# The Galois Connection between Syntax and Semantics

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1 The Galois Connection between Syntax and Semantics Peter Smith University of Cambridge Note: The material in Chs. 2 and 3 now appears, in revised form, as (stand-alone) chapters in my Notes on Category Theory, freely available at

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3 Contents Preface page ii 1 Partially ordered sets Posets introduced Partial orders and strict orders Maps between posets Compounding maps Order similarity Inclusion posets as typical 10 2 Galois connections Galois connections defined An alternative definition The relation between adjoints Fixed points and closures One way a Galois connection can arise 20 3 The Galois connection between syntax and semantics Some reminders Making the connection Drawing the consequences Triviality? More consequences Posets as sets 28 i

4 Preface In his Dialectica paper Adjointness in foundations (1969), F. William Lawvere writes of the familiar Galois connection between sets of axioms and classes of models, for a fixed [signature]. But even if long familiar folklore to category theorists, the idea doesn t in fact seem to be that widely known. The ideas here are pretty enough, elementary enough, and illuminating enough to be worth rehearsing briskly in an accessible stand-alone form. Hence these brisk notes. If you are already happy with the idea of a poset and ideas about mappings from one poset to another, proceed immediately to Chapter 2! ii

5 1 Partially ordered sets This chapter introduces a class of structures the posets which crop up in all sorts of contexts. Of course, given that this type of structure is indeed exemplified all over the place, there won t be a great deal we can say about the general case. But still, our discussion will give us a chance to introduce some key ideas like the idea of an order-isomorphism. 1.1 Posets introduced Definition A partially ordered set P = P, henceforth, a poset is a set P, carrying an ordering relation which is reflexive, anti-symmetric and transitive. That is to say, for all x, y, z P, (i) x x; (ii) if x y and y x, then x = y; (iii) if x y and y z, then x z. Why talk here of a partial order? Because there is no requirement that a given element of the poset P be order-related to any element other than itself. But do note equally that our definition doesn t actually preclude a poset s ordering relation being total: that is to say, it may be the case that for every x, y in the carrier set, either x y or y x. Perhaps we should really call our orders at least partial : but plain partial is absolutely standard. Let s immediately draw out a simple consequence of this definition, one which will help to fix ideas about the kind of ordering we are talking about here: 1

6 2 Partially ordered sets Theorem A poset contains no cycles. By a cycle, we mean a chain of elements p 0, p 1, p 2,..., p α P, all different, such that p 0 p 1 p 2... p α p 0 (to use an obvious notational shorthand). Proof Suppose there is such a cycle. By transitivity, p 0 p 1 and p 1 p 2 implies p 0 p 2. By transitivity again, p 0 p 2 together with p 2 p 3 implies p 0 p 3. Keep on going in the same way until you ve derived p 0 p α (if α is infinite, our informal argument here is infinitary!). But we are already given p α p 0. So by anti-symmetry we can conclude p α = p 0, contradicting the assumption that the elements of the cycle are all distinct. To put this graphically, suppose we draw a diagram of a poset. Put a dot for each distinct element, and a directed arrow from the dot corresponding to p to the dot corresponding to p just when p p. Then, because of transitivity, dots connected by a chain of arrows will also be directly connected. But the diagram will have no loops where we can follow a linked chain of directed arrows around and end up where we started except for the trivial self-loops that correspond to the truth of each p p. Here, then, are just a few examples of posets, starting with some based on familiar number systems before broadening out to give some non-numerical examples. (1) The natural numbers N with read as the natural less than or equals to order relation. (2) N with the converse ordering i.e. we take the order relation to be greater than or equals to. (3) N with the reordering defined as follows in terms of the natural ordering : (a) if m is even and n is odd, then m n; (b) if m and n are both odd, then m n iff m n; (c) if m and both n are both even, then m n iff m n. In other words, we put all the even numbers in their natural order before all the odds in their natural order. (4) N, = is also a poset, since identity is trivially the limiting case of a partial ordering (the smallest partial ordering on any set, where each element is order-related only to itself).

7 1.1 Posets introduced 3 (5) N \ {0} ordered by divisibility: so we put m n iff m divides into n without remainder. (Here N \ {0} is of course the set of natural numbers with 0 removed.) (6) N \ {0, 1} ordered by the same divisibility relation. (7) The real numbers R with the natural less than or equals to ordering. (8) R with the converse ordering. (9) The set R 2 of ordered pairs of reals r, s, taken with the lexicographic ordering, i.e. putting r, s r, s just if either r < r or r = r s s. (10) Any set Σ of functions f: R R, putting f g iff for all x R, f(x) g(x). (11) Choose an interpreted formal language L. Let α be the equivalence class containing all the L-wffs logically equivalent to α, and let E be the set of all such equivalence classes. Let be the relation that holds between the equivalence classes α, β E if α β. Then it is easy to check that E, is a poset. (If L comes equipped with a proof-system, then we could similarly define a poset in terms of a syntactic consequence relation rather than semantic consequence.) (12) A downward-branching binary tree has a top node, and every node has zero, one or two children (think of truth-trees ). Take N to be set of nodes. Let s say that node n is a descendant of n if n is a child of n, or is a child of a child of n, or is a child of a child of a child.... Put n n iff n = n or n is a descendant of n (in other words, n is on the same branch of the tree as n and n is at least as high on the branch as n ). Then N, is a poset. (13) Just for modal logicians: let W be a set of possible worlds in a Kripke frame for intuitionistic logic, and let be the accessibility relation between worlds. Then W, is a poset. And so it goes. Posets are ubiquitous. We ll mention just two more cases for the moment: (14) The powerset P(A) of some set A, with the members of P(A) ordered by set-inclusion, is a poset. (15) In fact, take any set M P(A), i.e. take any arbitrary collection of subsets of a set A. Then M, too is a poset. Call these inclusion posets. A bit later, Theorem will tell us that inclusion posets are in a good sense typical of all posets.

8 4 Partially ordered sets Note immediately that some of our example posets in fact have total orderings, while others have incomparable elements, i.e. elements x, y in the carrier set such that neither x y nor y x. Exercise: which examples are which? Note too the elementary point illustrated by the relationship between examples 5 and 6, and by the relationship between examples 14 and 15. Having worked out that the divisibility relation indeed gives us a partial ordering on the naturals starting from 1, we don t have to do more work to show that the restriction of that relation to the numbers from 2 onwards is also a partial ordering. Likewise, having worked out that is a partial order on the carrier set P(A), we immediately see it remains a partial order when restricted to some M P(A). This observation evidently generalizes to give us a trivial mini-theorem: Theorem If P = P, is a poset then so is P = P, for any P P, where is the relation restricted to P. In such a case, we can say that P is a sub-poset of P. A further, but equally elementary, point is illustrated by the relationship between examples 1 and 2, and between 7 and 8. This too we could dignify as a mini-theorem (whose proof is another trivial exercise): Theorem If P = P, is a poset then so is P op = P, op, where op is the opposite or converse of, i.e. for all x, y P, x op y iff y x. We ll say that P op here is the order-dual of P. Of course, if a poset s order relation is naturally represented by e.g., we ll want to represent the converse relation by : so the order dual of R, is R,. Likewise, if a poset s order relation is naturally represented by, the converse relation which orders its dual will be represented by. We ll use this sort of obvious notation in future. 1.2 Partial orders and strict orders What we ve just defined are sometimes also called weakly ordered sets. By contrast, Definition A strictly ordered set Q = Q, is a set Q, carrying a relation which is irreflexive, asymmetric and transitive. That is to say, for all x, y, z Q,

9 1.2 Partial orders and strict orders 5 (i) x x; (ii) if x y then y x; (iii) if x y and y z, then x z. For example, both N, < and R, <, with < interpreted in the natural way both times, are strictly ordered sets. So is N \{0}, where m n iff m n and m divides n without remainder. That latter example emphasizes that a strict order need not be a total order (so you need to watch the jargon here). Should we care about developing a theory of strictly ordered sets as well as a theory of partially ordered sets? Yes and no. For note we have the following three-part theorem, whose proof can again be left as an exercise. Theorem (i) Suppose P = P, is a partially ordered set, and for x, y P put x y = def x y x y. Then P = P, is a strictly ordered set. (ii) Conversely, suppose Q = Q, is a strictly ordered set, and for x, y Q put x y = def x y x = y. Then Q + = Q, is a partially ordered set. (iii) The result of going from a partially ordered set to its strictly ordered correlate then back again returns us to the original structure: i.e. (P ) + = P. Similarly (Q + ) = Q. When we have a partially ordered set we can therefore immediately define a counterpart strictly ordered set (by ignoring identities); and likewise when we have a strictly ordered set we can define a counterpart partially ordered set (by adding identities). So, yes, if we are interested in partially ordered sets at all, we ll probably care equally about their strictly ordered counterparts. But no, because of the simple definitional link, we don t really need a separate theory to treat the strictly ordered sets. We can develop the theory of partially and strictly ordered sets treating either notion as the more basic and the other as derived. It is conventional and convenient to take the idea of posets as the fundamental one. Henceforth, it will occasionally be notationally convenient to use x y as an abbreviation for x y x y (and likewise, x y as an abbreviation for x y x y, etc.).

10 6 Partially ordered sets 1.3 Maps between posets Next, we ll define three different, increasingly constrained, types of map that can relate two posets. Here s the first pair of notions we want: Definitions Suppose that P = P, and Q = Q, are two posets. And let f: P Q be a map between their carrier sets. Then (i) f is monotone iff, for all p, p P, if p p then f(p) f(p ); (ii) f is an order-embedding iff for all p, p P, p p iff f(p) f(p ). So a monotone mapping f preserves partial orderings. But note, it can so to speak squeeze them up. In other words, if p p we won t get the reversal f(p) f(p ), but we might have f(p) = f(p ). An order-embedding however preserves more: it puts a faithful copy of P s ordering inside Q. Let s make that last point more precise, via a theorem, a definition, and another theorem. First, then, we remark that: Theorem An order-embedding f: P Q between P = P, and Q = Q, is an injective (one-one) map. Proof Suppose f(p) = f(p ). Then, by the reflexivity of, we have both f(p) f(p ) and f(p ) f(p). Since f is an order-embedding, it follows by definition that p p and p p. And since is anti-symmetric, it follows that p = p. So, contraposing, if p p then f(p) f(p ), so f is one-one or injective. That theorem prompts a further key definition: Definition As before, suppose that P = P, and Q = Q, are two posets, and let f: P Q be a map between their carrier sets. Then f is an order-isomorphism iff f is a surjection (is onto) and is an order-embedding. Note that, given our last theorem, an order-isomorphism f between P and Q is one-one and onto. Hence, isomorphism is indeed an apt label. An order-isomorphism f makes one poset a copy of the other, as far as its order-structure is concerned.

11 1.3 Maps between posets 7 And now we can sharpen our claim that, if f is an order-embedding between P and Q, then f maps P to a copy embedded inside Q. Stating the result with a slight abuse of notation, we have: Theorem If f: P Q is an order-embedding between P = P, and Q = Q,, then f is an order-isomorphism between P and the poset f(p ),. f(p ) is, of course, the image of P under f, i.e. {q ( p P )f(p) = q}. And the (very) slight abuse of notation comes in not explicitly marking that the order-relation in f(p ), is the strictly speaking the relation in Q, restricted to f(p ). But here and henceforth we ll allow ourselves that degree of notational slack. The proof of our theorem is trivial: for f is, of course, onto f(p ), and is still an order-embedding. Let s use our definitions, then, in a few examples: (1) Take the pair of posets R +,, N,, where R + is the set of positive reals, and the order relations are the natural ones. Let f: R + N map a real r in R + to the natural number corresponding to r s integral part. Then f is monotone, but not an embedding (since e.g. the reals 10 and 11 get mapped to the same natural number, i.e. 3). (2) That first example reminds us that a monotone map typically doesn t preserve all order information. For an extreme example to hammer home the point, take R +, again and the one-element poset {0}, = (check that that is a poset!). Then the function f that maps every element of R + to 0 is monotone, but it s a forgetful map that throws away all non-trivial order information. (3) Take the posets R +,, N again, but this time consider a map g: N R + going in the opposite direction which takes a natural number to the corresponding integral real. Then g is an order-embedding but not a full isomorphism: g just finds a copy of the naturals embedded inside R +. (4) Consider the poset R,, and its order-dual R,. Then the map h: R R defined by h: x x is an order-isomorphism. (5) But don t be misled by that last example there isn t always an order-isomorphism between a poset and its order-dual. Consider, for example, N, and its order-dual N,. An orderisomorphism would have to map 0 which comes first in the order to some element which similarly comes first in the order.

12 8 Partially ordered sets But there is no such element in N. So there can be no orderisomorphism between N, and N,. 1.4 Compounding maps Another key result is that monotone maps can be composed together to get another monotone map: Theorem If f: P Q is a monotone map from P = P, to Q = Q, and g: Q R is a monotone map from Q to R = R,, then g following f, i.e. the composite map g f: P R, is a monotone map from P to R. Proof Need we spell this out? If p p then f(p) f(p ), since f is monotone. And if f(p) f(p ) then g(f(p)) g(f(p )), since g is monotone. So if p p then g(f(p)) g(f(p )), hence g f is monotone. It is also worth remarking on the following easy result: Theorem Composition of monotone maps is associative: in other words, if f: P Q, g: Q R, h: R S are monotone, then (f g) h and f (g h) are monotone and equal. Which leads to a brief aside. For it is, of course, trivial that the identity map Id P : P P, which maps every p P to itself, is a monotone map from P = P, to itself. Hence if you happen to know what a category is you ll see from our last two theorems that it s immediate that, if we take posets as objects and monotone mappings between them as the associated arrows/morphisms, then we get a category Pos of posets. However, although we ll sometimes be in the close neighbourhood of some category-theoretic ideas, e.g. in the next chapter, we ll not explicitly discuss categories in this book. End of aside! To continue: Theorem tells us that monotone maps compose. But so do order-embeddings and (crucially) isomorphisms: Theorem If f: P Q is an order-embedding (isomorphism) from P to Q, and g: Q R is an order-embedding (isomorphism) from Q to R, then g f: P R, is an order-embedding (isomorphism) from P to R.

13 1.5 Order similarity 9 Proof To show order-embeddings compose, just change the if s to if and only if s in the proof of Theorem And to show order-isomophisms compose, we just need to remark that if f and g are onto, so is g f. 1.5 Order similarity To repeat, an order-isomorphism between posets is indeed a one-to-one correspondence between their carrier sets, one that preserves all the basic facts about the way that the elements are ordered. So let s say that Definition The posets P and Q are order-similar (in symbols, P = Q), iff there is an order-isomorphism from P to Q. And we can readily check that order-similarity is properly so-called: Theorem Order-similarity is an equivalence relation. Proof It is trivial that order-similarity is reflexive. Theorem tells us that order-isomorphisms compose, and hence that order-similarity is transitive. So we just need to check for symmetry. In other words, we just need to confirm that if f: P Q between P and Q, then there is an order-isomorphism between Q and P. To show that, (i) define f 1 : Q P to be such that f 1 (q) = p iff f(p) = q, and then (ii) check this is indeed an order-isomorphism between Q and P. Let s have just a very small handful of examples: (1) We have already seen that N, and N, are of different order-types. Likewise there can be no order-isomorphism between N, and N,, where is the evens-before-odds ordering. For in N,, 1 has an infinite number of predecessors, and there is no element in N, with the same order property. Hence N,, N,, N, and indeed N, = are posets with the same carrier set but of different order-types. However, N,, N \ {0}, and E, are all order-similar (where N \ {0} is the set of naturals with 0 deleted, and E is the set of even numbers). (2) If Q + is the set of positive rational numbers, and their natural ordering, then Q +, is plainly not order-similar to N,. For, in their natural ordering, the rationals are dense between any two different ones there is another and not so for the natural

14 10 Partially ordered sets numbers. However we can also put a different order on the positive rationals. Map each q Q + \{0} one-to-one to the pair of naturals m, n where q = m/n and the fraction is in lowest terms. Do a standard kind of zig-zag enumeration of such pairs m, n where m, n have no common divisors other than 1. In other words, say m, n precedes m, n if m+n m +n or else if m+n = m +n and m m. And now put q q in Q + when either q = 0 or the counterpart pair for q is no later than the counterpart pair for q in the zig-zag enumeration. Then Q +, = N,. (3) The examples just mentioned involve posets whose ordering relation is total (i.e., for every x, y in the carrier set, either x y or y x). Plainly, no poset which isn t totally ordered can be order-similar to one that is. For an example of a pair of posets which are not totally ordered but which are still order-similar, consider the following trivial example. (i) The inclusion poset P({0, 1, 2}),. (ii) The poset whose carrier set is {1, 2, 3, 5, 6, 10, 15, 30}, with the numbers ordered by divisibility. It can again be left as an exercise to specify an order-isomorphism from (i) to (ii) there is more than one! 1.6 Inclusion posets as typical Posets, as we said before, are ubiquitous. Evidently they can come in all manner of sizes and shapes. And they can be built up from all manner of elements, related in a huge variety of ways. It might seem quite hopeless to try to put any organization at all on the mess here. But we can in fact tidy things up just a bit. For remember, any collection of subsets of some set A, ordered by, is a poset. We dubbed this sort of poset an inclusion poset. And we can now show: Theorem Every poset is order-isomorphic to an inclusion poset. NB, an inclusion poset can be based on any collection of subsets of a set A. The case where the poset is based on the set of all subsets of some A is a very special one. The theorem does not say that every poset is similar to one of the very special cases! Proof Take the poset P = P,. For each p P, now form the set π p = {x x P x p}. Let Π be the set of all π p for p P. Then Π is a set of subsets of P, and P π = Π, is of course an inclusion poset.

15 1.6 Inclusion posets as typical 11 Further, P is order-isomorphic to P π. For consider the function f: P Π which maps p P to π p Π. This map is evidently a oneone correspondence between P and Π. Moreover, by definition p p iff π p π p. So we are done. Hence, if what we really care about is just the order-type of various posets i.e. we don t care about what sort of elements the carrier-set has, and are just highlighting the kind of ordering that is being put on a set of elements then, without loss of generality, we can concentrate on inclusion posets as typical.

17 2.1 Galois connections defined 13 There are plenty of serious mathematical examples (e.g. from number theory, abstract algebra and topology) of two posets with a Galois connection between them. But we really don t want to get bogged down in unnecessary mathematics at this stage; so for the moment let s just give some relatively simple cases, including a couple of logical ones. (1) Suppose f is an order-isomorphism between P and Q, so the inverse function f 1 is one too. Then f, f 1 is a Galois connection. For, trivially, f(p) q iff f 1 (f(p)) f 1 (q) iff p f 1 (q). (2) Let P = N, and Q be Q +,, where in each case the order relation is standard. Put f : N Q + to be the standard embedding of the natural numbers into the rationals; and let f : Q + N map a positive rational q to the natural corresponding to its integral part. Then (G) trivially holds, and f, f is a Galois connection between the integers and the rationals in their natural orderings. (3) Take an arbitrary poset P = P,, and put I = {0}, = (so I is a singleton poset with the only possible order relation). Let f be the trivial function i that maps each and every member of P to 0. And let f be some function that maps 0 to a particular member of P. Then f is right adjoint to f, i.e. the pair f, f form a Galois connection between P and I, just if, for every p P, f (p) = 0 iff p f (0), i.e. just if for every p, p f (0), i.e. just if f (0) is the maximum of P in the obvious sense. Dually, if we put f to be the same flattening function i, then f is left adjoint to it just if f (0) is the minimum of P. (4) Our next example is from elementary logic. Recall example 11 from Section 1.1. Let L be a formal language equipped with a proof-system (it could be classical or intuitionistic, or indeed any other logic with well-behaved conjunction and conditional connectives). Then α is the class of L-wffs interderivable with α; E seem to have borrowed the term from the old theory of Hermitian operators, where in e.g. a Hilbert space with inner product,, the operators A and A are said to be adjoint when we have, generally, Ax, y = x, A y. The formal analogy is evident. The first discussion of such a connection and hence the name is to be found in Evariste Galois s work in what has come to be known as Galois theory, a topic far beyond our purview here. Though you might like to find out about Galois s short life from the historical sketch in Stewart s Galois Theory (1989).

18 14 Galois connections is the set of all such equivalence classes; and α β iff α β. E = E, is a poset. Now consider the following two functions between E and itself. Fix γ to be some L-wff. Then let f be the function that maps the equivalence class α to the class (γ α), and let f be the function that maps the equivalence class α to (γ α). Now, we have (γ α) β if and only if α (γ β). So (γ α) β if and only if α (γ β). So f, f is a Galois connection between E and itself. ( Conjunction is left adjoint to conditionalization. ) (5) Our last example for the moment is again from elementary logic. Let L now be a first-order language, and Form( x) be the set of L-wffs with at most the variables x free. We ll write ϕ( x) for a formula in Form( x), ϕ( x) for the class of formulae interderivable with ϕ( x), and E x for the set of such equivalence classes of formulae from Form( x). With as in the last example, E x, is a poset for any x. Now we ll consider two maps between the posets E x, and E x,y,. In other words, we are going to be moving between (classes of) formulae with at most x free, and formulae with at most x, y free (where y isn t among the x). First, since every wff with at most the variables x free also has at most the variables x, y free, there is a trivial map f that sends ϕ( x) E x to the same element ϕ( x) E x,y. Second, we define the companion map f that sends ϕ( x, y) E x,y to yϕ( x, y) E x. Then f, f is a Galois connection between E x and E x,y. That is to say f ( ϕ( x) ) ψ( x, y) iff ϕ( x) f ( ψ( x, y) ). For that just reflects the familiar logical rule that ϕ( x) ψ( x, y) iff ϕ( x) yψ( x, y) so long as y is not free in ϕ( x). Hence universal quantification is right-adjoint to a certain trivial operation. And we can similarly show that existential quantification is left-adjoint to the same operation (exercise: spell out that last claim). Our first family of cases, then, shows that Galois connections exist

19 2.2 An alternative definition 15 and are at least as plentiful as order-isomorphisms. The second and third cases show that posets that aren t order-isomorphic can in fact still be connected. The third and fifth cases gives a couple of illustrations of how a more contentful notion (taking maxima, universal quantification) can be regarded as adjoint to some relatively trivial operation. The fourth case shows that even when the Galois-connected posets are isomorphic (in this case trivially so, because they are identical!), there can be functions f and f which aren t isomorphisms but which also go to make up a connection between the posets and since γ was arbitrary, this case also shows that there can be many different connections between the same posets. And the fourth and fifth cases already show that Galois connections are of interest to logicians. Finally in this section, let s show that connections are well-behaved, in the sense that connections compose to give new connections. In other words, we have: Theorem Let P = P,, Q = Q,, and R = R, be posets. Let suppose f, f is a Galois connections between P and Q, and g, g is a Galois connections between Q and R. Then g f, f g is a Galois connection between P and R. Proof By assumption, we have (G 1 ) f (p) q iff p f (q) and (G 2 ) g (q) r iff q g (r). So (G 2 ) gives us g (f (p)) r iff f (p) g (r). And (G 1 ) gives us f (p) g (r) iff p f (g (r)). Whence, as we want, g (f (p)) r iff p f (g (r)). 2.2 An alternative definition So far, so good. We ll meet a rather more telling example of a logically significant Galois connection in the next chapter. But first let s explore the general case in a bit more detail. Here s a simple but illuminating result: Theorem Suppose P = P, and Q = Q, are posets, and f : P Q and f : Q P are a pair of functions between their carrier sets. Then f, f is a Galois connection if and only if (i) f, f are both monotone, and (ii) for all p P, q Q, p f (f (p)) and f (f (q)) q.

20 16 Galois connections Proof (Only if ) Suppose f, f is a Galois connection. As a particular case of (G) from Defn we have f (p) f (p) iff p f (f (p)). Since is reflexive, the l.h.s. of that biconditional holds. So p f (f (p)). Similarly for the other half of (ii). Now, suppose also that p p. Since we ve shown p f (f (p )), we have p f (f (p )). But by (G) we have f (p) f (p ) iff p f (f (p )). Whence, f (p) f (p ) and f is therefore monotone. Similarly for the other half of (i). (If ) Now assume (i) and (ii) hold, and suppose f (p) q. Since by (i) f is monotone, f (f (p)) f (q). But by (ii) p f (f (p)). So, since is transitive, p f (q). Which establishes that if f (p) q then p f (q). The proof of the other half of the biconditional (G) is dual. This theorem can evidently be used to justify an alternative definition of a Galois connection as, simply, a pair of maps for which conditions (i) and (ii) hold. 2.3 The relation between adjoints Our next theorem tells us that the relation between the adjoint members of a Galois connection is rigid in the sense that if f, f is to be a connection, then f fixes what f uniquely has to be, and conversely f fixes what f has to be. Theorem If f, f 1 and f, f 2 are Galois connections between P, and Q,, then f 1 = f 2. Likewise, if f 1, f and f 2, f are Galois connections between the same posets, then f 1 = f 2. Because (i) holds, connections defined our way are sometimes called monotone Galois connections. But now recall that if Q = Q, is a poset, so is its dual Q op = Q, op. Suppose then that f, f is a (monotone) connection between P = P, and Q = Q, so that f (p) q iff p f (q). Then, trivially, there is an antitone Galois connection between P and Q op, meaning a pair of functions f, f such that q op f (p) iff p f (q). The only point of footnoting this here is to remark that, (a) you might well encounter the idea of a connection defined the antitone way (which indeed it how it appears in Galois s work); indeed (b) some authors only call the antitone case a Galois connection, reserving the term adjunction for connections defined our way. But (c) since a monotone connection between P and Q is an antitone connection between P and Q s dual, we just don t need to fuss about the difference, and without loss of generality can we can concentrate entirely on connections presented the monotone way.

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