Γ-Ultrametric spaces Definition Let (Γ,,0) be a complete lattice with minimal element 0 A Γ-ultrametric space is a pair (M,d M ) such that M is a set and d M : M M Γ. (Reflexivity) ( x,y M)d M (x,y) = 0 x = y (Symmetry) ( x,y M)d M (x,y) = d M (y,x) (Strong Triangle Inequality) ( x,y,z M)d M (x,y) d M (y,z) d M (x,z)
Γ-Ultrametric spaces Definition Let (Γ,,0) be a complete lattice with minimal element 0 A Γ-ultrametric space is a pair (M,d M ) such that M is a set and d M : M M Γ. (Reflexivity) ( x,y M)d M (x,y) = 0 x = y (Symmetry) ( x,y M)d M (x,y) = d M (y,x) (Strong Triangle Inequality) ( x,y,z M)d M (x,y) d M (y,z) d M (x,z) A non-expanding map between Γ-ultrametric spaces (M,d M ) and (N,d N ) is a function f : M N such that ( a,b M)d(f (a),f (b)) d(a,b) This gives us a category Γ-UltMet whose objects are Γ-ultrametric spaces and whose morphisms are the non-expanding maps.
Closed Balls Definition For each x M,γ Γ we define the (closed) ball of radius γ around x to be B M (x,γ) = {y : d M (x,y) γ}. We will omit mention of M when it is clear from context. Definition If A M the diameter of A is diam(a) = {d M (x,y) : x,y M}.
Closed Balls The following lemmas are immediate. Lemma ( γ Γ,x,y M)x B(y,γ) B(x,γ) = B(y,γ) Lemma ( γ x,γ y Γ)( x,y M)γ x γ y B(x,γ x ) B(y,γ y ) or B(x,γ x ) B(y,γ y ) =.
Intersection of Balls Theorem Suppose α = {γ i : i I } and {x i : i I } M B = i I B(x i,γ i ) Then ( x B)B = B(x,α). Suppose x B. Then ( i I)x B(x i,γ i ) and so B(x,γ i ) = B(x i,γ i ). Hence ( i I)B B(x,γ i ) and diam(b) γ i. Therefore diam(b) α and in particular B B(x,α). But we also have ( i I)B(x,α) B(x,γ i ) and so B(x,α) B.
Relative Distance Preserving Maps While much of our focus will be on Γ-ultrametric spaces for a specific Γ, it will at times be useful to be able to compare generalized ultrametric spaces with different sets of distances. Definition If (M,d M ) is a Γ-ultrametric space and (N,d n ) is an Λ-ultrametric space we say a map f : M N is relative distance preserving if ( a,b,c,d M)d M (a,b) d M (c,d) d N (f (a),f (b)) d N (f (c),f (d)) We let GenUltMet be the category whose objects are Γ-ultrametric spaces for some Γ and whose morphisms are the relative distance preserving maps.
Spherically Complete An important property that a Γ-ultrametric spaces may have is what is called spherical completeness. This is the Γ-ultrametric analog of completeness for metric spaces. Definition A Γ-ultrametric space (M,d M ) is spherically complete if whenever {γ i : i < κ} Γ and {x i : i < κ} M B(γ i,x i ) B(γ j,x j ) if i j. Then i<κ B(γ i,x i ). We define SComp(Γ) to be the full subcategory of Γ-UltMet whose objects are spherically complete.
Intersection in Spherically Complete Spaces Theorem Suppose (M,d M ) is a spherically complete Γ-ultrametric space and {γ i : i < κ} Γ and {x i : i < κ} M B(γ i,x i ) B(γ j,x j ) if i j. is a decreasing chain of closed balls. Then for all x i<κ B(γ i,x i )) B( γ i,x) = B(γ i,x i ) i<κ i<κ
Pruned Presheaves Definition We say a separated presheaf A on Ω is pruned if ( γ Ω op )( a A(γ))( a A(1))a γ = a. We define Pruned(Ω) to be the full subcategory of Sep(Ω) whose objects are the pruned presheaves. Likewise we define PrunedSheaf(Ω) to be the full subcategory of Sheaf(Ω) whose objects are pruned presheaves.
Equivalence of Pruned Presheaves Theorem There is an equivalence of categories between Pruned(Ω) and Ω op -UltMet.
Equivalence of Pruned Presheaves Theorem There is an equivalence of categories between Pruned(Ω) and Ω op -UltMet. In order to be consistent the names of all elements will be as in Ω. We will use a superscript op to signify the corresponding notion in Ω op. For example a b if and only if a op b and 1 = 0 op.
Equivalence of Pruned Presheaves:Claim 1 Claim If (A,d A ) is an Ω op ultrametric space let A (γ) = {B(a,γ) : a A} where B(a,γ) γ = B(a,γ ) for all γ γ. Then A is a separated pruned presheaf on Ω First we need to confirm that restriction is well defined. Suppose B(a,γ) = B(b,γ) A (γ), γ γ and x B(a,γ ) (i.e. d A (x,a) op γ ). d A (a,b) op γ op γ, so d A (x,b) op γ and x B(b,γ ). Hence B(a,γ ) = B(b,γ ) and restriction is well defined.
Equivalence of Pruned Presheaves:Claim 1 To see that A is pruned notice that if B(a,γ) A (γ) then B(a,0 op ) A (1) and B(a,0 op ) γ = B(a,γ). To see A is separated, suppose γ = i I λ i and B = {B(x i,λ i ) A (λ i ) : i I } is such that B(x i,λ i ) λi λ j = B(x i,λ i λ j ) = B(x j,λ i λ j ) = B(x j,λ j ) λi λ j.
Equivalence of Pruned Presheaves:Claim 1 Now if i I B(x i,λ i ) = then there is no element in A (γ) compatible with B as such an element would have be of the form B(x,γ) with B(x i,λ i ) = B(x,λ i ) for all i I and hence x i I B(x i,λ i ) =. On the other hand, if there exists x i I B(x i,λ i ) then by we have B(x,γ) = i I B(x i,λ i ) and hence B(x,γ) is the unique element of A (γ) compatible with B.
Equivalence of Pruned Presheaves:Claim 2 Claim If (A,d A ),(C,d C ) obj(ω op -UltMet) and f : (A,d A ) (C,d C ) is a non-expanding map, let f (B A (a,γ)) = B C (f (a),γ). Then f (Pruned(Ω))[A,C ] First we need to show that f (B(a,γ)) doesn t depend on our choice of a. Suppose B A (a,γ) = B A (b,γ) and x B C (f (a),γ). Then d C (f (a),f (b)) op d A (a,b) γ and d C (x,f (a)) op γ.
Equivalence of Pruned Presheaves:Claim 2 So d C (x,f (b)) op γ and x B C (f (b),γ). Hence f (B A (a,γ)) = f (B A (b,γ)). Next we need to show f is a natural transformation, i.e. that if λ γ then f λ (BA (a,γ) λ ) = f γ (B A (a,γ)) λ. But f λ (BA (a,γ) λ ) = f λ (BA (a,λ)) = B C (f (a),λ) = B C (f (a),γ) λ = f γ (BA (a,γ)) λ. So f is a natural transformation from A to C.
Equivalence of Pruned Presheaves: Claim 3 Let F be the functor where F(A,d A ) = A if (A,d A ) obj(ω op -UltMet) and F(f ) = f if f morph(ω op -UltMet). Claim If A is pruned separated presheaf on Ω let (A o,d A ) be such that A o = A(1) and d A (a,b) = {γ : a γ = b γ. Then (A o,d A ) is an Ω op -ultrametric space.
Equivalence of Pruned Presheaves: Claim 3 First notice a da (a,b) = b da (a,b) because {a γ : γ d A (a,b)} is a compatible collection of elements covering both a da (a,b) and b da (a,b). In particular this means that if d A (a,b) = 0 op then a = a 0 = b 0 = b. So (A o,d A ) satisfies (reflexivity). (symmetry) is immediate from the definition. To see the (strong triangle inequality) holds let a,b,c A(1) with d A (a,b) op γ and d A (b,c) op γ. Then a γ = b γ = c γ and hence d A (a,c) op γ.
Equivalence of Pruned Presheaves: Claim 4 Claim If A,C obj(pruned(ω)) and f Pruned(Ω))[A,C] then f 1 : A(1) C(1) is a non-expanding map f 1 : (A o,d A ) (C o,d C ). We know that for all a,b A(1), a da (a,b) = a da (a,b) and so f 1 (a) da (a,b) = f da (a,b)(a da (a,b)) = f da (a,b)(b da (a,b)) = f 1 (b) da (a,b) and hence d A (a,b) op d C (f 1 (a),f 1 (b)).
Equivalence of Pruned Presheaves: Claim 5 Let E be the functor where E(A) = (A o,d A ) if A obj(pruned(ω)) and E(f ) = f 1 if f morph(pruned(ω)) Claim For all A obj(pruned(ω)) there is an isomorphism η A : A F(E(A)) which is the identity on A(1). First notice that for all a,b A(1), a γ = b γ if and only if d A (a,b) op γ if and only if B Ao (a,γ) = B Ao (b,γ). So the maps (η A ) γ (a γ ) = B Ao (a,γ) is well defined and an injective natural transformation. Further, because A is pruned, (η A ) γ is also surjective and hence an isomorphism for all γ. So η A is a natural isomorphism.
Equivalence of Pruned Presheaves: Claim 6 Claim For any map f Pruned(Ω)[A,C] we have η C f = (f 1 ) η A. As all four maps are maps of pruned presheaves they are determined by their values on A(1). Hence η : 1 Pruned(Ω) F E is a natural isomorphism.
Equivalence of Pruned Presheaves: Claim 7 Claim For all (A,d A ) obj(ω op -UltMet) there is a natural isomorphism ε A : (A,d A ) E(F(A,d A )). If (A,d A ) = E(F(A,d A )) then A = {{a} : a A} and for all a,b A, d A ({a}, {b}) = op {γ : B A (a,γ) = B A (b,γ)} = d A (a,b). Hence the map ε A (a) = {a} is an isomorphism of Ω op -ultrametric spaces.
Equivalence of Pruned Presheaves: Claim 8 Claim For any map f Ω op -UltMet[(A,d A ),(C,d C )] we have ε C f = (f ) 1 ε A. Immediate. Hence ε : 1 Ω op -UltMet F E is a natural isomorphism. And in particular E,F are equivalences of categories.
Spherically Completeness and Pruned Sheaves Theorem (*) The equivalence of the theorem above restricts to an equivalence of categories between PrunedSheaf(Ω) and SComp(Ω op ). Claim If (A,d A ) is a spherically complete Ω op ultrametric space then F(A) = A is a sheaf.
Spherically Completeness and Pruned Sheaves Let I = γ j : j < κ Ω op with (a i,i) : i I be a compatible collection of elements of A (with I a sieve). So there are x i such that a i = B(x i,γ i ). We define B i by induction. B 0 = B(x 0,γ 0 ). B α+1 = B(x γα+1,γ α+1 ) B α B ω β = j<ω β B j. We now want to show ( l < κ,λ κ)b(x γl,γ l ) B λ. To get a contradiction assume j is least such that B(x γl,γ l ) B j = for some l < κ. We break into three cases.
Spherically Completeness and Pruned Sheaves Case 1: j = 0 This case can t happen because (a i,i) : i I is a compatible collection of elemetns and hence the a i s are closed under intersection. Case 2: j = ω β. Notice that B r B s if s r < ω β. So B(x γl,γ l ) B ω β = B(x γl,γ l ) h<ω β B h = h<ω β (B(x γ l,γ l ) B h ).
Spherically Completeness and Pruned Sheaves But B(x γl,γ l ) B h by the inductive hypothesis and B(x γl,γ l ) B h : h < ω β is a decreasing sequence of balls. Hence h<ω β B(x γ l,γ l ) B h because (A,d A ) is spherically complete. So this case can t happen.
Spherically Completeness and Pruned Sheaves Case 3: j = α + 1. B(x γl,γ l ) B α+1 = B(x γl,γ l ) (B(x γα,γ α ) B α+1 ) = B(x γl γ α,γ l γ α ) B α. So this case can t happen and we have our contradiction. We therefore have B κ = i<κ B i and so B κ = B(x, I) for any x B κ. Hence B κ is the unique element of A ( I) which is covered by (a i,i) : i I. So, because (a i,i) : i I was arbitrary, A is a sheaf.
Spherically Completeness and Pruned Sheaves Claim If A is a pruned sheaf on Ω then (A o,d A ) is a spherically complete Ω op -ultrametric space. Suppose {γ i : i < κ} Ω op and {x i : i < κ} A B Ao (γ i,x i ) B Ao (γ j,x j ) if i j.
Spherically Completeness and Pruned Sheaves Whenever j i we then have x i B Ao (γ j,x j ) and γ i γ j and so x j γi = x i γi. Hence ( i,j < κ)x j γi γ j = x i γi γ j and (x i ζ : i < κ,ζ γ i is a compatible collection of elements. But because A is a sheaf there is an y A( i<κ γ i) covered by (x i ζ : i < κ,ζ γ i.
Spherically Completeness and Pruned Sheaves Further, because A is a pruned sheaf we know there is an x A(1) = A o such that x i<κ γ i = y. So for all i,j < κ, x γi = x i γi hence x B Ao (γ i,x i ) and x i<κ BAo (γ i,x i ). So (A o,d A ) is spherically complete (as our sequence of balls was arbitrary).
Pruned Presheaves has a Right Adjoint Now that we have a way to describe the category of Γ-ultrametric spaces it is natural to ask what properties this category has.
Pruned Presheaves has a Right Adjoint Now that we have a way to describe the category of Γ-ultrametric spaces it is natural to ask what properties this category has. Theorem The inclusion map i : Pruned(Γ) Sep(Γ) has a right adjoint r. If A obj(sep(γ)), let r(a)(γ) = {x A : ( y A(1))y γ = x}. If f Sep(Γ)[A,B] then we know that the image of f restricted to r(a) is determined by f 1. As such if we let r(f ) X (x) = f X (x) for all x r(a)(x) then r(f ) X (x) r(b) X for all x r(a)(x). Hence we have r is a functor.
Let η X = id X : X r i(x) and let ε X : i r(x) X be the inclusion map. Then it is clear that η : 1 Sep(Γ) r i and ε : i r 1 Sep(Γ) are the unit and counit (respectively) of an adjunction.
Pruned Presheaves is Complete Corollary Pruned(Γ) is complete. Because r i = 1 Pruned(Γ), i r and Sep(Γ) is complete.
Pruned Presheaves is Complete Corollary Pruned(Γ) is complete. Because r i = 1 Pruned(Γ), i r and Sep(Γ) is complete. Notice though that r does not restrict to an adjunction for the inclusion map i : PrunedSheaf(Γ) Sheaf(Γ) because even if A is a sheaf, there is no guarantee that r(a) will be as well. In general, PrunedSheaf(Γ) is not a complete category because it does not necessarily have all equalizers.
Pruned Exponentials Theorem (*) If A is a pruned separated presheaf on Ω and B is any separated presheaf then A B is a pruned separated presheaf. We have for each γ, A B (γ) = A(γ) B(γ) (because we are only dealing with localic topoi up to this point). Suppose f A B (γ) and for each x A(γ), g(x) A(1) and g(x) γ = x (we know such a g exists as A is pruned and we can choose once because we have assumed the Axiom of Choice).
Pruned Exponentials Let x i : i < κ = A(γ). Define A i as follows. A 0 = {x 0 γ : γ γ}, A α = {x α γ : γ γ} i<α A i. If y A i let g(y) = g(x i ). So g takes an element of a A(η) (η γ) and returns a g(a) A(1) such that g(a) η = a. So in particular g(a ζ ) = g(a) ζ.
Pruned Exponentials Next let f A B (1) be define so that if b B(λ) then [f λ ](b) = g([f λ γ ](b λ γ )) λ. We then have that [f λ ](b) η = g([f λ γ ](b λ γ )) η = g([f η γ ](b η γ )) η = [f η ](b η ). Hence f yields a map of presheaves. So [f ] A B (1) and f γ = f and hence A B is pruned.
Pruned Presheaves is Cocomplete Theorem Pruned(Γ) is cocomplete. It suffices to show that the colimit of pruned presheaves (in the category of presheaves) is a pruned presheaf. But this follows immediately from the fact that in the category of presheaves colimits are taken pointwise.
Generating Set and Hom Sets We will now show that in many categories the Hom sets (i.e. sets of the form C[A,B] where A,B obj(c)) can be viewed as generalized ultrametric spaces. Definition Let C be a category with G obj(c) and let P(G,A) = Powerset( X G C[X,A]). For every A,B obj(c) define d G : C[A,B] C[A,B] P(G,A) as follows: d G (f,g) = {h : X A such that f h = g h}. For every k : B D we define a map k! : (C[A,B],d G ) (C[A,D],d G ) by k! (g) = k g
Maps, Generating Sets and Hom Sets Theorem If G is a generating set of objects for C then (C[A,B],d G ) is a P(G,A)-ultrametric space (where α β if and only if β α and 0 = X G C[X,A]). Further if f : B D then each map f! is non-expanding. First lets show that (C[A,B],d G ) is a P(G,A)-ultrametric space. (Symmetry) is immediate. For (Reflexivity) notice that because G is a generating set of objects if f,g C[A,B] and f g then there is some a : X A with X G such that f a g a. In particular if f g then d G (f,g) 0.
Maps, Generating Sets and Hom Sets To show the strong triangle inequality suppose f, g, h C[A, B]. Whenever a : D A where a d G (f,g) d G (g,h) we have f a = g a = h a and hence a d G (f,h). Therefore d G (f,g) d G (g,h) d G (f,h) or equivalently d G (f,g) d G (g,h) d G (f,h). To see that any k! is a non-expanding map notice that if a d G (f,g) then f a = g a and hence k f a = k g a and a d(k f,k g) = d(k! (f ),k! (g)). So d(k! (f ),k! (g)) d(f,g) or equivalently d(k! (f ),k! (g)) d(f,g).
Maps, Generating Sets and Hom Sets Corollary If C is a category and G is a generating set then for each A obj(c) there are functor F G,A : C P(G,A)-UltMet given by ( B obj(c))f G,A (B) = (C[A,B],d G ) ( k C[B,D])F G,A (k) = k! This follows from the fact that k! j! = (k j)!.
Maps, Generating Sets and Hom Sets Lemma F G,A preserves products that exist in C. Notice that the product Π i I (C[A,B i ],d Gi ) = (Π i I C[A,B] C[A,D],d G ) where d G ((a i : i I),(b i : i I)) = i I d G i (a i,b i ). The result then follows from the fact that C[A,Π i I B i ] = Π i I C[A,B i ] and for any two maps f,g : A Π i I B i and any map x : X A, x f = x g if and only if ( i I)π i x f = π i x g (where π i is the projection onto B i ).
Subobjects and Γ-Ultrametric Spaces If our categories have epi-mono factorizations then, up to isomorphism in GenUltMet, our choice of generating set doesn t matter in determining the generalized ultrametric space structure put on Hom sets of the category. Definition Suppose C is a category with epi-mono factorizations. Let P(Sub C ) = Powerset(Sub C (A)). For every A,B obj(c) define d S : C[A,B] C[A,B] P(S) as d S (f,g) = {[h] : f h = g h}. For every k : B D we define a map k! : (C[A,B],d S ) (C[A,D],d S ) by k! (g) = k g
Equivalence of Structure on Hom Sets Theorem If G obj(c) then for each A,B obj(c), and f,g,x,y C[A,B] we have (a) d S (x,y) d S (f,g) d G (x,y) d G (f,g). (b) If G is a generating set for C then d G (x,y) d G (f,g) d S (x,y) d S (f,g).
Equivalence of Structure on Hom Sets Part (a): Suppose a : D A with D G. a then has an epi-mono factorization a = a m a e. Now a d G (x,y) if and only if x (a m a e ) = y (a m a e ) if and only if x a m = y a m if and only if [a m ] d S (x,y) (the second to last equivalence follows because a e is an epimorphism). So if [a m ] d S (x,y) [a m ] d S (f,g) then a d G (x,y) a d G (f,g). Hence d S (x,y) d S (f,g) implies d G (x,y) d G (f,g).
Equivalence of Structure on Hom Sets Part (b): To get a contradiction suppose G is a generating set for C d G (x,y) d G (f,g) a : D A with [a] d S (f,g) [a] d S (x,y) Then x a y a and hence there must be a map c : C B such that x a c y a c and c G. So a c d G (x,y) and hence by assumption a c d G (f,g) and f a c g a c. But f a = g a (by definition of [a] d S (f,g)) and so f a c = g a c.
Equivalence of Structure on Hom Sets Corollary If G is a generating set for C and A,B obj(c) then (C[A,B],d S ) is isomorphic to (C[A,B],d G ) in GenUltMet. Corollary If C is a category then for each A obj(c) there is a functor F Sub,A : C P(Sub,A)-UltMet given by ( B obj(c))f Sub,A (B) = (C[A,B],d S ) ( k C[B,D])F Sub,A (k) = k! Further F Sub,A preserves all products which exist in C. This follows from the fact that k! j! = (k j)!.
Equivalence of Structure on Hom Sets Theorem If F Sub,1, E and r are as above then there exists η : E r F Sub,1 a natural isomorphism. This follows immediately from the definition of E r and F along with the fact that P(1) = Ω as a lattice.
Translation of Results Now that we have characterized the category of Γ-ultrametric spaces in terms of presheaves on Γ op and shown how the Hom sets of Sep(Γ op ) can be viewed as Γ-ultrametric spaces, we want to use these facts to translate results about Γ-ultrametric spaces into results about pruned presheaves.
Translation of Results Definition We say that an object A is D-pruned if ( e : E D)( f : E A)( h : D A) such that h e = f (where e is a monic). This is a generalization of the notion of a pruned presheaf. Specifically a presheaf is pruned if and only if it is 1-pruned.
Translation of Results Definition Suppose C has all colimits and A,B obj(c). We say B is A-complete if For all complete lattices (Γ, ) and all sequences e i : E i A : i Γ with e i,j : E i E j such that e j e i,j = e i whenever i j For all sequences functions f i C[A,B] : i Γ with f i e j = f j e j whenever j i. There is a map f : A B with f e i = f i e i for all i Γ.
Spherically Complete and A-Complete Theorem (*) Suppose C has all colimits and A,B obj(c). Then (C[A,B],d S ) is spherically complete if and only if B is A-complete. : Let Γ, e i : E i A : i Γ, f i C[A,B] : i Γ be as in the definition. Further let α s : s < κ Γ be an increasing sequence of elements of Γ such that α s : s < κ = Γ. Let B s = {g C[A,B] : g e αs = f αs }. Then B s is a closed ball and B s B t if s t. So, by spherical completeness, there exists f s<κ B s
Spherically Complete and A-Complete : For each i < κ let B i = B(f i,e i ) where E i Sub C (A) and if B(f i,e i ) = B(f i,e ) then E E i. Further suppose B i B j if i j and so E j E i if j i. Now B i = B(f i,colimit(e i )) and if e i Colimit(E i ) then f i : i < κ and e i : i < κ satisfy the conditions of the theorem. Hence there must be a f κ such that B(f κ,colimit(e i )) = B i and so f κ i<κ B i. Therefore C[A,B] is spherically complete.
Contracting Maps Theorem Suppose F,F,G,G : B D are maps in C and suppose x,y C[A,B]. Then the statement d S (F(x),G(y)) d S (F (x ),G (y )) holds if and only if ( z : X A)G y z = F x z G y z = F x z This follows immediately from the definitions.
Contracting Maps Theorem Suppose F,F,G,G : B D are maps in C and suppose x,y C[A,B]. Then the statement d S (F(x),G(y)) d S (F (x ),G (y )) holds if and only if ( z : X A)G y z = F x z G y z = F x z This follows immediately from the definitions. Definition We say a map a : A B in C is contracting for D if ( f,g C[D,A])d S (a(f ),a(g)) < d S (f,g)
Contracting Maps Theorem A map a : A B is contracting for D if and only if ( f,g : D A)( h : E D) such that f h g h but a f h = a f h. We call h a witness that a is contracting for f and g. This follows immediately from the fact that d S (a(f ),a(g)) < d S (f,g) if and only if d G (a(f ),a(g)) < d G (f,g)
Contracting Maps Theorem If a : A B is contracting for D then ( f,g : D A)( h : E D) such that f h g h but a f h = a f h with E G. Suppose f,g : D A and h : E D are such that f h g h but a f h = a g h. Then there is a map i : E E such that E X and f h i g h i. Hence h i : E D is the desired witness.
Contracting Maps Theorem If a : A B is contracting for all D G then a is contracting for all E obj(c). Suppose f,g : E A are such that f g. Then there is a Y X and h : Y E such that f h g h. But by assumption a is contracting for Y and so there is a k : Z Y such that f h k g h k but a f h k = a f h k. Hence h k witnesses that a is contracting for E (with respect to f,g).
Contracting Maps Theorem If a : A B is a contracting map for D and b : B C is any map, then b a : A C is a contracting map for D. Suppose f,g : D A are such that f g. Then there is a k : Z D such that f k g k but a f k = a g k. But then we also have b a f k = b a g k. Hence b a is contracting for D.
Contracting Maps Theorem If a : A B is a contracting map for D and c : C A is any map, then a c is a contracting map for D. Suppose f,g : D C. We then have two cases. In the first case c f = c g and hence a c f = a c g. But then id D witnesses a c is contracting for f,g.
Contracting Maps In the second case we have c f c g. However both of these are maps from D to A and so there must be a k : E D such that (c f ) k (c g) k and a (c f ) k = a (c g) k. Hence k witnesses that c a contracts f,g.
Contracting Maps In the second case we have c f c g. However both of these are maps from D to A and so there must be a k : E D such that (c f ) k (c g) k and a (c f ) k = a (c g) k. Hence k witnesses that c a contracts f,g. Notice though that if f : A B is a contracting map of pruned presheaves in Sep(Ω) that doesn t mean that f is a contracting map in the category Pruned(Ω). In general in Pruned(Ω) there are no contracting maps (other than those from the 1).
Fixed Point Theorem For this section fix a Grothendieck Topos G. Given any map a : A A define Fix(a) to be subobject of A corresponding to the equalizer of a and 1 A. Theorem (*) Suppose a : A A is a contracting map for D and A is D-pruned. Then there is a unique map f : D A such that a f = a. We call f the D-fixed of a. First notice that because 0 : 0 A and [0] Sub G (D) we know there is a map h : D A. We will prove the existence of such a map f by repeated applying the contracting map a. Define f α : D A as follows
Fixed Point Theorem f 0 = h, and Fix 0 = f 1 0 [im(f 0 ) Fix(a)] (which is a subobject of D) f α+1 = a f α and Fix α+1 = f 1 [im(f α+1 ) Fix(a)] (which is a subobject of D) Let Fix ω γ = i<ω γ Fix i and let fω γ : i<ω γ f 1 i [im(f i ) Fix(a)] A be a map agreeing with f i on Fix i (for all i < ω γ). Then define f ω γ to be any map from D into A which factors through fω γ (which we know exists as A is D-pruned).
Fixed Point Theorem Notice that Fix α Fix α+1 by construction and if Fix α id D then by the fact that a is contracting for D, we have Fix α Fix α+1. So for some α, f α : D A is a fixed point. Now suppose there are two f,g : D A such that a f = f and a g = g. Then d S (a(f ),a(g)) = d S (a f,a g) = d S (f,g). Hence, because a is contracting for D, we must have f = g. So there is a unique fixed point of a.