{1X } if X = Y otherwise.

Transcription

1 C T E G O R Y T H E O R Y Dr E. L. Cheng e.cheng@dpmms.cam.ac.uk elgc2 Michaelmas Categories, unctors and natural transormations 1.1 Categories DEFINITION category C consists o: a collection o objects, ob C; LECTURE 2 14/10/02 For every pair X, Y ob C, a collection C(X, Y) = HomC (X, Y) o morphisms : X Y, equipped with: or each X ob C, an identity map idx = 1 X C(X, X); or each X, Y, Z ob C, a composition map satisying: m XYZ : C(Y, Z) C(X, Y) C(X, Z) (g, ) g = g, unit laws i : X Y then 1 Y = = 1 X associativity i X Y g Z h W, then h(g) = (hg). category is said to be small i ob C and all o the C(X, Y) are sets, and locally small i each C(X, Y) is a set. REMRKS 1 I C(X, Y), we say that X and Y are the domain (or source) and the codomain (or target) o. 2 Morphisms are also reerred to as maps or arrows. 3 We can write Hom C or the collection o all morphisms. 4 It is convenient and customary to assume that the C(X, Y) are disjoint or distinct pairs (X, Y). 5 We don t worry ourselves with the niceties o set theory. DEFINITION category C is called discrete i the only morphisms are identities; i.e. { {1X } i X = Y C(X, Y) = otherwise. EXMPLES Large categories o mathematical structures: a Set o sets and unctions. b Categories derived rom or related to Set: 1

2 Pn o sets and partial unctions; Rel o sets and relations; Set o pointed sets and base point preserving unctions. c lgebraic structures and structure-preserving maps: Grp o groups and group homomorphisms; b o abelian groups and group homomorphisms; Ring o rings and ring homomorphisms; Vec o vector spaces over ; Mat o natural numbers and n m matrices. d Topological categories: Top o topological spaces and continuous maps; Haus o Hausdor spaces and continuous maps; Met o metric spaces and uniormly continuous maps; Htpy o topological spaces and homotopy classes o maps. 2 Mathematical structures as categories: a Posets: a poset (P, ) can be regarded as a category C with objects the elements o P and precisely one morphism x y when x y and none otherwise. b Monoids: a category with just one object is a monoid. c Groups: a group G can be regarded as a category with just one (ormal) object and whose morphisms are the elements o G. 3 Small categories can be presented by generators and relations. From a directed graph we can generate a category o paths through the graph and then add relations imposing equalities between some paths with the same domain and codomain. a There is a category 0 with no objects and no morphisms, generated by the empty graph. b There is a category 1 with one objects and one (identity) morphism, generated by the graph with just one vertex. c There is a category generated by the graph with one vertex and one edge. It is isomorphic to the additive monoid. d There is a category generated by the graph with one vertex and one edge s say, together with the relation s 2 = 1. It has one object and two morphisms and is isomorphic to the cyclic group o order 2. e There is a category generated by the graph with two vertices and one edge between them. It has two objects and three morphisms and is isomorphic to the poset 2 = {0 1}. 1.2 Universal properties DEFINITION morphism C(X, Y) is an isomorphism i g C(Y, X) such that g = 1 X and g = 1 Y. We say g is an inverse or. PROPOSITION I g 1 and g 2 are inverses or, then g 1 = g 2. g 1 = g 1 1 Y = g 1 ( g 2 ) = (g 1 ) g 2 = 1 X g 2 = g 2. 2

3 PROPOSITION The identity map is an isomorphism. 2 The composition o two isomorphisms is an isomorphism. 1 1 X is clearly sel-inverse. 2 Let C(Y, Z), g C(X, Y) be isomorphisms, with respective inverses h C(Z, Y), k C(Y, X). Then we claim that g C(X, Z) is an isomorphism, with inverse kh C(Z, X). For so we have the desired result. DEFINITION (g)(kh) = (gk)h = (1 Y )h = h = 1 Z (kh)(g) = k(h)g = k(1 Y )g = kg = 1 X terminal object in C is an element T ob C such that X C,! morphism X k T. EXMPLE In Set, every 1-element set is terminal. So sometimes we denote a terminal object by 1. PROPOSITION Suppose 1 and 1 are terminal in C. Then there exists a unique isomorphism C(1, 1 ). Since 1 is terminal, there is a unique morphism : 1 1. Similarly, 1 is terminal, so there is a unique morphism : 1 1. Now consider C(1, 1). Since 1 is terminal, there is a unique morphism 1 1, i.e. the identity. So = id 1 ; similarly = id 1. Hence is the desired unique isomorphism. DEFINITION Given, B ob C, a product o and B is an object B equipped with projections p B q B, such that or all : C, g: C B,! morphism (, g): C B such that p (, g) = and q (, g) = g; i.e. such that C (,g) B p q g B commutes. EXMPLE In Set, B = { (a, b) a, b B } with p, q the irst and second projections. 3

4 Note however, that we could also have taken p, q to be the second and irst projections, or the set to be { (b, a) b B, a }. PROPOSITION I p D q B and D p q are products o, B C, then! isomorphism k: D D such that q k = q and p k = p. Consider the diagrams p D D k q p k q D D p q p q B B. By our deinition o product, k is the unique morphism D D s.t. these diagrams commute; so q k = q and p k = p certainly. We claim that k is an inverse or k. For consider k k : D D. We have Hence p (k k ) = (p k) k = p k = p q (k k ) = (q k) k = q k = q B D p k k q D p q B commutes. But by the deinition o product, there is a unique morphism D D that makes this diagram commute, i.e. the identity. So k k = id D. Similarly k k = id D. So k is indeed an isomorphism, and is the unique one s.t. q k = q and p k = p. DEFINITION I, B C, there exists a product B, we say C has all binary products. PROPOSITION I C is a category with binary products, then given C(, C), g C(B, D), there exists a unique morphism g C( B, C D) such that 4

5 C B p 0 q 0 g C D p 1 q 1 B D g commutes. Immediate rom deinition o product. LECTURE 3 16/10/02 DEFINITION Suppose C is a category with binary products. Given B, C ob C, a unction space or exponential is an object C B equipped with an evaluation morphism ɛ: C B B C such that : B C,! : C B such that B C 1 B C B B ɛ commutes, i.e. ɛ ( 1 B ) =. In Set, C B = { : B C } = [B, C]. There is an evaluation map ɛ: C B B C (g, b) g(b). Given : B C, ix a to get a : B C b (a, b). So we have a unction : C B a a, such that (a, b) = a (b) = ɛ( a, b) = ɛ ( 1 B )(a, b). So ɛ ( 1 B ) = as required. 5

6 1.3 Categorical constructions DEFINITION subcategory D o C consists o subcollections ob D ob C; HomD Hom C, together with composition and identities inherited rom C. We say D is a ull subcategory o C i X, Y D, D(X, Y) = C(X, Y), and a llu subcategory o C i ob C = ob D. We can think o the data or a category as Hom C c 1 c 2 ob C We could have c 1 giving us the domain o a morphism and c 2 the codomain, or vice verse. This motivates the deinition: DEFINITION Given a category C, the dual or opposite category C op is deined by: ob C = ob C op ; C(X, Y) = C op (Y, X); identities inherited; op g op = (g ) op. THE PRINCIPLE OF DULITY Given any property, eature or theorem in terms o diagrams o morphisms, we can immediately obtain its dual by reversing all the arrows (this is oten indicated by the preix co- ). EXMPLES The dual notion o a terminal category object is an initial object. That is, an object I C such that or all Y C, there exists a unique : I Y. For example, the (unique) initial object in Set is ; we sometimes write 0 or an initial object. 2 The dual o a product is a coproduct: p B where p, q are coprojections such that, or any C(, C), g C(B, C),! h: B C such that C commutes. DEFINITION h B p morphism m B is monic i given any, g: C, we have m = mg = g. Dually, a morphism e B is epic i given any, g: B C, we have e = ge = g. 6 q q B g B

7 It is easy to see that any isomorphism is epic and monic. In Set, a morphism is monic i it is injective, and epic i it is surjective. DEFINITION Given C a category and X ob C, then the slice over X, C/X is the category with: objects (Y, ), where : Y X C; morphisms h: (Y1, 1 ) (Y 2, 2 ) such that Y 1 h Y 2 1 X 2 commutes, i.e. 2 h = 1. Dually, we have the slice under X, X/C, with: objects (Y, ), where : X Y C; morphisms h: (Y1, 1 ) (Y 2, 2 ) such that X 1 2 Y 1 h Y 2 commutes, i.e. h 1 = 2. We have a terminal object (X, 1 X ) in C/X and dually an initial object (X, 1 X ) in X/C. 1.4 Functors DEFINITION Let C and D be categories. unctor F: C D associates with each X ob C, an object FX ob D; with each C(X, Y), a morphism F D(FX, FY), such that F1X = 1 FX ; F(g) = Fg F. LECTURE 4 18/10/02 DEFINITION We deine the category Cat o small categories:- For any category C there is an identity unctor 1 C : C C X X 7

8 Composition o unctors C F D G E with GF deined in the obvious way. Similarly we have CT, the category o large categories and unctors. EXMPLES Cat has an initial object 0. 2 Cat has a terminal object 1. 3 Cat has products; given C, D ob Cat, we have the product C D with objects (c, d), c C, d D; morphisms (, g), : c c C, g: d d D. DEFINITION unctor F: C D is aithul/ull/ull and aithul i C(X, Y) D(FX, FY) is injective/ surjective/an isomorphism. EXMPLES Functors between collections o mathematical objects: a orgetul unctors: b ree unctors: c inclusion o subcategories: 2 Functors between mathematical structures: Gp Set Ring Set Ring b Haus Top; Set Gp Set Mnd; b Gp Haus Top. a posets : (P, ) (Q, ) is an order-preserving map; b groups : G H is a group homomorphism. 3 Presheaves a unctor C op Set is called a preshea on C. 4 Diagrams a unctor C Set is called a diagram on C. Note that a unctor will preserve any property that is expressible as a commutative diagram. For example, isomorphisms are preserved by all unctors; i is an isomorphism, then F is also. PROPOSITION I F is ull and aithul, then F isomorphic isomorphic. Let C(X, Y) such that F is an isomorphism. Then inverse g D(FY, FX) or F. Since F is ull, then g C(Y, X) such that g = Fg. But now F(g) = (F)(Fg) = 1 FY. 8

9 nd F(1 Y ) = 1 FY, so since F is aithul, we have g = 1 Y. Similarly g = 1 X. So g is an inverse or C(X, Y), i.e. is an isomorphism. 1.5 Contravariant unctors DEFINITION contravariant unctor C D is a unctor C op D. That is: on objects, X FX; on morphisms, X Y FY F FX; identities are preserved; F(g ) = F Fg. non-contravariant unctor is sometimes reerred to as a covariant unctor. 1.6 The Hom unctor REPRESENTBLES Let C be a locally small category. We have a contravariant unctor H U or C(, U): H U : C op Set X C(X, U) X Y C(X, U) C(1,g) C(Y, U) g g Dually, we have a covariant unctor H U or C(U, ): These are known as representables. H U : C Set X C(U, X) X Y C(U, X) C(,1) C(U, Y) g g THE Hom FUNCTOR gain, take C locally small. Then we have a unctor H: C op C Set (X, Y) C(X, Y) (X, Y) (,g) (X, Y ) C(X, Y) C(,g) C(X, Y ) h gh where : X X C op and g: Y Y C. 9

10 1.7 Natural transormations DEFINITION Let F, G: C D be unctors. natural transormation α: F G is a collection o morphisms (known as components) such that, : X Y C, { α X : FX GX X C }, FX α X GX F G FY α Y GY commutes (the naturality condition). DEFINITION Given categories C and D, we deine the (larger) category [C, D] where: objects are unctors F: C D; morphisms are natural transormations α: F G, such that: LECTURE 5 21/10/02 identities are natural transormations 1F : F F (or any F: C D with components FX 1 FX FX; α β or composition, given F G H, then β α is the natural transormation with components (β α) X : FX β X α X HX. F C α G β H D So, or example, [C, D](F, G) is a collection o natural transormations F G. DEFINITION natural isomorphism α: F G is an isomorphism in the unctor category; i.e. there exists β: G F such that α β = 1 G and β α = 1 F. Note that two natural transormations are equal i all their components are. PROPOSITION α: F G is a natural isomorphism i each component α X : FX GX is an isomorphism in D. Suppose α is a natural isomorphism, and let β be its inverse. Then α β = 1 G (α β) X = 1 GX α X β X = 1 GX 10

11 and β α = 1 F (β α) X = 1 FX β X α X = 1 FX. So β X is an inverse or α X or each X C. Thus each component is an isomorphism in D. Conversely, i each component α X is an isomorphism, then let β X be the corresponding inverses or each X C. Now, given C(X, Y), we have that F FX α X G GX FY α Y GY commutes; i.e. (G) α X = α Y (F). But now: hence β Y (G) α X β X = β Y α Y (F) β X so β Y (G) 1 GX = 1 FY (F) β X so β Y (G) = (F) β X ; GX β X FX G GY β Y F FY commutes; so we can legitimately deine the natural transormation β with components β X. nd clearly β is an inverse or α, so α is a natural isomorphism. We can prove similar results that tell us that α is epic/monic i all its components are. 1.8 The 2-category Cat DEFINITION We deine horizontal composition o natural transormations. We have seen vertical composition already: F F C G α β D = C β α D. But we can also compose: H H F H HF C α D β E = C β α E. We deine (β α) X : HFX KGX by G K KG HFX Hα X HGX β GX KGX 11

12 or HFX β FX KFX Kα X KGX. By the naturality o β, these deinitions are equivalent: HFX β FX KFX Hα X Kα X HGX βgx KGX so we can deine (β α) X = β GX Hα X = Kα X β FX. We consider the ollowing particular case: F H C α D 1 E 1 H H α: HF HG G H which we will (or convenience) write as: F C α D H E Hα: HF HG. Similarly we have: G H C F D β E βf: HF KF. PROPOSITION (THE MIDDLE-4 INTERCHNGE LW) Given K F J C G α (1) α (2) D K β (1) β (2) E, H L we have (β (2) β (1) ) (α (2) α (1) ) = (β (2) α (2) ) (β (1) α (1) ). 12

13 Consider components. We have and [(β (2) β (1) ) (α (2) α (1) )] X = (β (2) β (1) ) HX J(α (2) α (1) ) X = β (2) HX β (1) HX Jα (2) X Jα (1) X [(β (2) α (2) ) (β (1) α (1) )] X = β (2) HX Kα (2) X β (1) GX Jα (1) X. So it is suicient to prove that Kα (2) X β (1) GX = β (1) HX Jα (2) X. But we have that JGX β(1) GX KGX Jα (2) X JHX β (1) HX Kα (2) X KHX commutes (by the naturality o β (1) ), and so we are done. DEFINITION We can now deine the 2-category Cat, consisting o: objects, morphisms and two-cells; composition o morphisms; horizontal and vertical composition o 2-cells; axioms - unit, associativity and middle-4 interchange; any two ways o composing are the same. DEFINITION Given categories C and D, an equivalence consists o: unctors C F D, D G C; natural isomorphisms GF α 1 C, FG β 1 D. We call β the inverse up to isomorphism or the pseudo-inverse o α. DEFINITION unctor F: C D is essentially surjective on objects i Y D, X C such that FX = Y D. PROPOSITION F is an equivalence o categories i it is essentially surjective and ull and aithul. Omitted. 13

14 2 Representability 2.1 The Yoneda Embedding Recall that or each C, we have the unctor H : C op Set. So we have an assignation H. We can extend this to a unctor, known as the Yoneda embedding: H : C [C op, Set] H (: B) (H : H H B ), where H is the natural transormation with components (H ) X : H X H B X i.e. C(X, ) C(X, B) h h. We need to check that this is a well-deined natural transormation, i.e. that C(Y, ) (H ) Y = C(Y, B) H g= g H B g= g C(X, ) (H ) X = C(X, B) commutes. But along the two legs we just have: h h h and ( h) g h g (h g) so the naturality condition just says that composition is associative. LECTURE 6 23/10/ Representable Functors DEFINITION unctor F: C op Set is representable i it is naturally isomorphic to H or some C, and a representation or F is an object C together with a natural isomorphism α: H F. Dually, a unctor F: C Set is representable i F = H or some C, and a representation or F is an object with a natural isomorphism α: H F. 14

15 NOTE The naturality square says, that : V W C, C(W, ) α W FW H = F commutes. EXMPLES C(V, ) α V FV 1 The orgetul unctor U: Gp Set is representable. Take =, and α to be the natural transormation with components: α G : H G UG (1). Then we can check that α is natural, and it is an isomorphism, since any homomorphism : G is completely determined by (1). 2 ob: Cat Set is representable. For let be 1, the terminal category; then ob(c) = Cat(1, C) is a natural isomorphism. Now, we can make a ew suggestive observations about natural transormations α: H F. Consider the naturality square C(, ) α F F C(V, ) α V FV We know this commutes; in particular, or the element 1 C(, ), we have α V (1 ) = F(α (1 )), so that α is in act completely determined by α (1 ) F. So, we would like to deine a natural transormation α: H F by setting α(1 ) = x F, and α V () = (F)(x). I this is indeed a natural transormation, then we will have set up a bijection between F and the natural transormations H F. Hence we get The Yoneda Lemma THEOREM (YONED LEMM) Let C be a locally small category, F: C op Set. Then there is an isomorphism F = [C op, Set](H, F), 15

16 which is natural in and F; i.e. FB [C op, Set](H B, F) F [C op, Set](H, F) F H and θ θ F [C op, Set](H, F) G [C op, Set](H, G) commute, or all : B and or all θ: F G respectively. 1 Given x F, we deine x [C op, Set](H, F) by components: X V : C(V, ) FV F(x) We must check the naturality o x; given g: W V, we need C(V, ) x V FV g Fg to commute. On elements, we have C(W, ) x W FW F(x) and Fg ( F(x) ) g F( g)(x) But Fg(F(x)) = F( g)(x) by the (contravariant) unctoriality o F, so the square commutes as required. 2 Given α [C op, Set](H, F), we deine α F by α = α (1 ). 3 We check ( ) = ( ). Given x F, x = x (1 ) = F(1 )(x) = 1 F (x) = x. Given α [C op, Set](H, F), α is given by components α: C(V, ) FV F( α) = F(α (1 )). So we need only check that α V () = F(α (1 )). We have the ollowing naturality square 16

17 or α: C(, ) α F F C(V, ) α V FV so on the element 1 C(, ), we have α V (1 ) = F(α (1 )), as required. 4 We check naturality in, i.e. that given any B, F [C op, Set](H, F) F H FB [C op, Set](H B, F) commutes. On elements, we have: x x x and x H F(x) F(x). Now, the ormer has components C(V, B) (H ) V C(V, ) FV g g F( g)(x), x V and the latter C(V, B) g F(x) V FV Fg F(x). But (Fg F)(x) = F( g)(x) by the unctoriality o F; so the naturality square commutes as required. 5 Finally, we must check the naturality in F; given a natural transormation θ: F G, we show that F [Cop, Set](H, F) θ θ G [C op, Set](H, G) 17

18 commutes. We have x x x and θ x θ (x) θ (x) with respective components C(V, ) G θ V F(x) and C(V, ) G G θ (x) But these two are equal by the naturality o θ; so the naturality square commutes as required. Dually, or F: C Set, we have F = [C, Set](H, F). LECTURE 7 25/10/02 THEOREM The Yoneda embedding is ull & aithul. We need to show that C(, B) H [C op, Set](H, H B ) is an isomorphism. By the Yoneda lemma, with F = H B, we have H B () = [C op, Set](H, H B ). So we just need to check that H is the same isomorphism as that given by the Yoneda lemma; i.e. that = H or Ĥ =. But Ĥ = (H ) (1 ) =. Note that this shows that, given, g: B, then H = H g = g. lso, given H h HB, there exists : B such that H = h. PROPOSITION = B C implies C(X, ) = C(X, B) and C(, X) = C(B, X), each isomorphism being natural in X. H is ull and aithul, so = B H = HB, so C(X, ) = C(X, B) naturally in X. Similarly or the dual statement. 2.4 Parametrised representability Consider F: C op Set. For all, we get F(, ): C op Set X F(X, ). Suppose each F(, ) has a given representation, i.e. an object U ; 18

19 a natural isomorphism α : C(, U ) F(, ). So we have an assignation U. Can we extend it to a unctor? nd are the α the components o a natural transormation? PROPOSITION Given a unctor F: C op Set such that each F(, ): C op Set has a representation α : C(, U ) F(, ), then there is a unique way to extend U to a unctor U: C such that the α are components o a natural transormation H U F. First we construct U on morphisms; i.e. given : B, we seek U: U U B. In order to satisy the naturality condition on α, we need C(, U ) α F(, ) F(,) to commute. C(, U B ) αb F(, B) Since the horizontal morphisms are isomorphisms, we get a unique morphism on the let H U H UB making the diagram commute. Now, the Yoneda embedding is ull and aithul, so there exists a unique morphism U U B inducing it. Call this U. It only remains to check that U is unctorial; it will make α a natural transormation by construction. 1 Check U(1 ) = 1 U. Note that U(1 ) is the unique morphism making the naturality square commute, so it suices to check that 1 U makes the square commute. We have C(, U ) α F(, ) 1 U F(,1 ) which commutes as required. C(, U ) α F(, ) 2 We check U(g ) = Ug U given B g C. Consider C(, U ) α F(, ) H U F(,) C(, U B ) α B F(, B) H Ug F(,g) C(, U C ) αc F(, C) 19

20 Each square commutes, so the outside commutes. Now, the composite on the RHS is F(, g ), and by deinition it induces a unique map H U(g ) on the let such that the diagram commutes. So we must have H U(g ) = H Ug H U = H Ug U, by unctorality. But the Yoneda embedding is ull and aithul, so we have U(g ) = Ug U as required. DEFINITION Cartesian closed category is a category C equipped with: a terminal object T; binary objects; unction spaces. In act, in the light o the above results on representability, we can also characterise a Cartesian closed category as containing: a representation or the unctor F: X 1, since 1 = C(X, T) or T a terminal object; representations or the unctors F,B : X C(X, ) C(X, B), since C(X, ) C(X, B) = C(X, B) naturally in X; representations or the unctors FB,C : X C(X B, C), since C(X B, C) = C(X, C B ) naturally in X. We can do even better; using the parametrised representability result, we can: rom the unctor F: (X, (, B)) C(X, ) C(X, B), construct the unctor U: (, B) B; rom the unctor F: (X, (B, C)) C(X B, C) construct the unctor U: (B, C) C B. 3 Limits & colimits 3.1 Introduction CONES Consider any drawable diagram contained within some category D; or example Then a limit over this diagram is a universal cone: cone over a diagram consists o: a vertex - an object in D; projections - a morphism rom the vertex to each object o the diagram, such that all the resulting triangles commute: LECTURE 8 25/10/02 20

21 3.1.2 LIMITS S UNIVERSL CONES Inormally, something is universal with respect to a property i any other thing with that property actors through it uniquely. limit is a universal cone over a diagram; that is, a cone such that any other cone actors through it uniquely. For example: L s.t. given Y with Y ϕ there exists unique ϕ such that all the triangles commute. s beore, the limit is unique up to unique isomorphism LIMITS OVER d Let be a small category ( is a generalisation o our drawable diagram ), and let D be a unctor D. Then we have the cone over D: a vertex L D; or each object I, a morphism ki : L DI such that, or all u: I I, L k I k I L DI Du DI commutes. We write (L k X DI) I. limit is a universal cone, and the universal property says: given a cone (Y p X DI) I, there exists a unique morphism : Y L such that all triangles commute, i.e., or all I, Y L commutes. p I DI k I 3.2 Some speciic limits PRODUCTS product is a limit o shape with discrete. So, or example, we have L DI DI DI DI our cone, where DI, ob D. The universal property says, given any other cone rom L, say, then L L DI DI DI DI 21

22 has a unique morphism L L such that every triangle commutes. We write DI p I DI. I We have already seen the product over the empty set, i.e. a terminal object, and the product over {, }; that is, a binary product EQULISERS n equaliser is a limit o shape. diagram o this shape in D is o the orm g B. cone over this diagram is E e m g B. Note that m = e = ge as all triangles commute; so in act we can rewrite this more simply as E e g B such that e = ge. n equaliser is the universal such; so given any C then there exists a unique actorisation: h g B such that h = gh, E e g B!h C h such that h = eh PULLBCKS pullback is a limit o shape diagram o this shape in D is W U g V. 22

23 cone over this diagram is P g W a U V commuting (really, there is a projection c: Z V, but we must have c = a = gb). pullback is the universal such; so given any commutative square a Z U b W g V, we have Z b!h a P W g U g V a unique h such that g h = a, and h = b. We say that g is a pullback or g over, and that is a pullback or over g. LECTURE 9 30/10/ Limits ormally DEFINITION Given Y D, we deine the constant unctor Y: Y: D I Y 1 Y. From this we get a unctor: : D [, D] Y Y with every component o being. X X Y Y 23

24 DEFINITION limit or D: D is a representation or the unctor [, D](, D): D op Set. That is, an object L D and a natural isomorphism α with H L α = [, D](, D). We write L = lim D = I DI. So we have an isomorphism D(, I DI) = [, D](, D). Let us make explicit what the unctor on the right hand side does; call it F. Then: F: D op Set Y [, D]( Y, D) Y [, D]( X, D) F X [, D]( Y, D) θ θ. Now, what does a natural transorm Y k D look like? We have: or each I, a morphism ki : ( Y)I DI Y DI; or all u: I I in, ( Y)u ( Y)I DI Du ( Y)I DI commutes by naturality; i.e. Y k I k I commutes. DI Du DI So such a natural transormation is precisely a cone over D with Y as the vertex. Now, consider a representation as above, and let α be its natural isomorphism. Then we have α Y : D(Y, L) [, D]( Y, D) F(α L 1 L ); i.e., the natural transormation is completely determined by α L 1 L. Now, we have a cone given by α L 1 L = (k I ) I, say. So given any other Y and Y L on the let 24

25 hand side, we have F(α L 1 L ) with components k I ; hence we have a bijective correspondence morphisms Y L cones over D (k I ) I i.e., starting on the right hand side, given any cone (p I ) I, there exists a unique morphism : Y L such that p I = k I or all I; thus (k I ) I is a universal cone over D. Note that any isomorphism on the let hand side will give rise to a universal cone. DEFINITION I a limit exists or all unctors rom D: D, we say D has all limits o shape. I D has all limits o shape or all small/inite categories, we say D has all small/inite limits or that D is (initely) complete. 3.4 Limits in Set THEOREM Set has all small limits. We seek a limit or F: Set. We deine L, a set o tuples FI by taking all tuples (x I ) I satisying: I I, xi FI; I u I, Fu(x I ) = x I. We have projections L p I FI (x I ) I x I or each I. We now show that this is a minimal cone: 1 It is a cone; we need to show, or all u: I I, that L p I p I FI Fu FI commutes. On elements we have (x I ) I (x I ) I and Fu(x I ) x I x I so we are done here, since Fu(x I ) = x I. 2 It is universal: we show that every cone actors through it uniquely. So consider a cone (Z q I FI) I ; so Z q I q I FI Fu FI 25

26 commutes; that is, or all y Z, Fu(q I (y)) = q I (y). We seek a unique actorisation making the ollowing diagram commute or all I: L h Z p I FI q I On elements, this would give h(y) y q I (y) So, writing h(y) = (a I ) I, we must have a I = q I (y). So deine h by h(y) = (q I (y)) I. It remains to check that h(y) L, so that or all u: I I, Fu(a I ) = a I ; i.e. Fu(q I (y)) = q I (y), which ollows since (Z q I FI) I is a cone. LECTURE 10 01/11/ Limits in other categories THEOREM I a category D has all small products and equalisers, then D has all small limits. Given a diagram D: D, small, we seek a limit in D. The idea o the proo is to construct it as an equaliser E e P g 1 Put Q, where P and Q are certain products over the DI. P = I D I with projections P p I 2 Put DI; this is a small product, so exists. Q = DJ u: I J with projections Q q U DJ; again, a small product, so exists. 3 Induce by the universal property o Q as ollows: or all u: I J, we have p J : P DJ inducing a unique : P Q such that u, q U = p J. (1) 26

27 P! Q p J q U DJ 4 Induce g by the universal property o product Q (dierently) as ollows: or all u: I J, we have Du p I : P DJ inducing a unique g: P Q such that, or all u, q u g = Du p I. (2) P!g Q p I q U DI Du DJ 5 Take equaliser E e P g Q; so in particular e = ge. (3) Claim that (E p I e DI)I gives a universal cone over D. 6 First we show it is a cone; i.e. or all u: I J, Du p I e = p J e (4) This is true, since Du p i e = q u g e by (2) = q u e by (3) = p J e by (1) It remains to show that this cone is universal; i.e. given any cone (V v I unique x: V E such that or all I, p I e x = v I. V DI) I, we seek a x E v I p I e DI 27

28 We will construct a diagram v I x V E e k P m g Q p I DI So suppose we are given such a cone (V v I DI) I. So or all u: I J, Du v I = v J. (5) 7 Induce k: V P by the universal property o P: or all I, we have V v I unique k: V P such that, or all I, DI inducing a p I k = v I. (6) 8 Induce x: V E by the universal property o the equaliser; in order to do this, we must irst show that k = gk. Now, or all u: I J, we have V vj DJ inducing a unique m: V Q such that q u m = v J. (7) But k and gk both satisy this condition, since, or all u, and q u k = p J k by (1) = v J by (6) q u g k = Du p I k by (2) = D u v I by (6) = v J by (5) Hence k = gk; so we can induce a unique x: V E such that e x = k. (8) 9 We now check that x is a actorisation or the cones. So given I, so we have the desired result. p I e x = p I k by (8) = v I by (6) 10 Finally, we show that x is unique with this property; suppose we have a morphism y: V E such that, or all I, p I e y = v I. (9) Now by construction x is unique such that ex = k, so we seek to show also ey = k. By construction, k is unique such that or all I, p I k = v I (by (6)); but (9) says that ey also satisies this. Hence ey = k, so y = x and we are done. 28

29 3.6 Colimits DEFINITION colimit or a diagram D: D is a representation D( I DI, ) = [, D](D, ). So a colimit or D: D is essentially a limit o D op : op D op. I D has all small colimits, we say it is cocomplete. LECTURE 11 04/11/ Parametrised limits Recall two results: 1 Given a diagram D: D, a limit or D is a representation D(, I DI) = [, D](, D) 2 Given a unctor X: C op Set such that each X(, ) has a representation α : C(, U ) = X(, ) then there is a unique way to extend U to a unctor such that C(Y, U ) = X(Y, ) naturally in Y and, with components o the implied natural transormation given by α. PROPOSITION Deine F: D such that each F(, ): D has a speciied limit in D: D(, I F(I, )) = [, D](, F(, )). Then there is a unique way to extend I F(I, ) to a unctor D such that naturally in Y and. D(Y, I F(I, )) = [, D]( Y, F(, )) Simple application o parametrised representability. PPLICTION Suppose D has chosen limits o shape. Consider the evaluation unctor E: [, D] D (I, D) DI Then E(, D) has a limit or each D, I DI. By parametrised limits, we get a unctor I : [, D] D D I DI such that D(Y, I DI) = [, D]( Y, D) naturally in Y and D. PPLICTION We can restate the deinition o a limit to get D(Y, I DI) = I D(Y, DI). 29

30 What does this mean? 1 The right hand side is the limit o the unctor D(Y, D ): Set I D(Y, DI) I ui D(Y, DI) Du D(Y, DI ) Set is complete, so this certainly has a limit. What does I D(Y, DI) look like? Well, it is all tuples (α I ) I such that I, α I D(Y, DI) and u: I I, Du α I = α I. So this is precisely a cone over D; i.e. I D(Y, DI) = [, D]( Y, D) 2 Observe that by parametrised limits, we have a unctor So naturally in Y and D. 3.8 Preservation, relection and creation o limits Y I D(Y, DI) I D(Y, DI) = [, D]( Y, D) = D(Y, DI) Let D D F E. We can consider limits over D and limits over FD. DEFINITION Suppose we have a limit cone or D We say F preserves this limit i ( I DI k I (F I DI Fk I DI) I FDI) I is a limit cone or FD in E. Note that it must preserve projections. DEFINITION Suppose FD: E has a limit cone. We say F relects this limit i any cone that goes to a limit cone was already a limit cone itsel. That is, given a cone (Z I DI) I such that (FZ F I FDI) I is a limit cone or FD, then (Z I DI) I is also a limit cone. 30

31 DEFINITION Suppose FD: E has a limit cone. We say F creates this limit i there exists a cone (Z I DI) I such that (FZ F I FDI) I is a limit cone or FD, and additionally F relects limits. That is, given a limit or FD, there is a unique-up-to-isomorphism lit to a limit or D. 3.9 Examples o preservation, relection and creation PROPOSITION Representable unctors preserve limits. We consider Given a limit cone or D, we need to show that D C HU Set I DI C(U, DI) ( I DI k I C(U, I DI) k I DI) I, C(U, DI) LECTURE 12 06/11/02 is a limit cone or C(U, D ). Certainly, C(U, I DI) = I C(U, DI). nd or projections so we are done. Dually, we have C(U, I DI) = [, C]( U, D) = I C(U, DI) k I C( I DI, U) = I C(DI, U) so H U takes a colimit in C to a limit in Set; and hence takes a limit in C op to a limit in Set. Thus H U also preserves limits. PROPOSITION ull and aithul unctor preserves limits. Consider D C F E, with F ull and aithul, and let (Z I DI) I be a cone such that F o it is a limit cone or FD. We need to show that this cone itsel is a limit. Now, given any other cone (W g I DI) I, we seek a unique h such that g I = I h or all I. So 1 Since F(Z I DI) is a limit, there exists unique m such that Fg I = F i m or all I. 2 Since F is ull, there exists h: W Z such that Fh = m. 3 Check commuting condition: we know that, or all I, Fg I = F i Fh, i.e. Fg I = F( i h). Hence I h = g I since F is aithul. 4 Suppose there is a k such that or all I, I k = g I. Then F I Fk = Fg I or all I; but we have that m is the unique morphism such that F i m = Fg I ; hence Fk = m = Fh, so k = h (as F aithul), and we are done. 31

32 4 Ends and coends 4.1 Dinaturality DEFINITION Given unctors F, G: C op C D, a dinatural transorm α: F G consists o, or each U C, a component α U : F(U, U) G(U, U) such that or all : U V, F(U, V) F(,1) F(U, U) α U G(U, U) G(1,) G(U, V) F(1,) F(V, V) α V G(V, V) G(,1) commutes. Note that there is no sensible composition o dinatural transormation, and hence Dinat(F, G) is just a set. 4.2 Ends and coends Recall that a limit or D: DEFINITION n end or F: op so that D is a representation or [, D](, D) = Nat(, D), such that D(Y, I DI) = Nat( Y, D) naturally in Y. D is a representation or the unctor Dinat(, F): D op Set D(Y, I F(I, I)) = Dinat( Y, F) naturally in Y. Dually, a coend or F is just a representation or Dinat(F, ): D Set so REMRK 4.3 Ends in Set D( I F(I, I), Y) = Nat(F, Y) naturally in Y. Ends are in act just a special sort o limit; any end can be expressed as a limit. Recall a limit in Set or D: n end in Set or X: op Set is given by { (x I ) I I, x i DI, u: I I, Du(x I ) = x I }. Set is given by { (x I ) I I, x i X(I, I), : I I, X(1, )(x I ) = X(, 1)(x I ) }. 32

33 4.4 Key observations OBSERVTION Parametric results ollow, so we can use ends in Set to restate the deinition o (co)ends. Consider We have an end in Set X V : op (I, J) Set D(V, F(I, J)) I X V(I, I) = I D(V, F(I, I)) = Dinat( V, F) So we get: End: D(V, I F(I, I)) = I D(V, F(I, I)) Coend: D( I F(I, I), V) = I D(F(I, I), V) OBSERVTION The set [, D](F, G) is an end in Set. For consider Then U X(U, U) = U D(FU, GI) is just But now X: op Set (U, V) D(FU, GV) { (α U ) U α U : FU GU and : U U, X(1, )(α U ) = X(, 1)(α U ) }. G α U = X(1, )(α U ) = X(, 1)(α U ) = α U F so this is just a naturality condition on the α U s; and hence we have U X(U, U) = U D(FU, GI) = [, D](F, G). OBSERVTION pplications We can restate the Yoneda lemma. Recall that i X: op Set, we have LECTURE 13 08/11/02 X(U) = [ op, Set](H U, X) = V [H U(V), X(V)] where [, ] means morphisms in Set = V [ (V, U), X(V)] Consider a unctor F: [, D]. What does a limit cone or this look like? We have (L α I FI) I with L a unctor and α I a natural transormation L FI with components (α I ) C : LC Now, given C, we can evaluate the whole cone at C: FI(C). (LC α IC FI(C)) I 33

34 Now i this is a limit cone in D or F C : I D FI(C) then we say that the limit or F is computed pointwise. PROPOSITION Suppose F: [, D] is such that or all C, has a limit cone Then F has a limit computed pointwise; i.e. We have a unctor F C : I D FI(C) ( I FI(C) (pc ) I FI(C) ) I. ( I FI k I FI ) I ( I FI) (C) = I FI(C) and (k I ) C = (p C ) I F: D (I, C) FI(C) and each F(, C) = F C has a limit, so by parametrized limits, we get a unctor C I FI(C) Call it L, and claim this gives the limit as required. So we need to show naturally in Y, and to check projections. Now, [, D](Y, L) = C D(YC, LC) [, D](Y, L) = [, [, D]]( Y, F) set o nat trans is end in Set = C D(YC, I F(I, C)) rewriting LC = C [, D]( (YC), F(, C)) by deinition o limit = [, [, D]]( (Y ), F(, )) end in Set is set o nat trans = [, [, D]]( Y, F) where the last isomorphism holds since [, [, D]] = [, D] = [, [, D]]. Note that each line is natural in Y; and the third line gives the projections as required. We have the same result or colimits, ends and coends. However, it may be possible or nonpointwise limits to exist i not all the F C s have limits. 34

35 THEOREM The Yoneda embedding preserves limits. Consider D H [ op, Set]. Suppose we have a limit cone or D, ( I DI k I DI) I We need to show that ( (, I DI) H k I (, DI))I. is a limit or H D. By the previous result, it suices to do this pointwise; so or all C, we need that ( (C I DI) k I (C, DI)) I is a limit or I (C, DI), i.e. H C D. But we have already shown this, since representables preserve limits, and the given cone is just H C o ( I DI DI) I. k I THEOREM (FUBINI) LECTURE 14 11/11/02 Suppose F: D is such that F J : D has a limit I F(I, J) or all J. Then we have a unctor I F(I, ): J I F(I, J) such that J I F(I, J) = (I,J) F(I, J) in the sense that i one exists, then so does the other, and they are isomorphic with corresponding limit cones. The right-hand side is a representation o [, D](, F); the let-hand side is a representation o [, D](, I F(I, )). Now, Hence representations give the result. COROLLRY [, D]( V, F) = [, [, D]]( V, F(, )) = I [, D]( V, F(I, )) = [, D]( V, I F(I, )). Suppose F: D such that I F(I, ): D and J F(, J): D exist. Then J I F(I, J) = I J F(I, J) in the same sense as above. Both are isomorphic to (I,J) F(I, J). Note that also we have colimits, ends and coends commuting with themselves; also (co)ends commute with (co)limits. THEOREM (DENSITY) For X: op Set, we have X(U) = W (U, W) X(W), 35

36 naturally in U. We aim to show that and deduce result by above. So: [ op, Set](X, Y) = [ op, Set]( W (, W) X(W), Y) RHS = U [ W (U, W) X(W), Y(U)] set o nat trans is end in Set = U W [ (U, W) X(W), Y(U)] restate deinition o colimit = W U [ (U, W) X(W), Y(U)] Fubini interchange = W U [X(W), [ (U, W), Y(U)]] deinition o unction space = W [X(W), U [ (U, W), Y(U)]] restate deinition o end = W [X(W), Y(W)] Yoneda restated = [ op, Set](X, Y) end in Set is set o nat trans Hence, since the Yoneda embedding is ull and aithul, we have the desired natural isomorphism X = W (, W) X(W). THEOREM Every preshea is a colimit o representables. By previous result, we have XU = W (U, W) X(W) The idea o the proo is that this is almost a colimit o representables. We would like to say that it is W,x X(W) (U, W). Can we do this in any way? We can, by deining the Grothendieck Fibration. Given X: op (X) with objects being pairs (W, x), W C, x XW. morphisms (W, x) (W, x ) being : W W such that X(x ) = x. There is a orgetul unctor P: (X) (W, x) W Set, we deine a category So we get (X) P H [ op, Set], and X(U) = α G(X) (U, P(α)) Hence we get X = α G(X) H P(α), a colimit o representables. THEOREM preshea category [ op, Set] is Cartesian closed. 36

37 Limits and colimits are computed pointwise, so we get the terminal object and binary products rom those in Set. So we need to ind unction spaces. So, given Y, Z [ op, Set], we seek Z Y [ op, Set] such that naturally in X and Y. So put Then [ op, Set](X, Z Y ) = [ op, Set](X Y, Z) Z Y (U) = [ op, Set](H U Y, Z) = V [ (V, U) Y(V), Z(V)] end in Set, products ptwise. [ op, Set](X, Z Y ) = U [X(U), ZY (U)] end in Set = U [X(U), V [ (V, U) Y(V), Z(V)]] write in deinition = U V [X(U), [ (V, U) Y(V), Z(V)]] restate den o limit = V U [X(U), [ (V, U)[Y(V), Z(V)]]] c.c. o Set, Fubini = V U [X(U) (V, U), [Y(V), Z(V)]] c.c. o Set = V [ U X(U) (V, U), [Y(V), Z(V)]] restate den o colimit = V [X(V), [Y(V), Z(V)]] Density = V [X(V) Y(V), Z(V)] c.c. o Set = [ op, Set](X Y, Z) end in Set, products ptwise. Thus Z Y is a unction space as required. 5 djunctions 5.1 Deinitions DEFINITION LECTURE 15 13/11/02 Let F: C D, G: D C be unctors. n adjunction F G consists o an isomorphism D(FX, Y) = C(X, GY) that is natural in X and Y. We say F is let adjoint to G, and G is right adjoint to F. So, we have a correspondence morphisms FX Y morphisms X GY NOTTION We write FX g Y D X g GY C and X GY C FX Y D We write ( ) or the adjunction operation, and call it transpose. Note =, g = g. 37

38 What do the naturality conditions mean? Naturality in X says that, or any h: X X, D(FX, Y) ( ) C(X, GY) Fh h D(FX, Y) ( ) C(X, GY) commutes. Similarly, naturality in Y says that or any k: Y Y, D(FX, Y) ( ) C(X, GY) k Gk D(FX, Y ) ( ) C(X, GY) commutes. That is, X FX h Fh X FX Y GY and FX g Y k Y X g GY Gk GY h = Fh k g = Gk g Now, this is actually the Yoneda lemma in disguise: D(FX, Y) = C(X, GY) is H FX = C(X, G ) and C(X, GY) = D(FX, Y) is H GY = D(F, Y) Yoneda tells us that each o these natural transorms is completely determined by where the identity goes: FX X Then by naturality, 1 FX η X FX and GX GFX FGY 1 GY ε Y GY Y and FX 1 FX FX Y g = Gg η X η X X GFX Gg GY = ε X F X FX F FGY 1 GY GY GY g ε Y Y nd in act, the η X, ε Y are components o a natural transormation. PROPOSITION Given F G, we have natural transormations η and ε with components given by η X, ε Y. 38

39 Check naturality. For η, given : X X, X η X GFX GF X η X GFX must commute. Now, we have:- X FX η X GFX GF GFX 1 FX FX X F FX 1 FX X η X FX GFX But we have transposed twice, and hence we have equality as required. Similarly or ε. DEFINITION Examples Given F G, we call η: 1 C GF the unit and ε: FG 1 D the counit o the adjunction. EXMPLES Free orgetul. For example: 1 U: Gp Set has a let adjoint F U, where F(S) gives the ree group on S; so we have Gp(FS, G) = Set(S, U(G)) 2 U: lg Vect which orgets the multiplicative structure; we have F U, where F(V) is the ree algebra on V. 3 U: Ring Monoid has a let adjoint : M M = {ormal inite combinations λ i m i, λ i, m i M.} 4 U: b Gp has a let adjoint ree abelianization : G B = G/[G, G]. 5 U: lg k Lie k has let adjoint L U(L) = universal enveloping algebra o L. EXMPLES Relections inclusions corelections. I C D has a let adjoint, it is called a relector and exhibits C as a relective subset o D. 1 s above, b Gp; b is relective in Gp. 2 { complete metric spaces, uniormly cts unctions 3 has let adjoint completion. { compact Hausdor spaces, uniormly cts unctions has let adjoint Stone- Cech compactiication. } { } { metric spaces, uniormly cts unctions topological spaces, uniormly cts unctions } } 39

40 4 Gp Monoid. Gp is relective and corelective in Monoid, via M { m M m is invertible } EXMPLE Closedness. Let C be a cartesian closed category. Then or all B C, we have B ( ) B i.e. C( B, C) = C(, C B ) naturally in and C. EXMPLE djoints or representable unctors are powers and copowers. Recall given an object C and a set I, we can orm the I-old power: I = i I = [I, ] and dually the I-old copower: I = i I. By parametrised limits, we get unctors: [, ]: Set C op : Set C Now, Set(I, C(U, )) = C(U, [I, ]) = C op ([I, ], U). So [, ] C(, ) = H. Similarly C(, ) = H, since Set(I, C(, U)) = C(I, U). So H has an adjoint i C has all small powers o i C op has all small copowers o. I C has all small powers and copowers o, we get via Set(I, C(, U)). So I [I, ]: C C. C(I, U) = C(, [I, U]) LECTURE 16 15/11/ Triangle identities PROPOSITION Given an adjunction F G, then the unit η: 1 GF and the counit ε: FG 1 satisy the triangle identities; that is, the ollowing diagrams commute: GY η GY GFGY FX Fη X FGFX 1 GY Gε Y and 1 FX ε X GY FX 40

41 η GY GY GFGY Gε Y GY FX FGY 1 FGY FGY GY ε Y Y and 1 GY GY X FX Fη X η X 1 FX FGFX GFX FX THEOREM n adjunction F G is completely determined by natural transormations η: 1 GF ε: FG 1 satisying the triangle identities. Suppose we are given such ε, η. We need to show that D(FX, Y) = C(X, GY) naturally in X and Y. So, given : X GY, put ε X FX 1 GFX GFX and given g: FX Y, put : FX F FGY ε Y Y g: X η X GFX Gg GY We need to check naturality. For naturality in X, we need, given h: X X, that h = Fh. Now, h = ε Y F(h) = (ε Y F) Fh = Fh. For naturality in Y, we need, or all k: Y Y, kg = Gk g. Now, kg = G(kg) η Y = Gk (Gg η Y ) = Gk g. Now we need to check that these are inverse: given : X have = FX F FGY ε Y Y. So = X η X GFX GF GFGY Gε Y GY, we need that =. We GY GY η GY 1 GY 41

42 Note that the let hand circuit commutes by the naturality o η, and the right hand circuit commutes by the irst triangle identity, so =. Similarly, given g: FX Y, g = FX Fη X FGFX FGg FGY ε Y Y 1 FX FX ε FX Here, the let circuit commutes by the second triangle identity, and the right circuit commutes by the naturality o ε; hence g = g, as required. REMRK djunctions can be composed: g C F 1 D F2 G 1 G 2 E giving C F2F1 rom E(F 2 F 1 X, Y) = D(F 1 X, G 2 Y) = C(X, G 1 G 2 Y). G 1 G 2 E 5.4 djunctions as parametrised representations To give a let adjoint to G: D C, it is suicient to give, or each X C, a representation or C(X, G ): D By parametrised representation, this extends uniquely to a unctor which is the let adjoint we are looking or. Dually, a right adjoint to F: C D is a representation or D(F, Y): C op Recall D has limits o shape means, or all D: [, D](, D): D op Set. Set. D, there exists a representation o i.e., D has limits o shape i : D [, D] has a right adjoint. Dually, D has colimits o shape i : D [, D] has a let adjoint. 5.5 djunctions as collections o initial objects DEFINITION Given G: D C and X C, we deine the comma category (X G): Set objects are pairs (, Y), X GY; morphisms (, Y) h (, Y ) are morphisms Y h Y such that commutes. PROPOSITION To give a let adjoint or G: D or the comma category (X G). GY X Gh GY C is equivalent to giving, or all X C, an initial object 42

43 n initial object in (X G) is a pair (u, V X ) with X u GV X such that, or all X GY, there exists a unique h: V X Y such that u X commutes. So GV X Gh GY D(V X, Y) = C(X, GY) Gh u We need to check naturality in Y. So, or all g: Y D(V X, Y) Y, we have C(X, GY) and so on elements h D(V X, Y ) C(X, GY ) Gh u g h G(g h) u = Gg Gh u and so this is a representation as required. 5.6 Duality We note that there are a lot o duality relations going on with adjunctions: let adjoint right adjoint unit counit natural in X natural in Y irst triangle identity second triangle identity Why is this? Consider F G, C F G D also G F, C op F G D op D(FX, Y) = C(X, GY) D op (Y, FX) = C op (GY, X) F G: D C G F: C op D op unit η X : X GFX counit η X : GFX X counit ε Y : FGY Y unit ε Y : Y FGY 43

44 6 djoint unctor theorems 6.1 Preservation THEOREM Suppose F G: D C. Then G preserves limits, and F preserves colimits. LECTURE 17 18/11/02 Consider D: cone or GD: D with limit cone ( I DI k I C. The cone becomes DI) I. We need to show that G o it is a limit (G I DI Gk I GDI) I. We need a natural transormations C(, G I DI) = [, C](, GD) with components C(V, G I DI) = [, C]( V, GD) Now, nd on projections: as required; and dually or F. (Gk I ) I C(V, G I DI) = D(FV, I DI) = I D(FV, DI) = I C(V, GDI) = [, C]( V, GD). k I Gk I 6.2 General adjoint unctor theorem DEFINITION Given a category, a collection is weakly initial i or all, there exists a morphism I or some I. EXMPLE {initial object} is a weakly initial set. THEOREM (GENERL DJOINT FUNCTOR THEOREM) Suppose we have a unctor G: D C that preserves small limits, and that D is locally small and complete. Then G has a let adjoint i or all X C, the category (X G) has a weakly initial set. This last condition is known as the solution set condition. 44

45 Here is the general structure o the proo: D locally small D complete G preserves small limits (X G) locally small P: (X G) D creates small limits (X G) complete (X G) has w.i.s. i (X G) has initial object G has a let adjoint i or all X, (X G) has initial object GFT where we deine P: (X G) D to be the obvious orgetul unctor. So: LEMM 1 P: (X G) D creates small limits. Let D: (X G) be a diagram. We need to show that, i PD has a limit cone, then there is a cone (V c I DI) I in (X G) such that (PV Pc I itsel a limit or D in (X G). PDI) I is a limit or PD in D, and that any such cone is 1 Suppose PD: D has a limit cone, say (L c I L c I c I PDI) I : c I PDI PDI PDI 2 G preserves small limits, so (GL Gc I GPDI) I is a limit or GPD in C. GL Gc I Gc I Gc I GPDI GPDI GPDI 45

46 3 (DI) I gives a diagram in (X G) X GPDI GPDI GPDI which is precisely a cone (X GPDI) I in C. Hence we induce a unique morphism u: X GL making everything commute: X GL GPDI GPDI GPDI (1) 4 Since everything in the diagram commutes, it orms a cone over D in (X G), with vertex V = (X u GL). Moreover, by construction is it unique such that applying P to it gives the original cone (L c I PDI) I. So we have shown that, given a limit cone or PD there is a unique cone in (X G) that maps to it, given by (1) above. It remains to show that this cone is universal. 5 Given any cone ( (X GY) DI) I in (X G), we seek a unique actorisation (X GY) V: GY Gh X GL GPDI GPDI pplying P, we get a cone (Y PDI) I in D, and since L is a limit, this induces a unique morphism h: Y L making everything commute in D. But now, by the uniqueness o u we have Gh = u, since Gh satisies the conditions making u unique. So h is a morphism in (X G): GY X u and so is the unique actorisation as required. So the cone (1) is indeed universal and P creates limits as required. So now we can quickly deduce LEMM 2 For each X C, (X G) is locally small and complete. Since D is locally small, so too is (X G). Now, let D be a diagram in (X G). pply P to get a diagram PD in D. This has a limit, since D is complete. nd by lemma 1, P creates it rom a limit in (X G); i.e. D has a limit in (X G). So (X G) is complete. Now, we need only prove 46 Gh GL