1 The Hyland-Schalke functor G Rel. 2 Weakenings

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1 1 The Hyland-Schalke functor G Rel Let G denote an appropriate category of games and strategies, and let Rel denote the category of sets and relations. It is known (Hyland, Schalke, 1996) that the following define a functor F : G Rel: F() = P ( ) F B = {(s, s B ) : s is a P -play} It can be shown that this functor F is faithful (see Calderon and McCusker, 2010 for a proof). This surprising fact will be very useful for us: if we want to show that a particular diagram commutes in G, we can normally do so more easily by applying the functor F to the diagram and showing that the resulting diagram commutes in Rel. Notation 1.1. We shall simplify our notation by writing P rather than F() and writing rather than F(). 2 Weakenings Let = (M, λ, P ) be a game. Then we have two notions of a weakening of - informally, a game that is easier than, from the player P s point of view. 1. n antecedent weakening of is a game = (M, λ, P ), where P P such that for every O-play s occuring in both P and P, the set of P -responses to s in is precisely the set of P -responses to s in and for every P -play t occurring in both P and P, the set of O-responses to t in is a subset of the set of O-responses to t in. 2. precedent strengthening of is a game = (M, λ, P ) where P P such that for every O-play s occuring in both P and P, the set of player responses to s in is a subset of the set of P -responses to s in, and such that for every P -play t occuring in both P and P the set of O-responses to t in is precisely the set of O-responses to t in. 1

2 In other words, a precedent strengthening of is a game in which the player has more choices at each move, but the opponent has the same choices, while an antecedent weakening of is a game in which the player has the same choices at each move, but the opponent has fewer choices. We can express the definition of an antecedent weakening more compactly as follows: (M, λ, P ) is an antecedent weakening of (M, λ, P ) if P P and the following condition holds: whenever s P is an O-play and a M is a P -move such that sa P, then sa P. This condition says that player P has all the moves available to her in that she had in P - the other conditions mentioned above follow automatically from the fact that P P. The important fact about both types of weakening is that we have copycat morphisms,. From here on, we will use the term weakening to refer to an antecedent weakening. If is a weakening of, we will write : for the copycat morphism. Proposition 2.1. Let be a weakening of. Then : has a right inverse. Proof. The inverse for is the partial strategy str:. This is also constructed as a copycat strategy; the only difference is that if O makes a move in that does not occur in P then the player P has no reply. We now claim that str = id. Formally: = {s P : if s is even, then s = s } str = { s P : if s is even, then s = s } We wish to show that str = id. In the style of section 1, we shall show that str = P in Rel. It is easy to see that = {(s, s) : s P } P P str = {(s, s) : s P } P P It is then easy to see that str = P P, as we wanted. s a sanity check, if we try to take the composition the other way round then we get P P, which is not the identity unless P = P. 2

3 3 The weakening monad of a game 3.1 Weakening pullback of a game Let be a game. Define Suppose B is a game and : B is a morphism. P B = {s B : s } P B is prefix closed, since is, so B = (M B, λ B, P B ) is a well-formed game. We claim that B is a weakening of B. Indeed, suppose s P B is an O-play and sa P B. By the definition of P B, there must be some s such that s B = s; choosing s minimal, we may assume that the last move in s coincides with the last move of s. Since s is an O-play, this last move is an O-move in B and therefore a P -move in B. Therefore, sa since is a strategy and so sa = sa B P B. We call B the weakening pullback of along the morphism. Next, we define a morphism : B. s sets, = (noting that, by definition, P B ). We claim that commutes. Once again, we check this in Rel. If we note that = PB P B P B and that = as sets, then it is clear that = in Rel, and so = in G. The weakening B and the morphism satisfy a universal property - namely, if B is a weakening of B and : B is a morphism such that commutes, then B is a weakening of B and B B (1) (2) 3

4 commutes. Indeed, working in Rel again, diagram (1) tells us that = as sets, which implies that P B P. Noting the definition of P B, it follows that B is the weakening pullback of along, and so diagram (2) commutes. 3.2 The weakening monad We shall want to prove various useful properties of these weakening pullbacks, notably a functoriality condition: if C B are morphisms, then we get a morphism : C B such that ( ) =. This and other useful facts are brought together in the following theorem: Theorem 3.1. Let be a game. Then we have a functor weak : G/ G/ that sends a morphism B to the pullback weakening B and that gives rise to an idempotent monad on G/. Proof. In order to complete the definition of the functor weak, we need to define its action on morphisms. Let C B be morphisms in G (so gives us a morphism from C to B in G/). We define a morphism : C B by = P C B. It is easy to see that is a strategy, so we need to check that = ( ). s usual, we perform this check in Rel. Since ( ) is created according to the pullback recipe given above, we have ( ) = as sets. So, as subsets of P C P, we have: ( ) = = = {(s, t) P C P : u P B. (s, u), (u, t) } Meanwhile, as sets we have = and = (P C P B ). So, as subsets of P C P, we have: = {(s, t) P C P : u P B. (s, u), (u, t) } - in other words, the same thing, but with P B instead of P B. It is clear then that ( ). In the other direction, suppose (s, t) P C P and that there exiss u P B such that (s, u) and (u, t). But now, 4

5 since (u, t), we must have u P B by the definition of P B and therefore (u, t). So far what we have done is show that the morphism does indeed give us a morphism in G/ from C ( ) to B. We still need to prove that what we have defined respects composition and so gives us a functor G/ G/. This means proving the following statement: suppose D υ C B are morphisms in G. This gives us morphisms υ, in G/ as pictured in the diagram below: υ D C B υ We have υ : D C given by υ = υ P D C and : C B given by = P C B, as before. We also have a morphism ( υ) : D B given by ( υ) = ( υ) P D B. We need to show that ( υ) = υ. We have ( υ) = ( υ) P D B, so: ( υ) = ( υ) (P D P B ) = ( υ) (P D P B ) = {(s, t) P D P B : u P C. (s, u) υ, (u, t) } Meanwhile, = P D and υ = υ P B, so υ = {(s, t) P D P B : u P C. (s, u) υ (u, t) } which is the same thing, but with P C rather than P C. Clearly, υ ( υ). In the other direction, suppose that s P D, t P B and that u P C such that (s, u) υ, (u, t). It will suffice to show that u P C. Now note that C is pullback of along and that B is pullback of along. Therefore: P C = {s C : s } P B = {t B : t } Since t P B, there must be some t such that t B = t. Put another way, there must be some v P such that (t, v). We have (u, t), so therefore (u, v), and it follows that u P C. 5

6 Therefore, we have that υ = υ = ( υ). Since the Hyland-Schalke functor is faithful, this means that υ = ( υ). This all means that we can define a functor weak : G/ G/ by C B B B C B We now define the monad structure on G/. The unit transformation η : id G/ weak is given by the weakening morphisms : B B for each object B ; we have already shown that the diagrams commute. To show that this is indeed a natural transformation, we need to show that if we are given some morphism in G/ C B then the following square commutes: To show this, we work in Rel again. On the one hand, we have: = {(s, t) P C P B : u P C. (s, u), (u, t) } = {(s, t) P C P B : (s, t) } = = (P C P B ) 6

7 On the other hand: = {(s, t) P C P B = (P C P B ) : u P B. (s, u), (u, t) } We certainly have. In the other direction, suppose that s P C and t P B such that (s, t). It will suffice to show that s P C. Indeed, by the definition of P B, there is some v P such that (t, v). Therefore, (s, v), and so s P C, as required. Before defining the multiplication µ : weak weak weak in the monad, we look at the identities that this unit transformation η will have to satisfy. If B is an object of G/ then we want the diagrams B B µ (B) and B B B µ (B) B to commute. The arrow : B B in the first diagram is the weakening arrow obtained from the object B of G/ using the natural transformation η, while the arrow : B B in the second diagram is obtained by applying the functor weak to the morphism : B B. In practice, however, there is no need for this distinction: B is the same game as B and the arrows and are the identity. To see this, recall the definitions of B and : P B = {s B : s } = (as sets of plays) Then we should have P B = {s B : s }, but since is the same set as, this is exactly the same as the definition of P B. Since, are identity maps, we observe that we have the strict equality weak weak = weak and we had better define µ to be the identity transformation. This µ is automatically associative and the unit diagrams above hold (since all maps involved are identity maps). We end up with an idempotent monad on G/. 7

8 4 Lax weakening morphisms We now introduce a lax version of our notion of weakening morphism that is stable under composition with isomorphisms: Definition 4.1. Let, B be games. We say that a morphism : B is a lax weakening if it is the composition of a weakening morphism and an isomorphism; in other words, if there exists a weakening B of B and an isomorphism ˆ : B such that the diagram commutes. Proposition 4.2. i) If B is a weakening of B, then : B B is a lax weakening. ii) If : B is an isomorphism, then is a lax weakening. iii) If C B are morphisms such that and are lax weakenings, then : C is a lax weakening. Proof. (i) and (ii) are both obvious. For part (iii), note that we have the diagram ˆ ˆ It will suffice to find a weakening C of C and an isomorphism ˆ : C B making the following diagram commute: C ˆ ˆ ˆ ˆ 8

9 Therefore, it will suffice to prove the lemma which follows. Lemma 4.3 (Lifting of weakenings along isomorphisms). Let B, C be games and let : C B be an isomorphism. Let B be a weakening of B. Then there is a weakening C of C and an isomorphism : C B such that the following diagram commutes: We shall give two proofs of this lemma - one direct proof using the Hyland- Schalke functor and one proof using the idempotent monad structure described in the previous section. First proof. We know that : C B is an isomorphism; therefore, its image P C P B under the Hyland-Schalke functor is an isomorphism in Rel. It is well-known (and easy to show) that the isomorphisms in Rel are precisely those relations that are the graphs of bijective functions. Therefore, we have a bijection : P C P B with the property that i) For all s P C there exists a P -play s such that s C = s and s B = (s) ii) For all P -plays s, s B = (s C ). Let P C = 1 (P B ) P C. We claim that P C is prefix-closed. Let s P C and let t s. By the definition of P C, (s) P B. Since P B is prefix-closed, it will suffice to show that (t) (s). By (i), we know that there exists a P -play s such that s C = s and s B = (s). Since t s, there is some prefix t s such that t C = t. Then, by (ii), we must have (t) = t B s B. Therefore, C = (M C, λ C, P C ) is a well-formed game. We claim that it is a weakening of C. Indeed, suppose s P C is an O-play and that sc is a P -reply in P C. We need to show that sc P C. By the definition of P C, we know that (s) P B. By (i) we know that there exists some P -play s such that s C = s and s B = (s); since s is a P -play, we can deduce that s and (s) have the same sign, and so (s) is an O-play in P B. Examining (i) and (ii) above, we can see that 9

10 (sc) = (s)b for some b P B. Therefore, since B is a weakening of B, we have (sc) P B and so sc P C as desired. Lastly, we define to be given by the set P C B, and we claim that the diagram commutes. This is most easily checked by applying the faithful functor G F Rel Rel op where : Rel Rel op is the identity on objects and flips a relation R B to give a relation R B. If : is a weakening morphism in G then its image under this functor is the graph of the inclusion P P. Meanwhile, the image of under this functor is the graph of the function 1 : P B P C. By the definition of P C, the image of under this functor is the graph of the function 1 PB : P B P C. It is then easy to check that the resulting diagram commutes in Rel: P C P C 1 1 PB P B P B The last thing that we need to do is to check that is an isomorphism. If we apply exactly the same argument to C C and the isomorphism 1 : B C then we get a weakening B of B and an isomorphism ( 1 ) : B C making the following diagram commute: 1 ( 1 ) Now note that P B is defined to be (P C ). Since is a bijection and P C 10

11 was defined to be 1 (P B ), we get that B = B. We now get: B B ( 1 ) C B = B 1 C C B = B 1 C B B = B B Since is an epimorphism, it follows that ( 1 ) = id B. n identical argument shows that ( 1 ) = id C. So is an isomorphism. Second proof. We start off with the diagram C We use the monad on the category G/B. Let C be the pullback of B along the morphism : C B. This gives us the diagram ( ) It remains to show that ( ) is an isomorphism. To show this, let us rearrange the original diagram a bit to make the monadic structure explicit: C B When we apply the functor weak B B to this diagram, we get C B ( ) By definition, = η B : B B, and so = (η B ). Since our monad is idempotent, (η B ) is an isomorphism. is an isomorphism by functoriality and therefore ( ) is an isomorphism, as desired. B 11