Notes on Homological Algebra

Transcription

1 Notes on Homological Algebra Marisz Wodzicki December 1, 2016

2

3 x 1 Fondations 1.1 Preliminaries A tacit assmption is that A, B,..., are abelian categories, i.e., additive categories with kernels, cokernels, every epimorphism being a kernel, and every monomorphism being a cokernel In an abelian category every morphism has an image-coimage factorization fl ι [^ j [ θ jfl α (1) in which ι is an image of α, i.e., a kernel of a cokernel of α, and θ is a coimage of α, i.e., a cokernel of a kernel of α, cf. Notes on Category Theory Exact composable pairs A composable pair of arrows α β (2) is said to be exact if β factorizes κ β with κ being a kernel of α and β being an epimorphism If λ is a cokernel of β, then λ is a cokernel of κ, i.e., a cokernel of β and a kernel of α form an extension λ. κ (3) Unlike the original condition this last condition is self-dal. 3

4 v x Note that a cokernel of β and a kernel of α form an extension precisely when κ is an image of β and λ is a coimage of α fl α [^ j [ λ jfl α β [^ [ j κ ^ jfl β in which case coimage factorization factorizes α throgh a cokernel of β with α being a monomorphism while image factorization factorizes β throgh a kernel of α with β being an epimorphism. Let s record this fact for ftre se. Lemma 1.1 For any composable pair of arrows, the following three conditions are eqivalent (a) β factorizes throgh a kernel of α (b) α factorizes being a cokernel of β β [^ [ ff κ ^ fffl β fl α [^ ff [ λ fffl (c) a cokernel of β and a kernel of α form an extension Condition α β = 0 A weaker condition, α β = 0, is eqivalent to existence of a factorization of β β κ α λ µ 4 (4)

5 x with µ being a monomorphism. Note that µ satisfying identity α = κ µ λ is niqe in view of κ being a monomorphism and λ being an epimorphism Homology A cokernel χ of µ is called a homology of composable pair (2) and it forms an extension χ µ. (5) The target of χ is sally referred to as the homology (grop) of the composable pair. This terminology originally was applied to homomorphisms of abelian grops bt the tradition is so well established that it contines to be sed also in the context of general abelian categories Note that a homology morphism χ is zero or, eqivalently, in view of the fact that χ is an epimorphism, its target is zero, precisely when µ is an isomorphism, i.e., when (2) is exact Composable pairs (2) satisfying the condition α β = 0 form a fll sbcategory of the category of composable pairs A 2,0. Exercise 1 Show that any assignment of a morphism χ on the class of sch composable pairs admits a niqe extension to a fnctor χ : A 2,0 Arr A The compositions Z s χ and H t χ 5

6 with the sorce and, respectively, the target fnctors s : Arr A A and t : Arr A A are referred to as a cycles and, respectively, homology fnctors. Exercise 2 Express the condition α β = 0 in terms of a certain factorization of α A dal approach to homology Based on the corresponding factorization of α instead of β one can develop a dal approach to homology Chain complices We shall se the term chain complex in two ways: either in a loose sense, to denote any seqence of composable morphisms of any length sch that the composition of any two composable arrows is zero or, in a strict sense, as a Z-graded object C = (C l ) l Z eqipped with an endomorphism of degree 1 sch that = 0. The graded endomorphism is referred to as the bondary morphism or operator Notation We may se notation (C, ), (C, ), C or, simply, C, to denote a chain complex (C l, l ) l Z. 1.2 Projective and injective objects Projective objects in an abelian category An object P is said to be projective if it possesses the Lifting Property for epimorphisms, more precisely, if for any diagram P ɛ 6

7 with ɛ being an epimorphism, there exists an arrow P \ \] sch that the diagram P \ \\] commtes. ɛ The above definition makes sense in any category, in general categories, however, there may be too many epimorphisms making the Lifting Property with respect to the class of all epimorphisms very restrictive An example: the category of topological spaces In the category of topological spaces and continos mappings, epimorphisms are mappings with dense image. For any nonsrjective epimorphism f : Y Z and any nonempty space X the maping that sends all points in X to a point in Z \ f (Y) cannot be lifted to Y. Ths, the only topological space that has Lifting Property for epimorphisms is the empty space Projective objects in a general category For the above reasons, projective objects in general categories are defined by the Lifting Property with respect to a specific class of epimorphisms, e.g., the class of all strong epimorphisms, see Notes on Category Theory. 7

8 v v w w Injective objects in an abelian category An object I is said to be injective if it possesses the Extension Property for monomorphisms, more precisely, if for any diagram I with µ being a monomorphism, there exists an arrow µ sch that the diagram commtes. I I \^ \ \^ \ \ µ Exercise 3 Show that P is projective if and only if for any commtative diagram with an exact row P there exists an arrow 0 j j φ jfl α P \ \] φ β sch that the diagram commtes. φ P \ \ φ \] α β 8

9 x Dally, I is injective if and only if for any commtative diagram with an exact row I [^ [ 0 ψ [ there exists an arrow sch that the diagram α ψ fffffi I β commtes. I ψ fffffi ψ ff α β Characterization of projectivity in terms of extensions Lemma 1.2 An object P is projective if and only if every extension of P, π P ι, splits. Proof. Consider the diagram P id P P π. If P is projective, then there exists ι P sch that π ι P = id P. 9

10 x x x Vice-versa, sppose that a diagram P α A ɛ A is given. Consider a pllback by α of the extension where κ is a kernel of ɛ, ɛ A A A ɛ P Ā A α ᾱ ɛ A A A κ κ κ ι If P w Ā splits the pllback extension, then ᾱ ι lifts α. Indeed, ɛ (id Ā ι ɛ) = 0 implies that for a niqe π and id Ā ι ɛ = κ π α ɛ = ɛ ᾱ = ɛ ᾱ id Ā = ɛ ᾱ ι ɛ + ɛ ᾱ κ π = ɛ ᾱ ι ɛ which in trn implies that α = e (ᾱ ι) since ɛ is an epimorphism. 1.3 Categories of chain complices The (strict) category of graded objects The strict category of Z-graded objects has Z-seqences (A q ) q Z as its objects and Z -seqences of morphisms (φ q ) q Z, A q φ q, (6) A q as morphisms. In order to avoid confsion with morphisms in the category of Z-graded chain complices Ch(A), we shall denote sch seqences graded maps. 10

11 1.3.2 Ch(A) Morphisms in the category of chain complices are graded maps that commte with the bondary operators, q φ q = φ q 1 q (q Z) The graded category of graded objects We shall also consider graded maps of degree d, i.e., seqences of morphisms (φ q ) q Z in A, φ q (q Z), (7) A q+d A q and often will refer to them simply as maps of degree d. The reslting graded category will be referred to as the graded category of graded objects of A Sbcategories of Ch(A) The category of chain complices Ch(A) of an abelian category A has several natrally defined sbcategories, for example, for any pair of sbsets J I Z, let Ch I,J (A) denote the fll sbcategory of chain complices with chain objects being zero in degrees otside I, C l = 0 for l I, and homology objects being zero in degrees otside J, H l = 0 for l J. Ths, Ch Z, (A) denotes the sbcategory of acyclic chain complices. When I = J we shall se notation Ch I (A) Ch + (A) The sbcategory Ch + (A) of complices vanishing at, i.e., satisfying C l = 0 for l 0, is the nion of the sbcatories Ch n (A) of complices satisfying C l = 0 for l < n. 11

12 1.3.6 Ch - (A) The sbcategory Ch - (A) of complices vanishing at, i.e., satisfying C l = 0 for l 0, is the nion of the sbcatories Ch n (A) of complices satisfying C l = 0 for l > n. 1.4 Fnctors associated with chain complices The category of chain complices is stdied together with with a nmber of associated fnctors The shift fnctors The shift fnctors where and [n] : Ch(A) Ch(A), C [n]c, (8) ([n]c) l C l n, ([n] ) l ( 1) n l n, (9) ([n] f ) q f q n, (10) provide an action of the additive grop of integers Z on Ch(A) and those sbcategories of Ch(A) that are invariant nder shifts The q-chain fnctors q-cycles fnctors Given a fnctor C l : Ch(A) A, C C q (q Z). (11) ζ l : Ch(A) Arr A, C ζ q, that assigns to a chain complex C a kernel of q, its composition with the sorce fnctor yields a q-cycles fnctor Z q : Ch(A) A, C s(ζ q ). (12) 12

13 1.4.5 q-bondaries fnctors Given a fnctor β q : Ch(A) Arr A, C β q, that assigns to a chain complex C an image β q of q+1, its composition with the sorce fnctor yields a q-bondaries fnctor B q : Ch(A) A, C s(β q ). (13) q-homology fnctors The condition q q+1 means that q+1 niqely factorizes throgh ζ q, q+1 = ζ q q+1. (14) Factorization (14) shold not be confsed with the image factorization of q+1 q+1 = β q q+1. (15) Assigning to a chain complex C the target of a cokernel of q+1 defines a q-homology fnctor H q : Ch(A) A. (16) The q-homology extensions Fnctors H q, Z q and B q form an extension χ q β q H q Z q x B q (17) becase the sorce of an image of q+1 is also the sorce of an image of q q-co-cycles fnctors Given a fnctor ζ q : Ch(A) Arr A, C ζ q, that assigns to a chain complex C a cokernel ζ q of q+1, its composition with the target fnctor yields a q-co-cycles fnctor Z q : Ch(A) A, C t(ζ q). (18) 13

14 x q-co-bondaries fnctors Given a fnctor β l : Ch(A) Arr A, C β q, that assigns to a chain complex C a coimage β q of q, its composition with the target fnctor yields a q-co-bondaries fnctor B q : Ch(A) A, C t(β q). (19) Since an image factorization of q is also its coimage factorization, fnctors B q and B q+1 are isomorphic by a niqe isomorphism of fnctors compatible with the corresponding kernel and cokernel fnctors q-co-homology fnctors The condition q q+1 means that q niqely factorizes throgh ζ q, q = q ζ q. Assigning to a chain complex C the sorce of a kernel of q defines a q-co-homology fnctor H q : Ch(A) A. (20) The q-co-homology extensions Fnctors H q, Z q and B q form an extension β q χ q B q Z q H q (21) becase the target of a coimage of q is also the target of a coimage of q. 14

15 v x x v The diagrams of fnctor extensions By definition the above seven seqences of fnctors from Ch(A)to A enter for extensions that form the following commtative diagram H q χ q β q ζ q β q Z q C q B q B q C q Z q β q ζ q β q (22) χ q Using the diagram chasing techniqes, one can constrct a canonical isomorphism of H q with H q, see the chapter on Diagram chasing in Notes on Category Theory for details By taking into accont this isomorphism, we may redraw diagram (22) in the form k h h h h h hk H q H q Z q Z q k [^ fl 447 h [ j 4 h jfl 4 h C q h A 4 h A 4 A hk A A 4 B q AD 7 B q (23) 15

16 w 2 Fndamental lemmata 2.1 First Fndamental Lemma Consider a diagram whose colmns are chain complices.. 2 Q 1 P Q 0 P 0 1 ɛ N f M ɛ 0 0 with the left colmn acyclic and P l projective for all l N Since P is projective and Q 0 ɛ N (24) 16

17 f ɛ is an epimorphism, P w N factorizes throgh (24),.. 2 Q 1 P φ 0 Q 0 P 0 ɛ N f M ɛ Since ɛ (φ 0 1 ) = f ɛ 1 = 0, and the left colmn is exact at Q 0, arrow φ 0 1 1, Q 1 w Q 0 factorizes throgh 17

18 φ 1 Q 1 P φ 0 Q 0 P 0 ɛ N f M ɛ Exercise 4 Show that φ 1 2 factorizes throgh Q 2 w Q This indctive procedre yields a morphism from the right colmn to the left colmn. If we denote by P = (P l, l ) l N and Q = (Q l, l ) l N the corresponding chain complices, then we established the first fndamental fact of Homological Algebra. Lemma 2.1 (First Fndamental Lemma) Given a morphism in an abelian category A, and a diagram of chain complices in A N f M Q P ɛ [0]N [0] f ɛ [0]M 18

19 with chain complices P and Q concentrated in nonnegative degrees, all P l being projective, and the left arrow qasiisomorphism, there exists a morphism of chain complices making the diagram φ Q P ɛ [0]N [0] f ɛ [0]M (25) commte In this case we say that φ covers f. The vertical arrows in diagram (25) are represented by the agmentations ɛ of complices P and Q by M and N, respectively. 19

20 2.2 Second Fndamental Lemma Sppose that f = 0. In this case ɛ φ 0 = 0 and φ 0 factorizes niqely, in 0, view of projectivity of P 0, throgh Q 1 w Q Q 1 P 1 \\^ \ χ 0 \ 1 φ 1 Q 0 P 0 φ 0 1 ɛ N 0 M ɛ Exercise 5 Show that (φ 1 χ 0 1 ) = 0 χ 1 and that there exists an arrow P 1 w Q φ 1 = χ χ 1. sch that This indctive procedre yields a seqence of morphisms χ l P l w Q l+1 (l N) sch that and l (φ l χ l 1 l ) = 0 φ l = χ l 1 l l+1 χ l (26) 20

21 for all l > 0. Identity (26) expresses the fact that morphism φ is the bondary of χ, a map of degree +1, in the Hom (P, Q) chain complex φ = Hom χ = [χ, ] Chain homotopy Morphisms of chain complices B φ φ A are said to by homotopic if there exists a degree +1 map χ sch that We may denote this fact by employing notation φ φ = Hom χ = [χ, ]. (27) φ φ. A degree +1 map χ that satisfies dentity (27) is said to be a homotopy from φ to φ. We may denote this fact by employing notation and represent it diagrammatically φ χ φ φ B φ χ A or φ B χ A φ (note what side of the wavy line the label χ is located; the placement indicates that χ is a homotopy from φ to φ ). 21

22 2.2.4 Nll homotopic morphisms A morphism homotopic to the zero morphism is said to be nll homotopic. In this case, χ sch that φ χ 0, φ = Hom χ = [χ, ], is called a contracting homotopy. We shall also say that χ contracts φ or that χ is a contraction of φ, and will represent this diagramatically B φ χ A or B φ χ A Two morphisms are homotopic if and only if their difference is nll homotopic φ We proved above that a morphism of chain complices Q P covers the zero arrow N 0 M the following fact. that is nll homotopic. Its corollary is Lemma 2.2 (Second Fndamental Lemma) Any two morphisms of chain complices that make diagram (25) commte are homotopic. 2.3 Homotopy categories of an abelian category Properties of the homotopy relation The zero arrow of degree +1 from A to B is a homotopy from φ to φ If χ is a homotopy from φ to φ, then it is also a homotopy from φ to φ and χ is a homotopy from φ to φ. 22

23 x If φ χ φ and φ χ φ, then χ + χ is a chain homotopy from φ to φ In particlar, the chain homotopy relation is an eqivalence relation on Hom Ch(A) (A, B). Exercise 6 Show that if then Exercise 7 Sppose that B are morphisms of chain complices and Show that φ 1 χ1 φ 1 and φ 2 χ2 φ 2 β φ 1 + φ 2 χ1 +χ 2 φ 1 + φ 2. B and A α A φ χ φ. φ α χ α φ α and β φ β χ φ. In other words, the chain homotopy relation is a congrence on any fll sbcategory of Ch(A). The corresponding qotient categories are referred to as the homotopy categories of A and are denoted K(A), K + (A), K (A) and so on ( Ch being replaced by K). 2.4 Projective resoltion fnctors A category with sfficiently many projectives We say that A has sfficiently many projectives if, for any object M, there exists an extension ɛ κ M P M 1 (28) with P projective. The kernel arrow in (28) is called a 1-st syzygy of M and M 1 is referred to as a 1-st syzygy object of M. 23

24 2.4.2 In particlar, there exists an extension ɛ 1 κ 1 M 1 P 1 x M 2 and so on. There exists a seqence of extensions ɛ l κ l M l P l x M l+1 (l 0) with M 0 M, P 0 = P, and all P l projective. By splicing them we obtain an acyclic agmented complex where... ɛ M P 0 P 1... (29) l κ l 1 ɛ l (l > 0). It is referred to as an (agmented) projective resoltion of M Syzygy objects The objects M l are referred to as l -th (projective) syzygy objects of M. Exercise 8 Given a projective resoltion P of M, show that Hom A (M l, N) for l 0 Hom Ch(A) (P, [l]n) = 0 for l < 0 (30) Exercise 9 Show that nll-homotopic morphisms P [l]n correspond to morphisms f Hom A (M l, N) which extend to P l 1. In particlar, Hom A (M l, N) for l 0 Hom K(A) (P, [l]n) = { f Hom A (M l, N) which extend to P l 1 } 0 for l < 0 (31) 24

25 2.4.4 According to Lemmata 2.1 and 2.2, any assignment on the class of objects of A, M P of its projective resoltion together with an agmentation P ɛ [0]M admits a niqe extension to an additive fnctor P from A to the homotopy category K 0,{0} (A) eqipped with a natral transformation of fnctors P ɛ [0] which, for each object M, indces an isomorphism of the 0-th homology of PM with M, H 0 (PM) M Moreover any two sch agmented fnctors are isomorphic by a niqe isomorphism of fnctors P P ɛ [0] [0] ɛ (32) 25

26 x x x x 2.5 Third Fndamental Lemma Consider a commtative diagram α β A A A π A ι A B B B π B β ι B C C C π C γ ι C D D with chain complices in colmns, extensions in rows, and a monomorphism in the bottom row, sch that A is projective and the right colmn is exact at C. ι D α β γ In view of projectivity of A, arrow α factorizes throgh π B, α = π B α for some α. Since π C β α = β π B α = β α = 0, exactness of the C-row implies that β α = ι δ 26

27 x x x x x x for a niqe arrow δ A A \ π A A ι A 4 \ 4 \ 4 \ B 4 \ δ β α B 4 \ 4 \ 4 \\] 46 α 4 β C C C π C γ ι C D D ι D α β γ Noting that ι D γ δ = γ β α = 0 and ι D is a monomorphism, we dedce γ δ = 0. In view of projectivity of A and exactness of the right colmn at C, arrow δ factorizes throgh β, δ = β δ for some arrow δ, A A [ A 4 [ 4 [ δ 4 [ B 4 [ δ β α 4 [ B 4 [ 4 [[] 46 α 4 β π A ι A C C C π C γ D D 27 ι C ι D α β γ

28 x In view of projectivity of A, the top row extension is split. Let π a left inverse of ι A. It is a niqe arrow sch that id A ι π A = ι A π where ι is a right inverse of π A. Set α α π A + ι B α π. (33) Exercise 10 Show that the diagram α π A A A A α ι A B B x π B B ι B α commtes and β α = 0. (34) Lemma 2.3 (Third Fndamental Lemma) Given a diagram whose colmns 28

29 x are chain complices.. 2 P 1 Q P 0 Q 0 1 (35) ɛ ɛ π M M M ι with the right colmn acyclic and P l projective for all l N, there exists a complex (R, ) agmented by M which is an extension of agmented complex by agmented complex P ɛ M Q ɛ M 29

30 x x x i.e., diagram (35) can be extended to a commtative diagram of chain complices ɛ π 1 2 P 1 R 1 Q 1 π 0 1 P 0 R 0 Q 0 π M M M ɛ ι 1 ι 0 ι 2 1 ɛ with an extension in each row Exercise 11 Prove the Third Fndamental Lemma. 30

31 3 The cone of a morphism 3.1 Direct sm and matrix calcls A direct sm of a family of objects A direct sm of a family (a i ) i I is, by definition, a coprodct ( c, (ιj ) j I ), (36) eqipped with a second family of morphisms niqely defined by the identities a j ι j c, π i ι j = { idai when i = j 0 when i j (37) Matrix morphisms between direct sms Given a matrix (α ij ) i I,j J of morphisms b i α ij a j sch that the set {i I α ij 0} is finite for every j I, the family of morphisms from a j to d, ι i α ij, i I defines a niqe morphism α from a coprodct c of a family (a j ) j j to a coprodct d of a family (b i ) i I. For obvios reasons, we shall denote this morphism by ι i α ij π j i I and call it the morphism associated with the matrix (α ij ) i I,j J Composition and addition of matrix morphisms corresponds to mltiplication and addition of their matrices. Accordingly, we shall be representing sch morphisms by their matrices and performing all calclations involving the morphisms by employing those matrices. 31

32 3.1.4 From now on we shall adopt the direct sm notation a j (38) j I to denote a direct sm of (a j ) j J eqipped with the two families of morphisms (ι j ) j J and (π j ) j J. Note that (38) shold be treated as a generic notation. 3.2 Cone( f ) Given a map f of degree 0 from a chain complex A to a chain complex B, the direct sm of graded objects B and [1]A, Cone( f ) q B q A q 1, eqipped with the degree 1 morphism ( ) B Cone q f q 1 q 0 q 1 A is called a cone of f. Exercise 12 Show that Cone q Cone q+1 = ( 0 B q f q q 1 A f ) q (39) In particlar, Cone Cone = 0 if and only if [ f, ] = By definition, a cone of the zero morphism Cone(0) is a direct sm of B and [1]A in the category of chain complices. The cone of a morphism of chain complices is ths an infinitesimal deformation of B [1]A. 32

33 3.2.3 The Cone Extension Exercise 13 Show that the maps [1]A π [1]A B [1]A and B [1]A B ι B are morphisms of chain complices [1]A Cone( f ) and Cone( f ) B. Since the composable pair ι [1]A Cone( f ) B B (40) π [1]A is in every degree a split extension associated with a direct sm B q A q 1, it is an extension in the category of chain complices. We shall refer to (40) as the cone extension A particlarly important featre of the cone extension is that the morphisms H q ( f ) indced by f in homology coincide with the connecting homomorphisms δ q+1 of the homology long exact seqence associated with extension (40) The cone fnctor If we fix a binary direct sm fnctor on the nderlying category A, we obtain a fnctor, denoted Cone, from the category of arrows Arr Ch(A) to Ch(A). Exercise 14 Indeed, given a sqare of morphisms of complices A φ ss A (41) f f B φ tt B show that [φ, Cone ] = ( 0 φtt f f ) φ ss

34 where φ ( ) φtt 0 0 φ ss (42) In particlar, (42) is a chain complex morphism if and only if sqare (41) commtes Note that where [1]A = Cone ( 0 A) and B = Cone ( 0 B ), 0 A 0 A and 0 B B 0, and the commtative diagram A A 0 0 B B is an extension in the category of arrows Arr Ch(A), f 0 A f x 0 B (43) which yields the Cone extension, cf. (40), when one applies the cone fnctor to (43) If we denote by 0 and 0 the fnctors Ch(A) Arr Ch(A) that send a complex C to 0 C and, respectively, 0 C, then extensions (43) give rise to an extension of fnctors on Arr Ch(A), 0 s id Arr Ch(A) x 0 t where s and t denote the sorce and the target fnctors from Arr Ch(A) to Ch(A). 34

35 w w A morphism of extensions in Ch(A) π A ι A A A A f f f B B B π B ι B indces an extension of the corresponding cones π ι, Cone( f ) Cone( f ) Cone( f ) the cone extension, (40), for example, being indced by the morphism of trivial extensions A A 0 f 0 B B It follows that Cone indces a fnctor from the category of arrows of the category of extensions of chain complices to the category of extensions of chain complices Arr Ext Ch(A) Cone Ext Ch(A). In particlar, Arr Ch(A) Cone Ch(A) is an exact fnctor. We shall see an explanation of this fact later Consider the extension [1]A Cone(ι B ) Cone ( ) id B (44) 35

36 indced by the morphism of extensions 0 B B (45) ι B [1]A Cone( f ) π [1]A Exercise 15 Find an explicit isomorphism ι B B Cone(ι B ) Cone ( id B ) [1]A by finding first a splitting of extension (44) in the category of chain complices Consider the extension Cone ( id [1]A ) Cone ( ) π [1]A [1]B (46) indced by the morphism of extensions π [1]A [1]A Cone( f ) ι B B (47) π [1]A [1]A [1]A 0 Exercise 16 Find an explicit isomorphism Cone ( π [1]A ) [1]B Cone ( id[1]a ) by finding first a splitting of extension (46) in the category of chain complices. 3.3 The cone diagram associated with an anticommtative sqare An anticommtative sqare of chain complices A 01 g 1 A 11 (48) f 0 + f 1 A 00 A 10 g 0 36

37 is a 3-complex with A pqr = A pq,r (p, q, = 0, 1, r Z), whose bondary operators in p-direction are provided by g p, in q-direction by f q, and in r-direction by the bondary operators of the corresponding complices Its total complex in degree n is the direct sm A 00,n A 10,n 1 A 01,n 1 A 11,n 2 eqipped with the bondary operators 00 n g 0,n 1 f 0,n 1 10 n 1 01 n 1 f 1,n 2 g 1,n 2 11 n 2 (49) By exchanging A 10 and A 01 in the direct sm, one obtains an alternate form of the matrix representation of the bondary operator 00 n f 0,n 1 g 0,n 1 10 n 1 g 1,n 2 (50) 01 n 1 f 1,n n 2 Let s denote the total complex of (48) as Cone( ). Formlae (49) (50) mean that it is canonically identified with Cone(f) and Cone(g) where f ( f0 = [1] f1) ( f0 1 f 1 ) 37

38 is the indced morphism between the g-cones Cone(g 0 ) Cone(g 1 ) and ( g0 g = [1]g1) ( g0 is the indced morphism between the f -cones 1g 1 ) Cone( f 0 ) Cone( f 1 ). Here and below l f denotes ( 1) l [l] f. This, as we shall soon see, is a proper way to define the shift fnctors on the category of arrows Arr Ch(A). The main advantage at this point of sing verss [ ], is that Cone ( l f ) = [l] Cone( f ). A conseqence of this observation is the existence of the following diagram [1]A 11 Cone(g 1 ) A 01 g 1 A 11 (51) [1] f 1 f f 0 + f 1 [1]A 10 g 0 Cone(g 0 ) A 00 A 10 [1] Cone( f 1 ) Cone( ) Cone( f 0 ) Cone( f 1 ) g [1]g 1 [2]A 11 [1] Cone(g 1 ) [1]A 01 [1]A 11 38

39 or, sing the notation, [1]A 11 Cone(g 1 ) A 01 g 1 A 11 (52) 1 f 1 f f 0 + f 1 [1]A 10 g 0 Cone(g 0 ) A 00 A 10 [1] Cone( f 1 ) Cone( ) Cone( f 0 ) Cone( f 1 ) g g 1 [2]A 11 [1] Cone(g 1 ) [1]A 01 [1]A 11 All sqares commte except the original one that generated the whole pictre, and located in the right top corner. That single sqare anticommmtes The diagram contains 8 extensions. In the left 3 2 sbdiagram all 4 rows are the cone extensions associated with the 4 morphisms on the right. In the bottom 2 3 sbdiagram all 4 colmns are the cone extensions that are similarly associated with the 4 morphisms at the top of the diagram One shold think of the 3 3 sbdiagram of extensions located in the bottom left corner as being the 2-dimensional version of the cone extension. 39

40 The latter relates the cone of an arrow (a 1-dimensional cell of the category of complices) to its target and sorce (0-dimensional faces ). The former relates the cone of an anticommtative sqare Cone( ) to the cones of its 1- and 0-dimensional faces, if we agree to consider the cone fnctor on 0-diemsional cells, i.e., objects of Ch(A), to be the identity fnctor Cone 0 id Ch(A) n-dimensional cone fnctors Totalization of the n-dimensional cbe involving 2 n chain complices with anticommting 2-dimensional faces can be regarded as the n-dimensional cone fnctor. For n = 1, we obtain the original cone fnctor on arrows of Ch(A), for n = 0, we obtain the identity fnctor on Ch(A) The role of the n-dimensional cone extension is played by the n-dimensional diagram involving 3 n complices forming 3 n 1 n one-dimensional cone extensions Cone(g f ) Given a composable pair of morphisms of chain complices C g B f A the above 4 4-diagram associated with the anticommtative sqare A id A A g f + f C g B 40

41 has the following form [1]A Cone( id A ) A id A A (53) [1] f f g f + f [1]B Cone(g) C g B [1] Cone( f ) Cone( ) Cone(g f ) Cone( f ) g [2]A Cone ( id [1]A ) [1]A id [1]A [1]A where ( ) ( g f g f f = [1] f is a morphism between the cones 1 f ) (54) and Cone(g) Cone( id A ) ( g g id [1]A is a morphism between the cones ) ( g = 1 ida) Cone(g f ) Cone( f ). 41

42 One can consider diagram (53) as expressing the relation between the cone of a composable pair of arrows (a 2-simplex of the category Ch(A)), to the cones of its 1-dimensional faces f, g and g f. In this respect the most important are the second row from the bottom and the second colmn from the left. This is the essence of the so called Octahedron Axiom of trianglated categories. Exercise 17 Find an explicit isomorphism Cone( ) Cone ( id [1]A ) Cone(g) by finding first a splitting of the cone extension of f, given by (54), Cone(g) Cone( ) in the category of chain complices. Cone ( id [1]A ) 3.4 Morphisms Cone( f φ ) Cone( f ) Given a pair of arrows in the category of chain complices A A f B a morphism Cone( f φ ) Cone( f ) φ = B ( ) φtt φ ts φ st φ ss f is represented by a matrix (55) 42

43 or, diagramatically, by the sqare with 6 arrows A φ ss A (56) φ st f f B φ ts φ tt B where φ st is of degree 1, arrows φ tt and φ ss are of degree 0, and φ ts is of degree +1. Exercise 18 Show that [ φ, Cone ] = [φ tt, ] f φ st φ tt f f φ ss [φ ts, ] (57) [φ st, ] φ st f [φ ss, ] where [, ] denotes the spercommtator of graded maps The meaning of the condition [ φ, Cone] = 0 The integrability condition [ φ, Cone ] = 0 translates into 4 separate conditions The st-condition says that φ st is a morphism of chain complices [1]A φ st. B The tt-condition says that φ tt is a contracting homotopy for f φ st while the ss-condition says that φ ss is a contracting homotopy for φ st f. Dia- 43

44 grammatically, A φ ss and A A f B φ tt φ st B φ st f B It follows that both φ tt f and f φ ss contract the triple composite f φ st f. Finally, the ts-condition says that φ ts is a homotopy from f φ ss to φ tt f. Diagrammatically, A φ ss A f B φ tt φ ts B f 3.5 The matrix homotopy category of arrows M(A) Define M(A) to be the category whose objects are arrows of Ch(A) and a morphism φ from f to f consists of a morphism from the target of the sorce arrow to the sorce of the target arrow (shifted by 1), υ : t( f ) [1]s( f ), a pair of homotopies ϕ t and ϕ s contracting f υ and υ f, respectively, and a frther homotopy ψ between the homotopies f ϕ s B ψ ϕ t f A 44

45 3.5.2 We shall refer to ϕ t and ϕ s as the primary homotopies, and to ψ as the secondary homotopy Fixing A, B, A and B, yields a fll small sbcategory M(A) B A BA which besides being preadditive has an abelian grop strctre on the set of objects The category of arrows UT(A) The category of arrows and commtative sqares UT(A) is a sbcategory of M(A) with morphisms being precisely the qadrples (0; ϕ t, ϕ s ; ψ) Note that the set of morphisms f φ f with fixed (υ; ϕ t, ϕ s ) is a torsor over the grop Hom Ch(A) (A, [ 1]B) while the set of morphisms in M(A) with fixed υ is a torsor over Hom UT(A) ( f, f ) The shift fnctors If we define the shift fnctors l on M(A) by the correspondence f A B [l]a [l]b ( 1) l f (58) on objects, and the correspondence (υ; ϕ t, ϕ s ; ψ) ([l]υ; [l]ϕ t, [l]ϕ s ; [l]ψ) (59) 45

46 on morphisms, then the correspondence f Cone( f ) on objects and (υ; ϕ t, ϕ s ; ψ) on morphisms, defines an epifnctor, ( ) ϕt ψ υ ϕ s Cone : M(A) Ch(A), (60) i.e., a fnctor that is srjective on the class of objects and on the class of arrows, which commtes with the shift fnctors, Cone(l f ) = [l] Cone( f ). This fnctor extends the cone fnctor from sbcategory Arr Ch(A) Caveat Note that l f = ( 1) l [l] f. On the left hand side f is an object of M(A), on the right hand side f is a morhism of Ch(A). The sign difference is necessitated by the following considerations. Each of the shift fnctors mst preserve the identity morphisms, as a conseqence the shift fnctors cannot reverse the sign of morphisms of shifted objects. At the same time shifts of morphisms in M(A) remain morphisms only if the shifts of arrows consider as objects of M(A), change sign simltanesly with the bondary operators. This is compatible with the fact that objects of M(A) are constitent parts of the bondary operators of their cones The category of arrows and homotopy commtative sqares UT(A) Morphisms with υ = 0 are triples (φ t, φ s ; ψ) that describe homotopy commtative sqares A φ s A f B φ t ψ B f 46

47 They correspond to those matrices (55) that are pper trianglar (note that morphisms in Arr Ch(A) correspond to diagonal matrices) The special case A A (61) f B corresponds to ψ being a homotopy from f to f or, eqivalently, ψ being a homotopy from f to f. ψ B f Exercise 19 Show that A A f B ψ B f is the inverse of (61) It follows that the cones of homotopic morphisms of chain complices are isomorphic in Ch(A), an isomorphism fnctorially depending on a particlar homotopy from f to f. 3.6 Morphisms between 0 A and 0 B We shall investigate morphisms in M(A) and in UT(A) between arrows whose sorce or target is 0. From now on we adopt the convention that the component arrows of a morphism not indicated in the morphism diagram are tacity assmed to be zero. 47

48 3.6.2 Note that A 0 υ 0 B is a morphism 0 B 0 A in M(A) precisely when υ is a morphism B [1]A in Ch(A). In particlar, Hom M(A) (0 B, 0 A ) Hom Ch(A) (B, [1]A) and Hom UT(A) (0 B, 0 A ) = Similarly, 0 A ψ B 0 is a morphism 0 A 0 B in M(A) precisely when ψ is a morphism [1]A B in Ch(A). In particlar, Hom M(A) (0 A, 0 B ) = Hom UT(A) (0 B, 0 A ) Hom Ch(A) ([1]A, B) In the category of arrows one has, of corse, Hom Arr Ch(A) (0 B, 0 A ) = 0 and and Hom Arr Ch(A) (0 B, 0 A ) = 0. 48

49 3.6.5 Note that Hom M(A) (0 A, 0 A ) = Hom Arr Ch(A) (0 A, 0 A ) Hom Ch(A) (A, A ) and Hom M(A) (0 B, 0 B ) = Hom Arr Ch(A) (0 B, 0 B ) Hom Ch(A) (B, B ). 3.7 Adjnctions The fnctor identities Cone 0 = id Ch(A) = [0] and Cone 0 = [1] mean that 0 is a right inverse of Cone while 0 is a right inverse of [ 1] Cone = Cone 1. In particlar, 0 and 1 0 are both right inverses of the cone fnctor The canonical identification Hom M(A) ( f, f ( ) w Hom Ch(A) Cone( f ), Cone( f ) ) means that 0 and 1 0 are both right and left adjoint to Cone This, in trn, means that 0 and 1 0 are isomorphic as fnctors Ch(A) M(A). There exists a canonical isomorphism

50 It is provided by the families of mtally inverse natral transformations given by the diagrams [ 1]C id 0 0 C and 0 [ 1]C id C 0 (62) Exercise 20 Show that the two morphisms in (62) are mtally inverse Isomorphism 0 Cone id M(A) Exercise 21 Show that the diagrams A 0 and 0 A (63) f π [1]A B Cone( f ) π B Cone( f ) ι [1]A ι B B f define morphisms in M(A) and that these morphisms are mtally inverse. It follows that the cone fnctor provides an eqivalence of categories between M(A) and Ch(A), with the fnctor 0 (or isomorphic to it fnctor 1 0 ) providing an inverse eqivalence Noting that Hom UT(A) ( f, 0 C ) = Hom M(A) ( f, 0 C ), bt Hom UT(A) (0 C, f ) Hom M(A) (0 C, f ), and Hom UT(A) (0 C, f ) = Hom M(A) (0 C, f ) bt Hom UT(A) ( f, 0 C ) Hom M(A) ( f, 0 C ), we observe that if we consider Cone as a fnctor UT(A) Ch(A), then 0 is its right bt not left adjoint while [ 1] 0 is its left bt not right adjoint. 50

51 x 3.8 Graded-split extensions An extension of chain complices π C C C ι (64) is said to be graded-split if it is split in the category of Z-graded objects, i.e., in every degree is a split extension in A If ι : C C is a map of degree 0 that spplies a right inverse to π in the category of graded objects A Z, then let π : C C be the corresponding graded projection onto C. Recall that it is defined by the identity The qartet id C ι π = ι π. C ι C π ι π C represents C as a direct sm of C and C in the category of Z-graded objects Graded-split epimorphisms and monomorphisms Morphisms that occr as π in a graded-split extension (64) are called graded-split epimorphisms. Morphisms that occr as ι are called graded-split monomorphisms. The former are precisely those morphisms that admit a graded right inverse ι, while the latter are those morphisms that admit a graded left inverse π Exercise 22 Show that In particlar, π [ι, ] = 0. [ι, ] = ι φ 51

52 x for a niqe morphism of chain complices C φ w [1]C. (65) We can represent (65) as [1] f = 1 f for f [ 1]C [ 1]φ w C. Exercise 23 Show that [ 1]C 0 and 0 [ 1]C (66) f C π π C ι C C ι f define morphisms in M(A) and that these morphisms are mtally inverse It follows that extension (64) is isomorphic to the cone extension of f π C C C ι [1] ( [ 1]C ) Cone( f ) x C A graded splitting σ [1]A w Cone( f ) of the cone extension is represented by the 2 1 matrix ( ) h id [1]A for some map h : [1]A B of degree 0. 52

53 Exercise 24 Show that [σ, ] is represented by the matrix ( ) [h, ] f 0 (67) i.e., σ is a morphism of complices if and only if h is a homotopy that contracts f In particlar, the cone extension of f is split if and only if f is nllhomotopic, and the set of splittings is in a natral bijective correspondence with the set of homotopies contracting f. 53

54 4 Contractible complices 4.1 Basic properties A complex C is said to be contractible if id C is nll-homotopic, i.e., if id C = [h, ] (68) for some degree +1 graded endomorphism h of C. The latter is referred to as a contracting homotopy or a contraction of C It follows from the discssion in Section that a complex C is contractible if and only if the cone extension splits. π [1]C [1]C Cone(id C ) x C Exercise 25 If h is a contracting homotopy, then the following identities hold (h ) 2 = h, ( h) 2 = h and h 2 = 0. (69) Direct sms of contractible complices A direct sm of a family (C i ) i I of contractible complices is contractible with ι i h i π i i I being a contracting homotopy where h i is a homotopy that contracts complex C i Direct smmands of contractible complices An object a in a preadditive category is said to be a direct smmand of an object a if there exist morphisms ι C a ι π a sch that π ι = id a. We shall refer to ι as the inclsion into a and to π as the projection onto a. 54

55 Exercise 26 If C is a direct smmand of a contractible complex C with a contracting homotopy h, then h ι h π contracts C. It is integrable if h is integrable. 4.2 Characterizations of contractible complices Example: Cone(id C ) Exercise 27 Show that contracts Cone(id C ). h ( 0 ) 0 id C 0 (70) Note that the id C in (70) occrs as a degree +1 map from C to [1]C The first characterization of contractible complices. By combining this with the reslt of Section 4.1.2, we obtain the following characterization of contractible complices in any additive actegory. Lemma 4.1 A complex is contractible if and only if it is a direct smmand of Cone(id A ) for some complex A The second characterization of contractible complices. All the considerations of the chapter devoted to the cone constrction so far were valid in any additive category. Under an additional hypothesis we shall establish that a contractible complex C is isomorphic to Cone(id ZC ) for the complex of cycles of C, the latter being eqipped with zero bondary operators. Proposition 4.2 Let C be a complex sch that kernels and coimages of C exist. If C is contractible, then it is isomorphic to Cone(id ZC ). Proof. Under the hypothesis, C is an extension of the complex B C of co-bondaries by the complex of cycles B β ζ C C x ZC, (71) 55

56 both eqipped with zero bondary operators. Given a contraction h of C, the graded map h niqely factorizes throgh β, and h = σ β β σ β = β h = β (h + h) = β id C = id B C β demonstrates, in view of β being epi, that β σ = id B C, i.e., σ is a graded splitting of extension (71) By calclating [σ, ] β = (σ B C C σ) β = h = σ β, we determine that extension (71) is isomorphic to the cone extension B C Cone( f ) x CZ for f being the niqe morphism from [ 1]B C to ZC sch that Note that ζ [1] f = σ. ζ [1] f ( β h ζ) = h h ζ = h ζ = ( h + h ) ζ = ζ implies, in view of ζ being mono, that Similarly, [1] f ( β h ζ) = id [1]ZC. ( β h ζ) [1] f β = β h σ β = β h h = β h = β (h + h) = β 56

57 implies, in view of β being epi, that ( β h ζ) [1] f = id B C. Ths, [1] f is an isomorphism between B C and [1]ZC. This, in trn, indces an isomorphism between Cone( f ) and Cone(id ZC ) ZC f [ 1]B C ZC f ZC which fits into the following diagram of isomorphisms of extensions B β ζ C C x ZC B C Cone( f ) x ZC [1]ZC Cone(id ZC ) x ZC [1] f 4.3 Characterization of nll-homotopic morphisms Exercise 28 If a morphism f factorizes throgh a complex C with a contractible homotopy h, B f A f C f 57

58 then is a contracting homotopy for f. f h f Vice-versa, if h is a homotopy that contracts a morphism f, then the diagram of composable morphisms in M(A) 0 A 0 (72) B f h A id A A represents a canonical factorization of f throgh Cone(id A ) Alternatively, the diagram 0 id [ 1]B B [ 1]B A h represents a canonical factorization of f throgh Cone(id B ). f 0 (73) Exercise 29 Explain why diagram (73) represents morphisms in M(A) Note that factorization of f throgh Cone(id A ) is realized via morphisms in UT(A). The two factorizations are eqivalent in the sense that they are obtained from a single triple factorization in M(A) 0 id h [ 1]B A 0 B [ 1]B A h f id A A (74) 58

59 by composing the first two, or the last two morphisms in M(A) We arrive at the following characterization of nll-homotopic morphisms in Ch(A). Lemma 4.3 A morphism of complices is nll-homotopic if and only if it factorizes throgh a contractible complex A very important conseqence is that the homotopy category of A is a qotient of Ch(A) by a sbcategory of contractible complices. 4.4 Graded-split projective and graded-split injective complices We say that a complex P is graded-split projective if it has the Lifting Property for the class of graded-split epimorphisms Graded-split injective complices are defined dally as those complices that have the Extension Property for the class of graded-split monomorphisms Consider a diagram Cone(id C ) g [1]A Cone( f ) π [1]A 59

60 The two arrows are represented by the following morphisms in M(A) A A and A ϕ C f υ 0 B 0 C and A A ϕ C f υ 0 B C f ϕ represents a canonical factorization of g throgh Cone( f ) Dally, consider a diagram Cone( f ) i B B υ Cone(id C ) The two arrows are represented by the following morphisms in M(A) A f 0 and C 0 υ B B C ϕ B and ϕ f C A 0 C ϕ υ f represents a canonical factorization of g throgh Cone( f ). B B 60

61 4.4.5 According to Section every graded-split epimorphism is isomorphic to [1]A π [1]A Cone( f ), for some morphism of complices f, and every graded-split monomorphism is isomorphic to Cone( f ) i B, ths we arrive at the following characterization of graded-split projective and graded-split injective complices. Lemma 4.4 The following three properties of a chain complex are eqivalent (a) C is graded-split projective; (b) C is graded-split injective; B (c) C is contractible. 4.5 Contractions and higher homotopies Integrable contracting homotopies Let h be a contracting homotopy of a contractible complex C. Exercise 30 Show that also is a contracting homotopy. h h h (75) Exercise 31 Show that h defined in (75) satisfies the identity h h = 0. (76) A contracting homotopy satisfying (76) will be said to be integrable. 61

62 4.5.2 The set of integrable contracting homotopies In Exercises 30 and 31 we established that, for every contractible complex C, the set of integrable contracting homotopies is nonempty. The set of all contracting homotopies is a torsor over the grop Hom Ch(A) (C, [ 1]C), i.e., the difference between any two contracting homotopies is a morphism from C to [ 1]C. h h = φ Exercise 32 Sppose that h is an integrable contracting homotopy. Show that h + φ is integrable if and only if the morphism φ : C [ 1]C satisfies the Marer-Cartan eqation [φ, h] + φ 2 = 0. (77) The integrability property of a contracting homotopy is eqivalently expressed ia the identity ( + h) 2 = id C. (78) When performing calclations involving spercommtators one is freqently sing the spercommtator Leibniz Rle. Exercise 33 (Spercommtator Leibniz Rle) Prove the following identity valid in any associative ring-type strctre involving even and odd elements [ab, c] = a[b, c] + ( 1) b c [a, c]b (79) where ã is the parity of a, nderstood to be an element of Z/2Z. Exercise 34 Show that [h n, ] = 0 if n is even. (80) h n 1 if n is odd 62

63 4.5.5 Ths, for any contracting homotopy, h 2 : C [ 2]C is a morphism of chain complices. Exercise 35 Show that h 2 is nll-homotopic and find a homotopy contracting h 2. Exercise 36 If h 3 is a graded endomorphism of C of degree 3 sch that then [h 3, ] + h 2 = 0, [h 3, h] is a morphism of complices C [ 4]C The spercommtator [h 3, h] is, in fact, again nll-homotopic Consider a formal series h h 1 + h 3 + h 5 + (81) where h i is a a degree i graded endomorphism of C, i.e., a map C C of degree i. The eqation expresses the infinite seqence of eqations id C = [h 1, ] ( + h) 2 = id C (82) 0 = [h 3, ] + h = [h 5, ] + [h 3, h 1 ] (83) 0 = [h 2n+1, ] + h i h j i+j=2n 1 i, j odd 63

64 or, eqivalently, of chain homotopies id C h1 0 0 h3 h h5 [h 3, h 1 ] (84) 0 h2n+1 h i h j i+j=2n 1 i, j odd A contraction h of C is integrable precisely when is a soltion of eqation (82) h = h We shall demonstrate that, for any contraction h of C, there exists a seqence of higher homotopies h 3, h 5,..., sch that formal series (81), with h 1 = h, is a soltion of eqation (82). We shall seek the soltion in the form h = F(h) where F(h) is a formal power series in h F(h) = h + i>2, odd a i h i (a i Z). The seqence of eqations (83), for i > 1, becomes the seqence of eqations in nknown coefficients a i, 0 = a 2n+1 + a i a j (n > 0) (85) i+j=2n 1 i, j odd which are eqivalent to a single fnctional eqation 0 = F h h 64 + F 2

65 or, eqivalently, The sole soltion of (82) of the form hf 2 + F h = 0. (86) F(h) = h + is given by the Taylor power series expansion at 0 of the fnction F(x) x (2x) 2 2x = x + n>1 ( 2) n 1 (2n 3)!! x 2n 1 n! = x x 3 + 2x 5 5x x 9 +. (87) A degree 3 graded map h 3 that satisfies the second eqation in (83) differs from h 3 by a morphism of chain complices h 3 : C [ 3]C, Exercise 37 Show that satisfies the third eqation in (83) (2n 1)-contractions Let s call h 3 = h 3 + h 3. h 5 2h 5 + h h 3 h h 2n 1 = h 1 + h h 2n 1 a (2n 1)-contraction of a complex C if the first n eqations (83) are satisfied A homotopy contraction of C is the same as a 1-contraction. We showed above that any 1-contraction extends to an -contraction Qestion Does any (2n 1)-contraction extend to a (2n + 1)-contraction? Above we showed that the answer is positive for 1- and 3-contractions. 65

66 5 Contractions of Cone( f ) and homotopy eqivalences 5.1 Nll-homotopic morphisms Cone( f φ ) Cone( f ) A degree +1 graded map Cone( f χ ) Cone( f ) a matrix χ = ( χtt χ ts χ st χ ss with entries being degree +1 graded maps ) is represented by. (88) B χ tt B, B [1]A χ ts, and [1]A χ ss. χ st B, [1]A [1]A Given a nll-homotopic morphism Cone( f φ ) Cone( f ) by a homotopy χ, the identity contracted φ = [χ, ] translates into 4 identities: φ tt = f χ st + [χ tt, ], φ ts = f χ ss χ tt f + [χ ts, ] (89) and φ st = [χ st, ], φ ss = χ st f + [χ ss, ]. (90) In those 4 identities we take into accont that, in terms of the inpt complices A, B, A and B, the off-diagonal entries of (88) have degrees 0 and 2, and ths are even, while the diagonal entries have degree 1, and ths are odd. This affects the signs in the correspoding sppercommtators. 66

67 5.2 Nll-homotopic morphisms in M(A) The above identities translate into the following notion of homotopy in the matrix homotopy category M(A). A morphism (υ; ϕ t, ϕ s ; θ) is nllhomotopic precisely when the following occrs: υ is nll-homotopic in Ch(A), i.e., υ η 0, (91) for some map η of degree 0 from B to A, the primary homotopies ϕ t and ϕ s are homotopic to f η and η f, respectively, ϕ t χt f η and ϕ s χs η f, (92) and the secondary homotopy ψ is homotopic to f χ s χ t f, ψ θ f χ s χ t f. (93) A qartet (η; χ t, χ s ; θ) will be referred to as a contracting homotopy for a morphism (υ; ϕ t, ϕ s ; ψ) in M(A) Homotopy split commtative sqares We shall say that a strictly commtative sqare A ϕ s A (94) f f B ϕ t B is homotopy split if there exists a morphism of chain complices η sch that both triangles in A f B ϕ s A (95) ϕ t are homotopy commtative. A homotopy splitting of the commtative sqare consists of the morphism η and the corresponding homotopies ϕ t χt f η and ϕ s χs η f. η B f 67

68 5.2.3 The reslting 2 homotopies between ϕ t f = f ϕ s and f η f may or may not be homotopic. If they are, and χ t f θ f χ s, we say that a homotopy splitting is strong and the qartet (η; χ t, χ s ; θ) is then referred to as a strong homotopy splitting data. If the sqare admits a strong homotopy splitting, we say that it is strongly homotopy split Commtative sqares nll-homotopic in M(A) A commtative sqare is nll-homotopic in M(A) precisely when it is strongly homotopy split Providing a contracting homotopy for (0; ϕ t, ϕ s ; 0) in M(A) is eqivalent to spplying the splitting morphism η, a pair of primary homotopies χ t and χ s between ϕ t and f η, and between ϕ s and η f and, finally, a secondary homotopy θ between f χ s and χ t f. 5.3 Nll-homotopic morphisms in UT(A) In the category of arrows and homotopy commtative sqares UT(A), a morphism (ϕ t, ϕ s ; θ) is nll-homotopic if ϕ t and ϕ s are nll-homotopic, ϕ t χt 0 and ϕ s χs 0, (96) and the homotopy ψ is homotopic to f χ s χ t f, exactly as in the case of M(A), cf. (93) In UT(A) a commtative sqare is nll-homotopic precisely when both ϕ t and ϕ s are nll homotopic and the reslting 2 homotopies that contract ϕ t f = f ϕ s are themselves homotopic. In particlar, a morphism nll-homotopic in M(A) is generally not nll-homotopic in UT(A). 68

69 5.4 Arrows contractible in M(A) Let s call an object f of M(A) contractible if id f when the sqare A A is nll-homotopic, i.e., (97) f B is strongly homotopy split. Identities (89) (90) become in this case and where [ f, χ] denotes B f id B f η = [χ t, ], 0 = [ f, χ] + [θ, ] (98) 0 = [η, ], id A η f = [χ s, ]. (99) f χ s χ t f A homotopy eqivalence data in Ch(A) A pair of morphisms of chain complices together with a pair of homotopies f A B (100) g id B hb f g and id A ha g f will be referred as a homotopy eqivalence (data) between complices A and B. By definition, A and B appear in it on eqal footing. If complices A and B admit sch data, we say that they are homotopy eqivalent. 5.5 Homotopy eqivalence of chain complices We say that a morphism f is a homotopy eqivalence if there exist g, h B and h A, that form a homotopy eqivalence data. In this case g is said to be a 69

70 x homotopy inverse of f. These are precisely the morphisms in Ch(A) that correspond to isomorphisms in the homotopy category K(A). In contrast with the definition of a homotopy eqivalence data where f and g appear on eqal footing, f is primary data while the homotopy inverse g and the two homotopies are secondary. One needs to be aware of these two related bt different ses of the term homotopy eqivalence This is very similar to the two ses of the term eqivalence of categories If f is contractible in M(A), then f is a chain homotopy eqivalence. More precisely, if (η; χ t, χ s ; θ) contracts f in M(A), then η is a homotopy inverse of g while χ t and χ s are the corresponding homotopies h B and h A. A homotopy eqivalence data with fixed f is the same as a homotopy splitting of (97) while a contraction of f in M(A) is a strong homotopy splitting. Exercise 38 Show that the monomorphism A B ι B A is a homotopy eqivalence if A is a contractible complex We shall now investigate the strctre of a homotopy eqivalence in greater detail. We start by making a few observations. Exercise 39 Show that Ths, [ f h A h B f, ] = 0 and [g h B h A g, ] = 0. (101) [ f, h] f h A h B f is a morphism of complices. A given homotopy eqivalence can be extended to a contraction of f in M(A) precisely when [ f, h] is nllhomotopic in Ch(A). 70