Order Polarities

Home , Join and meet

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Stone’s representation theorem [31] to Boolean algebras with operators (BAOs). The approach there was to first (non-constructively) dualize to relational structures, then construct the canonical extension from these. Early generalizations to distributive lattices used Priestly duality [26, 27] in a similar way (see for example [14, 29, 30]). More recent approaches using polarities bypass the dual construction, which is significantly more complicated outside of the distributive setting, and have the additional advantage of being constructive [12, 7]. Indeed, an innovation of [7] is to use the canonical extension of a poset to construct a dual, which can then play the same role in providing completeness results for substructural logics as the canonical frame does in the modal setting (see e.g. [5, Chapters 4 and 5]). For more on the development of the theory of canonical extensions see, for example, [17] or the introduction to [19]. We note that for operations that are not operators in the sense of [20], the canonical extension construction is ambiguous, as there are often several non-equivalent choices for the lifts of each operation, each of which may be ‘correct’ depending on the situation (see for example the epilogue of [15] for a brief discussion of this). Moreover, for posets, what is meant by the canonical extension is even less clear than it is in the lattice case. This is a consequence of ambiguities surrounding the notions of ‘filters’ and ‘ideals’ in the more general setting. See [24] for a thorough investigation of this issue. For canonical extensions in their various guises to play a role in ‘completeness via canonicity’ arguments, general results concerning the preservation of equations and inequalities are extremely useful. Some results of this sort can be found in [32, 33], where arguments from [18] are extended to more general settings. One component of these arguments is the exploitation of the so called intermediate structure, an extension of the original poset intermediate between it and the canonical extension. The idea is that operations are, in a sense, lifted first to the intermediate structure, and then to the canonical extension. More generally, the class of ∆1-completions [13] includes canonical extensions (however we define them), and also others such as the MacNeille (aka Normal) completion. Given a poset P , the ∆1-completions of P are, modulo suitable concepts of isomorphism, in one-to-one correspondence with certain kinds of polarities constructed from P [13, Theorem 3.4]. Here also the intermediate structure appears. Indeed, a ∆1-completion is the MacNeille completion of its intermediate structure [13, Section 3].

1.2. What is done here. In the existing literature, the intermediate structure emerges almost coincidentally from the construction of a completion. Given a polarity (X,Y, R), first a complete lattice G(X,Y, R) is constructed using the antitone Galois connection between ℘(X) and ℘(Y ) induced by R, as we explain in more detail in Section 2.3. The intermediate structure is then found sitting inside it as a subposet. There are natural maps from X and Y into the intermediate structure, and, if these are injective, partial orderings are thus induced on X and Y . When G(X,Y, R) is an extension of a poset P , it will also follow that X and Y are extensions of P . It turns out that the preorder on X ∪ Y induced by the intermediate structure agrees with R on X × Y . The broad goal of this paper is to take the idea of a polarity involving order extensions eX : P → X and eY : P → Y as primitive, and develop a theory from this. More explicitly, we are interested in the interaction between the relation R ORDER POLARITIES 3 and the orders on X and Y , and, in particular, under what circumstances something corresponding to the ‘intermediate structure’ can be defined on X ∪ Y . This issue raises several questions, depending on exactly what properties we think an ‘intermediate structure’ should have. Based on our answers to these questions, we define a sequence of so-called coherence conditions for polarities. The choice of name here comes from the idea that we have three things, ≤X , ≤Y and R, all giving us information about a possible preorder on X ∪ Y with certain properties, and we want this information to not contradict itself. The bulk of this work is done in Section 3, where the main definitions are made, and in Section 5, where, among other things, we prove our defined conditions are strictly increasing in strength. In Section 4 we define a Galois polarity to be a triple (eX ,eY , R) satisfying the strongest of our coherence conditions, and with the additional property that eX is a meet-extension, and eY is a join-extension. The ‘aptness’ of this definition is partly demonstrated by the fact that, if (eX ,eY , R) is a Galois polarity, there is one and only one possible preorder structure on X ∪ Y that agrees with the orders on X and Y , agrees with R on X × Y , and also preserves meets and joins from the base poset P (see Theorem 4.12 for a more precise statement). Galois polarities are studied further in Section 7. First we justify the choice of terminology by demonstrating that, for Galois polarities, the unique preorder structure described above can be defined in terms of a Galois connection between any join-preserving join-completion of Y and any meet-preserving meet-completion of X. This requires some technical results on extending and restricting polarity relations, which we provide in Section 6. Here we investigate the ‘simplest’ way we might hope to extend a relation between posets to a relation between meet- and join-extensions of these posets, and conversely the simplest way we might restrict a relation between extensions to a relation between the original posets. In particular we prove that it is rather common for coherence properties of a polarity to be preserved by extension and restriction as we define them. By defining suitable morphisms, we can equip the class of Galois polarities with the structure of a category. This can be seen as a generalization of the concept of a δ-homomorphism from [16, Section 4]. We define an adjunction between this category and the category of ∆1-completions (see Theorem 7.29). This produces the correspondence between ∆1-completions of a poset and certain kinds of polarities from [13, Theorem 3.4] via the categorical equivalence between fixed subcategories. In the long term we imagine handling lifting of operations, and the preservation of inequalities and so on, to ‘intermediate structures’ induced by polarities, and Galois polarities in particular. This is, of course, not an entirely new idea. Indeed, we have mentioned previously that lifting operations to canonical extensions is often done by first lifting to the intermediate structure. The hope is that, by shifting the focus a little from intermediate structures as they emerge in the construction of completions, to intermediate structures as algebraic objects of interest in their own right, some new insight might be gained. However, to control the length of this document, we leave the pursuit of this rather vague goal to future work.

2. Orders and completions 2.1. A note on notation. We use the following not entirely standard notations: 4 ROB EGROT

• Given a poset P and p ∈ P , we define p↑ = {q ∈ P : q ≥ p} and p↓ = {q ∈ P : q ≤ p}. • Given a function f : X → Y , and given S ⊆ X, we define f[S]= {f(x): x ∈ S}. • With f as above and with y ∈ Y and T ⊆ Y we define f −1(y)= {x ∈ X : f(x)= y} and f −1(T )= {x ∈ X : f(x) ∈ T }. • If P is a poset, then P ∂ is the order dual of P . • If X and Y are sets, then we may refer to a relation R ⊆ X × Y as being a relation on X × Y . 2.2. Extensions and completions. We assume familiarity with the basics of order theory. Textbook exposition can be found in [6]. In this subsection we provide a brisk introduction to some more advanced order theory concepts. This serves primarily to establish the notation we will be using. Definition 2.1. Let P and Q be posets. We say an order embedding e : P → Q is a poset extension, or just an extension. If Q is also a complete lattice we say e is a completion. If for all q ∈ Q we have q = e[e−1(q↑)], then we say e is a meet-extension, or a meet-completion if Q is a complete lattice. Similarly, if q = e[e−1(q↓)] for all q ∈ Q, then e is a join-extensionV ,or a join-completion when Q is complete. W Note that it is common in the literature to refer to completions using the codomain of the function. For example, we might say “Q is a completion of P ” when talking about the completion e : P → Q. This has the disadvantage of obfuscating the issue of what it means for two extensions to be isomorphic, as an isomorphism between codomains is not sufficient for extensions to be isomorphic in the sense used here (see below). This rarely causes significant problems in practice, as it is usually clear from context what kind of isomorphism is required. However, we find the identification of extensions with maps to be more elegant, and will generally use this approach. Definition 2.2 (Morphisms between order preserving maps and extensions). Given posets P1, P2,Q1,Q2, and order preserving maps f1 : P1 → Q1 and f2 : P2 → Q2, a map, or morphism, from f1 to f2 is a pair of order preserving maps gP : P1 → P2 and gQ : Q1 → Q2 such that the diagram in Figure 1 commutes. If gp and gQ are both order isomorphisms, then we say f1 and f2 are isomorphic. If f1 : P → Q1 and f2 : P → Q2 are extensions of a poset P , then f1 and f2 are isomorphic as extensions of P if they are isomorphic in the sense described above and the map gP is the identity on P . Definition 2.2 equips the class of order preserving maps between posets, and in particular the subclass of poset extensions, with the structure of a category. We will make frequent use of the idea of extensions being isomorphic, and we will return to the idea of a category of extensions in Section 7.4. Definition 2.3. Given a poset P , the MacNeille completion of P is a map e : P → N (P ) that is both a meet- and a join-completion. ORDER POLARITIES 5

f1 P1 / Q1

gP gQ P2 / Q2 f2

Figure 1.

The MacNeille completion was introduced in [23] as a generalization of Dedekind’s construction of R from Q. It is unique up to isomorphism. The characterization used here is due to [1]. See e.g. [6, Section 7.38] for more information. Note that a MacNeille completion e is completely join- and meet-preserving. Definition 2.4. The canonical extension of a lattice L is a completion e : L → Lδ such that: (1) e[L] is dense in Lδ. I.e. Every element of Lδ is expressible both as a join of meets, and as a meet of joins, of elements of e[L]. (2) e is compact. I.e. for all S,T ⊆ L, if e[S] ≤ e[T ], then there are finite S′ ⊆ S and T ′ ⊆ T with S′ ≤ T ′. V W Canonical extensions are also uniqueV upW to isomorphism. This characterization, and the proof that such a completion exists for all L, is due to [12]. It generalizes the definitions of the canonical extension for Boolean algebras [20], and distributive lattices [14]. The construction used in [12] can, as noted in Remark 2.8 of that paper, also be used for posets, and will again result in a dense completion. However, the kind of compactness obtained is weaker. This idea is expanded upon in [7]. The differences between the lattice and poset cases arise from the fact that definitions for filters and ideals which are equivalent for lattices are not so for posets. This issue is discussed in detail in [24]. One way to address this systematically is to talk about the canonical extension of P with respect to F and I, where F and I are sets of ‘filters’ and ‘ideals’ of P respectively. By making the definitions of ‘filter’ and ‘ideal’ weak enough, this allows all notions of the canonical extension of a poset to be treated in a uniform fashion. This is the approach taken in [25], for example.

Deﬁnition 2.5. Given a poset P , a ∆1-completion of P is a completion e : P → D such that e[P ] is dense in D.

∆1-completions, introduced in [13], include both MacNeille completions and canonical extensions. As such they are not usually unique up to isomorphism, so it doesn’t make sense to talk about the ∆1-completion. Deﬁnition 2.6. Let P and Q be posets. Then a monotone Galois connection, or just a Galois connection, between P and Q is a pair of order preserving maps α : P → Q and β : Q → P such that, for all p ∈ P and q ∈ Q, we have α(p) ≤ q ⇐⇒ p ≤ β(q). The map α is the left adjoint, and β is the right adjoint. An antitone Galois connection between P and Q is a Galois connection between P and the order dual, Q∂ , of Q. 6 ROB EGROT

Definition 2.7. A preorder on a set is a binary relation that is reflexive and transitive. Every preorder induces a canonical partial order by identifying pairs of elements that break anti-symmetry. 2.3. Polarities for completions. Following [3], we define a polarity to be a triple (X,Y, R), where X and Y are sets, and R ⊆ X × Y is a binary relation. For convenience we will assume also that X and Y are disjoint. See the section on polarities in [9] for several examples. Polarities have also been called polarity frames [35]. Given any polarity (X,Y, R), there is an antitone Galois connection between ℘(X) and ℘(Y ). This is given by the order reversing maps (−)R : ℘(X) → ℘(Y ) and R(−): ℘(Y ) → ℘(X) defined as follows: (S)R = {y ∈ Y : x R y for all x ∈ S}. R(T )= {x ∈ X : x R y for all y ∈ T }. The set G(X,Y, R) of subsets of X that are fixed by the composite map R(−) ◦ (−)R is a complete lattice. Indeed, this is a closure operator on ℘(X). Polarities in the special case where X and Y are sets of subsets of some common set Z, where the relation R is that of non-empty intersection, and which also satisfy some additional conditions, have been referred to as polarizations in the literature [36, 25]. Polarizations play an important role in the construction of canonical extensions. There are maps Ξ : X → G(X,Y, R) and Υ : Y → G(X,Y, R) defined by: Ξ(x)= R({x}R) for x ∈ X, and Υ(y)= R{y} for y ∈ Y. Ξ[X] and Υ[Y ] join- and meet-generate G(X,Y, R) respectively [11, Proposition 2.10]. Moreover, the (not usually disjoint) union Ξ[X] ∪ Υ[Y ] inherits an ordering from G(X,Y, R). Thus the inclusion of the poset Ξ[X] ∪ Υ[Y ] into G(X,Y, R) can be characterized as the MacNeille completion of Ξ[X] ∪ Υ[Y ]. The order on Ξ[X] ∪ Υ[Y ] can be defined without first constructing G(X,Y, R). We expand on this in Proposition 2.8 below. Proposition 2.8. Define a preorder on Ξ[X] ∪ Υ[Y ] as follows:

(1) ∀x1, x2 ∈ X Ξ(x1) Ξ(x2) ↔ ∀y ∈ Y x2 R y → x1 R y . (2) ∀y1,y2 ∈ Y Υ(y1) Υ(y2) ↔ ∀x ∈ X x R y1 → x R y2 . (3) (∀x ∈ X)(∀y ∈ Y ) Ξ(x) Υ(y) ↔ x Ry . (4) (∀x ∈ X)(∀y ∈ Y ) Υ(y) Ξ(x) ↔ (∀x′ ∈ X)(∀y′ ∈ Y ) (x′ R y & x R y′) → x′ R y′ . The partial ordering of Ξ[X] ∪ Υ[Y ] inherited from G (X,Y, R) is the canonical partial order induced by . Proof. This is essentially [11, Proposition 2.7]. Proposition 2.9 below provides another perspective on the conditions from Propo- sition 2.8. Proposition 2.9. Let (X,Y, R) be a polarity. Then the following are equivalent: 1. is the least preorder deﬁnable on Ξ[X] ∪ Υ[Y ] such that: ORDER POLARITIES 7

(a) Ξ(x) Υ(y) ⇐⇒ x R y for all x ∈ X and y ∈ Y . (b) The restrictions of to Ξ[X] and Υ[Y ] agree with the orders on these sets inherited from G(X,Y, R). 2. satisﬁes the conditions from Proposition 2.8 Proof. Suppose is any preorder on Ξ[X] ∪ Υ[Y ] satisfying conditions 1(a) and 1(b). Then, by Proposition 2.8 we have

Ξ(x1) Ξ(x2) ⇐⇒ Ξ(x1) ⊆ Ξ(x2) ⇐⇒ (∀y ∈ Y ) x2 R y → x1 R y , and thus 2.8(1) is satisﬁed. A similar argument works for 2.8(2), and 2.8(3) holds automatically. Finally, as is transitive, we must have Υ(y) Ξ(x) =⇒ (∀x′ ∈ X)(∀y′ ∈ Y ) Ξ(x′) Υ(y) & Ξ(x) Υ(y′) → Ξ(x′) Υ(y′) .

Thus, any such preorder satisfies 2.8(1)-(3), and the ‘forward implication only’ version of 2.8(4). To complete the proof it is sufficient to show that the ‘minimal’ defined from R using conditions 2.8(1)-(4) defines a preorder on Ξ[X] ∪ Υ[Y ] satisfying conditions 1(a) and 1(b). But this is what Proposition 2.8 tells us.

Lemma 2.10. The following are equivalent: (1.a) The map Ξ: X → G(X,Y, R) is injective. (1.b) Whenever x1 6= x2 ∈ X there is y ∈ Y such that either (x2,y) ∈ R and (x1,y) ∈/ R, or vice versa. (1.c) Whenever x1 6= x2 ∈ X we have either x1 ∈/ Ξ(x2) or x2 ∈/ Ξ(x1). The following are also equivalent: (2.a) The map Υ: Y → G(X,Y, R) is injective. (2.b) Whenever y1 6= y2 ∈ Y there is x ∈ X such that either (x, y2) ∈ R and (x, y1) ∈/ R, or vice versa. Proof. Observe that Ξ(x) = {z ∈ X : x R y → z R y for all y ∈ Y } for all x ∈ X. Let x1 6= x2 and suppose without loss of generality that there is z ∈ Ξ(x1) \ Ξ(x2). ′ ′ Then (z,y) ∈ R for all y ∈ Y with (x1,y) ∈ R, but there is y ∈ Y with (x2,y ) ∈ R ′ ′ ′ ′ and (z,y ) ∈/ R. For this y we must have (x2,y ) ∈ R and (x1,y ) ∈/ R. Thus (1.a) =⇒ (1.b). That (1.b) =⇒ (1.c) and (1.c) =⇒ (1.a) is automatic. The proof for Υ is similar, but even more straightforward.

What if X and Y are not merely sets but also have a poset structure? We make the following deﬁnition. Deﬁnition 2.11. A polarity (X,Y, R) is an order polarity if X and Y are posets. In this situation we might, for example, want the maps Ξ and Υ to be order embeddings, which places constraints on R. Building on Lemma 2.10 we have the following result. Proposition 2.12. Let (X,Y, R) be an order polarity. Then the map Ξ: X → G(X,Y, R) is an order embedding if and only if

∀x1, x2 ∈ X x1 ≤X x2 ↔ ∀y ∈ Y x2 R y → x1 R y . 8 ROB EGROT

The map Υ: Y → G(X,Y, R) is an order embedding if and only if

∀y1,y2 ∈ Y y1 ≤Y y2 ↔ ∀x ∈ X x R y1 → x R y2 . Proof. We could appeal to Proposition 2.8, but the direct argument is also extremely simple. Explicitly, Ξ is an order embedding if and only if x1 ≤X x2 ⇐⇒ Ξ(x1) ⊆ Ξ(x2), and a little consideration reveals that Ξ(x1) ⊆ Ξ(x2) if and only if x2 R y → x1 R y for all y ∈ Y . Again, the argument for Υ is even more straightforward. Propositions 2.9 and 2.12, while essentially trivial in themselves, contain, in a sense, the seed of inspiration for the rest of the paper. In broad terms, we want to investigate the conditions for the existence of preorders on X ∪ Y such that similar results can be proved. This we do in the next section and onwards. First, a little more notation.

Deﬁnition 2.13 (X ∪ Y , X ⊎ Y ). Given disjoint sets X and Y , we sometimes write X ∪ Y to specify that we are talking about X ∪Y ordered by a given preorder . We use X ⊎ Y to denote the canonical partial order arising from X ∪ Y .

3. Coherence conditions for order polarities 3.1. The basic case. In the previous section we discussed polarities and order polarities from the perspective of G(X,Y, R), and the inherited order structure on Ξ[X] ∪ Υ[Y ]. In this situation the maps Ξ and Υ may fail to be order preserving, order reflecting, or even injective. In this section we forget about G(X,Y, R), and ask instead, given an order polarity (X,Y, R), under what circumstances can we define preorders on X ∪ Y that agree with R on X × Y , and also extend the order structures of X and Y ? In other words, when are there preorders on X ∪Y agreeing with R on X × Y such that the natural inclusions of X and Y into X ∪ Y are order preserving? What about if we require the inclusions to be order embeddings, or to have stronger preservation properties? We will address these questions, but first some definitions. Definition 3.1. Let (X,Y, R) be an order polarity. Define a 0-preorder for (X,Y, R) to be a preorder on X ∪ Y with additional properties as follows: (P1): (∀x ∈ X)(∀y ∈ Y ) x y ↔ x R y . (P2): ∀x , x ∈ X(x ≤X x → x x ). 1 2 1 2 1 2 (P3): ∀y1,y2 ∈ Y (y1 ≤Y y2 → y1 y2). Definition 3.2. Let (X,Y, R) be an order polarity. We say (X,Y, R) is 0-coherent if it satisfies the following conditions:

(C1): (∀x1, x2 ∈ X)(∀y ∈ Y ) x1 ≤X x2 → (x2 R y → x1 R y) . (C2): (∀y1,y2 ∈ Y )(∀x ∈ X) y1 ≤Y y2 → (x R y1 → x R y2) . Deﬁnition 3.3 (0). Let O = (X,Y, R) be an order polarity. Deﬁne the relation 2 0 ⊆ (X ∪ Y ) by 0 =≤X ∪≤Y ∪ R .

Technically, 0 depends on the choice of O, so one could make the case that O the notation should be something like 0 to reflect that. We choose not do this, as we find it rather unwieldy, and the choice of O will always be obvious from the ORDER POLARITIES 9 context. Similar issues arise later at several points, for example in Definitions 3.8 and 3.17, and we take the same approach. Theorem 3.4. Given an order polarity O = (X,Y, R), the following are equivalent: (1) O is 0-coherent. (2) 0 is a 0-preorder for O. (3) There exists a 0-preorder for O. Moreover, the set of 0-preorders for (X,Y, R) is closed under non-empty intersections and, if it is non-empty, has 0 as its smallest member. Proof. Suppose first that (X,Y, R) is 0-coherent. It is immediate from its definition that 0 is reflexive, so it remains only to check transitivity. To do this we consider 3 triples (z1,z2,z3) ∈ (X ∪ Y ) , with z1 0 z2, and z2 0 z3. A simple counting argument reveals there are eight cases, depending on the containment of each zi in X or Y . The cases where the z values are either all in X or all in Y follow from the fact that 0 agrees with the orders on X and Y . The cases that require y 0 x are ruled out by the definition of 0, so the only remaining cases are (x1, x2,y), where x1, x2 ∈ X and y ∈ Y , and (x, y1,y2) where x ∈ X and y1,y2 ∈ Y . These cases are covered by the definition of 0-coherence, so we have transitivity, and thus (1) =⇒ (2). That (2) =⇒ (3) is automatic, so suppose now that is a 0-preorder for (X,Y, R). Then that (X,Y, R) is 0-coherent is an almost immediate consequence of the transitivity of along with the additional conditions from Definition 3.1. Thus (3) =⇒ (1). Finally, the set of 0-preorders for (X,Y, R) is obviously closed under non-empty intersections. Moreover, 0 must be a member of this set (if it is non-empty), by what we have proved already. The proof that it must be the smallest member is routine.

Conditions (C1) and (C2) are, in a sense, dual to each other, as are several other pairs of conditions to come. We will often appeal to this duality in proofs. Infor- mally, we mean something like “by switching some conditions to their (intuitively obvious) duals we could prove this using essentially the same argument”. It is possible to formalize this intuition, but we omit the details for reasons of space. The ad hoc understanding suﬃces to reconstruct proofs as necessary.

3.2. Extension polarities. Suppose in addition that X and Y are both extensions of some poset P . In other words, that there are order embeddings e1 : P → X and e2 : P → Y . What conditions must R satisfy in order for there to be a 0-preorder such that the diagram in Figure 2 commutes (recall Definition 2.13)? In this figure ιX and ιY stand for the compositions of the natural inclusion functions into X ∪ Y with the canonical map from X ∪ Y to X ⊎ Y . As this situation will be the focus of most of the rest of the document, we make the following definition.

Deﬁnition 3.5. An extension polarity is a triple (eX ,eY , R), where eX : P → X and eY : P → Y are order extensions of the same poset P , and (X,Y, R) is an order polarity. When both eX and eY are completions, we say (eX ,eY , R) is complete. We sometimes say an extension polarity (eX ,eY , R) extends P . The concepts of 0-preorders and 0-coherence, from Deﬁnition 3.1 and 3.2, also apply, mutatis mutandis, to extension polarities. 10 ROB EGROT

e P Y / Y

eX ιY X / X ⊎ Y ιX

Figure 2.

Note that an order polarity is an extension polarity in the special case where P is empty.

Deﬁnition 3.6. Let E = (eX ,eY , R) be an extension polarity. Deﬁne a 1-preorder for E to be a 0-preorder for E with the additional property that the diagram in Figure 2 commutes.

Deﬁnition 3.7. Let E = (eX ,eY , R) be an extension polarity. We say E is 1- coherent if it is 0-coherent and also satisﬁes the following conditions:

(C3): ∀p ∈ P eX (p) R eY (p) .

(C4): (∀p ∈ P)(∀x ∈ X)(∀y ∈ Y ) (x R eY (p)& eX (p) R y) → x R y . Deﬁnition 3.8 (m). Let (eX ,eY , R) be an extension polarity. Deﬁne the relation 2 m ⊆ (X ∪ Y ) to be the union of R with the sets 2 ZX ={(x1, x2) ∈ X : ∃p ∈ P x1 R eY (p)& eX (p) ≤X x2 }∪ ≤X 2 ZY ={(y1,y2) ∈ Y : ∃p ∈ P y1 ≤Y eY (p)& eX (p) R y2 }∪ ≤Y ZY X ={(y, x) ∈ Y × X : ∃p, q ∈ P y ≤Y eY (p)& eX (p) ReY (q)& eX (q) ≤X x }.

Lemma 3.9. Let E = (eX ,eY , R) be a 1-coherent extension polarity, and let be a 0-preorder for E. Then the following conditions are equivalent:

(1) ∀p ∈ P eY (p) eX (p) . (2) ZY X ⊆ .

Proof. Assume (1) holds and that there are p, q ∈ P such that we have y ≤Y eY (p), and eX (p) R eY (q), and eX (q) ≤X x. Then

y eY (p) eX (p) eY (q) eX (q) x, and so y x by transitivity, and thus (1) =⇒ (2). Conversely, if we assume ZY X ⊆, then setting y = eY (p) and x = eX (p) produces eY (p) eX (p), and thus (1) and (2) are equivalent as claimed.

Lemma 3.10. Let E = (eX ,eY , R) be a 1-coherent extension polarity, and let be a 0-preorder for E. Then is a 1-preorder for E if and only if satisfies the conditions from Lemma 3.9. Moreover, if these conditions are satisfied, then ZX ∪ ZY ⊆, where ZX and ZY are as in Definition 3.8. Proof. It is clearly necessary that (1) hold in order for to be a 1-preorder, as otherwise the diagram in Figure 2 will not commute. Conversely, if satisfies (1), ORDER POLARITIES 11 then, from (C3) it follows that the diagram in Figure 2 commutes, and thus is a 1-preorder. Now, suppose the conditions are satisfied, and let x1 ≤ x2 ∈ X. Then x1 x2 by (P2). This shows that ≤X ⊆. Suppose then that x1 R eY (p) and eX (p) ≤X x2 for some x1, x2 ∈ X and p ∈ P . Then x1 eY (p) eX (p) x2, and so we must have x1 x2 by transitivity. It follows that ZX ⊆, and that ZY ⊆ follows from a dual argument.

Theorem 3.11. Given an extension polarity E = (eX ,eY , R), the following are equivalent: (1) E is 1-coherent. (2) m is a 1-preorder for E. (3) There exists a 1-preorder for E. Moreover, the set of 1-preorders for E is closed under non-empty intersections and, if it is non-empty, has m as its smallest member.

Proof. Suppose ﬁrst that E is 1-coherent. That m is reﬂexive is automatic, so we show now that it is transitive. As in the proof of Theorem 3.4, we consider the 3 eight relevant cases of the triples (z1,z2,z3) ∈ (X ∪ Y ) . Unfortunately we must proceed case by case, and each case may have several subcases.

• (x1, x2, x3): Here x1 m x2, and x2 m x3. This case breaks down into subcases, depending on the reason m holds for each pair. − If x1 ≤X x2 and x2 ≤X x3 in X, then we have x1 ≤X x3, and thus x1 m x3, by transitivity of ≤X . − Suppose instead that x1 ≤X x2, and that there is p ∈ P with x2 R eY (p) and eX (p) ≤X x3. Then x1 R eY (p) by (C1), and so x1 m x3 by definition of m. − Alternatively, if x1 R eY (p), eX (p) ≤X x2 and x2 ≤X x3, then eX (p) ≤X x3, and so x1 m x3 by definition of m. − Finally, suppose there are p, q ∈ P with x1 R eY (p), with eX (p) ≤X x2, with x2 R eY (q) and with eX (q) ≤X x3. Then eX (p) R eY (q) by (C1), and so x1 R eY (q) by (C4), and thus x1 m x3 by definition of m. • (y1,y2,y3): This case is dual to the previous one. • (x1, x2,y): Here we have x2 m y, and thus x2 R y. We also have x1 m x2, which breaks down into two cases. − First suppose x1 ≤X x2. Then x1 R y by (C1), and so x1 m y as required. − Suppose instead that there is p ∈ P with x1 R eY (p) and eX (p) ≤X x2. Then eX (p) R y by (C1), and so x1 R y by (C4), and thus x1 m y as required. • (y, x1, x2): Here we have y m x, and so there are p, q ∈ P with y ≤Y eY (p), with eX (p) R eY (q), and with eX (q) ≤X x1, and x1 m x2. There are two subcases. − Suppose first that x1 ≤X x2. Then eX (q) ≤X x2 by the transitivity of ≤X , and the result then follows immediately from the definition of m. − Suppose instead that there is r ∈ P with x1 R eY (r) and eX (r) ≤X x2. Then an application of (C1) produces eX (q) R eY (r). Using this with (C4) provides eX (p) R eY (r). Thus we get y m x2 from the definition of m. • (x1,y,x2): We have x1 R y, and, by the definition of m, there are p, q ∈ P with y ≤Y eY (p), with eX (p) R eY (q), and with eX (q) ≤X x2. Then (C2) gives us x1 R eY (p), and consequently (C4) produces x1 R eY (q). Thus x1 m x2 by the definition of m. • (y1,x,y2): Dual to the previous case. 12 ROB EGROT

• (x, y1,y2): Dual to the (x1, x2,y) case. • (y1,y2, x): Dual to the (y, x1, x2) case.

From the above argument we conclude that m is transitive, and thus defines a preorder. To see that m is a 1-preorder first note that it is obviously a 0- preorder (just examine Definitions 3.1 and 3.8). That m is a 1-preorder then follows immediately from Lemma 3.10, as we have ZY X ⊆m by definition. Thus (1) =⇒ (2). That (2) =⇒ (3) is automatic, so suppose (3) holds, and let be a 1-preorder for E. Then (C3) must hold as otherwise the diagram in Figure 2 would not commute. Similarly, if this diagram commutes, then we must have eY (p) eX (p) for all p ∈ P . So, given p ∈ P , x ∈ X and y ∈ Y with x R eY (p) and eX (p) R y, we have x eY (p) eX (p) y, as is a 1-preorder, and thus x y by transitivity of . By the definition of a 1-preorder this implies x R y. So (C4) also holds, and thus (3) =⇒ (1). Finally, that the set of 1-preorders for E is closed under non-empty intersections is easily seen. It follows immediately from Lemma 3.10 that, if non-empty, the smallest member is m.

Deﬁnition 3.12. Let E = (eX ,eY , R) be an extension polarity. Deﬁne a 2- preorder for E to be a 1-preorder for E with the additional property that the maps ιX and ιY from the diagram in Figure 2 are both order embeddings.

Deﬁnition 3.13. Let (eX ,eY , R) be an extension polarity. We say (eX ,eY , R) is 2-coherent if it is 1-coherent and also satisﬁes the following conditions:

(C5): (∀x1, x2 ∈ X)(∀p ∈ P ) (x1 R eY (p)& eX (p) ≤X x2) → x1 ≤X x2 . (C6): (∀y1,y2 ∈ Y )(∀p ∈ P ) (y1 ≤Y eY (p)& eX (p) R y2) → y1 ≤Y y2 . Theorem 3.14. Given an extension polarity E = (eX ,eY , R), the following are equivalent: (1) E is 2-coherent. (2) m is a 2-preorder for E. (3) There exists a 2-preorder for E. Moreover, the set of 2-preorders for E is closed under non-empty intersections and, if it is non-empty, has m as its smallest member. In this case we have ≤X = ZX and ≤Y = ZY , where ZX and ZY are as in Definition 3.8. Proof. Suppose first that E is 2-coherent. As we know from Theorem 3.11 that m is a 1-preorder, we need only show that the maps ιX and ιY from the diagram in Figure 2 are order reflecting. But this is immediate from (C5) and (C6), which amount to stating that ZX ⊆≤X and ZY ⊆≤Y , respectively. Thus (1) =⇒ (2). That (2) =⇒ (3) is automatic, so suppose now that is a 2-preorder for E. Then, given x1, x2 ∈ X and p ∈ P with x1 R eY (p) and eX (p) ≤X x2, as is a 2-preorder (so necessarily a 1-preorder), we have x1 eY (p) eX (p) x2, and thus x1 x2 by transitivity. It follows immediately from the assumption that ιX is an order embedding that x1 ≤X x2. This proves that (C5) holds for E, and that (C6) also holds follows by a dual argument. That the set of 2-preorders is closed under non-empty intersections is easy to see, and that m is its smallest member (assuming it has any) follows immediately from the corresponding statement about 1-preorders made as part of Theorem 3.11. ORDER POLARITIES 13

Finally, we have noted that (C5) and (C6) imply that ZX ⊆≤X and ZY ⊆≤Y , and the opposite inclusions are automatic, so we are done. We will provide examples showing that the strengths of the coherence conditions deﬁned so far are strictly increasing, but we defer this till Section 5.2.

Deﬁnition 3.15. Let E = (eX ,eY , R) be an extension polarity. Deﬁne a 3- preorder for E to be a 2-preorder for E such that the maps ιX and ιY from the diagram in Figure 2 satisfy the following conditions:

(P4): For all S ⊆ P , if eX [S] exists in X, then ιX ( eX [S]) = ιX ◦eX [S]. (P5): For all T ⊆ P , if eY [T ] exists in Y , then ιY ( eY [T ]) = ιY ◦ eY [T ]. V V V Deﬁnition 3.16. Let E =W (eX ,eY , R) be an extensionW polarity. WeW say E is 3-coherent if it is 2-coherent and also satisﬁes the following conditions: (C7):

(∀x ∈ X)(∀y1,y2 ∈ Y )(∀S ⊆ P )

eX [S]= x & x R y2 & ∀p ∈ S y1 ≤Y eY (p) → y1 ≤Y y2 . (C8): ^

(x1, x2 ∈ X)(∀y ∈ Y )(∀T ⊆ P )

eY [T ]= y & x1 R y & ∀q ∈ T eX (q) ≤X x2 → x1 ≤X x2 . _ Definition 3.17 (g). Let (eX ,eY , R) be an extension polarity. Define the relation 2 g ⊆ (X ∪ Y ) to be the union of 0 with the sets ZS and ZT defined below.

ZS = {(y, x) ∈ Y × X : ∃S ⊆ P eX [S] exists in X, eX [S] ≤X x ^ ^ and ∀p ∈ S y ≤Y eY (p) }.

ZT = {(y, x) ∈ Y × X : ∃T ⊆ P eY [T ] exists in Y,y ≤Y eY [T ] _ _ and ∀q ∈ T eX (q) ≤X x }. Lemma 3.18. Let E = (eX ,eY , R) be a 3-coherent extension polarity, and let be a 2-preorder for E. Then is also a 3-preorder for E if and only if ZS ∪ ZT ⊆.

Proof. Suppose first that is a 3-preorder for E, let S ⊆ P , and suppose eX [S] exists. Suppose also that y ≤Y eY (p) for all p ∈ S. Then, by the assumption that V is a 3-preorder (and so necessarily a 1-preorder) it follows that y eX (p) for all p ∈ S, and thus that ιY (y) is a lower bound for ιX ◦ eX [S]. It then follows from the meet-preservation property of 3-preorders that y eX [S] as claimed. Thus ZS ⊆. That ZT ⊆ follows from a dual argument. V For the converse, let be a 2-preorder for E and suppose first that ZS ⊆. Let S ⊆ P and suppose eX [S] is defined in X. Then ιX ( e[S]) is obviously a lower bound for ιX ◦ eX [S]. Let z ∈ X ⊎ Y and suppose z is also a lower bound for V V ιX ◦ eX [S]. If z ∈ ιX [X], then we must have z ≤ ιX ( e[S]), as ιX is an order embedding. Moreover, if z = ιY (y) for some y ∈ Y , then we have y eX (p) for V all p ∈ S, and it follows from the fact that is a 2-preorder that y ≤Y eY (p) for 14 ROB EGROT all p ∈ S. Consequently, that y eX [S], and thus that z ≤ ιX ( eX [S]), follows from the definition of ZS. So satisfies (P4), and thus also (P5) by duality. V V Theorem 3.19. Given an extension polarity E = (eX ,eY , R), the following are equivalent: (1) E is 3-coherent. (2) g is a 3-preorder for E. (3) There exists a 3-preorder for E. Moreover, the set of 3-preorders for E is closed under non-empty intersections and, if it is non-empty, has g as its smallest member.

Proof. Suppose first that E is 3-coherent. It’s apparent from the definition of g that it is reflexive and that the ι maps will be order embeddings. Now, we have eX (p) g eY (p) from (C3) and the definition of g, and by setting S = {p} and x = eX (p), we can get eY (p) g eX (p) from the fact that (eY (p),eX (p)) ∈ ZS. Thus the diagram in Figure 2 will commute. So, if g is transitive, then it is a 2-preorder, and thus a 3-preorder, for E by Lemma 3.18. The main work now is showing that g is transitive. Again this breaks down into eight cases of form (z1,z2,z3). The cases where a y value does not appear before an x value are covered by the proof of Theorem 3.4, so the proofs need not be repeated. There are four remaining cases.

• (y, x1, x2): We have x1 ≤X x2, and two subcases. − Suppose ﬁrst that (y, x1) ∈ ZS. So there is S ⊆ P with eX [S] ≤X x1 and y ≤Y eY (p) for all p ∈ S. Then, since x1 ≤X x2 we also have eX [S] ≤X x2, V and so (y, x2) ∈ ZS too. V − Suppose instead that (y, x1) ∈ ZT . Then there is T ⊆ P with eY [T ] ≥Y y and eX (q) ≤X x1 for all q ∈ T . Then, as x1 ≤X x2 we also have eX (q) ≤X x2 W for all q ∈ T , and so (y, x2) ∈ ZT too. • (y1,y2, x): Dual to the previous case. • (x1,y,x2): We have x1 R y and two subcases. − Suppose ﬁrst that (y, x2) ∈ ZS. Then there is S ⊆ P with eX [S] ≤X x2 and y ≤Y eY (p) for all p ∈ S. So, given p ∈ S, as x1 R y by assumption, we V have x1 R eY (p) by (C2). It then follows from (C5) that x1 ≤X eX (p), and so x1 ≤X eX [S] ≤X x2 as required. − Suppose now that (y, x2) ∈ ZT . Then there is T ⊆ P with y ≤Y eY [T ] V and eX (q) ≤X x2 for all q ∈ T . Then we have x1 R eY [T ] by (C2), and so W x1 ≤X x2 by (C8). • Dual to the previous case. W

This proves that g is a 3-preorder, and thus (1) =⇒ (2). Again, that (2) =⇒ (3) is automatic, so suppose now that is a 3-preorder for E. If (C7) were to fail for some y1,y2 ∈ Y , then we would have y1 y2 (via an appeal to (P4) and other properties of ), but not y1 ≤Y y2, which would contradict the fact that is a 3-preorder. That we also have (C8) follows from a dual argument. Thus (3) =⇒ (1). Finally, it is again easy to show that the set of 3-preorders on E will be closed under non-empty intersections, and that g is its smallest element whenever it is non-empty follows immediately from Lemma 3.18.

Note that, when (eX ,eY , R) is 2-coherent, given x ∈ X and y ∈ Y , and given p, q ∈ P such that ORDER POLARITIES 15

(1) y ≤Y eY (p), (2) eX (p) R eY (q), and (3) eX (q) ≤X x, by setting S = {q} we have eX [S] ≤X x, and also y ≤Y eY (q) by (C5). Recalling Deﬁnitions 3.8 and 3.17, it follows that ZY X ⊆ ZS, and we also have ZY X ⊆ ZT by a dual argument. ExampleV 5.6, later, demonstrates that these inclusions may be strict, as, even when E = (eX ,eY , R) is 3-coherent, there may be a 2-preorder for E that is not also a 3-preorder for E. In that example we have ZS ∩ ZT 6⊆, but we must have ZY X ⊆ as, by Theorem 3.14, we have m ⊆. Thus we cannot have either ZS ⊆ ZY X or ZT ⊆ ZY X .

4. Galois polarities 4.1. Entanglement. In applications of polarities to completion theory, the orders on the sets X and Y of an order polarity (X,Y, R) are related to R via a property we present here as Deﬁnition 4.1.

Deﬁnition 4.1. If (eX ,eY , R) is an extension polarity, we say it is entangled if the following conditions are satisﬁed:

(E1): For all x1 6≤ x2 ∈ X there is y ∈ Y with (x2,y) ∈ R and (x1,y) ∈/ R. (E2): For all y1 6≤ y2 ∈ Y there is x ∈ X with (x, y1) ∈ R and (x, y2) ∈/ R.

In this situation we also say that (eX ,eY , R) is an entangled polarity. For entangled polarities we can refine Definition 3.2 using the following lemma. Lemma 4.2. Let (X,Y, R) be an entangled order polarity. Then (X,Y, R) is 0- coherent if and only if: ′ (C1 ): ∀x1, x2 ∈ X x1 ≤X x2 ↔ ∀y ∈ Y x2 R y → x1 R y . ′ (C2 ): ∀y1,y2 ∈ Y y1 ≤Y y2 ↔ ∀x ∈ X x R y1 → x R y2 . Proof. We claim that (C1′) and (C2′) here are equivalent, respectively, to (C1) and (C2) when (eX ,eY , R) is entangled. This is immediate from the definitions.

In the case of entangled polarities, using (C1′) and (C2′) we could, if we were so inclined, restate things like the various coherence conditions to avoid explicit reference to the orders on X and Y . Lemma 4.2 also has the following corollary.

Corollary 4.3. Let E = (eX ,eY , R) be an entangled extension polarity. Then, if is a 0-preorder for E, for all x1, x2 ∈ X we have x1 ≤X x2 ⇐⇒ x1 x2, and for all y1,y2 ∈ Y we have y1 ≤Y y2 ⇐⇒ y1 y2. Similarly, if is a 1-preorder for E, then it is also a 2-preorder for E. Proof. Let be a 0-preorder for E. Appealing to Lemma 4.2 we assume that ′ ′ (C1 ) and (C2 ) both hold. Let x1 6≤X x2 ∈ X. Then, by entanglement, there is y ∈ Y with (x2,y) ∈ R and (x1,y) ∈/ R. So we cannot have x1 x2, as otherwise transitivity would produce x1 y, and consequently x1 R y. This proves the ﬁrst claim. The second claim also follows from this argument, as the diﬀerence between 1-preorders and 2-preorders is only that in the latter case the induced maps ιX and ιY from Figure 2 must be order embeddings. 16 ROB EGROT

Note that, if we treat order polarities as extension polarities where P is the empty poset, then given an order polarity O = (X,Y, R), every 0-preorder for O is automatically a 1-preorder. Thus if O is an entangled order polarity, the above result shows that its sets of 0-, 1-, and 2-preorders coincide. However, it is not the case that every 2-preorder for O is necessarily a 3-preorder, as the following example demonstrates. Example 4.4. Consider the order polarity O = ({x}, {y}, {(x, y)}) and the preorder 0 from Deﬁnition 3.3. Then 0 is trivially a 2-preorder for O, but it is not a 3-preorder for O as, for example,

eX [∅]= x 60 y = ∅ = ιX ◦ eX [∅], X X⊎ Y X⊎ Y ^ ^0 ^0 and thus 0 does not satisfy (P4). 4.2. Deﬁning Galois polarities.

Deﬁnition 4.5. A Galois polarity is a 3-coherent extension polarity (eX ,eY , R) such that eX : P → X is a meet-extension, and eY : P → Y is a join-extension. The motivation for the name Galois polarity will become clear in Section 7.1. Galois polarities have several strong properties, as we shall see. We will use the following lemma.

Lemma 4.6. Let (eX ,eY , R) be an extension polarity. Suppose (eX ,eY , R) satisfies (C3). Then, if (eX ,eY , R) satisfies (C1), it also satisfies (†1) below. Similarly, if (eX ,eY , R) satisfies (C2), then it also satisfies (†2).

(†1) (∀p ∈ P )(∀x ∈ X) x ≤X eX (p) → x R eY (p) . (†2) (∀p ∈ P )(∀y ∈ Y ) eY (p) ≤Y y → eX (p) R y .

Moreover, if a polarity (eX,eY , R) satisﬁes either (†1) or (†2), then it also satisﬁes (C3).

Proof. Suppose (eX ,eY , R) satisfies (C1) and (C3), and let x ≤X eX (p) for some x ∈ X and p ∈ P . Then eX (p) R eY (p) by (C3), and so x R eY (p) by (C1). Thus (eX ,eY , R) satisfies (†1). The case where we assume (C2) and (C3) to prove (†2) is dual. Suppose now that (eX ,eY , R) satisfies (†1), and let p ∈ P . Then, as eX (p) ≤X eX (p), we have eX (p) R eY (p) by (†1), and thus (eX ,eY , R) satisfies (C3). The case where we assume (†2) and prove (C3) is again dual. Lemma 4.7. Galois polarities are entangled.

Proof. Let (eX ,eY , R) be a Galois polarity, and let x1 6≤ x2 ∈ X. Then, as eX is a meet-extension there is p ∈ P with x1 6≤X eX (p), and x2 ≤X eX (p). Thus x2 R eY (p) by (†1) of Lemma 4.6. Moreover, if x1 R eY (p), then x1 ≤X eX (p) by (C5), which contradicts the choice of p. We conclude that (E1) holds. A dual argument works for (E2).

Corollary 4.8. If G = (eX ,eY , R) is a Galois polarity, then every 1-preorder for G is also a 2-preorder for G. Proof. This follows immediately from Lemma 4.7 and Corollary 4.3. ORDER POLARITIES 17

If G is a Galois polarity, then g from Deﬁnition 3.17 is the only 3-preorder for G as we show in Theorem 4.12. First, the following technical lemma will be useful.

Lemma 4.9. If E = (eX ,eY , R) is 3-coherent, then a) and b) below each imply c), for all x ∈ X and for all y ∈ Y . Moreover, if E is Galois, then a), b) and c) are all equivalent, for all x and y. a) There is S ⊆ P with eX [S] ≤X x and y ≤Y eY (p) for all p ∈ S. b) There is T ⊆ P with eY [T ] ≥ y and eX (q) ≤X x for all q ∈ T . V c) For all p, q ∈ P , if eY (p) ≤Y y and x ≤X eX (q), then p ≤P q. W Proof. As E is 3-coherent, we can let be a 3-preorder for E. Suppose ﬁrst that a) holds for x and y, and let p, q ∈ P with eY (p) ≤Y y and x ≤X eX (q). Then we have eY (p) y eX [S] x eX (q) for some S ⊆ P , using the fact that^ZS ⊆, by Lemma 3.18. By commutativity of the diagram in Figure 2 we must therefore have eX (p) eX (q), and so p ≤P q. This shows a) =⇒ c), and a dual argument shows b) =⇒ c). Suppose now that E is Galois and that c) holds for x and y. As E is Galois −1 ↑ −1 ↓ we have x = eX [S] for S = eX (x ), and y = eY [T ] for T = eY (y ). By c) we have q ≤P p for all q ∈ T and p ∈ S. Given a 3-preorder for E we thus V W have eX (q) eY (p) for all q ∈ T and p ∈ S, and, appealing to (P4) and (P5), we must have y = eY [T ] eX [S] = x, and thus y ≤Y eY (p) for all p ∈ S, and eX (q) ≤X x for all q ∈ T . It follows that c) implies both a) and b), and so we have the claimed equivalence.W V

Deﬁnition 4.10. Given an extension polarity (eX ,eY , R), deﬁne ′ −1 ↓ −1 ↑ ZY X = {(y, x) ∈ Y × X : (∀p ∈ eY (y )(∀q ∈ eX (x )) p ≤P q }.

Corollary 4.11. If G = (eX ,eY , R) is a Galois polarity, then we can deﬁne g ′ from Deﬁnition 3.17 to be 0 ∪ZY X . ′ Proof. It follows immediately from Lemma 4.9 that ZY X = ZS ∪ZT in this case.

Theorem 4.12. If G = (eX ,eY , R) is a Galois polarity, then:

(1) The maps ιX : X → X ⊎g Y and ιY : Y → X ⊎g Y are completely meet- and join-preserving respectively. (2) g is the only 3-preorder for G.

(3) X ⊎g Y is join-generated by ιX [X], and meet-generated by ιY [Y ].

Proof. We will start by showing that ιX is completely meet-preserving. Let x ∈ X −1 ↑ and suppose x = Z for some Z ⊆ X. For each z ∈ Z deﬁne Sz = eX (z ). Then for all z ∈ Z we have z = eX [Sz], as eX is a meet-extension. Moreover, V V x = eX [ Sz]. So, using (P4), z∈Z V ιX (Vx)= ιSX ( eX [ Sz]) = ιX ◦ eX [ Sz]= ιX ( eX [Sz]) = ιX [Z]. z∈Z z∈Z z∈Z ^ [ ^ [ ^ ^ ^ This shows ιX is completely meet-preserving, and that ιY is completely join- preserving follows from a dual argument. To see that g is the only 3-preorder for G note ﬁrst that it must be the smallest such preorder, by Theorem 3.19. Moreover, if is another 3-preorder for G, then is determined, by ≤X , ≤Y , and R, everywhere except on Y × X. So 6= g if 18 ROB EGROT and only if there is x ∈ X and y ∈ Y with y x and y 6g x. But, by Corollary 4.11, this is impossible, as for any p, q ∈ P with eY (p) ≤Y y and x ≤X eX (q) we are forced to have p ≤P q by the transitivity of and the commutativity of the diagram in Figure 2.

Finally, ιX [X] is a join-dense subset of X ⊎g Y because it’s a join-dense subset of itself, and it contains ιX ◦ eX [P ], which is a join-dense subset of ιY [Y ].

Given a 0-coherent extension polarity E = (eX ,eY , R) where eX and eY are meet- and join-extensions respectively, there is a simple necessary and suﬃcient condition for E to be Galois, as explained in the next proposition.

Proposition 4.13. Let E = (eX ,eY , R) be 0-coherent, and let eX and eY be, respectively, meet- and join-extensions of P . Then (eX ,eY , R) is Galois if and only if the following conditions are both satisﬁed:

(S1): (∀p ∈ P )(∀x ∈ X) x ≤X eX (p) ↔ x R eY (p) .

(S2): (∀p ∈ P )(∀y ∈ Y ) eY (p) ≤Y y ↔ eX (p) R y . Proof. Suppose first that E is Galois, and let p ∈ P and x ∈ X. Suppose x ≤X eX (p). Then x R eY (p) by Lemma 4.6. Conversely, if x R eY (p), then x ≤X eX (p) by(C5), as eX (p) ≤X eX (p). Thus (S1) holds, and (S2) holds by a dual argument. Suppose now that E is 0-coherent and satisfies (S1) and (S2), and also that eX and eY are meet- and join-extensions respectively. We will show that the conditions (C3)–(C8) are satisfied.

(C3): This follows immediately from (S1) as eX (p) ≤X eX (p) for all p ∈ P . (C4): If x R eY (p) and eX (p) R y, then x ≤X eX (p) by (S1), and so x R y by (C1). (C5): Let x1 R eY (p) and let eX (p) ≤X x2. Then x1 ≤X eX (p) by (S1), and so x1 ≤X x2 by transitivity of ≤X . (C6): This is dual to (C5). (C7): Let eX [S] = x, let x R y2, and suppose y1 ≤Y eY (p) for all p ∈ S. Let q ∈ P and suppose eY (q) ≤Y y1. Then q ≤P p for all p ∈ S, as eY is an V order embedding, and so eX (q) ≤X x. Thus eX (q) R y2 by (C1), and so eY (q) ≤Y y2 by (S2). As eY is a join-extension it follows that y1 ≤Y y2 as required. (C8): This is dual to (C7).

It follows from Proposition 4.13 that what we call a Galois polarity corresponds to what [13, Section 4] calls a ∆1-polarity. See also [13, Proposition 4.1], which tells us that the preorder g as deﬁned using Corollary 4.11 is the one arising naturally from G(X,Y, R). Theorem 4.12 says that this is in fact the only 3-preorder deﬁnable for a Galois polarity. Note that if (eX ,eY , R) is not Galois, then g may not be a preorder. Since Galois polarities have only one 3-preorder, to lighten the notation we will from now on write e.g. X ⊎ Y in place of X ⊎g Y when working with Galois polarities.

5. The satisfaction and separation of the coherence conditions 5.1. Sets of coherent relations. If X and Y are posets, it’s easy to see that the set of relations on X ×Y such that the induced order polarity is 0-coherent is closed ORDER POLARITIES 19 under arbitrary unions and intersections, and has ∅ and X ×Y as least and greatest elements respectively. The situation for extension polarities and more restrictive forms of coherence is a little more delicate, as illustrated by Proposition 5.2 below. First we introduce another deﬁnition.

Deﬁnition 5.1 (Rl). Let eX : P → X and eY : P → Y be poset extensions. Deﬁne the relation Rl ⊆ X × Y by −1 ↑ −1 ↓ x Rl y ⇐⇒ eX (x ) ∩ eY (y ) 6= ∅.

Proposition 5.2. Let eX : P → X and eY : P → Y be poset extensions. Then:

(1) The set of relations R such that (eX ,eY , R) is n-coherent is closed under non-empty intersections for all n ∈{0, 1, 2, 3}. (2) (eX ,eY , Rl) is 2-coherent. (3) If R ⊆ X × Y and (eX ,eY , R) is 1-coherent, then Rl ⊆ R. (4) If eX and eY are, respectively, meet- and join-extensions, then (eX ,eY , Rl) is 3-coherent (and thus Galois). (5) If (eX ,eY , Rl) is not 3-coherent, then there is no R such that (eX ,eY , R) is 3-coherent. Proof. First, that the sets in question are all closed under non-empty intersections can be proved by a routine inspection of the conditions (C1)–(C8) and we omit the details. Checking that (eX ,eY , Rl) is 2-coherent is a similarly straightforward check of conditions (C1)–(C6). Thus we have dealt with (1) and (2). For (3), If R is a relation such that (eX ,eY , R) is 1-coherent, then, in particular, R must satisfy −1 ↑ −1 ↓ (C1), (C2) and (C3). Given x ∈ X and y ∈ Y , if there is p ∈ eX (x ) ∩ eY (y ), then we have x ≤X eX (p) and eY (p) ≤Y y by choice of p, and eX (p) R eY (p) by (C3). Thus x R eY (p) by (C1), and so x R y by (C2). It follows that Rl ⊆ R as claimed. For (4), suppose that eX is a meet-extension and eY is a join-extension. We will check that Rl also satisfies (C7). Let S ⊆ P , and let x = eX [S] in X. Let y1,y2 ∈ Y and suppose that y1 ≤Y eY (p) for all p ∈ S, and that x Rl y2. Let q ∈ P V and suppose eY (q) ≤Y y1. Then eY (q) ≤Y eY (p), and thus q ≤P p, for all p ∈ S. It follows that eX (q) ≤X eX (p) for all p ∈ S, and so eX (q) ≤X x. Also, by definition ′ ′ ′ ′ of Rl, there is q ∈ P with x ≤X eX (q ) and eY (q ) ≤Y y2. But then q ≤P q , and −1 ↓ consequently eY (q) ≤Y y2. This is true for all q ∈ eY (y1 ), and so y1 ≤Y y2 as eY is a join-extension. Rl also satisfies (C8) by duality, and so the claim is proved. Finally, if (eX ,eY , Rl) is not 3-coherent, then, as we have shown it must be 2- coherent, it must fail to satisfy either (C7) or (C8). In either case, since by (3) any R making (eX ,eY , R) 3-coherent must contain Rl, inspection of (C7) and (C8) reveals that no such R can exist, which proves (5).

Note that when eY is not a join-extension (eX ,eY , Rl) may not satisfy (C7), as Example 5.3 demonstrates. By duality, when eX is not a meet-extension (eX ,eY , Rl) may not satisfy (C8).

Example 5.3. Let P be the poset in Figure 3, and let eX and eY be the extensions deﬁned in Figures 4 and 5 respectively. Here the embedded images of elements of P are represented using •, and the extra elements of X and Y using ◦. Note that eX is a meet-extension, but eY is not a join-extension. Let S = {p, q}. Then x = eX [S], and x Rl eY (r). But we also have y ≤Y eY (p) and y ≤Y eY (q), but y 6≤Y eY (r). So (C7) does not hold for Rl. V 20 ROB EGROT

•p •q •r •p •q •r ❇❇ ⑤ ⑥ ❇❇ ⑤⑤ ⑥⑥ ❇ ⑤⑤ ⑥⑥ •p •q •r ◦x ◦y

Figure 3. Figure 4. Figure 5.

5.2. A strict hierarchy for coherence. Example 5.3, taken with Proposition 5.2(2), also demonstrates that it is possible for an extension polarity to be 2- coherent but not 3-coherent (take (eX ,eY , Rl) from this example). Thus 3-coherence is a strictly stronger property than 2-coherence. However, this example only applies when either eY fails to be a join-extension, or, by duality, when eX fails to be a meet-extension. Example 5.4 below demonstrates that, even when eX and eY are meet- and join-extensions respectively, there may be choices of R for which (eX ,eY , R) is 2-coherent but not 3-coherent.

Example 5.4. Let eX and eY be as in Figures 6 and 7 respectively. Then it’s easy to see that eX and eY are meet- and join-extensions respectively. Moreover, if we define R = Rl ∪{(x, y2)}, then E = (eX ,eY , R) is 2-coherent, as can be observed by noting the preorder on X ∪ Y described in Figure 8. However, E is not 3- coherent, as we prove now. If E is 3-coherent, then it is Galois, by definition, and ′ the characterization of g from Corollary 4.11 is valid. Noting that (y1, x) ∈ ZY X , it follows that y1 g y2, and thus that g is not a 2-preorder for E, as it does not reflect the order on Y . But then E is not 3-coherent, by Theorem 3.19, and to avoid contradiction we must conclude that (eX ,eY , R) is not 3-coherent after all. 2-coherence is also a strictly stronger condition than 1-coherence, as witnessed by Example 5.5 below. Example 5.5. Let P be the two element antichain {p, q}. Define X =∼ Y =∼ P , and let eX and eY be isomorphisms. Define R = Rl ∪{(eX (p),eY (q))}. Let E = (eX ,eY , R). Then E is 1-coherent, as can be proved via Theorem 3.11, either by writing down a suitable 1-preorder (the one inducing the two element chain eX (p) = eY (p) ≤ eX (q) = eY (q)), or by formally proving that m is such a thing. But E is not 2-coherent, which we can intuit by noticing that there’s no way the ι maps from Figure 2 are going to be order reflecting, or prove formally via Theorem 3.14 by observing that ≤X is a strict subset of ZX here. 5.3. Separating the classes of preorders. We have seen that the classes of extension polarities defined by the coherence conditions are strictly separated. It

• • ◦ • • ◦ • • y2 ❈ y2 ❇❇ ⑥⑥ ✶ ❈❈ ④④ ④ ✶ ❇❇ ⑥⑥ ✶✶ ④❈④❈ ④④ ✶✶ ❇ ⑥⑥ ✶ ④④ ❈ ④④ ✶ ◦ ✶ ◦ ◦ ✶ ◦x y1 ✶✶ y1 x ✶✶ ⑤ ❈❈ ✶ ❊❊ ② ✶ ⑤⑤ ❈❈ ✶ ②❊②❊ ✶ ⑤⑤ ❈ ✶ ② ❊ ✶ • • •• • • • • • ② • • •

Figure 6. Figure 7. Figure 8. ORDER POLARITIES 21

•• • • ••❈ ❇❇ ⑥ ❈❈ ④④ ❇❇ ⑥⑥ ④❈④❈ ❇ ⑥⑥ ④④ ❈ ◦x ◦y ◦y ◦x ⑤ ❆❆ ❈❈ ④ ⑤⑤ ❆❆ ④❈④❈ ⑤⑤ ❆ ④④ ❈ • • • • • •

Figure 9. Figure 10. Figure 11.

• • • • ❇❇ ••❈ ⑥ ❇ ⑤⑤⑤ ❈❈ ④④ ⑥⑥ ◦ ④❈④❈ ⑥⑥ y ④④ ❈ ◦y ◦y ◦y ◦x ◦y ❄ 2 ❇ 2 ❄ ◦x ◦y2 ❇ ⑤ ❄❄ ❄❄ ⑤❇⑤❇ ❄ ⑧⑧⑧ ❄ ⑤⑤ ❇ • • ◦y1 • • ◦y1 • • ◦y1

Figure 12. Figure 13. Figure 14. is also true that, even for a Galois polarity G, the set of 3-preorders for G may be strictly contained in its set of 2-preorders. Moreover, for a 3-coherent polarity E, it may be the case that the set of 3-preorders for E is strictly contained in the set of its 2-preorders, which is itself strictly contained in its set of 1-preorders (from Corollary 4.8 we know this last statement is not true for Galois polarities). This is demonstrated in Examples 5.6 and 5.7 respectively.

Example 5.6. Let eX and eY be as in Figures 9 and 10 respectively. Then G = (eX ,eY , Rl) is Galois, by Proposition 5.2(4), and the order represented in Figure 11 is induced by a 2-preorder for G, which we call . However, is not a 3-preorder for G as (P4) and (P5) fail.

Example 5.7. Let P and eX be as in Example 5.6, and let eY be deﬁned by the diagram in Figure 12. Then E = (eX ,eY , Rl) is 3-coherent, because g induces the order illustrated in Figure 13. However, E has a 2-preorder that is not a 3- preorder (described in Figure 14), and a 1-preorder that is not a 2-preorder obtained by additionally setting y2 y1.

6. Extending and restricting polarity relations 6.1. Extension. If e : P → Q is an order extension, then given another order extension e′ : Q → Q′, the composition e′ ◦ e is also an order extension. It is natural to ask whether an extension polarity E = (eX ,eY , R) can be extended to ′ ′ ′ ′ something like E = (eX ◦ eX ,eY ◦ eY , R ), and under what circumstances the level of coherence of E transfers to E′. This is of particular interest, for example, if we wish to extend eX and eY to completions, as we shall do in Section 7.1. The next theorem provides some answers, but ﬁrst we need a deﬁnition.

Deﬁnition 6.1 (R). Let (eX ,eY , R) be an extension polarity, let iX : X → X and iY : Y → Y be order extensions with X ∩ Y = ∅. Let R be the relation on X × Y 22 ROB EGROT deﬁned by ′ ′ ′ ′ x R y ⇐⇒ (∃x ∈ X)(∃y ∈ Y ) x ≤X iX (x)& iY (y) ≤Y y & x R y . Theorem 6.2. Let E = (eX ,eY , R) be an extension polarity, let iX : X → X and iY : Y → Y be order extensions with X ∩ Y = ∅. Let E = (iX ◦ eX ,iY ◦ eY , R). Then: (1) E is 0-coherent. (2) For all x ∈ X and for all y ∈ Y we have

x R y → iX (x) R iY (y), and the converse is true if and only if E is 0-coherent. (3) If E is n-coherent, then E is n-coherent, for n ∈{1, 2}. (4) If E is Galois, and if iX : X → X and iY : Y → Y are meet- and join- extensions respectively, then E is also Galois. (5) Let (X, Y, S) be 0-coherent, and suppose

x R y → iX (x) S iY (y). Then R ⊆ S. (6) If E is not n-coherent, then there is no S ⊆ X × Y satisfying

x R y → iX (x) S iY (y)

such that (iX ◦ eX ,iY ◦ eY , S) is n-coherent, for n ∈{2, 3}. Proof. (1) We check that (X, Y, R) is 0-coherent using Definition 3.2. We need only ′ ′ ′ check (C1) as (C2) is dual. Let x1 ≤ x2 ∈ X, let y ∈ Y , and suppose ′ ′ ′ ′ x2 R y . Then there are x ∈ X and y ∈ Y with x1 ≤X x2 ≤X iX (x), with ′ ′ ′ iY (y) ≤Y y , and with x R y. But then x1 R y , by definition of R, so (C1) holds. (2) If x R y, then that iX (x) R iY (y) follows directly from the definition. Con- versely, suppose (eX ,eY , R) is 0-coherent, let x1 ∈ X, let y1 ∈ Y , and suppose iX (x1) R iY (y1). Then there is x2 ∈ X and y2 ∈ Y with x1 ≤X x2, with x2 R y2, and with y2 ≤Y y1. It follows from 0-coherence of (eX ,eY , R) that x1 R y1 as required. Moreover, (iX ◦eX ,iY ◦eY , R) is always 0-coherent by (1), so, if the converse holds E inherits 0-coherence from E. (3) Now suppose E is 1-coherent. We check that (C3) and (C4) hold for E. (C3): Let p ∈ P . Then eX (p) R eY (p) as E is 1-coherent, and it follows easily that iX ◦ eX (p) R iY ◦ eY (p). Thus (C3) holds for E as required. ′ ′ ′ (C4): Let x ∈ X, let y ∈ Y , and let p ∈ P . Suppose x R(iY ◦ eY (p)) ′ and (iX ◦ eX (p)) R y . Then there are x1 ∈ X and y1 ∈ Y , with ′ x ≤X iX (x1), with x1 R y1, and with iY (y1) ≤Y iY ◦ eY (p), and also x2 ∈ X and y2 ∈ Y with iX ◦ eX (p) ≤X iX (x2), with x2 R y2, and ′ with iY (y2) ≤Y y . As iX and iY are order embeddings we have y1 ≤Y eY (p) and eX (p) ≤X x2. As E is 1-coherent it follows from Theorem 3.11 that m is a 1-preorder for E, and thus

x1 m y1 m eY (p) m eX (p) m x2 m y2.

So x1 R y2 by transitivity of m and the fact that it agrees with R on X × Y . It follows immediately that x′ R y′, and so (C4) holds for E. ORDER POLARITIES 23

Thus E is 1-coherent. Suppose now that E is 2-coherent. We check that ′ ′ ′ (C5) holds for E. Let x1, x2 ∈ X, and let p ∈ P . Suppose x1 R(iY ◦ eY (p)), ′ ′ and iX ◦eX (p) ≤X x2. Then there are x ∈ X and y ∈ Y with x1 ≤X iX (x), with iY (y) ≤Y iY ◦ eY (p), and with x R y. As E is 2-coherent we know from Theorem 3.14 that m is a 2-preorder for E, and we have

x m y m eY (p) m eX (p).

So x ≤X eX (p), as m is a 2-preorder, and consequently ′ ′ x1 ≤X iX (x) ≤X iX ◦ eX (p) ≤X x2. ′ ′ Thus x1 ≤X x2, and so (C5) holds. By duality (C6) also holds, and so E is 2-coherent as claimed. (4) Suppose now that E is Galois, and that the iX and iY are meet- and join- extensions respectively. First, that iX ◦ eX and iY ◦ eY are, respectively, meet- and join-extensions follows from the corresponding properties of iX , eX , iY and eY . It remains only to check that (C7) and (C8) hold for E. ′ ′ ′ ′ Let x ∈ X, let y1,y2 ∈ Y , and let S ⊆ P . Suppose (iX ◦ eX [S]) = x . ′ ′ ′ Suppose also that x R y2, and that y1 ≤Y iY ◦ eY (p) for all p ∈ S. Then ′ V ′ there are x ∈ X and y ∈ Y with x ≤X iX (x) and iY (y) ≤Y y2, and with ′ ′ x R y. We aim to prove that y1 ≤Y y2. ′ −1 ↓ Let y0 ∈ Y be such that iY (y0) ≤Y y1, and let q ∈ eY (y0 ). Then

eY (q) ≤Y y0 ≤Y eY (p) for all p ∈ S, ′ and so iX ◦ eX (q) ≤X x ≤X iX (x), and consequently eX (q) ≤X x. Since E is Galois, we know from Theorem 3.19 that g is a 3-preorder for E, and

we have y0 g x as the map ιY : Y → X ⊎g Y preserves joins of sets in −1 ↓ eY [P ] and y0 = eY [yY (y0 )]. So we have

W y0 g x g y, ′ and thus y0 ≤Y y for all y0 with iY (y0) ≤Y y1. But, as iY is a join- extension, we have

′ −1 ′↓ y1 = iY [iY (y1 )], ′ ′ _ and so y1 ≤Y iY (y) ≤Y y2, which is what we are trying to prove. It follows that (C7) holds for E, and thus by duality (C8) also holds. ′ ′ ′ (5) Suppose x R y . Then there is x ∈ X and y ∈ Y with x ≤X iX (x), x R y, ′ and iY (y) R y . Let S ⊆ X × Y satisfy the conditions from (5). Then ′ ′ iX (x) S iY (y), and so x S y by (C1) and (C2), and the result follows. (6) From (5) we know that any relation on X ×Y that ‘extends R’ must contain R. Examination of the conditions (C5)–(C8) reveals that if they fail for R they will also fail for any relation containing R.

Theorem 6.2(6), tells us that if we want to ﬁnd a 2- or 3-coherent polarity extending (eX ,eY , R), then it suﬃces to look at R, as if this does not produce the desired result then nothing will. Note that this does not apply for 1-coherence. To see this, let P = {p} =∼ X =∼ Y =∼ X =∼ Y , and let R = ∅. Then (C3) fails for E, but if S = {(iX ◦ eX (p)),iY ◦ eY (p)}, then (iX ◦ eX ,iY ◦ eY , S) is obviously 1-coherent. 24 ROB EGROT

For 0-coherent polarities we can add converses to some of the statements in Theorem 6.2, but we will leave this till Corollary 6.12. Note that for E to be 3- coherent it is not sufficient for E to be 3-coherent, or even Galois. The additional restrictions on the extensions iX and iY from Theorem 6.2(4) are necessary, as Example 6.3 demonstrates below. Example 6.3. Let P be the three element antichain from Figure 3, and let X =∼ Y =∼ P . Let iX : X → X and iY : Y → Y be the poset extensions illustrated in Figures 4 and 5 respectively. Define R on X × Y by x R y ⇐⇒ there is p ∈ P with x = eX (p) and y = eY (p). We can put a poset structure on X ∪ Y just by identifying copies of elements of P appropriately, in which case we end up with something isomorphic to P . Clearly the natural maps ιX and ιY are meet- and join-preserving order embeddings here, and so (eX ,eY , R) is Galois. However, (iX ◦eX ,iY ◦eY , R) is not 3-coherent. Indeed, it follows from Example 5.3 that, if we define Sl ⊆ X × Y analogously to Definition 5.1, the polarity (iX ◦ eX ,iY ◦ eY , Sl) is not 3-coherent. Thus there is no relation S such that (iX ◦ eX ,iY ◦ eY , S) is 3-coherent, by Proposition 5.2(5). The following lemma says, roughly, that the extension of the ‘minimal’ polarity relation Rl is again the minimal polarity relation.

Lemma 6.4. Let (eX ,eY , Rl) be an extension polarity, where Rl is as in Deﬁnition 5.1, and let iX : X → X and iY : Y → Y be order extensions with X ∩ Y = ∅. Then Rl = Sl, where Sl ⊆ X × Y is deﬁned analogously to Rl. Proof. Let x′ ∈ X and let y′ ∈ X. Then ′ ′ ′ ′ x Rly ⇐⇒ x ≤X iX (x), x Rl y and iY (y) ≤Y y for some x ∈ X and y ∈ Y ′ ′ −1 ↑ −1 ↓ ⇐⇒ x ≤X iX (x), iY (y) ≤Y y and eX (x ) ∩ eY (y ) 6= ∅ −1 ′↑ −1 ′↓ ⇐⇒ (iX ◦ eX ) (x ) ∩ (iY ◦ eY ) (y ) 6= ∅ ′ ′ ⇐⇒ x Sl y .

6.2. Restriction. If iX : X → X and iY : Y → Y are order extensions, then a polarity (X, Y, S) can be restricted in a natural way to a polarity (X,Y, S). The following deﬁnition makes this precise.

Definition 6.5. Let X and Y be posets, and let iX : X → X and iY : Y → Y be order extensions with X ∩ Y = ∅. Let S be a relation on X × Y . Define the relation S ⊆ X × Y by x S y ⇐⇒ iX (x) S iY (y). It turns out the coherence properties behave quite well under restriction. We make this precise in Theorem 6.7, but first we need another definition.

Deﬁnition 6.6 (φ, φ). Let P be a poset, let eX : P → X, iX : X → X, eY : P → Y and iY : Y → Y be order extensions, where X ∩ Y = ∅ = X ∩ Y . Deﬁne φ′ : X ∪ Y → X ∪ Y by i (z) if z ∈ X. φ′(z)= X (iY (y) if z ∈ Y. ORDER POLARITIES 25

Let be a preorder on X ∪ Y , and deﬁne the preorder φ on X ∪ Y by setting ′ ′ z1 φ z2 ⇐⇒ φ (z1) φ (z2).

Let X ⊎φ Y and X ⊎ Y be the posets induced by X ∪φ Y and X ∪ Y respectively, and let ιX : X → X ⊎φ Y , ιY : Y → X ⊎φ Y , ιX : X → X ⊎ Y and ιY : Y → X ⊎ Y be the maps induced by the inclusion functions. Deﬁne

φ : X ⊎φ Y → X ⊎ Y by ι ◦ i (x) if z = ι (x) for some x ∈ X. φ(z)= X X X (ιY ◦ iY (y) if z = ιY (y) for some y ∈ Y . It should be reasonably clear that φ′ is well defined. It may not be immediately obvious that φ is a preorder, but a quick check reveals that this is indeed the case. We show that φ is well defined as part of the next theorem. Theorem 6.7. With a setup as in Definition 6.6, the map φ is a well defined order embedding. Moreover, suppose S ⊆ X × Y and define E = (iX ◦ eX ,iY ◦ eY , S). Then:

(1) If is a 0-preorder for E, then the maps ιX and ιY are order preserving. (2) If is a 1-preorder for E, then the diagram in Figure 15 commutes. (3) If is a 2-preorder for E, the maps ιX and ιY are order embeddings. (4) Suppose is a 3-preorder for E, and suppose also that iX preserves meets in X of subsets of eX [P ] whenever they exist, and that iY likewise preserves joins in Y of subsets of eY [P ]. Then φ satisﬁes (P4) and (P5).

Proof. To see that φ is well defined and order preserving, suppose ιX (x) ≤ ιY (y) for some x ∈ X and y ∈ Y . Then x φ y, and thus, by definition of φ, we have iX (x) iY (y). It follows that ιX ◦ iX (x) ≤ ιY ◦ iY (y), and thus φ(x) ≤ φ(y). By a similar argument, if ιY (y) ≤ ιX (x), then we also have φ(y) ≤ φ(x), and so φ is well defined and order preserving as claimed.

To see that φ is an order embedding, let z1,z2 ∈ X ⊎φ Y , and suppose that φ(z1) ≤ φ(z2). There are four cases. Suppose first that z1 = ιX (x) and z2 = ιY (y) for some x ∈ X and y ∈ Y . Then iX (x) iY (y), and so x φ y, from which it follows immediately that ιX (x) ≤ ιY (y), and thus that z1 ≤ z2. The other cases are more or less exactly the same. Suppose is a 0-preorder for E. Then that ιX is order preserving follows immediately from the fact that iX and ιX are order preserving, and ιY is order preserving by duality. Suppose now that is a 1-preorder for E, and let p ∈ P . Then iX ◦ eX (p) iY ◦ eY (p). It follows immediately from this that eX (p) φ eY (p), and thus that ιX ◦eX (p) ≤ ιY ◦eY (p). By a similar argument we also have ιY ◦eY (p) ≤ ιX ◦eX (p). This shows the upper left square of the diagram in Figure 15 commutes, and that the rest of the diagram commutes follows immediately from the definition of φ and the assumption that is a 1-preorder for E. Suppose now that is a 2-preorder for E, let x1, x2 ∈ X, and suppose that ιX (x1) ≤ ιX (x2). Then φ(x1) ≤ φ(x2), and so ιX ◦ iX (x1) ≤ ιX ◦ iX (x2), by definition of φ. As is a 2-preorder for E, the map ιX is an order embedding, so, since iX is also an order embedding, we must have x1 ≤X x2. This shows ιX is an order embedding, as we have already proved it is order preserving. The argument for ιY is dual. 26 ROB EGROT

Finally, suppose is a 3-preorder for E, and that iX and iY have the preservation properties described above. Let S ⊆ P , and suppose eX [S] exists in X. Then iX ◦ eX [S] = iX ( eX [S]) in X. Let z ∈ X ⊎ Y and suppose z is a lower φ V bound for ιX ◦ eX [S]. Then φ(z) is a lower bound for φ ◦ ιX ◦ eX [S], and so by Vcommutativity of theV diagram in Figure 15 it follows that φ(z) is a lower bound for

ιX ◦ iX ◦ eX [S]. So, as is a 3-preorder for E, we have

φ(z) ≤ ιX ◦ iX ◦ eX [S] ^ = ιX ◦ iX ( eX [S])

= φ ◦ ιX ( ^eX [S]), and so z ≤ ιX ( eX [S]), as φ is an order embedding.^ It follows that ιX ◦ eX [S]= ιX ( eX [S]), and thus φ satisﬁes (P4). The argument for (P5) is dual. V V V

e i P Y / Y Y / Y

eX ιY ι X / X ⊎φ Y Y ιX ▲ ▲▲ φ i ▲▲ X ▲▲▲ ▲% X / X ⊎ Y ιX

Figure 15.

Corollary 6.8. Let X and Y be disjoint posets, and let iX : X → X and iY : Y → Y be order extensions with X ∩ Y = ∅. Let S be a relation on X × Y . Then: (1) If (X, Y, S) is 0-coherent, then so is (X,Y, S).

Moreover, if P is a poset, and if eX : P → X and eY : P → Y are order extensions, then both E = (iX ◦ eX ,iY ◦ eY , S) and E = (eX ,eY , S) are extension polarities, and: (2) If E is n-coherent, then so is E for n ∈{1, 2}. (3) Suppose iX preserves meets in X of subsets of eX [P ] whenever they exist, and let iY likewise preserve joins in Y of subsets of eY [P ]. Then, if E is 3-coherent, so is E, and the same is true if we replace ‘3-coherent’ with ‘Galois’. Proof. This all almost follows immediately from Theorems 3.4, 3.11, 3.14, 3.19 and 6.7, as we have almost proved that if is n-preorder for E, then φ is an n- preorder for E for all n ∈ {0, 1, 2, 3} (modulo some extra conditions for n = 3). To complete the proof we need only show that, for all x ∈ X and y ∈ Y , we have x φ y ⇐⇒ x S y. Now,

x φ y ⇐⇒ iX (x) iY (y) ⇐⇒ iX (x) S iY (y) ⇐⇒ x S y, so we are done. ORDER POLARITIES 27

Converses for the implications in Corollary 6.8 do not hold, as Example 6.9 demonstrates. Example 6.9. Let P be the poset represented by the • elements in Figure 16, let P =∼ X =∼ Y =∼ X, and let Y be represented by Figure 16. Then the implicit maps iX and iY are obviously meet- and join-extensions respectively, and are also, respectively, trivially completely meet- and join-preserving. Let S = Sl ∪{(p,y)}, where Sl ⊆ X × Y is deﬁned analogously to Deﬁnition 5.1. Then S = Rl, and so (eX ,eY , S) is Galois by Proposition 5.2(4). However, (iX ◦ eX ,iY ◦ eY , S) is not even 0-coherent, as we have (p,y) ∈ S but (q,y) ∈/ S, and so (C1) fails.

◦y ❄ ❄❄ ❄❄ • • •p

•q

Figure 16.

Using the notation of Definitions 6.1 and 6.5, we can define a map (−) from the complete lattice of relations on X ×Y to the complete lattice of relations on X ×Y , by taking R to R. Similarly, we can define a map (−) going back the other way by taking S to S. These maps are obviously order preserving. We also have the following result.

Lemma 6.10. Let X and Y be disjoint posets, let iX : X → X and iY : Y → Y be order extensions with X ∩ Y = ∅. Then: (1) Let R ⊆ X × Y . Then R ⊆ (R). Moreover, if (X,Y, R) is 0-coherent, then R = (R). (2) Let S ⊆ X × Y . If (X, Y, S) is 0-coherent, then (S) ⊆ S. Proof. We start with (1). Let x ∈ X, let y ∈ Y and suppose x R y. Then iX (x) R iY (y) by definition of R, and so x(R)y by definition of (R). Suppose now that (X,Y, R) is 0-coherent and let x(R)y. Then iX (x) R iY (y) by definition of (R), and thus x R y by Theorem 6.2(2). For (2), suppose first that (X, Y, S) is 0-coherent, and let x′ ∈ X and y′ ∈ Y with ′ ′ ′ x (S)y . Then, by definition of (S) there are x ∈ X and y ∈ Y with x ≤X iX (x), ′ with iY (y) ≤Y y , and with xSy. But then iX (x) S iY (y) by definition of S, and so ′ ′ ′ ′ x ≤X iX (x) S iY (y) ≤Y y , and thus x S y by 0-coherence of (X, Y, S). Note that the opposite inclusion to that in Lemma 6.10(2) may fail, even when (iX ◦ eX ,iY ◦ eY , S) is Galois, as is demonstrated in Example 6.13 below. Note also that the polarity (X, Y, S) from Example 6.9 is not 0-coherent, but, appealing to Lemma 6.4, we have (S) ⊆ S. Thus (X, Y, S) being 0-coherent is strictly stronger than having (S) ⊆ S. Corollary 6.11. Using the notation of Lemma 6.10, let L be the complete lattice of relations between X and Y , and let M be the complete lattice of relations S ⊆ X ×Y 28 ROB EGROT

• • • ❈ • • ◦y′ • ❈❈ ④④ ⑤ ❇❇ ❈ ④ ⑤⑤ ❇❇ ❈ ④④ ⑤⑤ ❇ •• • ◦x′ • • •

Figure 17. Figure 18. Figure 19. such that (X, Y,S) is 0-coherent. Then the maps (−): L → M and (−) : M → L are, respectively, the left and right adjoints of a Galois connection. Proof. First, recall the discussion at the start of Section 5.1 for the lattice structure of M. Moreover, (−): L → M is well deﬁned by Theorem 6.2(1). By Lemma 6.10 we have R ⊆ (R) for all R ∈ L, and (S) ⊆ S for all S ∈ M, which is one of the equivalent conditions for two order preserving maps to form a Galois connection (see e.g. [6, Lemma 7.26]). Using Corollary 6.8 and Lemma 6.10 we get partial converses for Theorem 6.2.

Corollary 6.12. With notation as in Theorem 6.2, let E = (eX ,eY , R) and suppose E is 0-coherent. Let E = (iX ◦ eX ,iY ◦ eY , R). Then: (1) If E is n-coherent, then so is E for n ∈{1, 2}. (2) Suppose iX preserves meets in X of subsets of eX [P ] whenever they exist, and let iY likewise preserve joins in Y of subsets of eY [P ]. Then, whenever E is 3-coherent, so is E, and this is also true if we replace ‘3-coherent’ with ‘Galois’.

Proof. For (1), given n ∈ {1, 2}, if E is n-coherent, then so is (eX ,eY , (R)), by Corollary 6.8, and as E is assumed to be 0-coherent we have R = (R), by Lemma 6.10. The proof of (2) is essentially the same.

Example 6.13. Let P be the poset in Figure 17, and let X =∼ Y =∼ P . Let X and Y be the posets in Figures 18 and 19 respectively. Deﬁne S ⊆ X × Y so that ′ ′ iX ◦ eX (p) S eY ◦ iY (p) for all p ∈ P , and also x S y . Then (iX ◦ eX ,iY ◦ eY , S) is Galois, as can be seen by considering the poset in Figure 20, and deﬁning ιX and ′ ιY in the obvious way. However, there is no x ∈ X and y ∈ Y with x ≤X iX (x), ′ ′ ′ with iY (y) ≤Y y , and with xSy. Thus (x ,y ) ∈/ (S).

• ❆ ◦y′ • ❆❆⑥ ❆❆⑥⑥ ⑥⑥❆❆ ⑥⑥❆❆ ⑥⑥ ❆ ⑥⑥ ❆ • ◦x′ •

Figure 20.

7. Galois polarities revisited 7.1. Galois polarities via Galois connections. Galois polarities are so named because their associated (unique) 3-preorder can be described in terms of a Galois connection. This idea is precisely articulated in Corollary 7.7 below. ORDER POLARITIES 29

When G = (eX ,eY , R) is Galois, we know from Theorem 4.12 that G has only a single 3-preorder, g. As mentioned previously, to lighten the notation we write e.g. X ⊎ Y in place of X ⊎g Y when working with Galois polarities. The next proposition collects together some useful facts, but ﬁrst we need a deﬁnition.

Definition 7.1 (γ). Let (eX ,eY , R) be a Galois polarity. Define γ : P → X ⊎ Y by γ = ιX ◦ eX = ιY ◦ eY . Noting Figure 2, it’s easy to see that γ is well defined.

Proposition 7.2. Let G = (eX ,eY , R) be a Galois polarity, let iX : X → X be a completely meet-preserving meet-extension, and let iY : Y → Y be a completely join-preserving join-extension. Then:

(1) G = (iX ◦ eX ,iY ◦ eY , R) is Galois. (2) The map γ is an order embedding. Moreover, if S,T ⊆ P and S and T exist in P , then V W (a) γ( S)= γ[S] ⇐⇒ eX ( S)= eX [S], and (b) γ( T )= γ[T ] ⇐⇒ eY ( T )= eY [T ]. V V V V (3) γ[P ]= ιX [X] ∩ ιY [Y ]. W W W W Proof. (1) That G is Galois is Theorem 6.2(4). (2) That γ is well deﬁned follows from 1-coherence of (eX ,eY , R), and that γ is an order embedding follows from 2-coherence of (eX ,eY , R), as γ is the composition of two order embeddings, ιX ◦ eX . That (a) and (b) hold follows from 3- coherence of (eX ,eY , R), as, for example, γ = ιX ◦ eX and ιX preserves meets in X of subsets of eX [P ]. (3) We obviously have γ[P ] ⊆ ιX [X] ∩ ιY [Y ], so let z ∈ ιX [X] ∩ ιY [Y ]. Then there are x ∈ X and y ∈ Y with z = ιX (x) = ιY (y). Thus, as ιX (x) ≤ ιY (y) −1 ↑ −1 ↓ −1 ↑ −1 ↓ we have eX (x ) ∩ eY (y ) 6= ∅. Suppose p ∈ eX (x ) ∩ eY (y ), and that eX (p) 6≤X x. Then, as eX is a meet-extension, there is q ∈ P with x ≤X eX (q) and eX (p) 6≤X eX (q). But this is a contradiction, as, since ιY (y) ≤ ιX (x), appealing to Corollary 4.11 we see that that p ≤P q. Thus x = eX (p), and so z = γ(p). It follows that ιX [X] ∩ ιY [Y ] ⊆ γ[P ] as claimed.

Note that we could slightly relax the preservation properties of iX and iY in the above proposition (and also later) to be the same as those in Corollary 6.12(2), but this isn’t necessary for what we want to do with it. The following fact will be useful.

Proposition 7.3. Let P and Q be posets, let e1 : P → J be a join-completion, and let e2 : Q → M be a meet-completion. Then any Galois connection α : P ↔ Q : β extends uniquely to a Galois connection α′ : J ↔ M : β′. Proof. This is [28, Corollary 2].

Deﬁnition 7.4 (F, G). Let P be a poset, and let eX : P → X and eY : P → Y be meet- and join-completions respectively. Deﬁne maps F : Y → X and G : X → Y as follows: F −1 ↓ (y)= eX [eY (y )]. _ 30 ROB EGROT

G −1 ↑ (x)= eY [eX (x )]. F and G are well deﬁned as X and^Y are complete.

Lemma 7.5. Let P be a poset, and let eX : P → X and eY : P → Y be meet- and join-completions respectively. Then there is a unique Galois connection F : Y ↔ X : G such that eX = F ◦ eY and eY = G ◦ eX . Here F and G are as in Deﬁnition 7.4.

Proof. Using the fact that eX and eY are, respectively, meet- and join-completions, we have, for all x ∈ X and for all y ∈ Y , F −1 ↓ (y) ≤X x ⇐⇒ eX [eY (y )] ≤X x _ −1 ↓ −1 ↑ ⇐⇒ (∀p ∈ eY (y ))(∀q ∈ eX (x )) p ≤P q

−1 ↑ ⇐⇒ y ≤Y eY [eX (x )]

⇐⇒ y ≤Y G^(x). To see that this is the only such Galois connection between X and Y we apply Proposition 7.3 with P = Q and the Galois connection produced by the identity function on P .

Definition 7.6. Let E = (eX ,eY , R) be an extension polarity, let iX : X → X be a completely meet-preserving meet-completion of X, and let iY : Y → Y be a completely join-preserving join-completion of Y . Let F : Y → X and G : X → Y be as in Definition 7.4, with respect to the maps iX ◦ eX , and iY ◦ eY . Define ′′ F ZY X ={(y, x) ∈ Y × X : (iY (y)) ≤X iX (x)} G ={(y, x) ∈ Y × X : iY (y) ≤Y (iX (x))}.

In the above deﬁnition, the maps F and G exist as iX ◦ eX : P → X and iY ◦ eY : P → Y are meet- and join-completions respectively.

′ Corollary 7.7. Recall ZY X from Deﬁnition 4.10. With a setup as in Deﬁnition ′ ′′ 7.6 we have ZY X = ZY X . Proof. This is an immediate consequence of the following equivalence: F (iY (y)) ≤X iX (x) −1 ↓ −1 ↑ ⇐⇒ iX ◦ eX [(iY ◦ eY ) (iY (y) )] ≤X iX ◦ eX [(iX ◦ eX ) (iX (x) )] _ −1 ↓ −1 ↑ ^ ⇐⇒ (∀p ∈ eY (y ))(∀q ∈ eX (x )) iX ◦ eX (p) ≤X iX ◦ eX (q)

−1 ↓ −1 ↑ ⇐⇒ (∀p ∈ eY (y ))(∀q ∈ eX (x )) p ≤ q .

Corollary 7.7 justiﬁes the terminology ‘Galois polarity’, as the unique 3-preorder for any Galois polarity (eX ,eY , R) is deﬁned by R, the orders on X and Y , and the Galois connection given by F and G for any suitable choices of iX and iY . ORDER POLARITIES 31

7.2. Polarity morphisms. Recall from Definition 3.5 that an extension polarity (eX ,eY , R) is complete if eX and eY are completions. Noting Proposition 4.13, we see that [13, Theorem 3.4] establishes a one-to-one correspondence between what we call complete Galois polarities and ∆1-completions of a poset. Theorem 7.26 below expands on the proof of this result, and in Section 7.4 we reformulate it in terms of an adjunction between categories. First we need to define a concept of morphism between Galois polarities. ′ ′ ′ Definition 7.8. Let P and P be posets, let G = (eX ,eY , R) and G = (eX′ ,eY ′ , R ) be Galois polarities extending P and P ′ respectively. Then a polarity morphism h : G → G′ is a triple of order preserving maps ′ ′ ′ h = (hX : X → X ,hP : P → P ,hY : Y → Y ) such that: (M1): The diagram in Figure 21 commutes. (M2): For all x ∈ X and y ∈ Y we have

ιY (y) ≤ ιX (x) → ιY ′ ◦ hY (y) ≤ ιX′ ◦ hX (x). (M3): For all x′ ∈ X′ and for all y′ ∈ Y ′, if (x′,y′) ∈/ R′, then there is x ∈ X and y ∈ Y such that: −1 ′↑ ↑ (i) hX (x ) ⊆ x . −1 ′↓ ↓ (ii) hY (y ) ⊆ y . ′ ′ (iii) hX (a) R y → a R y for all a ∈ X. ′ ′ (iv) x R hY (b) → x R b for all b ∈ Y . (v) (x, y) ∈/ R.

For a polarity morphism h, if hX , hP and hY are all order embeddings, and also ′ hX (x) R hY (y) → x R y for all x ∈ X and y ∈ Y , then h is a polarity embedding. If, in addition, all maps are actually order isomorphisms, then h is a polarity isomorphism, and we say G and G′ are isomorphic. Sometimes we want to ﬁx a poset P and deal exclusively with isomorphism classes of Galois polarities extending P . In this case we say Galois polarities G and G′ are isomorphic as Galois polarities extending P if there is a polarity ′ isomorphism (hX ,hP ,hY ): G → G where hP is the identity on P .

e e X o X P Y / Y

hX hP hY X′ o P ′ / Y ′ eX′ eY ′

Figure 21.

Note that if hX and hY are order embeddings, then hP will be too, but it is not the case that hX and hY being order isomorphisms implies that hP is too, as hP may not be surjective. Note also that Deﬁnition 7.8, while being similar in some respects, is largely distinct from the notion of a bounded morphism between polarity frames from [34]. It is also completely diﬀerent to the frame morphisms of [7, 11], which are duals to complete lattice homomorphisms, rather than ‘decomposed’ 32 ROB EGROT versions of certain maps X ⊎ Y → X′ ⊎ Y ′. We will make this clear in Theorem 7.13 later.

′ ′ ′ Lemma 7.9. If h = (hX : X → X ,hP : P → P ,hY : Y → Y ) is a polarity ′ morphism, then for all x ∈ X and for all y ∈ Y we have x R y → hX (x) R hY (y). ′ Proof. Suppose (hX (x),hY (y)) ∈/ R . Then, by (M3) there are x0 ∈ X and y0 ∈ Y −1 ↑ ↑ −1 ↓ ↓ with hX (hX (x) ) ⊆ x0, with hY (hY (y) ) ⊆ y0 and with (x0,y0) ∈/ R. From −1 ↑ ↑ hX (hX (x) ) ⊆ x0 it follows that x0 ≤X x, and similarly we have y ≤Y y0. Thus (x, y) ∈/ R, as otherwise (C1) and (C2) would force x0 R y0.

The following definition is due to Erné. This will be useful to us as it precisely characterizes those maps between posets that lift (uniquely) to complete homomorphisms between their MacNeille completions [8, Theorem 3.1]. Definition 7.10. An order preserving map f : P → Q is cut-stable if whenever −1 ↑ ↑ −1 ↓ ↓ q1 6≤ q2 ∈ Q, there are p1 6≤ p2 ∈ P such that f (q1 ) ⊆ p1 and f (q2 ) ⊆ p2. (M3) is related to cut-stability, as we shall see in Theorems 7.13 and 7.16. We can think of this as an adaptation of ideas from [16, Section 4]. We extend from what, according to our terminology, is the special case of (eX ,eY , R) where eX and eY are, respectively, the free directed meet- and join-completions and R = Rl, to Galois polarities in general. We will need the following definition.

′ ′ Definition 7.11 (ψh). Let G = (eX ,eY , R) and G = (eX′ ,eY ′ , R ) be Galois polarities extending posets P and P ′ respectively. Let γ : P → X ⊎ Y and γ′ : ′ ′ ′ ′ P → X ⊎ Y be the maps from Definition 7.1. Let h = (hX ,hP ,hY ): G → G be ′ ′ a polarity morphism. Define ψh : X ⊎ Y → X ⊎ Y by

ιX′ ◦ hX (z) when z ∈ ιX [X] ψh(z)= (ιY ′ ◦ hY (z) when z ∈ ιY [Y ]

We prove that ψh is well deﬁned as part of Theorem 7.13, below. We will also use the following deﬁnition.

′ ′ Definition 7.12. Let G = (eX ,eY , R) and G = (eX′ ,eY ′ , R ) be Galois polarities, let γ : P → X ⊎ Y and γ′ : P ′ → X′ ⊎ Y ′ be as in Definition 7.1, and let ψ : X ⊎ Y → X′ ⊎ Y ′. We say ψ is Galois-stable if it is order preserving, cut- stable, and satisfies: (G1): ψ ◦ γ[P ] ⊆ γ′[P ′]. ′ (G2): ψ ◦ ιX [X] ⊆ ιX′ [X ]. ′ (G3): ψ ◦ ιY [Y ] ⊆ ιY ′ [Y ]. Theorem 7.13. Let G, G′ γ and γ′ be as in Definition 7.12. Then, given a polarity ′ morphism h = (hX ,hP ,hY ): G → G , we have:

(1) ψh is the unique map such that the diagram in Figure 22 commutes (if we replace ψ with ψh). (2) ψh is Galois-stable. (3) ψh is an order embedding if and only if h is a polarity embedding. (4) If hX and hY are both surjective, then ψh is surjective. Proof. Let and ′ be the unique 3-preorders for G and G′ respectively. ORDER POLARITIES 33

(1) Given (hX ,hP ,hY ), the commutativity of the diagram in Figure 22 demands that ψh can only be defined as in Definition 7.11, if it exists at all. This deals with uniqueness. Now, if x ∈ X and y ∈ Y , then, using Lemma 7.9, we have ′ ′ x y ⇐⇒ x R y =⇒ hX (x) R hY (y) ⇐⇒ hX (x) hY (y). ′ If y x, then ιY (y) ≤ ιX (x) by definition, and so hY (y) hX (x) by (M2). This shows ψh is well defined, and combined with the fact that hX and hY are both order preserving, also shows ψh is order preserving. (2) We have already proved that ψh is order preserving. To see that it is cut-stable, ′ ′ ′ ′ let z1 6≤ z2 ∈ X ⊎ Y . Since by Theorem 4.12(3) we know ιX′ [X ] and ιY ′ [Y ] are, respectively, join- and meet-dense in X′ ⊎ Y ′, there are x′ ∈ X′ and y′ ∈ Y ′ with ′ ′ ′ ′ ′ ′ ′ ιX′ (x ) ≤ z1, with z2 ≤ ιY ′ (y ), and with ιX′ (x ) 6≤ ιY ′ (y ) (i.e. (x ,y ) ∈/ R ). Thus by (M3) there are x ∈ X and y ∈ Y with the five properties described in that definition. We will satisfy the condition of Definition 7.10 using the pair ιX (x) 6≤ ιY (y). −1 ↑ ↑ Let z ∈ ψh (z1 ). We must show that z ∈ ιX (x) . We have ψh(z) ≥ z1 ≥ ′ ιX′ (x ). There are two cases. If z = ιX (a) for some a ∈ X, then ψh(z) = ′ −1 ′↑ ↑ ιX′ ◦ hX (a), and so hX (a) ≥X′ x . Thus a ∈ hX (x ), and so a ∈ x , by ↑ (M3)(i). It follows that z = ιX (a) ∈ ιX (x) as claimed. Alternatively, suppose ′ z = ιY (b) for some b ∈ Y . Then ψh(z)= ιY ′ ◦ hY (b), and so ιY ′ ◦ hY (b) ≥ ιX′ (x ), ′ ′ and consequently x R hY (b). It follows from (M3)(iv) that x R b, and thus that −1 ↓ ↓ ιX (x) ≤ ιY (b) = z as required. That ψh (z2 ) ⊆ ιY (y) follows by a dual argument, and so ψh is cut-stable. To see that (G1) holds for ψh note that

ψh ◦ γ(p)= ψh ◦ ιY ◦ eY (p)

= ιY ′ ◦ hY ◦ eY (p)

= ιY ′ ◦ eY ′ ◦ hP (p) ′ = γ (hP (p)).

That (G2) and (G3) hold follows immediately from the deﬁnition of ψh. (3) If ψh is an order embedding, then that hX and hY , and thus also hP , are order embeddings follows directly from the commutativity of the diagram in Figure 22 ′ (substituting ψh for ψ). Moreover, that hX (x) R hY (y) → x R y for all x ∈ X and y ∈ Y holds for the same reason. Thus (hX ,hP ,hY ) is a polarity embedding. Conversely, suppose (hX ,hP ,hY ) is a polarity embedding. Since we already ′ ′ know ψh is order preserving, suppose z,z ∈ X ⊎Y and that ψh(z) ≤ ψh(z ). There ′ ′ ′ are four cases. If either z,z ∈ ιX [X], or z,z ∈ ιY [Y ], then that z ≤ z follows again from the commutativity of the diagram in Figure 22 and the assumption ′ that hX and hY are order embeddings. In the case where z = ιX (x) and z = ιY (y) for some x ∈ X and y ∈ Y , we have ′ ′ ′ ψh(z) ≤ ψh(z ) ⇐⇒ hX (x) R hY (y) ⇐⇒ x R y ⇐⇒ ιX (x) ≤ ιY (y) ⇐⇒ z ≤ z . ′ In the ﬁnal case we have z = ιY (y) and z = ιX (x) for some x ∈ X and y ∈ Y . Then ′ ψh(z) ≤ ψh(z ) ⇐⇒ ιY ′ ◦ hY (y) ≤ ιX′ ◦ hX (x).

If p, q ∈ P , and eY (p) ≤Y y and x ≤X eX (q), then

ιY ′ ◦ hY ◦ eY (p) ≤ ιY ′ ◦ hY (y) ≤ ιX′ ◦ hX (x) ≤ ιX′ ◦ hX ◦ eX (q), 34 ROB EGROT

and thus ιY ′ ◦ eY ′ ◦ hP (p) ≤ ιX′ ◦ eX′ ◦ hP (q), by the commutativity of the ′ diagram in Figure 21. It follows that eY ′ ◦ hP (p) eX′ ◦ hP (q), and thus that ′ ′ hP (p) ≤P ′ hP (q), as is a 3-preorder for G . Thus p ≤P q, as hP is an order embedding. So by Corollary 4.11 we have y x as required. ′ ′ ′ ′ (4) If hX and hY are onto, then, given z ∈ X ⊎ Y , we have either z = ιX′ (hX (x)) ′ for some x ∈ X, or z = ιY ′ (hY (y)) for some y ∈ Y . In either case it follows there is z ∈ X ⊎ Y with ψh(z)= z, and thus that ψh is onto.

Note that the converse to (4) in the above theorem may not hold, as illustrated by Example 7.14.

h Y Y / Y ′ ①< d❍❍ eY ①① ❍❍ eY ′ ①① ιY ιY ′ ❍❍ ①① ❍❍ ①① γ ψ γ′ ❍ P / X ⊎ Y / X′ ⊎ Y ′ o 1 P ′ ❋❋ O O ✈ ❋❋ hP ✈✈ ❋❋ ιX ιX′ ✈✈ eX ❋❋ ✈✈eX′ ❋# {✈✈ X / X′ hX

Figure 22.

Example 7.14. Let P = X be a two element antichain, and let Y be this two element antichain extended by adding a join for the two base elements. Let P ′ =∼ X′ =∼ Y ′ =∼ Y . Then the inclusion maps (modulo isomorphism) and the relations ′ ′ ∼ ∼ ′ ′ Rl and Rl deﬁne Galois polarities G and G , and X ⊎ Y = Y = X ⊎ Y . Let hX , ′ ′ ′ hP and hY be the maps induced by the inclusions P ⊂ P , X ⊂ X and Y = Y (modulo isomorphism). Then ψh is an isomorphism, and so is surjective, but hX is obviously not surjective.

′ ′ Definition 7.15. Let G = (eX ,eY , R) and G = (eX′ ,eY ′ , R ) be Galois polarities. Let γ : P → X ⊎ Y and γ′ : P ′ → X′ ⊎ Y ′ be as in Definition 7.1. Let ψ : X ⊎ Y → ′ ′ ψ ′ ψ ′ ψ ′ X ⊎ Y satisfy (G1)–(G3). Define hX : X → X , hY : Y → Y and hP : P → P as follows: ψ −1 hX =ιX′ ◦ ψ ◦ ιX ψ −1 hY =ιY ′ ◦ ψ ◦ ιY ψ ′−1 hP =γ ◦ ψ ◦ γ.

−1 −1 ′−1 ψ ψ ψ Here ιX′ , ιY ′ and γ are the partial inverse maps. Define hψ = (hX ,hP ,hY ). −1 −1 ′−1 The maps ιX′ , ιY ′ and γ are well defined as partial functions because ιX′ , ′ ψ ψ ψ ιY ′ and γ are all order embeddings. That hX , hY and hP are well defined will be proved as part of the following theorem. Theorem 7.16. Let G, G′, γ and γ′ be as in Definition 7.15. Let ψ : X ⊎ Y → ′ ′ ψ ψ ψ X ⊎ Y be a Galois-stable map, and let hX , hY and hP be as above. Then: ORDER POLARITIES 35

ψ ψ ψ (1) hX , hY and hP are the unique maps such that the diagram in Figure 22 ψ ψ ψ commutes (if we replace hX ,hY and hP with hX , hY and hP , respectively). ψ ψ ψ (2) hψ = (hX ,hP ,hY ) is a polarity morphism. Proof. ψ ψ −1 (1) If the diagram in Figure 22 is to commute, then hX and hY must be ιX′ ◦ ψ ◦ ιX −1 and ιY ′ ◦ ψ ◦ ιY respectively, if they are to exist at all. It follows from (G2) that −1 ψ ιX′ is total on ψ ◦ ιX [X], and thus that hX is well defined. A similar argument ψ with (G3) shows hY is also well defined. The commutativity of this diagram also ψ demands that, if hP exists, we have ψ ψ −1 −1 eX′ ◦ hP = hX ◦ eX = ιX′ ◦ ψ ◦ ιX ◦ eX = ιX′ ◦ ψ ◦ γ, and thus ψ −1 −1 ′−1 hP = eX′ ◦ ιX′ ◦ ψ ◦ γ = γ ◦ ψ ◦ γ, if this is well defined, which it is, by (G1). (2) To see that that hψ satisfies (M1), observe that we have ψ −1 hX ◦ eX = ιX′ ◦ ψ ◦ ιX ◦ eX −1 = ιX′ ◦ ψ ◦ γ −1 ′ ψ = ιX′ ◦ γ ◦ hP −1 ψ = ιX′ ◦ ιX′ ◦ eX′ ◦ hP ψ = eX′ ◦ hP , ψ ψ ψ ψ ψ and hY ◦ eY = eY ′ ◦ hP by a similar argument. That (hX ,hP ,hY ) satisfies (M2) also follows easily from the definitions of hX and hY combined with the fact that ψ is order preserving. It remains only to check (M3). As ψ is Galois-stable it is necessarily cut-stable. ′ ′ ′ ′ ′ ′ ′ ′ ′ So, let x ∈ X , let y ∈ Y , and suppose (x ,y ) ∈/ R . Then ιX′ (x ) 6≤ ιY ′ (y ), −1 ′ ↑ ↑ and thus by cut-stability there are z1 6≤ z2 ∈ X ⊎ Y with ψ (ιX′ (x ) ) ⊆ z1 , and −1 ′ ↓ ↓ ψ (ιY ′ (y ) ) ⊆ z2 . As ιX [X] and ιY [Y ] are, respectively, join- and meet-dense in X ⊎ Y , there are x ∈ X and y ∈ Y with ιX (x) ≤ z1, with z2 ≤ ιY (y), and with −1 ′ ↑ ↑ −1 ′ ↓ ↓ (x, y) 6∈ R. It follows that ψ (ιX′ (x ) ) ⊆ ιX (x) and ψ (ιY ′ (y ) ) ⊆ ιY (y) . We check the conditions required for (M3) are satisfied by the pair (x, y): ψ−1 ′↑ ψ ′ (i) Let a ∈ X and suppose a ∈ hX (x ). Then hX (a) ≥X′ x and so, by ψ −1 ′ ↑ ↑ ↑ definition of hX , we have ιX (a) ∈ ψ (ιX′ (x ) ) ⊆ ιX (x) , and so a ∈ x . (ii) Dual to (i). ψ ′ ′ ψ ′ (iii) Let a ∈ X and suppose hX (a) R y . Then ιX′ ◦ hX (a) ≤ ιY ′ (y ), and thus ′ −1 ′ ↓ ↓ ψ ◦ ιX (a) ≤ ιY ′ (y ). It follows that ιX (a) ∈ ψ (ιY ′ (y ) ) ⊆ ιY (y) , and so a R y as required. (iv) Dual to (iii). (v) By choice of (x, y).

′ ′ Theorem 7.17. Let G = (eX ,eY , R) and G = (eX′ ,eY ′ , R ) be Galois polarities. Let h : G → G′ be a polarity morphism. Then, recalling Deﬁnitions 7.11 and 7.15, we have

h = hψh . 36 ROB EGROT

Similarly, if ψ : X ⊎ Y → X′ ⊎ Y ′ is a Galois-stable map, then

ψ = ψhψ .

Proof. Given h, by Theorem 7.13 we know ψh is the unique map such that the diagram in Figure 22 commutes (replacing ψ with ψh). We also know that ψh is ψh ψh ψh Galois-stable. But by Theorem 7.16 it follows that hψh = (hX ,hP ,hY ) is the unique triple such that the same diagram commutes (replacing ψ with ψh, and ψh ψh ψh replacing hX , hP and hY hX , hP and hY respectively). From this it follows immediately that h = hψh , as we now have two ‘unique’ commuting diagrams involving ψh. The argument proving ψ = ψhψ is essentially the same. We can now easily prove a result analogous to Theorem 7.13(3). ′ Corollary 7.18. Let G, G , ψ and hψ be as in Theorem 7.16. Then ψ is an order embedding if and only if hψ is a polarity embedding.

Proof. We have ψ = ψhψ from Theorem 7.17, and from Theorem 7.13(3) we have that ψhψ is an order embedding if and only if hψ is a polarity embedding. The result follows immediately. The next lemma is needed to show polarity morphisms compose as expected.

Lemma 7.19. For each i ∈ {1, 2, 3}, let Gi = (eXi ,eXi , Ri) be a Galois polarity, and let ψ1 : X1 ⊎ Y1 → X2 ⊎ Y2 and ψ2 : X2 ⊎ Y2 → X3 ⊎ Y3 be Galois-stable maps.

Then the composition ψ2 ◦ ψ1 is also Galois-stable, and hψ2◦ψ1 = hψ2 ◦ hψ1 .

Similarly, let h1 = (hX1 ,hP1 ,hY1 ): G1 → G2 and h2 = (hX2 ,hP2 ,hY2 ): G2 → G3 be polarity morphisms. If we deﬁne

h2 ◦ h1 = (hX2 ◦ hX1 ,hP2 ◦ hP1 ,hY2 ◦ hY1 ), then h2 ◦ h1 is also a polarity morphism, and ψh1◦h2 = ψh2 ◦ ψh1 . Proof. It’s straightforward to show that the composition of maps satisfying (G1)– (G3) also satisﬁes these conditions, and compositions of order preserving maps are obviously order preserving. Moreover, cut-stability is preserved by composition [8, Corollary 2.10]. So compositions of Galois-stable maps are also Galois-stable.

Moreover, unpacking Deﬁnition 7.15 reveals that hψ2◦ψ1 = hψ2 ◦ hψ1 . For example, hψ2◦ψ1 = ι−1 ◦ ψ ◦ ψ ◦ ι X1 X3 2 1 X1 = ι−1 ◦ ψ ◦ ι ◦ ι−1 ◦ ψ ◦ ι X3 2 X2 X2 1 X1 = hψ2 ◦ hψ1 . X1 X1

Similarly, unpacking Deﬁnition 7.11 reveals that ψh2◦h1 = ψh2 ◦ ψh1 , and is consequently well deﬁned and Galois-stable, and so it follows from Theorem 7.16 that h is a polarity morphism. So h ◦ h is a polarity morphism, as, using ψh2 ◦ψh1 2 1 Theorem 7.17, we have h ◦ h = h ◦ h 2 1 ψh2 ψh1 = h . ψh2 ◦ψh1

Proposition 7.20. The class of Galois polarities and polarity morphisms forms a category, where polarity morphisms compose componentwise. ORDER POLARITIES 37

Proof. Identity morphisms obviously exist, so we need only check composition be- haves appropriately, and this follows immediately from Lemma 7.19. We will expand on this categorical viewpoint in Section 7.4.

7.3. Galois polarities and ∆1-completions. Recall that if P is a poset we write, for example, e : P → N (P ) for the MacNeille completion of P (see Deﬁnition 2.3).

Lemma 7.21. If (eX ,eY , R) is a Galois polarity, and if e : X ⊎ Y → N (X ⊎ Y ) is the MacNeille completion of X ⊎ Y , then e ◦ γ : P → N (X ⊎ Y ) is a ∆1-completion (where γ is as in Deﬁnition 7.1).

Proof. X ⊎ Y is join-generated by ιX [X], and meet-generated by ιY [Y ] (by The- orem 4.12(3)), and X and Y are meet- and join-generated by eX [P ] and eY [P ] respectively. N (X ⊎ Y ) is both join- and meet-generated by e[X ⊎ Y ]. Thus every element of N (X ⊎ Y ) is both a join of meets, and a meet of joins, of elements of e ◦ γ[P ] as required.

Deﬁnition 7.22. If G is a Galois polarity, deﬁne the ∆1-completion dG = e ◦ γ constructed from G as in Lemma 7.21 to be the ∆1-completion generated by G.

Now, given a ∆1-completion d : P → D, deﬁne XD and YD to be (disjoint isomorphic copies of) the subsets of D meet- and join-generated by d[P ] respectively.

Deﬁne eXD : P → XD and eYD : P → YD by composing d with the isomorphisms into XD and YD respectively. Abusing notation by identifying XD and YD with their images in D, deﬁne RD on XD × YD by x RD y ⇐⇒ x ≤ y in D. Then

(eXD ,eYD , RD) is a complete Galois polarity. To see this note that it is straightforward to show that the order on D induces an order on XD ∪ YD satisfying the necessary conditions. This leads to the following deﬁnition.

Deﬁnition 7.23. If d : P → D is a ∆1-completion, deﬁne the complete Galois polarity Gd = (eXD ,eYD , RD) constructed from d as described in the preceding paragraph to be the Galois polarity generated by d.

Lemma 7.24. Let G = (eX ,eY , R) be a Galois polarity. Let dG = e ◦ γ : P → N (X ⊎ Y ) be the ∆1-completion generated by G. Denote the Galois polarity generated by dG by GdG = (eXN ,eYN , RN ). Then there is a polarity embedding

(hX ,hP ,hY ) from G to GdG with hP = idP . Moreover, if G is complete, then this embedding is an isomorphism of polarities extending P , and is the only such isomorphism.

Proof. Recall that γ = ιX ◦ eX = ιY ◦ eY by definition. Define hP = idP . Note that we are using e.g. XN to denote XN (X⊎Y ). Let µX be the isomorphism used to define XN (recall that XN is an isomorphic copy of a subset of N (X ⊎ Y )). Define the map hX : X → XN by hX = µX ◦ e ◦ ιX . This is clearly a meet-completion. Similarly define a meet-completion hY : Y → YN by hY = µY ◦ e ◦ ιY . Note that eXN is just µX ◦ e ◦ γ = µX ◦ e ◦ ιX ◦ eX , and similarly eYN = µY ◦ e ◦ ιY ◦ eY . Thus we trivially have the commutativity required by (M1). To show that (M2) is also satisfied, let x ∈ X, let y ∈ Y , and suppose ιY (y) ≤

ιX (x). Then e ◦ ιY (y) ≤ e ◦ ιX (x). The unique 3-preorder = g for GdG can only be the order inherited from N (X ⊎ Y ), so µX ◦ e ◦ ιX (x) µY ◦ e ◦ ιY (y), and thus

ιYN ◦ hY (y) ≤ ιXN ◦ hX (x) as required. 38 ROB EGROT

′ ′ ′ ′ −1 ′ For (M3), let x ∈ XN , let y ∈ YN , and suppose (x ,y ) ∈/ RN . Then µX (x ) 6≤ −1 ′ µY (y ). As e ◦ ιX [X] and e ◦ ιY [Y ] are, respectively, join- and meet-dense in N (X ⊎ Y ), there is x ∈ X and y ∈ Y with e ◦ ιX (x) 6≤ e ◦ ιY (y), with e ◦ ιX (x) ≤ −1 ′ −1 ′ µX (x ), and with µY (y ) ≤ e ◦ ιY (y). We check the necessary conditions are satisﬁed for this choice of x and y: (i) Let a ∈ X. Then

′ −1 −1 ′ hX (a) ≥XN x ⇐⇒ µX ◦ µX ◦ e ◦ ιX (a) ≥ µX (x )

=⇒ e ◦ ιX (a) ≥ e ◦ ιX (x)

⇐⇒ a ≥X x,

−1 ′↑ ↑ and so hX (x ) ⊆ x as required. (ii) Dual to (i). ′ ′ (iii) Let a ∈ X, let y ∈ YN , and suppose hX (a) RN y . Then −1 −1 ′ µX ◦ µX ◦ e ◦ ιX (a) ≤ µY (y ) ≤ e ◦ ιY (y),

and so ιX (a) ≤ ιY (y), and thus a R y as required. (iv) Dual to (iii). (v) Since e ◦ ιX (x) 6≤ e ◦ ιY (y) we must have (x, y) ∈/ R.

Now let x ∈ X, let y ∈ Y , and suppose hX (x) RN hY (y). Then −1 −1 µX ◦ µX ◦ e ◦ ιX (x) ≤ µY ◦ µY ◦ e ◦ ιY (y), and so ιX (x) ≤ ιY (y), and thus x R y, and we conclude that (hX , idP ,hY ) is a polarity embedding as claimed. Finally, suppose eX and eY are completions. Then, ′ given x ∈ XN , there is S ⊆ P with

′ x = µX ( e ◦ γ[S])

= µX (^ e ◦ ιX ◦ eX [S])

= µX ◦^e ◦ ιX ( eX [S])

= hX ( eX [S])^, so hX is surjective, and thus an isomorphism.^ The same is true for hY by duality, and, as hP is the identity on P , we have a polarity isomorphism extending P as claimed. Finally, suppose g = (gX , idP ,gY ) is another such polarity isomorphism.

Then we must have gX ◦ eX = eXN = hX ◦ eX . Since eX is a meet-extension it follows that gX = hX . A dual argument gives gY = hY , so we are done.

Lemma 7.25. Let d : P → D be a ∆1-completion. Let Gd = (eXD ,eYD , RD) be the complete Galois polarity generated by d. Then d is isomorphic to dGd , the ∆1-completion generated by Gd.

Proof. Let µX and µY be the isomorphisms onto XD and YD, respectively, let γd : P → XD ⊎ YD, and let ed : XD ⊎ YD → N (XD ⊎ YD) be the MacNeille −1 −1 completion of XD ⊎ YD. Then µX (X) ∪ µY (Y ) has an order inherited from D, and we have the situation described in Figure 23. This gives the result, as ed ◦ γd is the ∆1-completion generated by Gd. ORDER POLARITIES 39

γd ed P / XD ⊎ YD / N (XD ⊎ YD) ▼▼▼ O O ▼▼▼ ▼▼▼ =∼ ∼= d ▼▼▼ & µ−1(X) ∪ µ−1(Y ) / D X Y ⊆

Figure 23.

Theorem 7.26. There is a 1-1 correspondence between (isomorphism classes of) ∆1-completions and (isomorphism classes of) complete Galois polarities. More- over, for a ﬁxed poset P this correspondence restricts to a 1-1 correspondence between (isomorphism classes of) ∆1-completions of P and (isomorphism classes of) complete Galois polarities extending P .

Proof. We define a map Θ from the class of all isomorphism classes of ∆1-completions to the class of all isomorphism classes of complete Galois polarities. Given a ∆1- completion d, let [d] be the associated isomorphism class, and define Θ([d]) by G ∈ Θ([d]) if and only if G is isomorphic to Gd, the complete Galois polarity generated by d. Now, if G ∈ Θ([d]) and G′ =∼ G, then G′ ∈ Θ([d]) by definition of Θ. Conversely, ′ if G and G are both in Θ([d]), then they are both isomorphic to Gd, and thus each other. So Θ([d]) is indeed an isomorphism class, provided it is well defined. Suppose then that d =∼ d′, let G ∈ Θ([d]) and let G′ ∈ Θ([d′]). Since d and d′ are isomorphic, it’s easy to construct an isomorphism between the complete Galois polarities they generate (since these polarities are based on subsets of the corresponding complete lattices). If we denote these generated Galois polarities by Gd and Gd′ , we have ′ ′ ′ G =∼ Gd =∼ Gd′ =∼ G , and thus G ∈ Θ([d ]) and G ∈ Θ([d]). It follows that Θ([d]) = Θ([d′]), and so Θ is well defined. ∼ By Lemma 7.24, if G is a complete Galois polarity, then G = GdG , so G ∈ ′ Θ([dG]), and so Θ is surjective. Now suppose that Θ([d]) = G = Θ([d ]). Then Gd =∼ G =∼ Gd′ . Now, it’s easy to see that isomorphic Galois polarities generate isomorphic ∆1-completions (the map ψh will be an isomorphism, by Theorem 6.7, and this will lift to the required isomorphism), so ∼ ∼ dGd = dG = dGd′ . ∼ ∼ ′ ∼ ′ By Lemma 7.25 we have dGd = d and dGd′ = d , and so d = d . This shows Θ is injective. Finally, appealing to Proposition 4.13, the second claim is essentially [13, Theo- rem 3.4], and we can also obtain this result with the proof above by working modulo ∆1-completions of a fixed poset P , and complete Galois polarities extending P .

Before moving on we pause to consider a technical question regarding the polarity extensions discussed in Section 6. Given a Galois polarity (eX ,eY , R) which is not complete, by Lemma 7.24 there is a polarity embedding from (eX ,eY , R) to

(eXD ,eYD , RD), where eXD and eYD are meet- and join-completions respectively. By the deﬁnition of polarity embeddings, there are order embeddings hX : X → XD and hY : Y → YD, and its easy to see these will be meet- and join-completions respectively. 40 ROB EGROT

Thus Theorem 6.2 applies and produces a Galois polarity (eXD ,eYD , R). We certainly have R ⊆ RD, by Theorem 6.2(5), but does the other inclusion also hold? The answer, in general, is no. To see this we borrow [13, Example 2.2], and lean heavily on the discussion at the start of Section 4 in that paper. The MacNeille completion of a poset P can be constructed from the Galois polarity (eFp ,eIp , Rl), where eFp : P →Fp and eIp : P → Ip are the natural embeddings into the sets of principal upsets and downsets of P respectively. Consider the poset P = ω ∪ ω∂. I.e. P is made up of a copy of ω below a disjoint copy of the dual ω∂ . Then N (P )= ω ∪{z} ∪ ω∂. I.e. P with an additional ∂ element above ω and below ω . Let X = Fp and let Y = Ip, so N (P ) is generated by (eX ,eY , Rl). Let the complete Galois polarity arising from Theorem 7.26 be ∼ ∂ (eXD ,eYD , RD), where XD = X ∪{zX } (a copy of ω ∪ ω with zX in between), and ∂ YD =∼ Y ∪{zY } (a copy of ω ∪ ω zY in between). Now, to produce N (P ) it is necessary that zX RD zY , but (zX ,zY ) ∈/ Rl, and thus RD 6= R in this case. It also follows from this that the ∆1-completion generated by (eX ,eY , R) may not be isomorphic to the one generated by (iX ◦ eX ,iY ◦ eY , R) from Theorem 6.2, even when iX and iY are meet- and join-completions respectively (just set iX = hX and iY = hY and use the example above). 7.4. A categorical perspective. Here we assume some familiarity with the basic concepts of category theory. The standard reference is [22], and an accessible introduction can be found in [21]. Definition 7.27. We define a pair of categories, Pol and Del as follows: Pol: Let Pol be the category of Galois polarities and polarity morphisms (from Definition 7.8). Del: Let Del be the category whose objects are ∆1-completions, and whose maps are commuting squares as described in Definition 2.2, with the additional property that gQ in that definition is a complete lattice homomorphism. Definition 7.28. Let Γ : Pol → Del and ∆ : Del → Pol be defined as follows: Γ: Let Γ : Pol → Del be the map that takes a Galois polarity G to the ∆1-completion it generates (recall Definition 7.22), and takes a polarity ′ morphism h : G → G to the map between ∆1-completions described in Figure 24, where N (ψh) is the unique complete lattice homomorphism lift of the map ψh from Definition 7.11 to the respective MacNeille completions, as described in [8, Theorem 3.1]. ∆: Let ∆ : Del → Pol be the map that takes a ∆1-completion d : P → D to the Galois polarity it generates (recall Definition 7.23), and takes a ∆1- completion morphism as in Figure 25 to the triple (g| ,f,g| ), where, XD1 YD1 for example, g| is (modulo isomorphism) the restriction of g to X . XD1 D1 That Γ and ∆ are well defined is proved as part of the next theorem. Theorem 7.29. Γ and ∆ are functors and form an adjunction Γ ⊣ ∆. Proof. For easier reading we will break the proof down into discrete statements which obviously add up to a proof of the claimed result. • “Γ is well defined”. Γ is certainly well defined on objects. For maps, as Theorem 7.13 says ψh will be cut-stable, [8, Theorem 3.1] says that N (ψh) will be a complete lattice homomorphism. Moreover, the conjunction of these theorems also implies that the diagram in Figure 24 commutes. ORDER POLARITIES 41

• “Γ is a functor”. To see that Γ lifts identity maps to identity maps note that the identity morphism on (eX ,eY , R) clearly lifts via Deﬁnition 7.11 to the identity on X ⊎ Y , and since taking MacNeille completions is functorial for cut-stable maps (see [8, Corollary 3.3]), that Γ maps identity morphisms appropriately follows immediately. Similarly, given polarity morphisms h1 and h2, if h2 ◦ h1 exists

we have ψh2◦h1 = ψh2 ◦ ψh1 , by Lemma 7.19, so Γ respects composition as the MacNeille completion functor does. • “∆ is well defined”. ∆ is also clearly well defined on objects. Consider now a ∆1- completion map as in Figure 25. Abusing notation by, for example, identifying X with its isomorphic copy inside D , we must show that g| : X → X , D1 1 XD1 D1 D2 that g| : Y → Y , and that (g| ,f,g| ) satisfies the conditions of YD1 D1 D2 XD1 YD1 Definition 7.8. First, to lighten the notation define g = g| , and g = g| . X XD1 Y YD1

Now, let x ∈ XD1 . Then there is S ⊆ P1 with x = d1[S]. So, using the commutativity of the diagram in Figure 25 and the fact that g is a complete lattice homomorphism, we have V

g(x)= g( d1[S]) = g ◦ d1[S]= d2 ◦ f[S]. This proves that g does^ indeed have^ codomain X^ , and the analogous result XD1 D2 for gY holds by duality. Since gX , f, and gY are all obviously order preserving, it remains only to check the conditions of Deﬁnition 7.8: (M1): This follows immediately from the deﬁnitions of gX and gY and the commutativity of the diagram in Figure 25. (M2): Without loss of generality, we can think of the ι maps as inclusion func-

tions, and so the claim is just the statement that y ≤D1 x → g(y) ≤D2 g(x), and thus is true as g is order preserving. ′ ′ (M3): Abusing notation in the same way as before, let x ∈ XD2 , let y ∈ YD2 , ′ ′ −1 ′↑ and suppose x 6≤ y ∈ D2. Using the completeness of D1, let z1 = g (x ), −1 ′↓ and let z2 = g (y ). It follows easily from the fact that g is a complete ′ ′ V lattice homomorphism that x ≤D2 g(z1) and g(z2) ≤D2 y , so if z1 ≤D1 z2, ′ W ′ then x ≤D2 g(z1) ≤D2 g(z2) ≤D2 y , as g is order preserving. Thus to avoid

contradiction we must have z1 6≤D1 z2.

As XD1 and YD1 are, respectively, join- and meet-dense in D1, there is x ∈ ↓ ↑ −1 ′↑ XD1 ∩ z1 and y ∈ YD1 ∩ z2 with x 6≤D1 y. Now, as x ≤D1 z1 = g (x ) we have g−1(x′↑) ⊆ z↑ ⊆ x↑. Thus (i) holds for this choice of x, and (ii) holds for X 1 V y by a dual argument. ′ −1 ′↓ Moreover, suppose a ∈ XD1 , and that g(a) ≤D2 y . Then a ∈ g (y ), and so

a ≤D1 z2, by deﬁnition of z2, and consequently a ≤D1 y. Thus (iii) holds, and (iv) is dual. That (v) holds is automatic from the choice of x and y. • “∆ is a functor”. ∆ obviously sends identity maps to identity maps, and almost as obviously respects composition. • “Γ ⊣ ∆”. The unit η is deﬁned so that its components are the embeddings of

G = (eX ,eY , R) into ∆Γ(G)= GdD = (eXN ,eYN , RN ) described in Lemma 7.24. From this lemma, if we again abuse notation by treating XN and YN as if they are subsets of N (X ⊎ Y ), we have

ηG = (e ◦ ιX , idP ,e ◦ ιY ).

We ﬁrst show that η is indeed a natural transformation. Let G1 = (eX1 ,eY1 , R1)

and G2 = (eX2 ,eY2 , R2) be Galois polarities, and let g = (gX ,gP ,gY ) be a 42 ROB EGROT

polarity morphism from G1 to G2. We aim to show that the diagram in Figure 26 commutes.

Deﬁne hX1 = e1 ◦ ιX1 to be the standard embedding X1 of into XN1 , and sim-

ilarly define hY1 = e1 ◦ ιY1 . Make analogous definitions for hX2 and hY2 . Define + gX : XN1 → XN2 to be the X component of ∆Γg, which is the restriction of + N (ψg) to XN1 . Define gY analogously. Consider the diagram in Figure 27. The inner squares commute by definition of g, and that the outer squares commute can be deduced from the commutativity of the diagram in Figure 28, the commutativity of whose right square follows from the commutativity of the diagram in Figure 22. + + Now, ∆Γg ◦ ηG1 is the polarity morphism (gX ◦ hX1 ,gP ,gY ◦ hY1 ), and ηG2 ◦ g

is the polarity morphism (hX2 ◦ gX ,gP ,hY2 ◦ gY ), and these are equal by the commutativity of the diagram in Figure 27. Thus η is a natural transformation. Let G = (eX ,eY , R) be a Galois polarity extending P . We will show that ηG has the appropriate universal property (see e.g. [21, Theorem 2.3.6]). Let d : Q → D be a ∆1-completion, and let h = (hX ,hP ,hY ): G → ∆(d) be a polarity morphism. We must find a map g : Γ(G) → d such that ∆g ◦ ηG = h, and show that g is unique with this property. By Lemma 7.24, the map η∆(d) is an isomorphism, as ∆(d) is a complete Galois polarity, and is the unique such polarity isomorphism that is the identity on Q. Consider the diagram in Figure 30. The upper triangle commutes as η is a natural transformation (see Figure 29). By Lemma 7.25 there is an isomorphism between Γ∆(d) and d (as extensions of Q), which we define to be τ = (idQ,t), and illustrate in Figure 31. As functors preserve isomorphisms, ∆τ is an isomorphism. Since ∆τ is also the identity on Q, it must be the inverse of η∆(d). Thus, noting Figure 30, τ ◦ Γh : Γ(G) → d has the property that −1 ∆(τ ◦ Γh) ◦ ηG = ∆τ ◦ ∆Γh ◦ ηG = η∆(d) ◦ ∆Γh ◦ ηG = h. We must show that τ ◦ Γh is unique with this property, so let f : Γ(G) → d be a Del morphism, and suppose f is defined by (f1,f2) for some f1 : P → Q and f2 : N (X ⊎ Y ) → D. Suppose also that ∆f ◦ ηG = h. Then ∆f =

(f2|XN ,f1,f2|YN ), and, continuing to identify XN and YN with the corresponding subsets of N (X ⊎ Y ), we have

∆f ◦ ηG = (f2|XN ◦ e ◦ ιX ,f1 ◦ idP ,f2|YN ◦ e ◦ ιY ).

Suppose τ ◦ Γh is deﬁned by the maps t1 : P → Q and t2 : N (X ⊎ Y ) → D. Then

∆(τ ◦ Γh) ◦ ηG = (t2|XN ◦ e ◦ ιX ,t1 ◦ idP ,t2|YN ◦ e ◦ ιY ).

Since we have assumed that both ∆f ◦ ηG and ∆(τ ◦ Γh) ◦ ηG are equal to h, we must have f1 = t1 = hP , and, moreover, f2 must agree with t2 on e[X ⊎ Y ]. But f2|e[X⊎Y ] is cut-stable, as it is the restriction of a complete lattice homomorphism between MacNeille completions, and so it extends uniquely to a complete lattice homomorphism (by [8, Theorem 3.1]). Thus f2 = t2, and so f = τ ◦ Γh as required. It follows that Γ ⊣ ∆ as claimed.

The components of counit of the adjunction between Γ and ∆ are provided by the isomorphisms produced in Lemma 7.25. Thus the subcategory, Fix(Γ∆), of ORDER POLARITIES 43

γ1 e 1 d1 P1 / X1 ⊎ Y1 / N (X1 ⊎ Y1) P1 / D1

hP ψh N (ψh) f g P2 // X2 ⊎ Y2 / N (X2 ⊎ Y2) P D γ2 e2 2 / 2 d2

Figure 24. Figure 25.

ηG 1 hX1 eX1 eY1 hY1 G1 / ∆Γ(G1) XN1 o X1 o P1 / Y1 / YN1

+ + g ∆Γg gX gP gY gX gY G2 / ∆Γ(G2) XN o X o P / Y / YN ηG 2 2 eX 2 eY 2 2 2 hX2 2 2 hY2

Figure 26. Figure 27.

ιX e1 1 ηG N (X1 ⊎ Y1) o X1 ⊎ Y1 o X1 G / ∆Γ(G)

g N (ψg) ψg X h ∆Γh N (X ⊎ Y ) o X ⊎ Y o X 2 2 e 2 2 ιX 2 ∆(d) / ∆Γ∆(d) 2 2 η∆(d)

Figure 28. Figure 29.

ηG G / ∆Γ(G) ❅❖❖❖ ❅ ❖❖ η ◦h ❅❅ ❖❖ ∆(d) ❅ ❖❖ ∆Γh ❅❅ ❖❖❖ ❅ ❖❖' ❅❅ Γ∆(d) = ed◦γd ❅ ∆Γ∆(d) Q / N (X ⊎ Y ) h ❅❅ ❯❯❯ D D ❅ O ❯❯❯❯ ❅❅ ∼ ❯❯❯ ❅ η∆(d)(=) ❯❯❯❯ t(=∼) ❅❅ d ❯❯❯❯ ❯❯❯❯ ∆(d) ❯* D

Figure 30. Figure 31.

Del is just Del itself. The canonical categorical equivalence between Fix(∆Γ) and Fix(Γ∆) produces a categorical version of the correspondence in Theorem 7.26. We end with a universal property for Galois polarities whose relation is the minimal Rl.

Proposition 7.30. Let G = (eX ,eY , Rl) be a Galois polarity extending P , let Q be a poset, and let f : X → Q and g : Y → Q be order preserving maps such that f ◦ eX = g ◦ eY . Let be the unique 3-preorder for G. Then the following are equivalent: (1) y x → g(y) ≤ f(x) for all x ∈ X and for all y ∈ Y . (2) There is a unique order preserving map u : X ⊎ Y → Q such that the diagram in Figure 32 commutes. 44 ROB EGROT

e P Y / Y

eX ιY g (f1,g1) (X, Y ) / Q X ι / X ⊎ Y 1 X ❋❋ ❋❋ ❋❋ u ❋❋ ❋❋ ❋❋ h ❋❋ (f2,g2) ❋❋ f ❋" # , Q Q2

Figure 32. Figure 33.

′ Proof. Suppose (1) holds. We define u : X ∪ Y → Q by f(z) if z ∈ X u′(z)= (g(z) if z ∈ Y We show that u′ is order preserving with respect to the preorder and the order on Q. Let z1 z2 ∈ X ∪ Y . If z1 and z2 are both in X, or both in Y , then ′ ′ that u (z1) ≤ u (z2) follows immediately from the definition of u and the fact that both f and g are order preserving. If z1 = x ∈ X, and z2 = y ∈ Y , then there is −1 ↑ −1 ↓ p ∈ eX (x ) ∩ eY (y ), and so f(x) ≤ g(y) by the assumption that f ◦ eX = g ◦ eY . If z1 = y ∈ Y and z2 = x ∈ X, then that g(y) ≤ f(x) is true by (1), and so ′ ′ u (y) ≤ u (x) as required. Define u by u(ιX (x)) = f(x) and u(ιY (y)) = g(y). Then u is well defined and order preserving by the monotonicity of u′, and that u is unique with these properties is automatic from the required commutativity of the diagram. Conversely, suppose (2) holds. Then

y x =⇒ ιY (y) ≤ ιX (x) =⇒ u ◦ ιY (y) ≤ u ◦ ιX (x) =⇒ g(y) ≤ f(x) as required.

Proposition 7.30 says that, if eX and eY are ﬁxed meet- and join-extensions ′ ′ respectively, the pair of maps (ιX , ιY ) arising from (eX ,eY , Rl) is initial in the category whose objects are pairs of order preserving maps (f : X → Q,g : Y → Q) such that f ◦eX = g◦eY and y x → g(y) ≤ f(x), and whose maps are commuting triangles as in Figure 33 (here h is order preserving, and commutativity means f2 = h ◦ f1 and g2 = h ◦ g1). In particular this category contains all (ιX , ιY ) arising from Galois polarities (eX ,eY , R) based on eX and eY . Acknowledgement

The author would like to thank the anonymous referees for their thorough reports, and for suggestions which considerably improved the clarity of the paper. References

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