AA categorical duality for algebras of partialfunctions
Brett McLean
Laboratoire J. A. Dieudonné UMR CNRS 7351, Université Nice Sophia Antipolis,06108 Nice Cedex 02 https://math.unice.fr/~bmclean/ [email protected]
Abstract
We prove a categorical duality between a class of abstract algebras of partial functions and a classof (small) topological categories. The algebras are the isomorphs of collections of partial functionsclosed under the operations of composition, antidomain, range, and preferential union (or ‘override’).The topological categories are those whose space of objects is a Stone space, source map is a localhomeomorphism, target map is open, and all of whose arrows are epimorphisms.
Primary: 06F05; Secondary: 20M30, 06E15, 20M35
Keywords and phrases partial function, duality, Stone space, finite state transducer
Funding
This project has received funding from the European Research Council (ERC) under theEuropean Union’s Horizon 2020 research and innovation program (grant agreement No. 670624).
Variants and extensions of Stone duality are pervasive in logic and computer science: forexample in modal [14], intuitionistic [8], substructural [7], and many-valued [13] logic, and insemantics [2], formal language theory [11], and logics for static analysis [5]. In its most basicform—between Boolean algebras and Stone spaces—it provides a duality for the isomorphsof algebras of unary relations equipped with the union and complement operations. Oneextremely prominent ‘real world’ example of such an algebra is the set of regular languages(for a fixed finite alphabet). Indeed, recently it has been shown how the presence of anextended Stone duality is the explanation behind many of the great successes of algebraiclanguage theory [11, 10].Another genre of algebras of relations that has been studied is algebras of partial functions [6, 16, 15, 18]. Here the algebras are formed of collections of partial functions closed undercertain natural operations such as composition or ‘preferential union’. Again, pre-existingexamples can be found in automata/formal language theory.
Transducers are finite statemachines that take words as input and produce words as output. In general they realiseword-to-word relations, but certain collections of word-to-word partial functions defined viatransducers are considered important—we think particularly of the rational functions andthe regular functions [9].In this paper, motivated by the utility of duality applied to regular languages, we givea description of a duality (Theorem 3.7) applicable to both the rational functions and theregular functions. Specifically, on one side is the category of isomorphs of algebras of partialfunctions equipped with four operations: composition , antidomain , range , and preferentialunion (see Section 2). On the other side is the category of (small) categories each equippedwith a topology and satisfying certain extra conditions (see Section 3). This duality is apartial function analogue of a duality due to Mark Lawson between a certain subclass ofinverse semigroups and certain topological groupoids [17]. Recall that the inverse semigroupsare the isomorphs of injective partial functions. In fact the duality presented here not onlybroadens the scope from injective to arbitrary partial functions, but also generalises which a r X i v : . [ m a t h . R A ] S e p A categorical duality for algebras of partial functions morphisms of algebras are handled. In our duality, the morphisms of algebras are exactlythe homomorphisms, whereas in [17] they are a restricted type of homomorphism only.Given that partial functions are just a special type of binary relation, we are compelled toacknowledge the numerous dualities applicable to, or even designed specifically for, algebras ofbinary relations. The algebras concerned by these dualities span a range from the very highlystructured Boolean algebras with operators [4, Chapter 5: Algebras and General Frames]down to the much more general cases of (bounded) distributive lattices with completelyarbitrary additional operations [12] and of posets with monotone operations [7]. Two remarksare in order here. Firstly, none of the mentioned dualities are applicable to our algebras,since on the one hand they are not distributive lattices, but on the other hand not all theiroperations are monotone. Secondly, the use of a general theory would not be optimal in anycase, since our algebras do possess a great deal of structure, reflecting their concrete origins. Structure of the paper
In Section 2 we define formally our algebras of functions and thecategory they constitute. In Section 3 we do the same for the small topological categories onthe other side of our duality. In Section 4, we describe one half of the duality: the functorfrom algebras to topological categories. In Section 5, we describe the remaining half of theduality: the functor from topological categories to algebras. In Section 6, we prove thatthese two functors do indeed form a contravariant equivalence of categories. In Section 7, wesay a little about the duality as it applies to the rational and regular functions.
Given an algebra A , when we write a ∈ A or say that a is an element of A , we mean that a is an element of the domain of A . Similarly for the notation S ⊆ A or saying that S is asubset of A . We follow the convention that algebras are always nonempty. If S is a subset ofthe domain of a map θ then θ [ S ] denotes the set { θ ( s ) | s ∈ S } . As is common in algebraiclogic, compositions denoted with the symbol ; are written with the first composee on theleft, that is, contrary to the usual mathematical convention. We will, however, also use theconventional ◦ notation with the conventional ordering in situations having no connection tothe composition of partial functions. If S and T are subsets of A , then we abuse notationby writing S ; T for { s ; t | s ∈ S and t ∈ T } and abuse further by writing S ; a and a ; S for S ; { a } and { a } ; S respectively.We begin by making precise what is meant by partial functions and algebras of partialfunctions. (cid:73) Definition 2.1.
Let X be a set. A partial function on X is a subset f of X × X satisfying ( x, y ) ∈ f and ( x, z ) ∈ f = ⇒ y = z. (cid:73) Definition 2.2.
Let σ ⊆ { ; , A , R , t} be a functional signature, where ; and t are binaryand A and R are unary. An algebra of partial functions of the signature σ is a universalalgebra A = ( A, σ ) where the elements of the universe A are all partial functions on some(common) set X , the base , and the interpretations of the symbols are given as follows.The binary operation ; is composition of partial functions.The unary operation A is the operation of taking the diagonal of the antidomain of apartial function: A( f ) := { ( x, x ) ∈ X |6 ∃ y ∈ X : ( x, y ) ∈ f } . . McLean 3 The unary operation R is the operation of taking the diagonal of the range of a partialfunction: R( f ) := { ( y, y ) ∈ X | ∃ x ∈ X : ( x, y ) ∈ f } .the binary operation t is preferential union : ( f t g )( x ) = f ( x ) if f ( x ) defined g ( x ) if f ( x ) undefined, but g ( x ) definedundefined otherwise Note that (despite the symmetry of the symbol t ) preferential union is not generally acommutative operation, though it is associative. (cid:73) Definition 2.3.
An algebra A of the signature σ is representable by partial functions ifit is isomorphic to an algebra of partial functions of the signature σ . An isomorphism from A to an algebra of partial functions is a representation of A . We begin by looking at representable { ; , A , R } -algebras, but we will soon see that thealgebras we are interested in are equivalent to the representable { ; , A , R , t} -algebras. (cid:73) Remark 2.4.
Note that the constants 0 (empty function) and 1 (identity function), theoperation D (domain), and the relation ≤ (subset) are all definable in the signature { ; , A , R } .That is, the term 0 := A( a ) ; a , the term 1 := A(0), and the term D( a ) := A(A( a )) are allnecessarily represented in the intended way by any representation, and the relation a ≤ b ⇐⇒ D( a ) ; b = a corresponds via any representation precisely to the subset relation on the image of therepresentation.Statements involving order will always be with respect to ≤ . (cid:73) Remark 2.5.
The representable { ; , A , R } -algebras form a proper quasivariety, axiomatisedby a finite number of quasiequations [15, Theorem 4.1]. (cid:73) Definition 2.6.
Two elements a and b of an algebra of the signature { ; , A , R } are com-patible if D( a ) ; b = D( b ) ; a . Clearly in any representable { ; , A , R } -algebra the compatibility relation expresses preciselythat for any representation the representing functions agree on their common domain. Insuch an algebra compatibility is necessary for the existence of a least upper bound of a pair a and b . If a and b have an upper bound, c say, they have a least upper bound (join) givenby A(A( a ) ; A( b )) ; c . From this term, we see that in concrete algebras any binary joins mustbe given by binary unions. (cid:73) Lemma 2.7.
Homomorphisms of { ; , A , R } -algebras preserve binary joins. Proof.
Let h : A → B be a homomorphism of { ; , A , R } -algebras, and suppose a, b ∈ A havean upper bound. Since the algebras are representable, we may assume they are algebras ofpartial functions. As ≤ is defined by an equation, h is order preserving, so h ( a ∨ b ) is anupper bound for { h ( a ) , h ( b ) } , that is, h ( a ) , h ( b ) ⊆ h ( a ∨ b ). On the other hand h ( a ∨ b ) = h ((D( a ) ∨ D( b )) ; ( a ∨ b ))= (D( h ( a )) ∪ D( h ( b ))) ; h ( a ∨ b ) . A categorical duality for algebras of partial functions (We know joins correspond to unions on the subalgebra of elements of the form D( − ), sincejoin is expressible there, as A(A( − ) ; A( − )).) Hence h ( a ∨ b ) ⊆ h ( a ) ∪ h ( b ), and so h ( a ∨ b ) isthe smallest possible upper bound for h ( a ) and h ( b )— h ( a ) ∪ h ( b ). (cid:74) Let A be the subclass of the representable { ; , A , R } -algebras consisting of those validatingthe first-order condition that every compatible pair has an upper bound. (cid:73) Corollary 2.8.
The category consisting of A with { ; , A , R } -homomorphisms is isomorphicto the category of representable { ; , A , R , t} -algebras with { ; , A , R , t} -homomorphisms. Proof.
In any representable { ; , A , R } -algebra in which compatible pairs have upper bounds,the operation t is definable as a t b := a ∨ A( a ) ; b . And there is an inverse interpretation ofany representable { ; , A , R , t} -algebra as a representable { ; , A , R } -algebra with compatiblejoins, since for compatible a and b we have a ∨ b = a t b . It remains to see that the morphismsare the same. Since { ; , A , R } -homomorphisms preserve binary joins, they must preserve t ,since t is then definable in terms of preserved operations. (cid:74) Let Σ be a finite alphabet. The rational functions are the partial functions from Σ ∗ toΣ ∗ realisable by a one-way transducer. The regular functions are the partial functions fromΣ ∗ to Σ ∗ realisable by a two-way transducer. The rational and the regular functions areboth closed under composition, antidomain, and range and also under the partial operationof compatible union. These classes of partial functions are not closed under other familiaroperations that we may be tempted to include in the signature, such as intersection andrelative complement. This is the reason for our interest in the class A .We mention one other important class of partial functions important to the theoryof transducers. The sequential functions are those partial functions realised by one-way input-deterministic transducers. However, the sequential functions do not fit within ourframework for the reason that they are not closed under compatible unions. (For example,the sequential functions a n a n and a n b b n have disjoint domains, hence are compatible,but their union is not sequential.)In view of Corollary 2.8, we can choose to work with the representable { ; , A , R , t} -algebras,in lieu of A , and henceforth that is what we will do. This pays off immediately: the classhas a syntactically simple finite axiomatisation (and therefore is algebraically well behaved). (cid:73) Theorem 2.9 (Hirsch, Jackson, and Mikulás [15, Corollary 4.2 + Lemma 3.6]) . The repres-entable { ; , A , R , t} -algebras form a proper quasivariety, axiomatised by the following finitelist of equations and quasiequations. a ; ( b ; c ) = ( a ; b ) ; c (1)A( a ) ; a = A( b ) ; b (2)1 ; a = a (3) a ; A( b ) = A( a ; b ) ; a (4)D( a ) ; b = D( a ) ; c ∧ A( a ) ; b = A( a ) ; c → b = c (5)D(R( a )) = R( a ) (6) a ; R( a ) = a (7) a ; b = a ; c → R( a ) ; b = R( a ) ; c (8)D( a ) ; ( a t b ) = a (9)A( a ) ; ( a t b ) = A( a ) ; b (10) . McLean 5 The category of representable { ; , A , R , t} -algebras and their homomorphisms is the firstof the two categories between which we will exhibit a duality.We now introduce a small running example by starting with an eight-element { ; , A , R , t} -algebra. Though finite algebras cannot inform very much about the topological aspect of ourduality—their duals all have discrete topologies—the example will be sufficient to grasp theessence of the duality. (cid:73) Example 2.10.
Let A be the following collection of partial functions on the set { , , } .The empty function, ∅ , the identity id { , } on { , } , the identity id { } on { } , the identityid { , , } on { , , } , the ‘swap’ s := { , } , the function { , , } ,the constant function c := { , } , and the constant function { , , } .Then one can check that A is closed under the operations of composition, antidomain, range,and preferential union and is therefore a { ; , A , R , t} -algebra of partial functions.Later we will take a particular interest in homomorphisms that are what we call locallyproper , though they are not essential to our duality. To define locally proper homomorphisms,we first need the notion of a prime filter in a representable { ; , A , R , t} -algebra. (cid:73) Definition 2.11.
A homomorphism of representable { ; , A , R , t} -algebras is locally proper if the inverse image of every prime filter (see Definition 4.1) is a prime filter. This is the condition restricting the morphisms in Lawson’s inverse semigroup duality[17]. It is evident, even before reading the definition of a prime filter, that locally properhomomorphisms are closed under composition and include all identity maps.
In this section we describe the other (large) category participating in our duality. It is acategory of small categories with extra structure. To reduce the potential for confusion, wewill call the ‘object level’ morphisms of the small categories arrows (which underlines theirabstract nature) and reserve morphism for the ‘meta level’ morphisms of the large category.Composition in the small categories is denoted · and like the ; notation, the first composeeappears on the left-hand side. (cid:73) Definition 3.1. A topological category is a (small) category whose sets of objects O and arrows M are both topological spaces and such thatthe source map d : M → O is continuous,the target map r : M → O is continuous,the composition map · : M × O M → M is continuous, the identity-assigning map x x sending each object to its identity arrow is continuous. Put concisely, for us a topological category is a category internal to the category
Top of topological spaces. Note that a topological category is a particular type of topologicalpartial algebra —a partial algebra on a topological space whose (possibly) partial operationsare continuous when considered as functions on their domains of definition (equipped withthe subspace topology). Unlike algebras, categories are allowed to be empty. Not to be confused with the various other (unrelated) usages of this term. The topology on the pullback M × O M is the initial topology with respect to the two projections. That is,the topology generated by sets of the form { ( x, y ) ∈ M × O M | x ∈ U } and { ( x, y ) ∈ M × O M | y ∈ U } for open subsets U of M . In other words, it is the subspace topology on M × O M ⊆ M × M . A categorical duality for algebras of partial functions (cid:73)
Definition 3.2. A local homeomorphism π : X → Y of topological spaces is a continuousmap such that for every x ∈ X there exists an open neighbourhood U of x such that π ( U ) is open, π | U : U → π ( U ) is a homeomorphism. (cid:73) Definition 3.3. An étale category is a topological category such thatthe source map d : M → O is a local homeomorphism,the target map r : M → O is an open map.An étale category is Stone if its space of objects is a Stone space (also known as a Booleanspace), that is, a compact and totally separated space.
The condition that d is a local homeomorphism and the condition, coming from thedefinition of a category, that d is surjective, together say that in an étale category, d gives M the structure of an étale space (of sets) over O (also known as a sheaf space ).One might expect to take functors given by continuous maps as the morphisms oftopological categories. However we require a more general definition in order to capture allthe duals of algebra homomorphisms. (cid:73) Definition 3.4.
Let C and D be categories. A multivalued functor F : C → D consistsof: a function from the objects of C to the objects of D ,a relation from the arrows of C to the arrows of D ,(both denoted F ) such that: if f : x → y is an arrow of C and g ∈ F ( f ) , then g : F ( x ) → F ( y ) , F ( x ) ∈ F (1 x ) for each object x of C , if f · f is defined, g ∈ F ( f ) , and g ∈ F ( f ) , for arrows f and f , then g · g ∈ F ( f · f ) .For F to be a multivalued functor between topological categories we also require that: (i) the object component of F is continuous, (ii) the arrow component of F is a continuous relation from the arrows of C to the arrowsof D (that is, inverse images of open sets are open). Note that we are using ‘multivalued’ in the sense ‘zero or more values’. We now pick outcertain special multivalued functors to account for the structure of the algebras for whosehomomorphisms they are to provide duals. (cid:73)
Definition 3.5.
A multivalued functor F : C → D between categories is star injective iffor every object x of C , it restricts to an injective relation on the ‘star’ Hom( x, − ) . That is: F ( f : x → y ) ∩ F ( f : x → y ) = ∅ = ⇒ f = f . The same functor is star surjective if its restrictions to stars are surjective relations.We call a multivalued functor F : C → D between topological categories pseudo starsurjective if whenever U is an open set of arrows of D , and f ∈ U ∩ Hom( F ( x ) , − ) , thenthere exists some f ∈ U that is in the image of F | Hom( x, − ) . The same functor is co-pseudostar surjective if F op is pseudo star surjective.We call a multivalued functor between topological categories star coherent if it is starinjective, star surjective and co-pseudo star surjective. Composing two multivalued functors in the evident way yields a multivalued functor.The identity functor provides a two-sided identity for this composition. . McLean 7 (cid:73)
Lemma 3.6.
Composition of multivalued functors between topological categories preservesstar coherency.
Proof.
We give the details showing co-pseudo star surjectivity is preserved. Let F : C → D and G : D → E be co-pseudo star surjective multivalued functors. Suppose U is open in E ,and f ∈ U ∩ Hom( − , G ( F ( y )). Then by co-pseudo star surjectivity of G , we can find some g ∈ G − ( U ) ∩ Hom( − , F ( y )). As G is continuous, G − ( U ) is open. Hence by co-pseudo starsurjectivity of F , we can find an h ∈ F − ( G − ( U )) ∩ Hom( − , y ). That is, there is some f ∈ U in the image of ( G ◦ F ) | Hom( x, − ) . Hence G ◦ F is co-pseudo star surjective. (cid:74) We see therefore that star-coherent multivalued functors can legitimately be used asmorphisms of topological categories.We are now finally ready to state our duality theorem. (cid:73)
Theorem 3.7.
There is a categorical duality between the following two categories.The category A with objects the { ; , A , R , t} -algebras representable by partial functions, morphisms the homomorphisms of { ; , A , R , t} -algebras.The category C with objects the Stone étale categories all of whose arrows are epimorphisms, morphisms the star-coherent multivalued functors of topological categories. We will also show that the duality restricts to the following sub-duality. (cid:73)
Theorem 3.8.
There is a categorical duality between the following two categories.The category A with objects the { ; , A , R , t} -algebras representable by partial functions, morphisms the locally proper homomorphisms of { ; , A , R , t} -algebras.The category C with objects the Stone étale categories all of whose arrows are epimorphisms, morphisms the star-coherent functors of topological categories. In this section, we will define a contravariant functor in the direction from algebras totopological categories that forms one half of our duality. Following [17], we present thefunctor in terms of certain filters. That is, given an algebra, the entities used to construct atopological category—the entities that will constitute the arrows—will be filters satisfying aprimality condition. We mention, however, that an alternative presentation is possible usingthe sort of algebraic distillation of germs of functions found in [3], for example.We start with some easily verifiable remarks. In a representable { ; , A , R , t} -algebra A ,the elements of the form A( − ) form a subalgebra. We call an element of this subalgebraa domain element (since it is equivalently of the form D( − )). This subalgebra, D[ A ], isa Boolean algebra, with least element 0, greatest element 1 , meet given by ;, complementgiven by A, and join given either by De Morgan or by t .In all the following lemmas, we will be working exclusively with isomorphism-invariantproperties of representable { ; , A , R , t} -algebras. Hence, whenever convenient, we may assumewe are working genuinely with partial functions and make free use of any property of partialfunctions that is both intuitively obvious and easily verifiable from definitions. A categorical duality for algebras of partial functions (cid:73)
Definition 4.1.
Let A be a representable { ; , A , R , t} -algebra. A filter of A is a nonempty,upward-closed, downward-directed subset of A . A filter F is prime if it is proper andwhenever a t b ∈ F , either a ∈ F or b ∈ F . Our definition of filters being the standard one, many basic facts are already known to us.For example there is a smallest filter including any given subset; that is, the notion of thefilter generated by a subset is well defined. Many of the properties of prime filters that weneed have been proven in [15, Section 4] (where a prime filter is called an ultrasubset ). Theproofs there apply to any representable { ; , A , R } -algebra, so in particular the representable { ; , A , R , t} -algebras. (cid:73) Lemma 4.2.
In representable { ; , A , R , t} -algebras, the prime and maximal filters coincide. Proof.
Take first a prime filter P . Let a be an arbitrary element not in P . Since P isnonempty, we can find p ∈ P . The filter generated by P ∪ { a } must contain a lower boundfor { a, p } , call this b . Since b ≤ p it follows, by reasoning about partial functions, that p = b ∨ (A( b ) ; p ). Hence either b ∈ P or A( b ) ; p ∈ P . But b is not in P , else a would be in P .Hence A( b ) ; p ∈ P , and so the filter generated by P ∪ { a } contains both b and A( b ) ; p andhence some lower bound for this pair—necessarily 0. Hence any extension of P is improper,so P is maximal.For the converse, we first establish:for any upward-closed set S , the set D[ S ] is upward closed in D[ A ] . If a ∈ S and D( b ) ≥ D( a ), then a t b ≥ a , so a t b ∈ S . It is a property of partial functionsthat D( b ) ≥ D( a ) = ⇒ D( a t b ) = D( b ), hence D[ S ] contains D( b ). We conclude that D[ S ] isupward closed.Now take a maximal filter M , and suppose a t b ∈ M . By [15, Lemma 4.5(ii)] andthe fact that D[ M ] is upward closed in D[ A ], the set D[ M ] is an ultrafilter of D[ A ]. Theneither D( a ) ∈ D[ M ] or A( a ) ∈ D[ M ]. Suppose the former, that is, there is some c ∈ M withD( c ) = D( a ). As M is downward directed, there is some d ∈ M with d ≤ a t b, c . It follows,by reasoning about partial functions, that d ≤ a , and hence a ∈ M . By a similar argument,if A( a ) ∈ D[ M ] then b ∈ M . Hence M is prime. (cid:74) By [15, Lemma 4.5(ii)] we now know that the following conditions are equivalent. P is a prime filter. P is a maximal filter. P = ( µ ; a ) ↑ for some a ∈ A and ultrafilter µ of D[ A ] such that 0 µ ; a . For some ultrafilter µ of D[ A ], for all a ∈ P , we have P = ( µ ; a ) ↑ .In the following lemmas, let A be a representable { ; , A , R , t} -algebra. The notation S ↑ denotes the upward closure of the set S in A or in D[ A ] if specified. (cid:73) Lemma 4.3.
Let P be a prime filter of A . Then D[ P ] and R[ P ] ↑ are both ultrafilters of D[ A ] , where the upward closure is taken in D[ A ] . Proof. As P is nonempty, D[ P ] is too. Now suppose D( a ) , D( b ) ∈ D[ P ], with a, b ∈ P . Thenas P is downward directed, it contains some c ≤ a, b . This inequality implies (for any partial This is equivalent to the condition that whenever a ∨ b ∈ F , either a ∈ F or b ∈ F . As is conventional, ‘maximal filter’ will always mean maximal proper filter. . McLean 9 functions) that D( c ) ≤ D( a ) , D( b ). Hence D[ P ] is downward directed. If D[ P ] contained 0,then P would have to too, since D( a ) = 0 implies a = 0 for partial functions. So D[ P ] mustbe a proper filter. It remains to show that for any a ∈ A , either D( a ) or A( a ) belongs toD[ P ]. Take any element b of P . Then b = (D( a ) ; b ) t (A( a ) ; b ), so either D( a ) ; b ∈ P orA( a ) ; b ∈ P . Then we can obtain an element of D[ P ] that is less than or equal to either D( a )or A( a ), respectively.As P is nonempty, R[ P ] ↑ is too. It is upward closed by definition. Suppose R( a ) , R( b ) ∈ R[ P ], with a, b ∈ P . Then as P is downward directed, it contains some c ≤ a, b . Theinequality implies that R( c ) ≤ R( a ) , R( b ). Hence R[ P ] is downward directed, so R[ P ] ↑ istoo. Since R[ P ] cannot contain 0, neither can R[ P ] ↑ , so it is proper. Given any a ∈ A ,take any b ∈ P . Then b = ( b ; D( a )) t ( b ; A( a )), so either b ; D( a ) ∈ P or b ; A( a ) ∈ P .But R( b ; D( a )) ≤ D( a ) and R( b ; A( a )) ≤ A( a ). So either D( a ) ∈ R[ P ] ↑ or A( a ) ∈ R[ P ] ↑ ,respectively. (cid:74)(cid:73) Lemma 4.4.
Let µ be an ultrafilter of D[ A ] . Then µ ↑ , where the upward closure is takenin A , is a prime filter of A . Proof.
We have µ ↑ = ( µ ; 1 ) ↑ , and by [15, Lemma 4.5(ii)] this is a maximal filter. ByLemma 4.2, it is a prime filter. (cid:74)(cid:73) Lemma 4.5.
Let µ be an ultrafilter of D[ A ] . Then D[ µ ↑ ] = µ , where the upward closure istaken in A . Proof.
We know µ ⊆ D[ µ ↑ ] because D fixes all domain elements. Conversely, take an elementof D[ µ ↑ ]: an element D( b ) such that b ≥ α for some α ∈ µ . Then D( b ) ≥ D( α ). But D( α ) = α and µ is upward closed; hence D( b ) ∈ µ . This proves the reverse inclusion. (cid:74)(cid:73) Lemma 4.6.
Let
P, Q be prime filters of A . Then ( P ; Q ) ↑ is a prime filter if and only if D[ Q ] = R[ P ] ↑ . Otherwise ( P ; Q ) ↑ contains and hence is all of A . Proof.
Let P = ( µ ; a ) ↑ and Q = ( ν ; b ) ↑ for µ, ν ultrafilters of domain elements. By [15,Lemma 4.6(i)], the set ( P ; Q ) ↑ is of the form ( µ ; a ; ν ; b ) ↑ . By [15, Lemma 4.6(ii)], the set( µ ; a ; ν ; b ) ↑ is a prime filter if and only if it does not contain 0. So it remains to show that0 ∈ ( P ; Q ) ↑ if and only if D[ Q ] = R[ P ] ↑ .First suppose 0 ∈ ( P ; Q ) ↑ , so 0 ∈ P ; Q , and so 0 = a ; b for some a ∈ P and b ∈ Q .It follows, by reasoning about partial functions, that R( a ) ; D( b ) = 0. Since R( a ) ∈ R[ P ] ↑ and D( b ) ∈ D[ Q ], and ; is meet on domain elements, these two ultrafilters cannot be equal.Conversely, suppose D[ Q ] = R[ P ] ↑ . Since they are ultrafilters, we can then find Booleancomplements α and A( α ) with α ∈ R[ P ] ↑ and A( α ) ∈ D[ Q ]. That is, there are a ∈ P withR( a ) ≤ α and b ∈ Q with D( b ) = A( α ). Then R( a ) ; D( b ) = 0, and it follows by reasoningabout partial functions that a ; b = 0; hence 0 ∈ ( P ; Q ) ↑ . (cid:74)(cid:73) Lemma 4.7.
Let
P, Q be prime filters of A . If ( P ; Q ) ↑ is proper, then R[ P ; Q ] ↑ = R[ Q ] ↑ . Proof.
First we show R[ P ; Q ] ⊆ R[ Q ], giving R[ P ; Q ] ↑ ⊆ R[ Q ] ↑ . Take an R( a ; b ) ∈ R[ P ; Q ],with a ∈ P and b ∈ Q . Now b = (R( a ) ; b ) t (A(R( a )) ; b ), so, since Q is a prime filter, eitherit contains R( a ) ; b or A(R( a )) ; b . If the latter, then P ; Q contains a ; A(R( a )) ; b , equal to0, contradicting the hypothesis that ( P ; Q ) ↑ is proper. Hence Q contains R( a ) ; b , so R[ Q ]contains R(R( a ) ; b ). But R(R( a ) ; b ) = R( a ; b ) is a property of partial functions (it is axiom(r.VII) in [15]). Hence R( a ; b ) ∈ R[ Q ], and we have our first inclusion.Conversely, suppose R( b ) ∈ R[ Q ], with b ∈ Q . Take any a ∈ P . As before, it must be thecase that R( a ); b ∈ Q . Then a ;(R( a ); b ) ∈ P ; Q , that is, a ; b ∈ P ; Q . Hence R( a ; b ) ∈ R[ P ; Q ]. By a property of partial functions, R( b ) ≥ R( a ; b ), hence R( b ) ∈ R[ P ; Q ] ↑ . Since R( b ) wasan arbitrary element of R[ Q ] we have R[ Q ] ⊆ R[ P ; Q ] ↑ and hence R[ Q ] ↑ ⊆ R[ P ; Q ] ↑ . (cid:74)(cid:73) Lemma 4.8. If P and Q are nondisjoint prime filters with D[ P ] = D[ Q ] , then P = Q . Proof.
This is [15, Lemma 4.5(iv)]. (cid:74)(cid:73)
Lemma 4.9.
Let
P, Q, R be prime filters of A . Suppose ( P ; Q ) ↑ and ( P ; R ) ↑ are properand equal. Then Q = R . Proof.
If ( P ; Q ) ↑ and ( P ; R ) ↑ are proper, then by Lemma 4.6, we have D[ Q ] = R[ P ] ↑ = D[ R ].So by Lemma 4.8 it is sufficient to show that Q and R are nondisjoint. Let a ∈ P and b ∈ Q . So a ; b ∈ ( P ; Q ) ↑ = ( P ; R ) ↑ , that is, a ; b ≥ a ; c for some a ∈ P and c ∈ R . Bydefinition this means D( a ; c ) ; a ; b = a ; c . But D( a ; c ) must belong to the ultrafilterD[ P ]—it cannot be that its Boolean complement A( a ; c ) is in D[ P ], since ( P ; R ) ↑ is proper.Hence D( a ; c ) ; a ∈ P . Pick some a ∈ P with a ≤ D( a ; c ) ; a and a ≤ a . It is a propertyof partial functions that if d ; b = d ; c and d ≤ d, d then d ; b = d ; c . Hence, fromD( a ; c ) ; a ; b = a ; c we obtain a ; b = a ; c . By axiom (8), this gives R( a ) ; b = R( a ) ; c .Now R( a ) is an element of D[ Q ] = R[ P ] ↑ = D[ R ], so R( a ) ; b ∈ Q and R( a ) ; c ∈ R . Thatis, we have found our element common to Q and R . (cid:74)(cid:73) Lemma 4.10.
Let µ be an ultrafilter of D[ A ] and a ∈ A be such that R( a ) ∈ µ . Then thereexists a prime filter P such that a ∈ P and R[ P ] ↑ = µ . Proof.
Suppose R( a ) ∈ µ . Consider the subset D[ a ; µ ] of the Boolean algebra D[ A ]. Thisset D[ a ; µ ] is nonempty (because µ is nonempty) and downward directed—because givenD[ a ; α ] and D[ a ; β ], for α, β ∈ µ , we know α ; β ∈ µ , and it is a property of partial functionsthat D( a ; α ; β ) is a lower bound for { D( a ; α ) , D( a ; β ) } (in fact it is the meet). Further,D[ a ; µ ] does not contain 0, since that would imply 0 ∈ a ; µ , which implies that A(R( a )) ∈ µ ,but this is prohibited, since µ is an ultrafilter containing R( a ).We have shown that D[ a ; µ ] ↑ is a proper filter of D[ A ]. Extend it to an ultrafilter ν . Now( ν ; a ) ↑ is a prime filter because 0 ν ; a , by the following reasoning. The filter D[ a ; µ ] ↑ contains D( a ; 1 ) = D( a ), and hence ν does too, meaning ν does not contain A( a ), which isa necessary condition for ν ; a to contain 0. Our prime filter ( ν ; a ) ↑ contains a , as desired.Finally, we claim that R[( ν ; a ) ↑ ] ↑ = µ . For any α ∈ µ , we know that D( a ; α ) ∈ ν and therefore R(D( a ; α ) ; a ) ∈ R[( ν ; a ) ↑ ] ↑ . But it is a property of partial functions thatR(D( a ; α ) ; a ) ≤ α (given that α is a domain element). Since R[( ν ; a ) ↑ ] ↑ is upward closedand α was an arbitrary element of µ , we obtain R[( ν ; a ) ↑ ] ↑ ⊆ µ . Since R[( ν ; a ) ↑ ] ↑ and µ areultrafilters, they are equal, as required. (cid:74) PF on objects We now define the functor PF : A → C used for one half of the duality. (The PF stands for‘prime filter’, not ‘partial function’!) For A a representable { ; , A , R , t} -algebra, let PF( A ) bethe following Stone étale category.The objects of PF( A ) are the ultrafilters of D[ A ].The arrows of PF( A ) are the prime filters of A .The source and target of an arrow P are D[ P ] and R[ P ] ↑ respectively. By Lemma 4.3these are objects.The identity arrow for an object µ is µ ↑ , where the upward closure is taken in A . ByLemma 4.4, this is an arrow. By Lemma 4.5, its source is µ . By Lemma 4.6, the targetof µ ↑ must also be µ since ( µ ↑ ; µ ↑ ) ↑ is proper. . McLean 11 For composable arrows P and Q , the composition is given by P · Q := ( P ; Q ) ↑ . ByLemma 4.7, this is an arrow, evidently with the same source as P . By Lemma 4.6, it hasthe same target as Q .The confirmation that the structure so defined validates the axioms for categories is thecontent of Lemma 4.11, which follows shortly. By Lemma 4.9, all arrows are epimorphisms.Let uf(D[ A ]) denote the ultrafilters of D[ A ], and let pf( A ) denote the prime filters of A .The topology on the objects is the topology generated by { b α | α ∈ D[ A ] } , where b α := { µ ∈ uf(D[ A ]) | α ∈ µ } .The topology on the arrows is the topology generated by { a θ | a ∈ A } , where a θ := { P ∈ pf( A ) | a ∈ P } .The confirmations that the source, target, composition, and identity-assigning maps arecontinuous with respect to these topologies is the content of the following Lemma 4.12. Theconfirmation that the source map is a local homeomorphism is the following Lemma 4.13,and the confirmation that the target map is an open map is Lemma 4.14. It is immediatethat the objects form a Stone space, since we have used for this space exactly the standardconstruction of the Stone dual of the Boolean algebra D[ A ]. (cid:73) Lemma 4.11.
Let A be a representable { ; , A , R , t} -algebra. Then PF( A ) satisfies theassociativity and identity axioms for categories. Proof.
First the identity laws: let P be an arrow (a prime filter) and let µ = D[ P ] be itssource, so P is of the form ( µ ; a ) ↑ for some a . Then the identity arrow at µ is µ ↑ = ( µ ; 1 ) ↑ and contains 1 . By definition µ ↑ · P = ( µ ↑ ; P ) ↑ and so contains 1 ; a = a . Then sinceD[ µ ↑ · P ] = D[ µ ↑ ] = µ = D[ P ], by Lemma 4.8 we conclude µ ↑ · P = P . Similarly, P · µ ↑ contains a and has source equal to D[ P ]. Hence we also have P = µ ↑ · P .For the associativity law, by similar reasoning, if the compositions ( P · Q ) · R and P · ( Q · R )are defined, then they are nondisjoint. And since both have source D[ P ], by Lemma 4.8 theyare equal. (cid:74)(cid:73) Lemma 4.12.
Let A be a representable { ; , A , R , t} -algebra. The source, target, composition,and identity-assigning maps on the category PF( A ) are continuous with respect to the topologiesgenerated by { b α | α ∈ D[ A ] } and { a θ | a ∈ A } . Proof.
First d : we take b α and show that d − ( b α ) is open. So let P be a prime filter with α ∈ D[ P ]. Take any a ∈ P . Then b := α ; a is also in P , so P ∈ b θ . Now for any Q ∈ b θ ,we have D( b ) ∈ D[ Q ] = d ( Q ), and D( b ) ≤ α , so α ∈ d ( Q ). That is, Q ∈ d − ( b α ). So P ∈ b θ ⊆ d − ( b α ). Since P was an arbitrary element of d − ( b α ) and b θ is by definition open,we are done.Next r : we take b α and show that r − ( b α ) is open. So let P be a prime filter with α ∈ R[ P ] ↑ .Take any a ∈ P . Then b := a ; α is also in P —because a = ( a ; α ) t ( a ; A( α )), but a ; A( α )cannot be in P , else R[ P ] ↑ (which contains α ) would contain 0. Hence P ∈ b θ . Now for any Q ∈ b θ , we have R( b ) ∈ R[ Q ] ↑ = r ( Q ), and R( b ) ≤ α , so α ∈ r ( Q ). That is, Q ∈ r − ( b α ). So P ∈ b θ ⊆ r − ( b α ). Since P was an arbitrary element of r − ( b α ) and b θ is by definition open,we are done.For composition: we take a θ and show that the inverse image under · is open. So let P and Q be two prime filters such that P · Q ∈ a θ , that is, a ∈ ( P ; Q ) ↑ . Then there are b ∈ P and c ∈ Q such that b ; c ≤ a . Now D( b ; c ) ; b must also be in P , for if A( b ; c ) ; b were in P then P ; Q would contain A( b ; c ) ; b ; c = 0. So we have open sets (D( b ; c ) ; b ) θ containing P and c θ containing Q , and for any two composable arrows R ∈ (D( b ; c ) ; b ) θ and S ∈ c θ , theircomposition R · S = ( R ; S ) ↑ contains D( b ; c ) ; b ; c = b ; c ≤ a and therefore lies in a θ . Since P and Q were arbitrary subject to P · Q ∈ a θ , this proves that a θ is open. For the identity-assigning map: we take a θ and confirm that the set { µ ∈ uf(D[ A ]) | a ∈ µ ↑ } is open. Take ν in this set; then a ≥ α for some α ∈ ν . It follows that the open set b α ofobjects contains ν and is included in { µ ∈ uf(D[ A ]) | a ∈ µ ↑ } , so we are done. (cid:74)(cid:73) Lemma 4.13.
The map d : P D[ P ] is a local homeomorphism from the arrows to theobjects of PF( A ) . Proof.
We know from Lemma 4.12 that d is continuous. Next we establish that d is anopen map by showing that d [ a θ ] = (cid:91) D( a ) for any a ∈ A . Clearly if a belongs to a prime filter P then D( a ) belongs to d ( P ) = D[ P ], hence d [ a θ ] ⊆ (cid:91) D( a ). Conversely, any µ ∈ (cid:91) D( a ) is theimage under d of the element ( µ ; a ) ↑ of a θ (for D( a ) ∈ µ ensures 0 µ ; a ). Hence d is anopen map.Now any open and continuous map f is a local homeomorphism if every point in itsdomain has an open neighbourhood U such that the restriction of f to U is injective. For d we take, for any P in its domain, any a ∈ P we wish and use the open neighbourhood a θ of P . The map d is injective on a θ by Lemma 4.8. (cid:74)(cid:73) Lemma 4.14.
The map r : P R[ P ] ↑ is an open map from the arrows to the objects of PF( A ) . Proof.
We argue that r [ a θ ] = (cid:91) R( a ) for any a ∈ A . Clearly if µ ∈ r [ a θ ], that is, µ = R[ P ] ↑ for some P containing a , then µ contains R( a ), so µ ∈ (cid:91) R( a ). Conversely, suppose µ ∈ (cid:91) R( a ),that is, R( a ) ∈ µ . By Lemma 4.10, there exists a prime filter P containing a and such thatR[ P ] ↑ = µ . That is, P ∈ a θ and µ = R[ P ] ↑ = r ( P ) ∈ r [ a θ ]. (cid:74)(cid:73) Example 4.15.
Let A be the example from Example 2.10. The Boolean subalgebra D[ A ]consists of the four elements ∅ , id { , } , id { } , and id { , , } . The ultrafilters of D[ A ] (theobjects of the dual) are { id { , } , id { , , } } and { id { } , id { , , } } , which we call ‘1 ,
2’ and ‘3’respectively. The prime filters of A (the arrows of the dual) are the up-sets of the minimalnonzero elements of A , and there are four of these: id { , } , id { } , s , and c . Those thatcorrespond to identity arrows in the dual are id { , } and id { } . We can calculate that bothid { , } and s have source ‘1 ,
2’ and target ‘1 , s · s = id { , } , and so on. A suggestivediagram of the dual PF( A ) of A follows. Note how the dual is smaller and easier to depictthan the algebra. Figure 1
The dual PF( A ) of A , PF on morphisms Let h : A → B be a homomorphism of representable { ; , A , R , t} -algebras. It is immediatethat h restricts to a Boolean algebra homomorphism from D[ A ] to D[ B ]. The action of PFon morphisms of A is given by inverse image. In more detail, the continuous multivaluedfunctor PF( h ) : PF( B ) → PF( A ) is given by:for an object µ ∈ uf(D[ B ]): µ ( h | D[ A ] ) − ( µ ) , for an arrow P ∈ pf( B ): P
7→ { Q ∈ pf( A ) | Q ⊆ h − ( P ) } . . McLean 13 That the object component of PF( h ) is a well defined and continuous function follows fromclassical Stone duality. The proof that PF( h ) is a multivalued functor from the underlyingcategory of PF( B ) to that of PF( A ) is Lemma 4.17. The proof that the arrow component ofPF( h ) is continuous is Lemma 4.18. The proof that PF( h ) is star coherent is Lemma 4.19.The proof that PF is itself functorial is Lemma 4.20. (cid:73) Lemma 4.16.
For any homomorphism h : A → B and any prime filter P of B , the set h − ( P ) is partitioned into prime filters of A . Proof.
Let µ be the ultrafilter D[ P ] and ν be the ultrafilter ( h | D[ A ] ) − ( µ ). The relation a ∼ b ⇐⇒ ∃ α ∈ ν : α ; a = α ; b is easily seen to be an equivalence relation on h − ( P ). Weclaim that each ∼ -equivalence class is a prime filter.Each equivalence class is by definition nonempty. Each equivalence class is a propersubset of A , for if h − ( P ) contained 0 then P would contain h (0) = 0, in contradiction to P being a prime filter.We now argue equivalence classes are upward closed. Take a ∈ h − ( P ) and a ≥ a . As h is a homomorphism, it is order preserving, so h − maps upward closed sets to upward closedsets. Hence, as P is upward closed, h − ( P ) is upward closed. So a ∈ h − ( P ). Further, h (D( a )) = D( h ( a )) ∈ D[ P ] = µ , so D( a ) ∈ ν . Hence a ∼ a , for D( a ) ; a = a = D( a ) ; a . Weconclude that ∼ -equivalence classes are upward closed.To show that equivalence classes are downward directed, take a ∼ b and α ∈ ν such that α ; a = α ; b . By elementary reasoning about partial functions, α ; a is a lower bound for a and b , and if α ; a ∈ h − ( P ) then α ; a ∼ a . Hence we only need to show α ; a ∈ h − ( P ).We know h ( a ) ∈ P and h ( α ) ∈ µ , and so h ( α ) ; h ( a ) ∈ P . But as h is a homomorphism, h ( α ) ; h ( a ) = h ( α ; a ), so α ; a ∈ h − ( P ), as desired.Finally, to show equivalence classes satisfy the primality condition, take b t c = a ∈ h − ( P ).Then as ν is an ultrafilter of D[ A ], either D( b ) ∈ ν or A( b ) ∈ ν . If D( b ) ∈ ν then by the sameargument appearing in the previous paragraph, D( b ) ; a ∈ h − ( P ). But D( b ) ; a = b , so then b ∈ h − ( P ). Similarly, if A( b ) ∈ ν then A( b ) ; a ∈ h − ( P ). But A( b ) ; a ≤ c and h − ( P ) isupward closed, so then c ∈ h − ( P ). (cid:74)(cid:73) Lemma 4.17.
For any homomorphism h : A → B , the function/relation pair PF( h ) is amultivalued functor between the underlying categories of PF( B ) and PF( A ) . Proof.
We must show that PF( h ) validates conditions 1–3 of Definition 3.4. Let P : D[ P ] → R[ P ] ↑ be an arrow in PF( B ) (that is, a prime filter of B ) and suppose Q ⊆ PF( h )( P ) is a prime filter of A . We want to show that the ultrafilters D[ P ] andR[ P ] ↑ are mapped to the source and target of Q respectively. That is, we want to show( h | D[ A ] ) − (D[ P ]) = D[ Q ] and ( h | D[ A ] ) − (R[ P ] ↑ ) = R[ Q ] ↑ . We prove the second equality;the proof of the first is similar (but simpler). Suppose α ∈ R[ Q ] ↑ , so α ≥ β = R( b )for some b ∈ Q . Then h ( b ) ∈ P , so h ( β ) = h (R( b )) = R( h ( b )) ∈ R[ P ]. That is, β ∈ ( h | D[ A ] ) − (R[ P ]) (since β ∈ D[ A ]). As β ∈ ( h | D[ A ] ) − (R[ P ]) is an ultrafilter, by upwardclosure α ∈ ( h | D[ A ] ) − (R[ P ]) also. We have shown that ( h | D[ A ] ) − (R[ P ] ↑ ) ⊆ R[ Q ] ↑ . Thereverse inclusion follows, since both sides are ultrafilters. We want to show that for every ultrafilter µ of D[ B ], the identity arrow on ν := PF( h )( µ ) =( h | D[ A ] ) − ( µ ) belongs to PF( h )( µ ↑ ) (upward closure in B ), that is, is a subset of h − ( µ ↑ ).The identity arrow on ν is ν ↑ . Suppose a ≥ α ∈ ν . Then h ( a ) ≥ h ( α ) ∈ µ , so h ( a ) ∈ µ ↑ .That is, a ∈ h − ( µ ↑ ), as required. Suppose P · P is defined, Q ⊆ h − ( P ), and Q ⊆ h − ( P ). We know Q · Q is definedand has the same source as the prime filters that, by Lemma 4.16, partition h − ( P · P ). Choose some a ∈ Q and b ∈ Q . Then a ; b ∈ Q · Q , and h ( a ; b ) = h ( a ); h ( b ) ∈ P · P . So a ; b also belongs to h − ( P · P ). Hence the prime filter Q · Q has nonempty intersectionwith one of the prime filters partitioning h − ( P · P ). By Lemma 4.8, Q · Q equals thatprime filter. So Q · Q ⊆ h − ( P · P ), that is, Q · Q ∈ PF( h )( P · P ), as required. (cid:74)(cid:73) Lemma 4.18.
For any homomorphism h : A → B , the arrow component of the multivaluedfunctor PF( h ) : PF( B ) → PF( A ) is continuous. Proof.
Since sets of the form a θ form a subbasis for the topology on A , it suffices to show thatgiven a , . . . , a n ∈ A , the set PF( h ) − ( T i a θi ) is open in B . Suppose P ∈ PF( h ) − ( T i a θi ).Then there exists a prime filter Q ⊆ h − ( P ) with Q ∈ T i a θi . So a , . . . , a n ∈ Q . Since Q isa filter, there is an a ∈ Q with a ≤ a , . . . , a n . Then h ( a ) ∈ P , that is, P ∈ h ( a ) θ . And forany prime filter P of B : P ∈ h ( a ) θ = ⇒ h ( a ) ∈ P = ⇒ a ∈ h − ( P )= ⇒ ∃ Q ∈ pf( A ) : a ∈ Q ⊆ h − ( P ) , so for such a Q := ⇒ a , . . . , a n ∈ Q = ⇒ Q ∈ a θ , . . . , a θn = ⇒ Q ∈ \ i a θi , and hence P ∈ PF( h ) − ( T i a θi ). So the open set h ( a ) θ contains P and is included entirelywithin PF( h ) − ( T i a θi ). Hence PF( h ) − ( T i a θi ) is open. (cid:74)(cid:73) Lemma 4.19.
For any homomorphism h : A → B , the multivalued functor PF( h ) :PF( B ) → PF( A ) is star coherent. Proof.
For star injectivity, suppose P , P ∈ PF( B ) are prime filters with the same source—D[ P ] = D[ P ]—and such that Q ∈ PF( h )( P ) ∩ PF( h )( P ). So Q ⊆ h − ( P ) , h − ( P ). As Q is nonempty, we can choose some a ∈ Q , and then h ( a ) ∈ P , P . By Lemma 4.8, we get P = P .For star surjectivity, suppose µ is an ultrafilter of D[ B ], and Q is a prime filter of A withD[ Q ] = ( h | D[ A ] ) − ( µ ). Choose some element a ∈ Q . Then D( a ) ∈ h − ( µ ), so h (D( a )) ∈ µ .Since h is a homomorphism, this gives D( h ( a )) ∈ µ . Then ( µ ; h ( a )) ↑ is a prime filter of B that we will denote P . We claim that Q ∈ PF( h )( P ), that is, h [ Q ] ⊆ P . Suppose b ∈ Q .Then by the proof of Lemma 4.16, there is some α ∈ ( h | D[ A ] ) − ( µ ) such that α ; a = α ; b .We then have h ( b ) ≥ h ( α ; b ) = h ( α ; a ) = h ( α ) ; h ( a ), giving us h ( b ) ∈ ( µ ; h ( a )) ↑ = P . As b was an arbitrary element of Q , we have h [ Q ] ⊆ P and Q ∈ PF( h )( P ). As µ was arbitrary,and Q was arbitrary subject to D[ Q ] = ( h | D[ A ] ) − ( µ ), the relation PF( h ) is star surjective.For co-pseudo star surjectivity suppose U is an open set of arrows in PF( A ). Sincethe topology on the arrows of PF( A ) is generated by sets of the form a θ , we may assume U = T i a θi for some a , . . . , a n . Now suppose µ is an ultrafilter of D[ B ] and Q is a prime filterof A with Q ∈ U and R[ Q ] ↑ = ( h | D[ A ] ) − ( µ ). Then a , . . . , a n ∈ Q , and as Q is downwarddirected we can choose some a ∈ Q with a ≤ a , . . . , a n . By the second hypothesis on Q , weknow R( a ) ∈ h − ( µ ), so h (R( a )) ∈ µ . Since h is a homomorphism, this gives R( h ( a )) ∈ µ .By Lemma 4.10, there exists a prime filter P containing h ( a ) and such that R[ P ] ↑ = µ .Then h − ( P ) contains a , so one of the prime filters that, by Lemma 4.16, partition h − ( P ), . McLean 15 contains a . Call this prime filter Q . By upward closure, a , . . . , a n ∈ Q , hence Q ∈ U .Since Q is in the image under PF( h ) of P ∈ Hom( − , µ ), this completes the proof. (cid:74)(cid:73) Lemma 4.20.
The map
PF : A → C preserves composition of morphisms and identitymorphisms and hence is a functor from A to C . Proof.
Let h : A → B and h : B → C be homomorphisms of representable { ; , A , R , t} -algebras. Since the object components of PF( h ), PF( h ), and PF( h ◦ h ) are given by inverseimages of the induced maps D[ A ] → D[ B ] → D[ C ], the object components of PF( h ) ◦ PF( h )and PF( h ◦ h ) coincide. By Lemma 4.16, we can make the same conclusion for the arrowcomponents. It is evident that PF applied to an identity map yields an identity map. (cid:74) The next lemma relates to the restricted duality of Theorem 3.8. (cid:73)
Lemma 4.21. If h is a locally proper homomorphism, the multivalued functor PF( h ) :PF( B ) → PF( A ) is a functor. Proof.
Conditions 1–3 of Definition 3.4 reduce to the conditions defining a functor in the casethat the relation component of the multivalued functor is (precisely) single valued. Hence weonly need to know that PF( h ) relates each prime filter P of B to precisely one prime filter of A . This is practically the definition of a locally proper homomorphism. For by the definitionof a locally proper homomorphism, the set h − ( P ) is a prime filter. So by definition PF( h )relates P to h − ( P ). Conversely, if PF( h ) relates P to a prime filter Q , then by definition Q ⊆ h − ( P ). Since Q and h − ( P ) are both prime filters, this gives Q = h − ( P ). (cid:74) In this section, we define the contravariant functor SecCl : C → A used for the second halfof the duality. The notation for the functor stands for ‘section on a clopen’. (cid:73) Definition 5.1.
Let π : E → X be a local homeomorphism of topological spaces. A (local)section of π is a continuous function f : U → E , for some open U ⊆ X , with π ◦ f = id U . Since a local section is completely determined by its image, we will often identify it withthis image, in which case an upper-case Roman letter will be used.
SecCl on objects
Let C be a Stone étale category with objects O and arrows M all of which are epimorphisms.Then we define SecCl( C ) to be the following { ; , A , R , t} -algebra.The universe of SecCl( C ) is the set of all local sections U → M of d : M → O with U clopen.The operation ; is defined by A ; B := { a · b | a ∈ A, b ∈ B, and r ( a ) = d ( b ) } . Theconfirmation that A ; B is a section on a clopen is Lemma 5.5.The operation A is defined by A( A ) := { x | x ∈ O \ d [ A ] } . The confirmation that A( A )is a section on a clopen is Lemma 5.2.The operation R is defined by R( A ) := { x | x ∈ r [ A ] } . The confirmation that R( A ) is asection on a clopen is Lemma 5.4.The operation t is defined by A t B := A ∪ (A( A ) ; B ). The confirmation that A t B is asection on a clopen is Lemma 5.6.Proving that SecCl( C ) is representable by partial functions is achieved by verifying that itvalidates all the equations and quasiequations of Theorem 2.9; this is done in Lemma 5.7. (cid:73) Lemma 5.2.
Let A be a section with clopen domain. Then A( A ) := { x | x ∈ O \ d [ A ] } isa section with clopen domain. Proof.
By definition, d [ A ] is clopen in O , hence O \ d [ A ] is clopen in O . So A( A ) defines afunction f : x x with clopen domain, clearly with left inverse d . It remains to show thisfunction is continuous. But this is immediate, since f is a restriction—given by a restrictionof the domain—of the (continuous) identity-assigning map, and all such restrictions ofcontinuous maps are continuous. (cid:74)(cid:73) Lemma 5.3.
Let π : E → X be a local homeomorphism and U ⊆ X be open, and suppose f : U → E has left inverse π . Then f is continuous (that is, is a local section) if and only if f [ U ] is open in E . Proof.
For the forward direction let e ∈ f [ U ]. Then as π is a local homeomorphism, thereis some V open in E and containing e such that π | V is a homeomorphism onto its image π | V [ V ] = π [ V ]. As f is continuous, the set f − ( V ), which equals f − ( V ∩ f [ U ]), is open in X .But f − ( V ∩ f [ U ]) ⊆ π [ V ], so as π V is a homeomorphism the inverse image of f − ( V ∩ f [ U ])under π | V , that is, V ∩ f [ U ], is open in V . Since V itself is open, V ∩ f [ U ], which contains e , is open in E . Since e was an arbitrary element of f [ U ], the set f [ U ] is open in E .Conversely, suppose f [ U ] is open. Let V be an open subset of E , hence an open subsetof f [ U ]. We want to show that f − ( V ) is open. It suffices to show that any x ∈ f − ( V )is contained in an open neighbourhood included in f − ( V ). But this is clear, since π is alocal homeomorphism, hence maps some open W containing f ( x ), and contained in V , tothe open π [ W ] = f − ( W ), which of course contains x and is included in V . (cid:74) Note that Lemma 5.3 immediately implies that the identity arrows form an open set,since the identity-assigning map is manifestly a section. (cid:73)
Lemma 5.4.
Let A be a section with clopen domain. Then R( A ) := { x | x ∈ r [ A ] } is asection with clopen domain. Proof.
Let A correspond to the function f : U → M . Then R( A ) is the identity-assigningmap restricted to ( r ◦ f )[ U ]. As R( A ) is a restriction of a continuous function it is continuous.By Lemma 5.3 (with local homeomorphism set to d ), the set f [ U ] is open. Then since r is anopen map, ( r ◦ f )[ U ] is open. It remains to show ( r ◦ f )[ U ] is closed. Now r ◦ f is a continuousmap from a compact space ( U ) to a Hausdorff space (the space of objects). It is a basic andeasy-to-prove result of general topology that such a map is a closed map (the ‘closed maplemma’). Hence ( r ◦ f )[ U ] is indeed closed. (cid:74)(cid:73) Lemma 5.5.
Let A and B be sections with clopen domains. Then A ; B := { a · b | a ∈ A, b ∈ B, and r ( a ) = d ( b ) } is a section with clopen domain. Proof.
Let A correspond to the function f : U → M and B to g : V → M . It is clear that A ; B corresponds to a function h on a subset of O and that d is a left inverse for this function.Since h can be expressed as a composition g ◦ r ◦ f of restrictions of the continuous functions f , r , and g , we see that h is continuous. It remains to show that the domain of h is clopen.By Lemma 5.4, the set ( r ◦ f )[ U ] is clopen; hence ( r ◦ f )[ U ] ∩ V is clopen. Now the domainof h is ( r ◦ f ) − (( r ◦ f )[ U ] ∩ V ), so clopen by continuity of f and r . (cid:74)(cid:73) Lemma 5.6. If A and B are sections on clopens, then so is A t B := A ∪ (A( A ) ; B ) . . McLean 17 Proof.
It is immediate that A t B defines a function with left inverse d . By Lemma 5.5and Lemma 5.2, we know A( A ) ; B is a section on a clopen, and hence the domain of A t B is clopen. It remains to argue that A t B is continuous. But this is a function given bythe union of two continuous functions with open domains, which always yields a continuousfunction. (cid:74)(cid:73) Lemma 5.7.
The { ; , A , R , t} -algebra SecCl( C ) validates the axioms for representability bypartial functions listed in Theorem 2.9. Proof.
We state each axiom anew before giving a justification for its validity. A ; ( B ; C ) = ( A ; B ) ; C Clear.A( A ) ; A = A( B ) ; B Both sides yield the empty set. A ; A( B ) = A( A ; B ) ; A Suppose c ∈ A ; A( B ). Then c ∈ A and there is no arrow in B whose source is r ( c ). Since c is the unique arrow in A with d ( c ), there is then no pair of composable arrows a ∈ A and b ∈ B with d ( a ) = d ( c ). Hence 1 d ( c ) is in A( A ; B ), so c belongs to the right-hand side.Conversely, suppose c ∈ A( A ; B ) ; A , then in particular there is no ( A, B )-path from d ( c )so in particular, no arrow in B whose source is r ( c ). Hence c ∈ A ; A( B ).D( A ) ; B = D( A ) ; C ∧ A( A ) ; B = A( A ) ; C → B = C For any partition
I, J of the identity arrows into two parts, any set D of arrows is the unionof I ; D and J ; D . Since D( A ) , A( A ) is such a partition, the validity of the quasiequationfollows.1 ; A = A We noted that A( A ) ; A yields the empty set. So 1 is by definition A( ∅ )—precisely theidentity arrows. The equation is then clear.0 ; A = A Clear.D(R( A )) = R( A )The set R( A ) is a set of identity arrows, and D := A is the identity operation on any suchset. A ; R( A ) = A By definition R( A ) is all identities on objects that are the target of some arrow in A . So thevalidity of the equation is clear. A ; B = A ; C → R( A ) ; B = R( A ) ; C This is the only case of note, for we must use the fact that all arrows of C are epimorphisms.Assume the antecedent holds, and suppose b ∈ R( A ) ; B . Then b ∈ B and there is some a ∈ A whose target is d ( b ). So a · b ∈ A ; B and hence, by the supposition, a · b ∈ A ; C .Hence a · b = a · c , for some a ∈ A and c ∈ C , though necessarily a = a , as A is a sectionand d ( a ) = d ( a · b ) = d ( a · c ) = d ( a ). Since a is an epimorphism, we obtain b = c , hence b ∈ C , and therefore b ∈ R( A ) ; C . The reverse inclusion is by a symmetric argument.D( A ) ; ( A t B ) = A If a ∈ D( A ) ; ( A t B ) then firstly there is an arrow in A with the same source as a . Secondly,by the definition of A t B on local sections, a is in either A or A( A ) ; B . But a cannot be inA( A ); B , since that implies there is not an arrow in A with the same source as a . Hence a ∈ A .Conversely, if a ∈ A then a ∈ A t B and 1 d ( a ) ∈ D( A ), so 1 d ( a ) · a = a ∈ D( A ) ; ( A t B ).A( A ) ; ( A t B ) = A( A ) ; B If a ∈ A( A ) ; ( A t B ) then firstly there is no arrow in A with the same source as a .Secondly, by the definition of A t B on local sections, a is in either A or A( A ) ; B —itmust be A( A ) ; B . Conversely, if a ∈ A( A ) ; B then immediately a ∈ A t B , using thedefinition of t on local sections again. That a ∈ A( A ) ; B also implies 1 d ( a ) ∈ A( A ). Hence a = 1 d ( a ) · a ∈ A( A ) ; ( A t B ). (cid:74)(cid:73) Example 5.8.
Returning again to our running example (Example 2.10 and Example 4.15),the reader may examine the (discrete) category PF( A ) described in Example 4.15 andcalculate its dual. Sections must contain at most one arrow with source ‘1 ,
2’ (there are threesuch arrows, so four possible choices) and at most one arrow with source ‘3’ (one arrow, sotwo choices). In total there are eight sections, so eight element of the dual of PF( A ). Onecan verify that this ‘double dual’ is isomorphic to the original algebra A . SecCl on morphisms
The action of SecCl on morphisms is given by inverse image. That is, given C , D Stoneétale categories whose arrows are epimorphisms and a star-coherent multivalued functor F : C → D , we defineSecCl( F ) : SecCl( D ) → SecCl( C ) A F − ( A ) . First it must be checked that F − ( A ) really is an element of SecCl( C ). If b , b ∈ F − ( A )have the same source, then as F is a multivalued functor, arrows in F ( b ) and F ( b ) all sharethe same source. As A is a section, b and b must be inverse images under F of the same a ∈ A . Then by star injectivity of F , we get b = b . Hence F − ( A ) defines a function on d [ F − ( A )]. As F is by assumption continuous, and d [ A ] is clopen, F − ( d [ A ]) = d [ F − ( A )](star surjectivity) is clopen. Since F − ( A ) is open, Lemma 5.3 implies that it correspondsto a continuous function. That SecCl( F ) is a homomorphism of { ; , A , R , t} -algebras isLemma 5.9. It is immediate from its definition by inverse image that SecCl is functorial,that is, respects compositions of star-coherent multivalued functors and acts as the identityon identity functors. (cid:73) Lemma 5.9.
The map
SecCl( F ) : SecCl( D ) → SecCl( C ) of { ; , A , R , t} -algebras is ahomomorphism. . McLean 19 Proof.
We start by showing that ; is preserved. Let A and B be sections of D . If c ∈ F − ( A ) ; F − ( B ) that means c = c · c for some c , c such that there exist a ∈ A ∩ F ( c )and b ∈ B ∩ F ( c ). Then by functoriality of F , we know a · b ∈ F ( c ), so a · b ∈ ( A ; B ) ∩ F ( c ).That is, c ∈ F − ( A ; B ). Hence F − ( A ) ; F − ( B ) ⊆ F − ( A ; B ). Conversely, if c ∈ F − ( A ; B ),with a · b ∈ F ( c ) say, then by star surjectivity applied at d ( a ) there is some c with the samesource as c and with a ∈ F ( c ). Applying star surjectivity again at r ( a ) = d ( b ) we obtainsome c with c · c = c and b ∈ F ( c ). Hence c ∈ F − ( A ) ; F − ( B ), and we conclude that F − ( A ) ; F − ( B ) = F − ( A ; B ).Next we show that A is preserved. Let B be a section of D . First suppose a ∈ F − (A( B )),with 1 y ∈ F ( a ) ∩ A( B ) say. Then by functoriality of F , the identity arrow 1 y belongsto F (1 d ( a ) ). By star injectivity of F , we find a = 1 d ( a ) . Since 1 y ∈ A( B ) there is nomember of B with source y , hence there is no member of F − ( B ) with source d ( a ). Soby definition, 1 d ( a ) ∈ A( F − ( B )), that is, a ∈ A( F − ( B )). We have our first inclusion: F − (A( B )) ⊆ A( F − ( B )). Conversely, suppose 1 x ∈ A( F − ( B )). We want to argue that1 F ( x ) is in A( B ). But if there were a member b of B with source F ( x ), then by star surjectivityof F , there would be an arrow with source x whose image under F contains b , contradictingthe fact that 1 x ∈ A( F − ( B )). Hence 1 F ( x ) is indeed in A( B ), and since 1 F ( x ) ∈ F (1 x ) thisgives 1 x ∈ F − (A( B )). We conclude that F − (A( B )) = A( F − ( B )).Showing that R is preserved is fairly similar. Let B be a section of D with a clopendomain. By Lemma 5.3, the set B is open. First suppose we have an element in F − (R( B )).By functoriality and star injectivity of F , our element of F − (R( B )) is an identity element,1 y say. By definition of R, there is some b ∈ B with target F ( y ). As B is open, by co-pseudostar surjectivity of F there is some b ∈ B belonging to F ( a ) for some a with target y .So a ∈ F − ( B ), hence 1 y ∈ R( F − ( B )). We conclude that F − (R( B )) ⊆ R( F − ( B )).Conversely, suppose 1 y ∈ R( F − ( B )). So there is an a with target y such that F ( a ) ∩ B = ∅ .Then 1 F ( y ) ∈ R( B ), so 1 y ∈ F − (R( B )). We conclude that F − (R( B )) = R( F − ( B )).Finally we note that t is preserved, because of the definition A t B := A ∪ A( A ) ; B andthe elementary fact that relation inverse images preserve unions. (cid:74) In this section we will first show that the double dual functor on the category A ofrepresentable { ; , A , R , t} -algebras is naturally isomorphic to the identity functor. Thenwe will do the same for the double dual functor on the category C of Stone étale categoriesall of whose arrows are epimorphisms. This will complete the proof that we have given aduality between the categories A and C . First we describe an isomorphism from A to SecCl(PF( A )) for an arbitrary representable { ; , A , R , t} -algebra. Then we will show this construction is natural.In fact, our isomorphism is already hidden in notation we have defined. Recall that for a ∈ A , the set a θ is defined to be { P ∈ pf( A ) | a ∈ P } . We define θ : A → SecCl(PF( A ))by a a θ . Note that a θ is indeed an element of the algebra SecCl(PF( A )), for it is clearlya section, and its domain is the set (cid:91) D( a ) := { µ ∈ uf(D[ A ]) | D( a ) ∈ µ } , which is open bydefinition and closed because (cid:91) A( a ) is open.To see that θ is injective, it suffices to show that when a b there exists a prime filtercontaining a but not b . This argument can be found in the proof of Lemma 4.9 in [15]. Tosee that θ is surjective, we need to argue that all sections on clopens of PF( A ) are of the form a θ . Let A be a section on a clopen of PF( A ). By a similar argument to that in theproof of Lemma 4.18, for each P ∈ A there is an a P with P ∈ a θP ⊆ A . By compactness ofthe space of objects of PF( A ), the domain of A can be covered by (cid:92) D( a P ) , . . . , (cid:92) D( a P n ) forsome finite n . Then A = ( a P t · · · t a P n ) θ . (cid:73) Lemma 6.1.
The map θ : A → SecCl(PF( A )) given by a a θ is a homomorphism of { ; , A , R , t} -algebras Proof.
To show ; is preserved, let a, b ∈ A . If P ∈ a θ ; b θ , that means there are P containing a and P containing b such that ( P ; P ) ↑ = P . Hence a ; b ∈ P , giving P ∈ ( a ; b ) θ . Conversely,suppose P ∈ ( a ; b ) θ . Then P is of the form ( µ ; a ; b ) ↑ , where µ is the ultrafilter D [ P ] ofD[ A ]. As P does not contain 0, neither does ( µ ; a ) ↑ , which is therefore a prime filter, P say. Let ν be the ultrafilter R[ P ] ↑ . Using again the fact that P = ( µ ; a ; b ) ↑ = ( µ ; a ; ν ; b ) ↑ does not contain 0, the filter ( ν ; b ) ↑ is a prime filter, P say. Then P ∈ a θ and P ∈ b θ with P · P = P , hence P ∈ a θ ; b θ .To show A is preserved, let a ∈ A . If P ∈ A( a θ ), then P = µ ↑ for some ultrafilter µ ofD[ A ], and there is no prime filter containing a with source µ . Hence the filter ( µ ; a ) ↑ is notproper. That is, there is some α ∈ µ such that α ; a = 0. It is a property of partial functionsthat this implies α ≤ A( a ), so, by upward closure of µ ↑ , we know A( a ) ∈ µ ↑ = P . Hence P ∈ A( a ) θ . Conversely, suppose P ∈ A( a ) θ . Then A( a ) ∈ P so 1 ∈ P . Hence P is of theform ( µ ; 1 ) ↑ = µ ↑ for some ultrafilter µ . Hence P is the identity arrow for µ in the categoryPF( A ). We know A( a ) ∈ µ . If there were a prime filter P containing a with source µ thenD( a ) would also be in µ , contradicting µ being proper. Hence there is no such P , and weconclude P ∈ A( a θ ).To show R is preserved, let a ∈ A . If P ∈ R( a θ ), then P = µ ↑ for some ultrafilter µ ofD[ A ], and there is some prime filter P containing a with target R[ P ] ↑ = µ . Then R( a ) ∈ µ ,so R( a ) ∈ µ ↑ = P . Hence P ∈ R( a ) θ . Conversely, suppose P ∈ R( a ) θ . Then R( a ) ∈ P so1 ∈ P . Hence P is of the form ( µ ; 1 ) ↑ = µ ↑ for some ultrafilter µ . Hence P is the identityarrow for µ in the category PF( A ). We know R( a ) = D(R( a )) ∈ D[ P ] = µ . By Lemma 4.10there exists a prime filter P containing a with target µ . We conclude P ∈ R( a θ ).To see that t is preserved, note that if a prime filter contains either a t b = a t (A( a ) ; b )it contains a or A( a ) ; b (by primality), and if it contains a or A( a ) ; b then it contains a t b (by upward closure). Thus ( a t b ) θ = a θ ∪ (A( a ) ; b ) θ . The latter, given we know ; and A arepreserved by θ , equals a θ ∪ (A( a θ ) ; b θ ), which by definition is a θ t b θ . (cid:74) We now show that our isomorphisms together give a natural transformation from theidentity functor to the double dual. For each A ∈ A , denote now the isomorphism justdescribed by θ A . Then given A , B ∈ A and a homomorphism h : A → B , we are required toshow that SecCl(PF( h )) ◦ θ A = θ B ◦ h . The right-hand side sends an element a ∈ A to theset h ( a ) θ of prime filters of B . Seeing that the left-hand side has the same effect just involvesunravelling the definitions. The element a is sent first to a θ , then SecCl(PF( h )) sends this to { P ∈ PF( B ) | PF( h )( P ) ∈ a θ } = { P ∈ PF( B ) | h − ( P ) ∈ a θ } = { P ∈ PF( B ) | a ∈ h − ( P ) } = { P ∈ PF( B ) | h ( a ) ∈ P } = h ( a ) θ as required. . McLean 21 Let
C ∈ C , and write Id(SecCl( C )) for the elements of SecCl( C ) consisting entirely of identityarrows. Define ϕ : C →
PF(SecCl( C )) by:for an object x : x x ϕ := { A ∈ Id(SecCl( C )) | x ∈ A } , for an arrow c : c c ϕ := { A ∈ SecCl( C ) | c ∈ A } . We first verify that x ϕ is an ultrafilter of D[SecCl( C )], and c ϕ is a prime filter of SecCl( C ),so ϕ indeed has codomain PF(SecCl( C )).For x ϕ , the identity-assigning map is a section on a clopen, so x ϕ is nonempty. If A, B ∈ Id(SecCl( C )) then A ∩ B ∈ Id(SecCl( C )), so x ϕ is downward directed. It is trivialthat x ϕ is upward closed and clear that for A ∈ Id(SecCl( C )) precisely one of A and A( A ) isin x ϕ .For c ϕ , as d is a local homeomorphism from the arrows to the objects of C , there existssome section s on an open that has c in its image. Since the space of objects of C has abasis of clopens, we can restrict s to a clopen still with c in its image. Hence c ϕ is nonempty.By a similar argument, c ϕ is down directed. It is trivial that c ϕ is upward closed andstraightforward that it satisfies the primality condition. Clearly ∅ 6∈ c ϕ , and ∅ is the 0 ofSecCl( C ), so c ϕ is proper. (cid:73) Lemma 6.2.
The map ϕ is a functor between the categories C and PF(SecCl( C )) . Proof.
Let ( c : x → y ) ∈ C . To see that c ϕ : x ϕ → y ϕ , suppose A ∈ c ϕ . Then c ∈ A , so x = d ( c ) ∈ d [ A ], so 1 x ∈ D( A ). Hence D( A ) ∈ x ϕ . We conclude that D[ c ϕ ] ⊆ x ϕ . Since bothare ultrafilters, they are equal. Hence d ( c ϕ ) = D[ c ϕ ] = x ϕ . Similarly, we have y = r ( c ) ∈ r [ A ],so 1 y ∈ R( A ). Hence R( A ) ∈ y ϕ . We conclude that R[ c ϕ ] ⊆ y ϕ . Since the ultrafilter y ϕ isupward closed, we get R[ c ϕ ] ↑ ⊆ y ϕ , then since both these are ultrafilters, they are equal.Hence r ( c ϕ ) = R[ c ϕ ] ↑ = y ϕ .Next we argue that for any object x , we have (1 x ) ϕ = 1 x ϕ . The right-hand side is bydefinition ( x ϕ ) ↑ . That is, an element A of 1 x ϕ is an upper bound for (that is, superset of)some identity section A that contains 1 x . Hence A itself contains 1 x , so A ∈ (1 x ) ϕ . Weconclude (1 x ) ϕ ⊇ x ϕ . Since both are prime filters, they are equal.Let ( c : x → y ) , ( c : y → z ) ∈ C . We know that c ϕ · c ϕ : x → z is defined, in particular isa prime filter. To see that c ϕ · c ϕ = ( c · c ) ϕ , we have A ∈ c ϕ and B ∈ c ϕ = ⇒ c ∈ A and c ∈ B = ⇒ c · c ∈ A ; B = ⇒ A ; B ∈ ( c · c ) ϕ . Hence c ϕ · c ϕ ⊆ ( c · c ) ϕ . Since both are prime filters, they are equal. (cid:74) To see that ϕ is injective (on arrows, therefore on objects) let c = d ∈ C . Choose anysection on a clopen A that contains c . If d ( c ) = d ( d ), then A cannot contain d , so c ϕ and d ϕ are not equal. If d ( c ) = d ( d ), then we can find a clopen set U of objects that contains d ( c )but not d ( d ). By restricting A to U , we obtain a section on a clopen containing c but not d ,so again c ϕ and d ϕ are not equal. To see that ϕ is surjective (on arrows, therefore on objects), take a prime filter P ofsections on clopens of C . Let S = T P . If c ∈ S then P ⊆ c ϕ , and so P = c ϕ . Hencewe only need to show S cannot be empty. Let A ∈ P . As P is a filter, T P = ∅ implies T { B ∈ P | B ⊆ A } = ∅ . As A is compact (being the continuous image of a compact set),this implies B ∩ · · · ∩ B n = ∅ for some B , . . . , B n ∈ { B ∈ P | B ⊆ A } . As P is downwarddirected, this implies ∅ ∈ P —the required contradiction.Since ϕ is bijective, it is certainly star-coherent. To show ϕ is an isomorphism in C itremains to show that ϕ and its inverse are continuous. First we need a lemma. (cid:73) Lemma 6.3.
In any Stone étale category, the (images of) sections on clopens provide abasis for the topology on the arrows.
Proof.
Let U be an open set of arrows, and suppose c ∈ U . As d is a local homeomorphism, c has an open neighbourhood V , which we may assume is a subset of U , such that d | V provides a homeomorphism onto its image, which is also open. That is, d | − V is a section onan open. As the set of objects has a clopen basis, we may restrict d | − V to a clopen containing d ( c ), giving the section on a clopen containing c and included in U that we seek. (cid:74) By the lemma, to show that ϕ is an open map, it suffices to consider an arbitrary sectionon a clopen A of C . We claim that ϕ [ A ] equals A θ and is therefore in particular open. If c ϕ ∈ ϕ [ A ] (for c ∈ A ), then as A is a section on a clopen, A ∈ c ϕ , so c ϕ ∈ A θ . We concludethat ϕ [ A ] ⊆ A θ . Now let P ∈ A θ , and suppose for a contradiction that P ϕ [ A ]. Sincefor prime filters (which are maximal filters) inclusion implies equality, P ϕ [ A ] implies forall c ∈ A we have c ϕ P . That is, we can find, for each c ∈ A , a section on a clopen B c that contains c , but is not in P . We may assume (by the same reasoning as in the proofof Lemma 6.3) that each B c is a subset of A . Then as A is compact, some finite collection B c , . . . , B c n cover A . That is, A = B c t · · · t B c n in SecCl( C ). As P is prime and contains A , it must contain some B c i —the required contradiction.Continuity of ϕ now follows straightforwardly. That ϕ [ A ] = A θ and ϕ is injective implies ϕ − ( A θ ) = A , which is open if A is a section on a clopen. The set of A θ ’s such that A is asection on a clopen provides a basis for PF(SecCl( C )), so we are done.We now show that our isomorphisms together give a natural transformation from theidentity functor to the double dual. For each C ∈ C , denote now the isomorphism justdescribed by ϕ C . Then given C , D ∈ C and a star-coherent multivalued functor F : C → D ,we are required to show that PF(SecCl( F )) ◦ ϕ C = ϕ D ◦ F (as multivalued functors). Theright-hand side sends an arrow c ∈ C to { d ϕ | d ∈ F ( c ) } . On the left-hand side, c is first sentto c ϕ , then PF(SecCl( F )) sends this to the set of prime filters of SecCl( F ) that partition { A ∈ SecCl( D ) | SecCl( F )( A ) ∈ c ϕ } = { A ∈ SecCl( D ) | F − ( A ) ∈ c ϕ } = { A ∈ SecCl( D ) | c ∈ F − ( A ) } = { A ∈ SecCl( D ) | F ( c ) ∩ A = ∅} . And as the prime filters of SecCl( F ) are of the form d ϕ , the set of prime filters partitioning { A ∈ SecCl( D ) | F ( c ) ∩ A = ∅} is { d ϕ | d ∈ F ( c ) } , exactly as required. This completes theproof of Theorem 3.7.We now complete the proof of the restricted duality of Theorem 3.8. (cid:73) Lemma 6.4. If F is a functor, the homomorphism SecCl( F ) : SecCl( D ) → SecCl( C ) of { ; , A , R , t} -algebras is locally proper. . McLean 23 Proof.
Take P an arbitrary prime filter in SecCl( C ). By the preceding discussion we know P is of the form c ϕ for some arrow c of C . SoSecCl( F ) − ( P ) = { A ∈ SecCl( D ) | c ∈ SecCl( F )( A ) } = { A ∈ SecCl( D ) | c ∈ F − ( A ) } = { A ∈ SecCl( D ) | F ( c ) ∈ A } = F ( c ) ϕ , which we know is a prime filter of SecCl( D ). (cid:74) Now to finish the proof of Theorem 3.8, note that by Lemma 4.21 and Lemma 6.4, thedouble dual of any locally proper homomorphism is locally proper , and the double dual of anystar-coherent functor is a functor . Then θ and ϕ provide the required natural isomorphisms,since isomorphisms of representable { ; , A , R } -algebras are locally proper, and isomorphismsof topological categories are functors. Let Σ be a finite alphabet. Recall that the rational functions over Σ, which we denoteRat f (Σ), are the partial functions from Σ ∗ to Σ ∗ realisable by a one-way transducer. The regular functions over Σ, which we denote Reg f (Σ), are the partial functions from Σ ∗ toΣ ∗ realisable by a two-way transducer. (See [9] for an overview of these concepts.) Therational and the regular functions are both closed under ;, A, R, and t and hence are both { ; , A , R , t} -algebras of partial functions with base Σ ∗ . ClearlyRat f (Σ) ⊆ Reg f (Σ) , and so Rat f (Σ) is a subalgebra of Reg f (Σ). For both Rat f (Σ) and Reg f (Σ), the subalgebraof subidentity functions is the set of identity functions on regular languages (by a simple‘forgetting the output’ argument). The Stone dual of the regular languages over Σ is knownto be (the underlying space of) the profinite completion c Σ ∗ of the monoid Σ ∗ [19]. Henceboth Rat f (Σ) and Reg f (Σ) have duals whose space of objects is c Σ ∗ .In the setting of languages , duality provides a powerful and, in principle, fully generalmethod for characterising any family of languages that forms a sublattice of the regularlanguages [11, 10]. For example, for the subalgebra of star-free languages this yields thecharacterisation by the profinite equation x ω x = x ω , equivalent to Schützenberger’s celebratedcharacterisation by aperiodicity of the syntactic monoid of the given language [1]. It wouldbe useful to have similar tools available for regular functions.Since the category of representable { ; , A , R , t} -algebras with homomorphisms is a concretecategory, the embedding Rat f (Σ) , → Reg f (Σ) is a monomorphism in that category. (cid:73) Problem 7.1.
Is the embedding
Rat f (Σ) , → Reg f (Σ) locally proper? The answer is no, by the following general result. (cid:73)
Proposition 7.2.
Let h : A → B be a locally proper homomorphism of { ; , A , R , t} -algebrassuch that the induced map D[ A ] → D[ B ] is an isomorphism. Then h is an isomorphism. Proof.
If the hypotheses hold, then the dual F : PF( B ) → PF( A ) of h is a functor anda bijection on objects. But any star-coherent functor that is bijective on objects must bebijective on arrows, and thus F is an algebraic isomorphism of categories. It follows in particular (surjectivity of F ) that h is injective, so we may assume h is an inclusion of asubalgebra, and the induced D[ A ] → D[ B ] is the identity.We know that F is continuous, so to show that F is an isomorphism of topological categories it only remains to show F is an open map. From there the conclusion that h is anisomorphism is immediate, by duality.The topology on PF( B ) is generated by sets of the form θ B ( b ) for b ∈ B . We must showthat each F [ θ B ( b )] is open in PF( A ). For P ∈ B , we have F ( P ) = h − ( P ) = P ∩ A . Let µ = D[ P ] = D[ F ( P )]. Choose some a ∈ F [ P ] ⊆ P . We know F [ P ] = ( µ ; a ) ↑ , with upwardclosure taken in A , and P = ( µ ; a ) ↑ , with upward closure taken in B . That is the algebraicinverse F − of F is given by taking the upward closure in B . So, if b ∈ P then there is some a ≤ b ∈ P ∩ A , that is, F ( P ) ∈ θ A ( a ), and θ A ( a ) is included in F [ θ B ( b )]. Hence F [ θ B ( b )] isopen, for arbitrary b ∈ B . (cid:74) Since Rat f (Σ) and Reg f (Σ) have the same Boolean subalgebra of domain elements—theregular languages encoded as subidentity functions—if the embedding Rat f (Σ) , → Reg f (Σ)were locally proper, Proposition 7.2 would apply. But Rat f (Σ) is strictly included in Reg f (Σ),hence the embedding cannot be locally proper.We conclude with a problem. (cid:73) Problem 7.3.
Give descriptions of the duals of
Rat f (Σ) and Reg f (Σ) and of the dual ofthe embedding Rat f (Σ) , → Reg f (Σ) . References On finite monoids having only trivial subgroups.
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