The basics of frame theory

Transcription

1 First version released on 30 June 2006 This version released on 30 June 2006 The basics o rame theory Harold Simmons The University o Manchester hsimmons@ manchester.ac.uk This is the irst part o a series o notes [3, 4, 5, 6, 7] on rames, with several more to come. In this part I set down the basic deinitions and properties, give a ew examples, and begin the discussion o nuclei. The last section looks at various categorical properties, and could be omitted at a irst reading. 1 Ater that you should read [4] and [5] in the order you preer. That will give you a good grounding in the subject. The whole collection can be ound at [1] I you have never seen this collection beore then perhaps you should read [2]. Contents 1 The category o rames The basic notions Topologies Complete boolean algebras The implication on a rame Morphisms as poset adjunctions The universal algebra o rames Quotients o rames Quotients in Set Quotients in Sup Quotients in Frm Nuclei on rames Various relections Relections in general Relection rom Meet Relection rom Dlt Relection rom Sup Relection rom Set Reerences Index The category o rames In this section we irst set up the deinition o a rame and a rame morphism, we then look at two important classes o rames, and inally we sort out some useul gadgetry. 1 It might make more sense to make Section 5 the bulk o another set o notes on relections in general or various kinds o structured posets. 1

2 1.1 The basic notions Is there a better way to start than with a deinition? 1.1 DEFINITION. A rame is a structure where (A, ) is a complete poset (A,,, ) is a -semilattice (A,,, ) is a -semilattice and where these satisy (A,,,,, ) (FDL) a X = {a x x X} or each a A and X A. This is the Frame Distributive Law. A rame morphism A B between rames A, B is a unction : A B which preserves the distinguished attributes. Let Frm be the category o rames and rame morphisms. We must understand exactly what this deinition says so it is worth looking at some o its details. A rame A is carried by a poset (A, ) which is complete, that is both X X exist or each subset X A. In particular, A has both extremes = = A A = = the top and the bottom. Three o these, the two extremes and are selected as distinguished attributes. However, is not a distinguished attribute o the rame, but the binary join is. This signature, the selected attributes, determines the notion o a rame morphism. This is a unction, as indicated, which preserves the comparison, in other words it is monotone, and satisies ( ) = (x y) = (x) (y) ( ) = ( X) = (X) 2

3 or each x, y A and X A (where A is the source o the morphism). In particular we can have ( X) (X) or X A. We will see a general reason or this in Subsection 1.2. For each unction : A B we write or the image o (X) = {(x) x X} (Y ) = {x A (x) Y } direct X A inverse Y B across, respectively. As well as carrying these attributes a rame is required to satisy FDL. As a particular case o this we may take a couple X = {x, y} o elements to obtain a (x y) = (a x) (a y) and hence the rame is a distributive lattice. Consequently the rame also satisies or a, x, y A. However, it can happen that a (x y) = (a x) (a y) a X {a x x x} or a A and X A. We will see an example o this in Subsection 1.2. The import o the FDL can be expressed in a dierent way. We look at this in Subsection 1.4. As we have seen, each rame is both a -semilattice and a -semilattice. More precisely, there is a pair Frm Meet Frm Sup o orgetul unctors. We take a closer look at these in Section 5. The precise deinition o Meet is given in Subsection 5.2, and that o Sup slightly earlier in Subsection 3.2. It s time to look at two substantial examples. We make each o these a subsection. 1.2 Topologies Historically, topologies ormed the motivating examples o rames. Beore we look at them let s ix some notation which we used throughout these notes, and other sets in the series. Let S be a topological space. Thus we have two amilies OS CS 3

4 o subsets o S, the open closed subsets o S, respectively. Here OS is the topology. We also have two operations ( ) ( ) on subsets o S, providing the interior closure o a subset o S, respectively. Any o these our gadgets uniquely determines the topology. We also write ( ) or the set theoretic complementation on S. This, o course, is independent o the particular topology on S. To be a topology the amily OS must contain the two extremes S and, and be closed under binary intersections and arbitrary unions. This almost shows that (OS,,, S,, ) is a rame. All that is needed is the observation that FDL holds since OS sits inside the power set PS and the relevant operations are computed set theoretically. This is a useul way i motivating the distinguished attributes o a rame. We know that or U OS the intersection U need not be open. The inimum o U in OS is given by U = ( U) using the interior operation o the space. Note also that or a topology OS we need not have W U = {W U u U} or W OS and U OS. Thus the opposite o FDL need not hold in a rame. Consider a continuous map T φ between a pair spaces S, T. By deinition o continuity we have φ (U) OT or each U OS. A couple o simple calculations shows that the assignment S OS φ OT is a rame morphism. In particular, a rame morphisms need not preserve arbitrary inima. This construction sets up a contravariant unctor Top O Frm rom spaces to rames. I you are new to this game you should go through the various calculations missing here. In [5] we show that O is one hal o a contravariant adjunction between Frm and Top. This connection enables many topological properties to be analysed using rame theoretic methods. That is what point-ree topology is about. 4

5 1.3 Complete boolean algebras In this subsection we isolate a large subcategory o Frm. A boolean algebra is a distributive lattice A or which each element is complemented. That is, or each a A there is a, necessarily unique, element b A such that a b = a b = hold. We write a or the element b, so that ( ) is a 1-placed operator on A. Notice that a = a (or i b is the complement o a, then a is the complement o b). Notice that any lattice morphism A B rom a boolean algebra A preserves complements, that is i a, b are complementary o A, then (a), (b) are complementary in B. The operation ( ) on a boolean algebra A has various properties. In particular we have a x y x a y or a, x, y A. The proo o this is more important than the result. To prove we remember a a = so that x = ( a a) x = ( a x) (a x) a y using a x y at the inal step. A similar argument starting rom a a = gives the converse implication. This kind o trick is used several times in this subsection. A complete boolean algebra is a boolean algebra which is complete as a poset. Such an algebra has all the attributes to be a rame, but what about FDL? 1.2 LEMMA. Each complete boolean algebra is a rame. Proo. Let A be a cba. Certainly A is a complete lattice, so it suices to show that A satisies the FDL. To this end let l = a X r = {a x x X} or a A and x A. The comparison r l is trivial, so it remains to veriy the converse. We use the equivalence given above. For x X we have a x l so that x a l 5

6 to give X a l and hence r l, as required. In any boolean algebra the interaction between meets, joins, and complementation is given by the De Morgan laws. There are similar laws or complete boolean algebras. We need one o these. 1.3 LEMMA. For an arbitrary subset X o a complete boolean algebra A we have X = ( Y ) where Y = { x x X} is the auxiliary subset. Proo. Let a = X b = Y so that it suices to show that a, b are complementary, that is a b = a b = hold. For each y = x Y we have a y x x = and hence, since A is a rame, to give the let hand equality. Let c = b. For each x X we have so that c x. Thus c a, and hence a b = {a y y Y } = c x c b = a b c b = to give the right hand equality. Complete boolean algebras orm the objects o two dierent categories, Cba and CBA. Given two such algebras A, B and arrow A B 6

7 in Cba is simply a lattice morphism, that is a unction that preserves the initary attributes. As we saw above, such a morphism preserves complements. An arrow in CBA, sometime called a complete morphism, preserves arbitrary inima and suprema, that is ( X) = (X) ( X) = (X) or each subset X o the source. Each object o CBA is a special kind o rame, and each arrow o CBA is a special kind o rame morphism. Thus we have a orgetul unctor CBA Frm this time with Frm as the target, not the source. You might think that there is a third category where the objects are complete boolean algebras, and the arrows are the rame morphisms between these algebras. However, the next result shows that this is just CBA. 1.4 LEMMA. Let A B be a rame morphism rom a complete boolean algebra A to a rame B. Then is a complete morphism, that is ( X) = (X) or each X A. Proo. Let b = ( X) d = (X) or the given subset X. The monotonicity o ensures that b d, so it suices to show the converse comparison. Let Y = { x x X} c = ( Y ) so that ( Y ) = X (by Lemma 1.3), and hence b and c are complementary in B. For each y = x Y we have d (y) (x) ( x) = since passes across. Also passes across so that and hence c = {(y) y Y } d c = {d (y) y Y } = by the FDL in B. Finally, remembering that c b = 7

8 we have as required. d = d (c b) = (d c) (d b) = d b b For each pair A, B o complete boolean algebras, we certainly have an inclusions CBA[A, B] Frm[A, B] o arrow sets (since each complete morphism is a rame morphism). Lemma 1.4 shows that, in act, this inclusion is an equality. Thus we have the ollowing. 1.5 THEOREM. The category CBA o complete boolean algebras and complete morphisms is a ull subcategory o the category Frm o rames. In [7] we investigate just how CBA sits inside Frm. 1.4 The implication on a rame A rame is a complete lattice which satisies the FDL. This equational requirement can be codiied in a dierent way, and produces a useul tool. 1.6 DEFINITION. An implication on a -semilattice A is a two placed operation ( ) such that x (b a) b x a or all a, b, x A. Trivially, at most one implication can be carried by a -semilattice. Carrying one enorces certain properties. 1.7 LEMMA. A complete lattice A carries an implication precisely when it is a rame. Proo. Suppose irst that A is a rame. For a, b A set so that (b a) = {x A b x a} b x a = x (b a) (or arbitrary x A). We require the converse. But b (b a) = {b x b x a} a by the FDL, and this leads to the required result. Conversely, suppose A carries an implication, and consider any a A and X A. It suices to show a X {a x x X} (since the converse comparison is trivial). Let y = {a x x X} 8

9 so that a x y or each x X. The implication property gives x (a y) or each such x, so that X (a y) and hence by a second use o the implication property. a X y This shows that a rame is exactly the same thing as a complete heyting algebra. However, that description can be misleading since rame morphisms need not preserve implication. It is worth looking at two examples o implication. 1.8 EXAMPLES. (a) For a topology OS on a space S the implication is given by (V U) = (V U) or U, V OS. To see this consider an arbitrary W OS. Then W (V U) W (V U) W V U to give the required result. We oten use a topology OS to illustrate various aspects o rame theory. Sometimes we use OS directly, but sometimes it is more enlightening to look at CS. This oten requires the insertion o a complementation at appropriate places. In this particular case, or X, Y CS we have (Y X ) = (Y X ) = (X Y ) which is clearly important in some situations. (b) For a complete boolean algebra A the implication is given by (b a) = b a using the negation on A. In other words we have x b a b x a or a, b, x A. This is the observation made just beore Lemma 1.2. We will return to the idea o a negation shortly. There are many identities involving the implication. Some o these are given in the next two results. It isn t worth remembering most o these, but the proo technique repeated use o the characterizing equivalence is important. 9

10 1.9 LEMMA. On a rame A we have (i) a (x a) (ii) x (x a) = x a (iii) x y = (y a) (x a) (iv) (( X) a) = {(x a) x X} or all a, x, y A and X A. Proo. (i). Since x a a, this is immediate. (ii). For each z A we have z x (x a) z x and z (x a) z x and x z a z a or the required result. The last step here requires just a ew moment s thought. (iii). Let z = (x a) where x y. Then so that as required. (iv). For each z A we have x z y z a z (x a) z ( X) a z X a {z x X} a ( x X)[z x a] ( x X)[z (x a)] z {(x a) x X} or the required result. notice the use o FDL at the second step. Part (iv) o this result is one o the ew identities worth remembering. Notice that by taking X = {x, y} we obtain ((x y) a) = (x a) (y a) as a particular case. The next deinition may look a little strange but, as we will learn, the operator produced is very important DEFINITION. For a rame A we set w a (x) = ((x a) a) or each a, x A to produce an operator w a on A. 10

11 We will use these operators quite a lot. Here are their simple properties LEMMA. For each element a o a rame A, the operator w a is inlationary, monotone, and idempotent, and satisies or each x, y A. (w a (x) a) = w a (x a) = (x a) w a (x y) = w a (x) w a (y) Proo. For each x A a use o Lemma 1.9(ii) gives so that x (x a) = x a a x ((x a) a) = w a (x) to show that w a is inlationary. Two uses o Lemma 1.9(iii) shows that w a is monotone. Beore we show that w a we veriy the let hand identity. To this end let z = (w a (x) a) = w a (x a) where the right hand equality is an immediate consequence o the deinition o w a. Since x w a (x), we have x z w a (x) z a so that or the required result. With this we have to show that w a is idempotent. Finally, or x, y A we have z (x a) w a )x a) = z w 2 a (x) = ((w a(x) a) a) = ((x a) a) = w a (x) w a (x y) w a (x) w a (y) since w a is monotone. This a converse comparison will give the remaining required result. To this end let z = w a (x) w a (y) so that to give z w a (x) z (x a) a and hence, using Lemma 1.9(iv), we have z w a (y) z (y a) a z ((x y) a) = z ((x a) (y a)) = (z (x a)) (z (y a)) a 11

12 which leads to the required comparison. A particular case o implication gives the negation a = (a ) o an element a A. This is characterized by z a a z = or z A. Now this terminology and notation may be conusing you, or in Subsection 1.3 we have used both in conjunction with a complete boolean algebra. Let s clear that up. We say an element a A o a rame A is complemented i a b a b = or some (necessarily unique) element b A. We then say that b is the complement o a in A. Notice that this terminology agrees with the boolean case LEMMA. Let A be an arbitrary rame, and consider a A. The element a is complemented precisely when a a =. Furthermore, i a is complemented, then its complement is its negation a. Proo. Suppose irst that a is complemented, that is a b a b = or some b A. The irst o these gives b a and the second gives a = a (a b) = ( a a) ( a b) = ( a b) b to show that a = b, and hence a a = b a = holds. Conversely, suppose then, with b = a, we have a a = a b = a a = a b = a a = to show that a is complemented. This result show that we can deine the notion o a complement element o a rame in terms o the behaviour o the negation operation on that rame. We extend this idea. 12

13 1.13 DEFINITION. An element a A o a rame A is, respectively, complemented regular dense i using the negation operation on A. a a a = a a = Notice that double negation ( ) is nothing more than the operation w o Deinition In particular, or each element a we have a a a = a by a particular case o a part o Lemma This shows that a is dense precisely when a =. This observation also shows that a is regular (no matter what element a we start with). Also, as in the proo o Lemma 1.11 we have (a b) = a b or all elements a, b. In particular, and element o the orm a a is always regular. Finally, since a = a (a a) we see that each element o a rame is the meet o a regular element and a denser element. This is a standard lattice theoretic observation. Deinition 1.13 uses standard terminology rom lattice theory, but the words are also used in topology. There is no conlict EXAMPLE. Consider a topology OS viewed as a rame. For each U, V OS we have V U = V U V U = U to show that U is the negation o U in OS. In particular, U is complemented precisely when U U = S U = U U = U in other words when U is clopen. 2 Since U = U = U we see that U is regular in the sense o Deinition 1.13 precisely when it is topologically regular. Finally, U is dense in the sense o Deinition 1.13 precisely when U = U = S equivalently when that is when U is topologically dense. U = S 2 I sometimes tell students that this is where the negation symbol comes rom. Some o them don t believe me. 13

14 In time we will develop the ideas in this example quite a bit urther. Every complemented element is regular, but in general the converse doesn t hold. In act, we have the ollowing LEMMA. A rame A is boolean precisely when each element is regular. Proo. Suppose A is boolean, so that a a = or each element A. On replacing a be a we have a a = so that a is the complement o a. But this complement is a, so that a = a. Conversely, suppose that ( ) is the identity unction on A. For each a A we have a a) = so that a a = (a a) = ( a a) = = to show that a is the complement o a. Most o the calculations in this section have been rather standard, only only occasionally have we needed the completeness o the rame. In due course we will need to go through several calculations which are more rame speciic. 1.5 Morphisms as poset adjunctions Each rame morphism is a -preserving unction between complete posets. In particular, as a poset map this unction has a right adjoint. This is a useul gadget, and is worth a bit o notation DEFINITION. For each rame morphism = rom rame A to rame B A B the right adjoint is the unique monotone map rom B to A such that (a) b a (b) or all a A and b B. Each o A and B is a poset. I we view each as a category that a unctor rom one to the other is just a monotone map. Two such maps are adjoint as unctors precisely when the are adjoint in the sense o Deinition Here we use upper and lower decorations 14

15 to distinguish between the let and the right component o the adjunction. (However, you are warned that some writers don t use this convention, perhaps because they don t understand it.) Finally, not that not every poset adjunction between a pair o rames gives a rame morphism. This let adjoint must also preserve initary meets. A simple exercise show that or each rame morphism (or any poset adjunction) we have ( Y ) = (Y ) or each Y B. However, need not be a rame morphism. In act, it need not preserve even binary joins. Each continuous map T φ S between a pair o spaces gives a rame morphism OS φ OT between the carried topologies, and hence produces an adjoint pair OS φ φ OT where φ = φ is the let adjoint. What is the right adjoint φ? It is reasonable to expect this to have something to do with direct images LEMMA. For each continuous map φ, as above, the right adjoint φ is give by or each W OT. φ (W) = φ (W ) Proo. Consider U OS and W OT. We have U φ (W ) φ (W ) U φ (W ) U ( t T)[t W = φ(t) U ] ( t T)[φ(t) U = t W] ( t T)[t φ (U) = t W] φ (U) W to veriy the required equivalence. We said earlier that the right adjoint o a rame morphism need not preserve even binary joins. We can illustrate this using two very simple spaces EXAMPLE. Consider the 2-point spaces T = {l, r} S = {0, 1} 15

16 with T discrete and with (so that S is sierpinski space). Let OS = {, {1}, S} φ(l) = 0 φ(r) = 1 to produce a continuous map rom T to S. With U = {l} V = {r} we have φ (U V ) = φ (T) = φ ( ) = = = S φ (U) = φ (U ) = φ(r) = {1} = S = φ (V ) = φ (V ) = φ(l) = {0} = {0} = {1} so that φ (U V ) = S φ (U) φ (V ) = {1} to show that φ is not a -morphism. We will, o course, have a lot more to say about rame morphisms in general. 2 The universal algebra o rames Here by universal algebra I don t mean anything very sophisticated; just some inormation about subrames, and a ew o the simple categorical properties o Frm. Quotient rame are let until the next section. Given a rame A = (A,,,,, ) a subrame is a subset B A which is itsel a rame under the restriction o the distinguished attributes o A. Thus, B and B is closed under binary meets and arbitrary suprema. The FDL then automatically transers to B. This is as must as we need to know about subrames, or they are not very interesting. However, let s not dismiss them just yet. Given rames B A we know that any computation involving and done in B agrees with the corresponding computation done in A. However, the implication and negation on B need not agree with those on A. 2.1 EXAMPLE. Consider two topologies on the same set S. Let us write OS S ( ) ( ) or the two respective interior operations. Thus E E or each subset E S, and these can be ar apart. For U, V OS the two implications are given by (V U) = (V U) (V U) = (V U) respectively, and these can be very dierent. 16

17 Recall that, respectively, an arrow B m A A e B o Frm is i or each parallel pair arrows monic epic C g B B g C the implication m = m g = = g e = g e = = g hold. Since the arrows in Frm are unctions (o a certain kind) we have m injective = m monic e surjective = m epic by the obvious cancellation argument. We show that the monics o Frm are precisely the injective arrows, but there are epics which are not surjective. To deal with monics we use the 3-element rame 1 3 = 0 where 0 < < 1. This can be used to separate the elements o an arbitrary rame. Consider any rame B and any element b B. consider the unction as on the let : 3 B (1)= ( )= b (0)= given by the equalities on the right. Almost trivially this is a rame morphism. Trivially, every rame morphism rom 3 arises in this way. 2.2 THEOREM. In the category Frm the monics are precisely the injective morphisms. Proo. On general grounds each injective morphism is monic. Conversely, suppose B m A is monic and consider any b, c B with m(b) = m(c). We require b = c. Let 3 g 17 B

18 be the pair o arrows determined by ( ) = b g( ) = c with (0), (1), g(0), g(1) as they must be. We have (m )( ) = m(b) = m(c) = (m g)( ) so that and hence since m is monic. This gives as required. m = m g = g b = ( ) = g( ) = c Not every epic in Frm is surjective. We use a particular example to show this. In act, the example illustrates a little bit more. Remember that a bimorphism is a morphism that is both monic and epic. Thus each isomorphism is a bimorphism. we give an example o a bimorphism that is not surjective, and not an isomorphism. 2.3 EXAMPLE. Let S be a T 1 topological space, with topology OS. The insertion OS PS is certainly monic, and is surjective only when S is discrete (so that OS = PS). We show that the insertion is epic. Since S is T 1, we know that points are closed. For each point p S let U p = {p} X p = {p} to obtain the set attached to p. We have or each subset E S. Consider any morphism open closed E = {X p p E} PS C to an arbitrary rame C. Since U p and X p are complements in PS, the values (U p ) (X p ) are complements in C. Thus the value (X p ) is uniquely determined by (U p ). 18

19 consider two morphism PS g C which agree on OS. We show that = g, and hence the insertion is epic. For each p S we have (U p ) = g(u p ) and hence (X p ) = g(x p ) by the remarks above. Consider any E S. Using the representation o E in terms o the X p, the preservation properties o and g give (E) = {(X p ) p X} = {g(x p ) p X} = g(e) or the required result. To conclude this section we make a simple observation which, as the story unolds, will become more and more important. Let ( ) 1 2 = 1 = ( ) 0 be the 2-element and 1-element rame, respectively. Almost trivially, these are the initial inal objects o Frm, respectively. In other words, or each rame A there are unique morphisms 2 A A 1 given in the obvious way. This uses 2 as a rame. We can also view it as a topological space. We don t use the discrete topology, rather we use O2 = {, { },2} the upper section topology. Notice that 3 = O2. When viewed in this way we reer to the space 2 as the sierpinski space. As we will ind out, this has a controlling interest in point-ree topology. 3 Quotients o rames A quotient o a rame A is a surjective morphism A B to some rame B. In particular, the structure o the target B is completely determined by the structure o the source A and the nature o. As in almost any algebraic situation, 19

20 the structure o B can be coded by a congruence on A. However, rames being what they are, this congruence on A can be replaced by another ar more useul and amenable gadget. These gadgets, the nuclei on A, are the central components in a amily o techniques which are a distinctive eature o the analysis o rames. We begin to develop these in Section 4, then more extensively in [4]. In this section we concentrate on characterizing quotients and thereby showing how nuclei irst appear. To do that (and to prepare or more sophisticated uses) we set the development in a broader context. Thus we look at quotient in the category Sup o -semilattices, and to do that we look briely at quotients in the category Set o sets. 3.1 Quotients in Set Almost all algebraic quotients can be obtained by reining a simple construction in the category Set o sets and unctons. Let A be a set, and let be an equivalence relation on A. Let A/ be the set o blocks o, the set o -equivalence classes, and let A η A/ be the canonical surjection. Thus or each a A η(a) = {x A a x} is the block to which a belongs. In particular, we have or x, y A. Conversely consider any unction η(x) = η(y) x y A B rom A. Setting x y (x) = (y) or x, y A produces an equivalence relation on A. Because o what comes later it is useul to think o this as the kernel o. These two construction are related by the ollowing, rather simple but undamental, result. 3.1 THEOREM. Let be an equivalence relation on the set A and let be a unction rom A, as above. Suppose is included in the kernel o, that is we have x y = (x) = (y) 20

21 or x, y A. Then there is a unique unction such that A B commutes. η A/ Proo. Consider any block α A/ and set (α) = (a) or any a α. The compatibility o with ensures this is a well-deined unction A/ B and trivially we have ( η)(a) = (a) or each a A. This shows there is at least one unction that makes the triangle commutes. Since η is surjective this is the only possible ill-in unction. There are several algebraic reinements o this result. When the two sets A, B carry similar algebraic structures and is a companion morphism we may imposed on A/ a similar algebraic structure such that both η and become morphisms. 3.2 Quotients in Sup The construction o Theorem 3.1 can be reined to obtain a similar actorization o morphisms in many algebraic situations. In the irst instance we replace the arbitrary equivalence relation by a congruence, a special kind o equivalence relation that respects the distinguished attributes. Then, i we are lucky, we can replace that congruence by another, more amenable, gadget. For instance, or groups or rings the congruence can be replaced by a particular one o its blocks, the normal subgroups or ideals, respectively. However, these two version still involve dealing with block representatives, which is always a nuisance. A similar thing happens with -semilattices and rames, but or these the replacement has a quite dierent character and does not require block representatives. Beore we get to these new gadgets we nee to sort out what a Sup-congruence is. And beore that I suppose I should tell you what the category is. 3.2 DEFINITION. The objects o the category Sup are the complete posets. When viewed in this way we reer to such an object as a -semilattice. An arrow o Sup, or -morphism, is a unction A B : A B 21

22 between to -semilattices such that ( X) = (X) or each X A. Each -semilattice is a complete poset, and so has all inima as well as all suprema. However, a -morphism need not preserve any inima, even binary meets. Note that since =, such a -morphism must preserve bottom, but it need not preserve top. You should think about this, and make sure you understand why. We can now begin to sort out what a Sup-congruence is. To do that it is convenient to introduce a bit o temporary notation. Let A be a -semilattice, and be an equivalence relation on A. When is a Sup-congruence on A? Let X = {x i i I} Y = {y i i I} be a pair o similarly indexed subsets o A (that is, over the same index set). We write i X Y ( i I)[x i y i ] that is i X and Y are point-wise equivalent. 3.3 DEFINITION. An equivalence relation on a -semilattice A is a Sup-congruence i we have X Y = X Y or each similarly indexed pair X, Y o subsets o A. Let A B be a -morphism. As in the Set-case, the kernel o is the equivalence relation on A given by x y (x) = (y) or x, y A. The morphism property translates into the ollowing. 3.4 LEMMA. The kernel o a -morphism is a Sup-congruence on the source algebra. Our job here is to produce a converse o this result, We show that every Supcongruence is obtained rom a -morphism. Consider how we might do that. Let A be a -semilattice and let be a Sup-congruence on A. As in the Set-case, consider the unction A η A/ 22

23 which sends each element to its block. The idea is to urnish A/ as a -semilattice in such a way that η becomes a -morphism. How might we urnish A/ with a supremum operation? Consider a subset X o A/. By choosing block representatives we can view this as X = η [X] or some subset X o A. The idea is to deine X = η( X) which automatically ensures that η is a -morphism. There is, o course, some work to be done. We must show that the deinition o this supremum operation in A/ is independent o the choice o block representatives. We also have to show that it is a supremum operation, which mean we should irst set up a partial ordering on A/ and show that η is monotone. All this can be done, but it s messy and, thankully, avoidable. The cause o the mess is the use o arbitrary block representatives. This can be cleaned up using special block representatives. 3.5 LEMMA. Let be a Sup-congruence on a -semilattice A. Then each -block has a unique largest member. Proo. Consider any a A. Let X = {x i i I} be an indexing o the block to which a belongs. Also let Y = {y i i I} where y i = a or each i I. By construction we have X Y so that (since is a Sup-congruence) X Y = a to show that X is the largest member o the block in question. In the standard congruence situation each block is handled by some block representative, some member o that block. In general any one representative is no better than any other. However, Lemma 3.5 shows that each block o a Sup-congruence has a special representative, and that is the one we use. We also use the operator which attaches to each element its largest mate. 3.6 DEFINITION. Let be a Sup-congruence on a -semilattice A. The selector or is the operator j : A A given by j(a) = {x A a x} or each a A. 23

24 By construction, i j is the selector o the congruence on A then we have ( ) a j(a) ( ) x a = x j(a) or all a, x A. These two properties characterize being the selector o, and enable us to see selectors in a dierent light. Recall that a closure operation on A is a unction j : A A which is inlationary, monotone, and idempotent, that is or all a, b A. (i) a j(a) (m) b a = j(b) j(a) (c) j(j(a)) = j(a) 3.7 LEMMA. Let A be a -semilattice. The selectors on A are precisely the closure operators. Each closure operation is the selector o precisely one congruence. There is a bijective correspondence between Sup-congruence relations and closure operators on A. Proo. Suppose that j is the selector o the congruence on A. We use the two properties ( ) and ( ) rom above to show that j is a closure operation on A. For each a A we have a a, so that ( ) gives a j(a), to show that j is inlationary. Consider a, b A with b a. We have a j(a) b j(b) by ( ), so that by the congruence property, and hence j(a) j(b) a b = a j(b) j(a) j(b) j(a) by ( ). This shows that j is monotone. Consider any a A and let b = j(j(a)). We have a j(a) b by two uses o ( ), so that b a, and hence b j(a) by ( ). The converse comparison j(a) b holds since j is inlationary. This shows that j is idempotent. These three small arguments show that the selector j is a closure operation. Suppose j is a closure operation on A. We show that j is the selector o at least one congruence on A. To this end consider the relation on A given by x y j(x) = j(y) or x, y A. Trivially, this is an equivalence relation on A. To show it is a congruence relation suppose X Y or two similarly indexed subsets X, Y o A. For each y Y there is some x X with x y, so that y j(y) = j(x) j( X) 24

25 by the inlationary and monotone properties o j. Thus Y j( X) to give j( Y ) j(j( X)) j( X) by the monotone and idempotent properties o j. By symmetry this gives j( X) = j( Y ) and hence X Y as required to show that is a congruence on A. The idempotent and inlationary properties o j ensure that ( ) and ( ) hold, and hence j is the selector o. Trivially (by deinition) each congruence has just one selector. Thus, to complete the whole proo it suices to show that each closure operation is the selector o just one congruence. We show that i j is the selector o the congruence then x y j(x) = j(y) or each x, y A. Consider x, y A with x y. By hypothesis, j is the selector o, and hence x j(y) y j(x) by two uses o ( ). Thus j(x) = j(y) by the monotone and idempotent properties o j. Consider x, y A with j(x) j(y). By hypothesis, j is the selector o, and hence x j(x) = j(y) y by two uses o ( ). Thus x y since is an equivalence relation. This result enables us to do away with any use o block representatives, and to replace the use o congruences by the use o closure operators. In time this will give us a collection o powerul techniques or analysing -semilattices, but or now we concentrate on producing a Sup-version o Theorem DEFINITION. Let A be a -semilattice. A subset F A is -closed i X F or each X F. For a closure operation j on A we set A j = j (A) = {x A j(x) = x} to obtain the set o ixed elements o j. Observe that i F A is -closed then, by considering F, we have = F, and hence F is non-empty. These -closed sets have a more important property. 3.9 LEMMA. Let A be a -semilattice. For each closure operation j on A the subset A j is -closed. For each -closed subset F A there is a unique closure operation j on A with F = A j There is a bijective correspondence between closure operations on A and -closed subsets. 25

26 Proo. Let j be a closure operation on A and consider X A j. For each x X we have X x so that j( X) j(x) = x and hence j( X) X to show that X A j. Let F be -closed in A. For each a A set j(a) = {x F a x} to obtain an operator j on A. Almost trivially, this j is inlationary and monotone. Furthermore, or each a A we have j(a) F a F = j(a) = a (where the -closed property gives the let hand condition). In combination these two conditions give j(j(a)) = j(a) (or each a A), to show that j is idempotent, and hence is a closure operation. The right hand condition gives F A j. Conversely, i a A j then a = j(a) F by the let hand condition, to show F = A j. Finally, suppose we have A j = A k or closure operations on A. Consider any a A. We have j(a) A j = A k, so that and hence k(j(a)) = j(a) k(a) k(j(a)) = j(a) by the inlationary property o j and the monotone property o k. A similar argument gives j(a) k(a), and hence j = k. We now have enough background to begin the Sup-analogue o Theorem 3.1. Let A be a -semilattice, and let j be a closure operation on A. Remember that we think o j as a more socially acceptable version o a Sup-congruence on A. Consider the unction A a j A j j(a) (and do not conuse this with the very similar unction j). Since A j is a subset o A it inherits a comparison rom A, so A j is at least a poset. We show a bit more LEMMA. Consider the situation above. For each subset X A j, the element j( X) is the supremum o X in A j. The poset (A j, ) is complete. The assignment j is a Sup-morphism. 26

27 Proo. Consider X A j. For each x X we have x X j( X) A j so that j( X) is an upper bound o X in A j. Let a A j be any upper bound o X in A j. We have X a (in A) so that j( X) j(a) = a to show that j( X) is the least upper bound o X in A j. This also shows that A j is complete (as a poset). To show that j is a Sup-morphism we require j ( X) = j( j (X)) or each X A. The element on the right hand side is the supremum o the subset j (X)) in A j. This required equality rephrases as and we veriy this via two comparisons. For each x X we have j( X) = j( {j(x) x X}) x j(x) {j(x) x X}) so that and hence X {j(x) x X}) j( X) j( {j(x) x X}) since j is monotone. Conversely, or each x X we have x X so that to give and hence j(x) j( X) {j(x) x X} j( X) j( {j(x) x X}) j 2 ( X) = j( X) since j is monotone and idempotent. By Lemma 3.4 each Sup-morphism A B 27

28 has a kernel given by x y (x) = (y) or x, y A. This, o course, views the kernel as a congruence. By Lemma 3.7 we may re-view this congruence as a closure operation k given by k(x) = k(y) x y or x, y A. These two characterizations enable us to move directly rom the morphism to the closure operation k DEFINITION. For each Sup-morphism, as above, the kernel o is the unique closure operation k on A (the source o ) such that k(x) = k(y) (x) = (y) or all x, y A. This deinition uniquely speciies the kernel o a morphism, but doesn t really tell us what it is. To discover that we remember that each Sup-morphism has a right adjoint A B which, as I promised you earlier, has its uses LEMMA. For each Sup-morphism, as above, the kernel is the composite. Proo. By the general properties o poset adjunctions we know that k = is a closure operation on the source A. Thus it suices to show that ( )(x) = ( )(y) (x) = (y) holds or all x.y A. The implication is immediate, and the converse holds since =. With a little bit more work we can characterize the kernel directly in terms o the morphism without using the right adjoint COROLLARY. For each Sup-morphism, as above, the kernel is the closure operation k such that x k(a) (x) (a) or each x, a A. Proo. We have x (y) x y or each x A and y B. Setting y = (a) = (a) gives the required result. Finally we can prove the -reinement o Theorem

29 3.14 THEOREM. Let j be a closure operation on the -semilattice A, and let A B be a Sup-morphism with kernel k. Suppose j k. Then there is a unique Sup-morphism such that A j B A j commutes. Proo. Since j is surjective, there can be at most one such ill-in morphism. For each a A we have k(k(a)) = k(a) so that (k(a)) = (a) by one o the characteristic properties o the kernel k o. We also have so that and hence by the previous observation. For each x A j set to obtain a unction : A j a j(a) k(a) (a) (j(a)) (k(a)) = (a) (j(a)) = (a) (x) = (x) B. We have just seen that ( j )(a) = (j(a)) = (a) or each a A, so it suices to show that is a Sup-morphism. We require ( X) = (X) or each X A j. Here is the supremum operation on A j. This condition unravels as (j( X)) = (X) which, since j =, reduces to the given morphism property o. As you can probably guess, we are going to reine this result even urther by replacing Sup by Frm. 29

30 3.3 Quotients in Frm The results o Subsection 3.2 are important in a wider context, but here we are primarily concerned with rames. Each rame morphism A B is a -morphism o a special kind. As such it has a kernel j given by x j(a) (x) (a) or x, a A. This j is a closure operation on A, o a special kind. Our job in this subsection is to isolate and begin to investigate these special closure operations 3.15 DEFINITION. A nucleus on a rame A is a closure operation j on A such that j(a b) = j(a) j(b) or all a, b A. I you think about it you have already seen one amily o examples o nuclei. We will return to those examples later LEMMA. The kernel o a rame morphism is a nucleus on the source. Proo. Consider a rame morphism with its kernel j, as above. For a, b A (the source o and carrier o j) we have (a b) = (a) (b) and we require a corresponding equality or j. For each x A the characterizing property o j gives x j(a b) (x) (a b) = (a) (b) (x) (a) and (x) (b) x j(a) and x j(b) x j(a) j(b) so that as required. j(a b) = j(a) j(b) O course, we want to show that every nucleus arises as the kernel o a rame morphism. To do that we reine the ideas o Deintion DEFINITION. A ixed set o a rame A is a -closed subset F A such that a A j = (x a) A j or all a, x A. 30

31 By Lemma 3.9 the closure operations on a rame A correspond to the -subsets. This correspondence reines as ollows LEMMA. A closure operation j on a rame A is a nucleus precisely when its set A j o ixed elements is a ixed set. Proo. Suppose irst that j is a nucleus and consider y = (x a) or a A j and arbitrary x A. We have x y a, so that to give x j(y) j(x) j(y) j(x y) j(a) = a j(y) (x a) = y and hence y A j. This shows that A j is a ixed set. Conversely, suppose A j is a ixed set, and consider arbitrary x, y A. It suices to show that j(x) j(y) j(x y) (since the converse comparison is a consequence o the monotonicity o j). Let a = j(x y), so that a A j. We have x y j(x y) = a so that to give and hence y (x a) A j j(y) (x a) x j(y) a holds. A repeat o this argument (with x and y playing dierent roles) gives j(x) j(y) a which is the required result. Each nucleus j on a rame A has a ixed set A j and, by Lemma 3.10, j induces a -morphism rom A ro Aj with j as its kernel. This reines as ollows LEMMA. Let j be a nucleus on the rame A. The ixed set A j is a rame, and the assignment A a j A j j(a) is a rame morphism. 31

32 Proo. By Lemma 3.10 the poset (A j, ) is complete. Thus, by Lemma 1.7, it suices to show that A j carries an implication. But Lemma 3.18 shows that A j is closed under the implication carried by A, and it is easy to check that this provides an implication on A j. By Lemma 3.10 the assignment j is a Sup-morphism, thus we require j (x y) = j (x) j (y) or x, y A. This is immediate since j is a nucleus. With this we can prove the reinement o Theorem THEOREM. Let j be a nucleusn on the rame A, and let A B be a rame morphism with kernel k. Suppose j k. Then there is a unique rame morphism such that commutes. A j B A j Proo. By Theorem 3.14 there is a unique Sup-morphism or which the triangle commutes. Thus it suices to show that this satisies (x y) = (x) (y) or x, y A j. This is an immediate consequence o the deinition o since A j A and is a rame morphism. The results o this section show that we ought to ind out a lot more about nuclei. We begin that in the next section. 4 Nuclei on rames As in Deinition 3.15, a nucleus j on a rame A is a closure operation that passes across binary meets. The results o Subsection 3.3 suggest that these nuclei have a role to play in the analysis o rames. In act, as we will see later, they have a considerable impact on the whole subject. In this section we look at a ew o the basic examples and results. 4.1 DEFINITION. For each element a o a rame A we set u a (x) = (a x) v a (x) = (a x) w a (x) = ((x a) a) or each x A, to obtain three operators on A. 32

33 The operators w a are the same as those introduced in Deinition 1.10, and we now see that Lemma 1.11 shows that each o these is a nucleus. That is the harder part o the proo o the ollowing. 4.2 LEMMA. For each rame A and a A, the three operators u a, v a, w a are nuclei on A. Several straight orward calculations show that u a and v a are nuclei, but let s look at another proo. The kernel o each rame morphism is a nucleus on the source. This can be a useul way o showing that an operator is a nucleus. Here is a simple example o this technique. 4.3 EXAMPLE. Let A be a rame and let a A be an arbitrary element. Consider the two principal sections [a, ] [, a] above and below a. Each o these is a complete poset in its own right. In act, ater a ew moment s thought we see that each is a rame in its own right (but neither is a subrame o A, unless a takes an extreme position). However, each o the two assignments A [a, ] A [, a] y a y y a y is a surjective rame morphism. (You should check this and notice how the FDL is used.) Each o these morphisms has a kernel k given by y k(x) a y a x y k(x) a y a x respectively. These give k = u a k = v a to show that u a and v a are nuclei on A. The ixed set A ua o u a is precisely the interval [a, ] we started rom. This ixed set A va o v a is not the interval [, a], but is is canonically isomorphic to this interval. In a spatial situation the u and v nuclei have a common generalization. 4.4 DEFINITION. Let S be a space with topology OS, and let E S. We set [E](U) = (E U) or each U OS, to produce an operator on OS. Observe that or an open set A OS we have [A](U) = A U or each U OS, and so [A] = u A on OS. Similarly, we have [A ](U) = (A U) = (A U) or each U OS, and so [A ] = v A on OS. Later we will see that u a and v a are complementary on an arbitrary rame. A proo o the ollowing is straight orward. 33

34 4.5 LEMMA. For each space S with topology OS and each E S, the operator [E] is a nucleus on OS. Proo. Trivially, the operator [E] is inlationary and monotone. For F, G S we have (F G) = F G and hence [E] is a pre-nucleus. Finally, or U OS we have [E] 2 (U) = [E]([E](U)) = (E (E U) ) (E E U) = (E U) = [E](U) to show that [E] is idempotent. As with the u and v nuclei, it is instructive to see [E] exhibited as the kernel o a rame morphism. To do this consider a continuous map T φ S rom a space T to a space S. From Lemma 1.17 this induces a rame morphism and its adjoint OS φ φ OT where φ (U) = φ (U) φ (W) = φ (W ) or each U OS and V OT. This morphism has a kernel which has a simpler description. φ φ 4.6 LEMMA. For the situation above the kernel o φ is [E] where E = T φ (S), the complement o the range o φ. Proo. Let k be the kernel o φ. Remembering the characterization k given by Corollary 3.13, or each U, V OS we have V k(u) φ (V ) φ (U) ( t T)[φ(t) V = φ(t) U] ( s φ (T))[s V = s U] φ (T) V U V E U V [E](U) to give the required result. This idea deserve a bit o terminology. 34

35 4.7 DEFINITION. A nucleus on a topology OS is spatially induced i it has the orm [E] or some E S. In general or a space there are nuclei on OS that are not spatially induced. In act, as we will see in [6] every nucleus on a topology is spatially induced precisely when the parent space has a certain amount o pathology. Thus we could say that it is those nuclei on a topology that are not spatially induced that are the interesting ones. We irst met the nuclei w in Subsection 1.4, where I said they are important gadgets. Here is one (but not the only) reason or saying that. 4.8 LEMMA. For each rame A and element a A, when viewed as a rame the ixed set is boolean. Proo. Observe that a = w a ( ) is the bottom o A wa. Consider any x A wa. We must produce some y A wa with A wa x y = a w a (x y) = where, o course, w a (x y) = is the join o x and y in A wa. Let y = (x a) so that w a (y) = w a (x a) = (w a (x) a) = (x a) = y By Lemma In particular, y A wa. Also x y = x a = a since a x. Thus it remains to deal with the join in A wa. Using Lemma 1.9(iv) we have ((x y) a) = (x a) (y a) = y (y a) = a since a y. Thus or the required result. w a (x y) = (a a) = Each nucleus w a gives a boolean quotient. What is more interesting is that ever boolean quotient arises in this way. 4.9 THEOREM. Suppose A B is a surjective rame morphism with a boolean target B. Then the kernel k o is w a where a = k( ). 35

36 Proo. In this proo it is convenient to write A A B B or the extremes o the two rames. Thus a = k( A ), and satisies (a) = B. We irst check that (x ) (x) (x a) = B or each x A. For (x ) we have (x ) (x) (x a) = B (x) (x a) = (x (x a)) = (x a) (a) = B to give the equality. This does not use any properties o B. For (x ) we know that (x) has a complement in B and this is the image o some element o A. Thus we have (x) (z) = B (x) (z) = B or some z A. The lower one o these gives so that (since k is the kernel o ) and hence which with the upper one leads to as required. From (x, x ) we have (x z) ( A ) x z k( A ) = a z (x a) (x) (x a) (x) (z) = B b (x) b (x a) = B or each b B. Thus, or each y A we have y k(x) (y) (x) (y) (x a) = B = ( A ) (y (x a)) ( A ) y (x a) k( A ) = a y (x a) a y w a (x) to give the required result. Each nucleus j on the rame A produces a quotient A j A j 36

37 to the rame A j o element ixed by j. Furthermore, each quotient o A arises in this way (up to a unique isomorphism over A). We have seen that u a and v a arise rom A [a, ] A [, a] respectively, where a little bit o care is needed with the right hand quotient. What about A j A wa or arbitrary a? To answer that we irst look at the particular case a =. For each x A we have w (x) = ((x ) ) x so that w is just double negation on A. In particular A w = A = {x A x = x} is the set o regular elements o A. This is converted into a rame using x y = (x y) = ( x y) as the join o x and y in A. This is well known in topological circles EXAMPLE. For the topology OS on a space S the rame (OS) is the complete boolean algebra o regular open sets o S. We know that a regular element o OS is just a regular open set. For two such sets U, V the join in (OS) is given by U V = (U V ) = (U V ) which is precisely the way we convert the regular open sets into a boolean algebra. For each a A the quotient A wa is a complete boolean algebra. We have just seen that A = A wbot is the analogue o the boolean algebra o regular pen sets. To deal with A wa or arbitrary a we look at the interval [a, ] as a rame. What is the negation o an element x [a, ] in [a, ]? It is that element y [a, ] such that z y z x a or z [a, ]. In other words it is (x a), which does live in [a, ]. Furthermore, the double negation o x in [a, ] is just w a (x). Thus to veriy the ollowing. [a, ] = A wa 37