, Ḡ) making the diagram (,, τ, κ) (f,f ) (U, U, τ U, κ U ) (g,g) ( 0, 0, τ 0, κ 0 ) (i 0,I 0 ) (ḡ,ḡ) ( 0, 0, τ 0, κ 0 ) commutative. Finally, let us show that (ḡ, Ḡ) is bicontinuous. For A 0 we have ḡ A = Ḡ A = ϕ[g ϕ 1 0 [A]] = ϕ[g A] by the