Derived Categories

Home , Derived category, Derived functor, Exact functor, Homotopy category, Injective object, Triangulated category

DERIVED CATEGORIES

05QI

Contents 1. Introduction 2 2. Triangulated categories 2 3. The deﬁnition of a triangulated category 2 4. Elementary results on triangulated categories 5 5. Localization of triangulated categories 13 6. Quotients of triangulated categories 19 7. Adjoints for exact functors 24 8. The homotopy category 26 9. Cones and termwise split sequences 26 10. Distinguished triangles in the homotopy category 33 11. Derived categories 37 12. The canonical delta-functor 39 13. Filtered derived categories 42 14. Derived functors in general 45 15. Derived functors on derived categories 52 16. Higher derived functors 55 17. Triangulated subcategories of the derived category 59 18. Injective resolutions 62 19. Projective resolutions 67 20. Right derived functors and injective resolutions 69 21. Cartan-Eilenberg resolutions 71 22. Composition of right derived functors 72 23. Resolution functors 73 24. Functorial injective embeddings and resolution functors 75 25. Right derived functors via resolution functors 77 26. Filtered derived category and injective resolutions 77 27. Ext groups 85 28. K-groups 89 29. Unbounded complexes 90 30. Deriving adjoints 93 31. K-injective complexes 94 32. Bounded cohomological dimension 97 33. Derived colimits 99 34. Derived limits 103 35. Operations on full subcategories 104 36. Generators of triangulated categories 106 37. Compact objects 108 38. Brown representability 110

This is a chapter of the Stacks Project, version fac02ecd, compiled on Sep 14, 2021. 1 DERIVED CATEGORIES 2

39. Admissible subcategories 112 40. Postnikov systems 114 41. Essentially constant systems 119 42. Other chapters 121 References 122

1. Introduction 05QJ We ﬁrst discuss triangulated categories and localization in triangulated categories. Next, we prove that the homotopy category of complexes in an additive category is a triangulated category. Once this is done we deﬁne the derived category of an abelian category as the localization of the homotopy category with respect to quasi-isomorphisms. A good reference is Verdier’s thesis [Ver96].

2. Triangulated categories 0143 Triangulated categories are a convenient tool to describe the type of structure inherent in the derived category of an abelian category. Some references are [Ver96], [KS06], and [Nee01].

3. The definition of a triangulated category 05QK In this section we collect most of the definitions concerning triangulated and pre- triangulated categories. 0144Definition 3.1. Let D be an additive category. Let [n]: D → D, E 7→ E[n] be a collection of additive functors indexed by n ∈ Z such that [n] ◦ [m] = [n + m] and [0] = id (equality as functors). In this situation we define a triangle to be a sextuple (X,Y,Z,f,g,h) where X,Y,Z ∈ Ob(D) and f : X → Y , g : Y → Z and h : Z → X[1] are morphisms of D.A morphism of triangles (X,Y,Z,f,g,h) → (X0,Y 0,Z0, f 0, g0, h0) is given by morphisms a : X → X0, b : Y → Y 0 and c : Z → Z0 of D such that b ◦ f = f 0 ◦ a, c ◦ g = g0 ◦ b and a[1] ◦ h = h0 ◦ c. A morphism of triangles is visualized by the following commutative diagram

X / Y / Z / X[1]

a b c a[1] X0 / Y 0 / Z0 / X0[1] Here is the deﬁnition of a triangulated category as given in Verdier’s thesis.

0145Deﬁnition 3.2. A triangulated category consists of a triple (D, {[n]}n∈Z, T ) where (1) D is an additive category, (2) [n]: D → D, E 7→ E[n] is a collection of additive functors indexed by n ∈ Z such that [n] ◦ [m] = [n + m] and [0] = id (equality as functors), and (3) T is a set of triangles called the distinguished triangles subject to the following conditions DERIVED CATEGORIES 3

TR1 Any triangle isomorphic to a distinguished triangle is a distinguished triangle. Any triangle of the form (X,X, 0, id, 0, 0) is distinguished. For any morphism f : X → Y of D there exists a distinguished triangle of the form (X,Y,Z,f,g,h). TR2 The triangle (X,Y,Z,f,g,h) is distinguished if and only if the triangle (Y,Z,X[1], g, h, −f[1]) is. TR3 Given a solid diagram

f g h X / Y / Z / X[1]

a b a[1] 0 0 f g h0 X0 / Y 0 / Z0 / X0[1] whose rows are distinguished triangles and which satisﬁes b ◦ f = f 0 ◦ a, there exists a morphism c : Z → Z0 such that (a, b, c) is a morphism of triangles. TR4 Given objects X, Y , Z of D, and morphisms f : X → Y , g : Y → Z, and distinguished triangles (X,Y,Q1, f, p1, d1), (X,Z,Q2, g ◦ f, p2, d2), and (Y,Z,Q3, g, p3, d3), there exist morphisms a : Q1 → Q2 and b : Q2 → Q3 such that (a) (Q1,Q2,Q3, a, b, p1[1] ◦ d3) is a distinguished triangle, (b) the triple (idX , g, a) is a morphism of triangles (X,Y,Q1, f, p1, d1) → (X,Z,Q2, g ◦ f, p2, d2), and (c) the triple (f, idZ , b) is a morphism of triangles (X,Z,Q2, g◦f, p2, d2) → (Y,Z,Q3, g, p3, d3). We will call (D, [], T ) a pre-triangulated category if TR1, TR2 and TR3 hold.1

The explanation of TR4 is that if you think of Q1 as Y/X, Q2 as Z/X and Q3 as Z/Y , then TR4(a) expresses the isomorphism (Z/X)/(Y/X) =∼ Z/Y and TR4(b) and TR4(c) express that we can compare the triangles X → Y → Q1 → X[1] etc with morphisms of triangles. For a more precise reformulation of this idea see the proof of Lemma 10.2. The sign in TR2 means that if (X,Y,Z,f,g,h) is a distinguished triangle then in the long sequence −h[−1] f g −f[1] −g[1] (3.2.1)05QL ... → Z[−1] −−−−→ X −→ Y −→ Z −→h X[1] −−−→ Y [1] −−−→ Z[1] → ... each four term sequence gives a distinguished triangle. As usual we abuse notation and we simply speak of a (pre-)triangulated category D without explicitly introducing notation for the additional data. The notion of a pre-triangulated category is useful in finding statements equivalent to TR4. We have the following definition of a triangulated functor. 014VDefinition 3.3. Let D, D0 be pre-triangulated categories. An exact functor, or a triangulated functor from D to D0 is a functor F : D → D0 together with given functorial isomorphisms ξX : F (X[1]) → F (X)[1] such that for every distinguished triangle (X,Y,Z,f,g,h) of D the triangle (F (X),F (Y ),F (Z),F (f),F (g), ξX ◦ F (h)) is a distinguished triangle of D0.

1 We use [] as an abbreviation for the family {[n]}n∈Z. DERIVED CATEGORIES 4

An exact functor is additive, see Lemma 4.17. When we say two triangulated categories are equivalent we mean that they are equivalent in the 2-category of triangulated categories. A 2-morphism a :(F, ξ) → (F 0, ξ0) in this 2-category is simply a transformation of functors a : F → F 0 which is compatible with ξ and ξ0, i.e., F ◦ [1] / [1] ◦ F ξ

a?1 1?a ξ0 F 0 ◦ [1] / [1] ◦ F 0 commutes. 05QMDeﬁnition 3.4. Let (D, [], T ) be a pre-triangulated category. A pre-triangulated subcategory2 is a pair (D0, T 0) such that (1) D0 is an additive subcategory of D which is preserved under [1] and [−1], (2) T 0 ⊂ T is a subset such that for every (X,Y,Z,f,g,h) ∈ T 0 we have X,Y,Z ∈ Ob(D0) and f, g, h ∈ Arrows(D0), and (3) (D0, [], T 0) is a pre-triangulated category. If D is a triangulated category, then we say (D0, T 0) is a triangulated subcategory if it is a pre-triangulated subcategory and (D0, [], T 0) is a triangulated category.

0 In this situation the inclusion functor D → D is an exact functor with ξX : X[1] → X[1] given by the identity on X[1]. We will see in Lemma 4.1 that for a distinguished triangle (X,Y,Z,f,g,h) in a pre- triangulated category the composition g ◦ f : X → Z is zero. Thus the sequence (3.2.1) is a complex. A homological functor is one that turns this complex into a long exact sequence. 0147Definition 3.5. Let D be a pre-triangulated category. Let A be an abelian category. An additive functor H : D → A is called homological if for every distinguished triangle (X,Y,Z,f,g,h) the sequence H(X) → H(Y ) → H(Z) is exact in the abelian category A. An additive functor H : Dopp → A is called cohomological if the corresponding functor D → Aopp is homological. If H : D → A is a homological functor we often write Hn(X) = H(X[n]) so that H(X) = H0(X). Our discussion of TR2 above implies that a distinguished triangle (X,Y,Z,f,g,h) determines a long exact sequence (3.5.1) H(h[−1]) H(f) H(g) H(h) 0148 H−1(Z) / H0(X) / H0(Y ) / H0(Z) / H1(X) This will be called the long exact sequence associated to the distinguished triangle and the homological functor. As indicated we will not use any signs for the morphisms in the long exact sequence. This has the side effect that maps in the long exact sequence associated to the rotation (TR2) of a distinguished triangle differ from the maps in the sequence above by some signs.

2This deﬁnition may be nonstandard. If D0 is a full subcategory then T 0 is the intersection of the set of triangles in D0 with T , see Lemma 4.16. In this case we drop T 0 from the notation. DERIVED CATEGORIES 5

0150Deﬁnition 3.6. Let A be an abelian category. Let D be a triangulated category. A δ-functor from A to D is given by a functor G : A → D and a rule which assigns to every short exact sequence

0 → A −→a B −→b C → 0

a morphism δ = δA→B→C : G(C) → G(A)[1] such that

(1) the triangle (G(A),G(B),G(C),G(a),G(b), δA→B→C ) is a distinguished triangle of D for any short exact sequence as above, and (2) for every morphism (A → B → C) → (A0 → B0 → C0) of short exact sequences the diagram

G(C) / G(A)[1] δA→B→C

δ 0 0 0 G(C0) A →B →C / G(A0)[1] is commutative.

In this situation we call (G(A),G(B),G(C),G(a),G(b), δA→B→C ) the image of the short exact sequence under the given δ-functor.

Note how a δ-functor comes equipped with additional structure. Strictly speaking it does not make sense to say that a given functor A → D is a δ-functor, but we will often do so anyway.

4. Elementary results on triangulated categories 05QN Most of the results in this section are proved for pre-triangulated categories and a fortiori hold in any triangulated category.

0146 Lemma 4.1. Let D be a pre-triangulated category. Let (X,Y,Z,f,g,h) be a distinguished triangle. Then g ◦ f = 0, h ◦ g = 0 and f[1] ◦ h = 0.

Proof. By TR1 we know (X,X, 0, 1, 0, 0) is a distinguished triangle. Apply TR3 to X / X / 0 / X[1]

1 f 1[1]

f g h X / Y / Z / X[1] Of course the dotted arrow is the zero map. Hence the commutativity of the diagram implies that g ◦ f = 0. For the other cases rotate the triangle, i.e., apply TR2.

0149 Lemma 4.2. Let D be a pre-triangulated category. For any object W of D the functor HomD(W, −) is homological, and the functor HomD(−,W ) is cohomological.

Proof. Consider a distinguished triangle (X,Y,Z,f,g,h). We have already seen that g ◦ f = 0, see Lemma 4.1. Suppose a : W → Y is a morphism such that DERIVED CATEGORIES 6

g ◦ a = 0. Then we get a commutative diagram W W 0 W [1] 1 / / /

b a 0 b[1] X / Y / Z / X[1] Both rows are distinguished triangles (use TR1 for the top row). Hence we can ﬁll the dotted arrow b (ﬁrst rotate using TR2, then apply TR3, and then rotate back). This proves the lemma. 014A Lemma 4.3. Let D be a pre-triangulated category. Let (a, b, c):(X,Y,Z,f,g,h) → (X0,Y 0,Z0, f 0, g0, h0) be a morphism of distinguished triangles. If two among a, b, c are isomorphisms so is the third. Proof. Assume that a and c are isomorphisms. For any object W of D write HW (−) = HomD(W, −). Then we get a commutative diagram of abelian groups

HW (Z[−1]) / HW (X) / HW (Y ) / HW (Z) / HW (X[1])

0 0 0 0 0 HW (Z [−1]) / HW (X ) / HW (Y ) / HW (Z ) / HW (X [1])

By assumption the right two and left two vertical arrows are bijective. As HW is homological by Lemma 4.2 and the ﬁve lemma (Homology, Lemma 5.20) it follows that the middle vertical arrow is an isomorphism. Hence by Yoneda’s lemma, see Categories, Lemma 3.5 we see that b is an isomorphism. This implies the other cases by rotating (using TR2). 09WARemark 4.4. Let D be an additive category with translation functors [n] as in Deﬁnition 3.1. Let us call a triangle (X,Y,Z,f,g,h) special3 if for every object W of D the long sequence of abelian groups

... → HomD(W, X) → HomD(W, Y ) → HomD(W, Z) → HomD(W, X[1]) → ... is exact. The proof of Lemma 4.3 shows that if (a, b, c):(X,Y,Z,f,g,h) → (X0,Y 0,Z0, f 0, g0, h0) is a morphism of special triangles and if two among a, b, c are isomorphisms so is the third. There is a dual statement for co-special triangles, i.e., triangles which turn into long exact sequences on applying the functor HomD(−,W ). Thus distinguished triangles are special and co-special, but in general there are many more (co-)special triangles, than there are distinguished triangles. 05QP Lemma 4.5. Let D be a pre-triangulated category. Let (0, b, 0), (0, b0, 0) : (X,Y,Z,f,g,h) → (X,Y,Z,f,g,h) be endomorphisms of a distinguished triangle. Then bb0 = 0.

3This is nonstandard notation. DERIVED CATEGORIES 7

Proof. Picture X / Y / Z / X[1]

0 b,b0 0 0 α β Ð Ð X / Y / Z / X[1] Applying Lemma 4.2 we ﬁnd dotted arrows α and β such that b0 = f ◦ α and 0 b = β ◦ g. Then bb = β ◦ g ◦ f ◦ α = 0 as g ◦ f = 0 by Lemma 4.1. 05QQ Lemma 4.6. Let D be a pre-triangulated category. Let (X,Y,Z,f,g,h) be a distinguished triangle. If Z / X[1] h c a[1]

h Z / X[1] is commutative and a2 = a, c2 = c, then there exists a morphism b : Y → Y with b2 = b such that (a, b, c) is an endomorphism of the triangle (X,Y,Z,f,g,h). Proof. By TR3 there exists a morphism b0 such that (a, b0, c) is an endomorphism of (X,Y,Z,f,g,h). Then (0, (b0)2 − b0, 0) is also an endomorphism. By Lemma 4.5 we see that (b0)2 − b0 has square zero. Set b = b0 − (2b0 − 1)((b0)2 − b0) = 3(b0)2 − 2(b0)3. A computation shows that (a, b, c) is an endomorphism and that 2 0 2 0 0 2 0 2 b − b = (4(b ) − 4b − 3)((b ) − b ) = 0. 014B Lemma 4.7. Let D be a pre-triangulated category. Let f : X → Y be a morphism of D. There exists a distinguished triangle (X,Y,Z,f,g,h) which is unique up to (nonunique) isomorphism of triangles. More precisely, given a second such distinguished triangle (X,Y,Z0, f, g0, h0) there exists an isomorphism (1, 1, c):(X,Y,Z,f,g,h) −→ (X,Y,Z0, f, g0, h0) Proof. Existence by TR1. Uniqueness up to isomorphism by TR3 and Lemma 4.3. 0FWZ Lemma 4.8. Let D be a pre-triangulated category. Let (a, b, c):(X,Y,Z,f,g,h) → (X0,Y 0,Z0, f 0, g0, h0) be a morphism of distinguished triangles. If one of the following conditions holds (1) Hom(Y,X0) = 0, (2) Hom(Z,Y 0) = 0, (3) Hom(X,X0) = Hom(Z,X0) = 0, (4) Hom(Z,X0) = Hom(Z,Z0) = 0, or (5) Hom(X[1],Z0) = Hom(Z,X0) = 0 then b is the unique morphism from Y → Y 0 such that (a, b, c) is a morphism of triangles. Proof. If we have a second morphism of triangles (a, b0, c) then (0, b − b0, 0) is a morphism of triangles. Hence we have to show: the only morphism b : Y → Y 0 such that X → Y → Y 0 and Y → Y 0 → Z0 are zero is 0. We will use Lemma 4.2 without further mention. In particular, condition (3) implies (1). Given condition (1) if the composition g0 ◦ b : Y → Y 0 → Z0 is zero, then b lifts to a morphism Y → X0 which has to be zero. This proves (1). DERIVED CATEGORIES 8

The proof of (2) and (4) are dual to this argument. Assume (5). Consider the diagram

X / Y / Z / X[1] f g h

0 b 0 0 0 0 f g h0 X0 / Y 0 / Z0 / X0[1] We may choose such that b = ◦ g. Then g0 ◦ ◦ g = 0 which implies that g0 ◦ = δ ◦ h for some δ ∈ Hom(X[1],Z0). Since Hom(X[1],Z0) = 0 we conclude that g0 ◦ = 0. Hence = f 0 ◦ γ for some γ ∈ Hom(Z,X0). Since Hom(Z,X0) = 0 we conclude that = 0 and hence b = 0 as desired. 05QR Lemma 4.9. Let D be a pre-triangulated category. Let f : X → Y be a morphism of D. The following are equivalent (1) f is an isomorphism, (2) (X,Y, 0, f, 0, 0) is a distinguished triangle, and (3) for any distinguished triangle (X,Y,Z,f,g,h) we have Z = 0. Proof. By TR1 the triangle (X,X, 0, 1, 0, 0) is distinguished. Let (X,Y,Z,f,g,h) be a distinguished triangle. By TR3 there is a map of distinguished triangles (1, f, 0) : (X,X, 0) → (X,Y,Z). If f is an isomorphism, then (1, f, 0) is an isomorphism of triangles by Lemma 4.3 and Z = 0. Conversely, if Z = 0, then (1, f, 0) is an isomorphism of triangles as well, hence f is an isomorphism. 05QS Lemma 4.10. Let D be a pre-triangulated category. Let (X,Y,Z,f,g,h) and (X0,Y 0,Z0, f 0, g0, h0) be triangles. The following are equivalent (1) (X ⊕ X0,Y ⊕ Y 0,Z ⊕ Z0, f ⊕ f 0, g ⊕ g0, h ⊕ h0) is a distinguished triangle, (2) both (X,Y,Z,f,g,h) and (X0,Y 0,Z0, f 0, g0, h0) are distinguished triangles. Proof. Assume (2). By TR1 we may choose a distinguished triangle (X ⊕ X0,Y ⊕ Y 0, Q, f ⊕ f 0, g00, h00). By TR3 we can ﬁnd morphisms of distinguished triangles (X,Y,Z,f,g,h) → (X ⊕ X0,Y ⊕ Y 0, Q, f ⊕ f 0, g00, h00) and (X0,Y 0,Z0, f 0, g0, h0) → (X ⊕ X0,Y ⊕ Y 0, Q, f ⊕ f 0, g00, h00). Taking the direct sum of these morphisms we obtain a morphism of triangles (X ⊕ X0,Y ⊕ Y 0,Z ⊕ Z0, f ⊕ f 0, g ⊕ g0, h ⊕ h0)

(1,1,c) (X ⊕ X0,Y ⊕ Y 0, Q, f ⊕ f 0, g00, h00).

In the terminology of Remark 4.4 this is a map of special triangles (because a direct sum of special triangles is special) and we conclude that c is an isomorphism. Thus (1) holds. Assume (1). We will show that (X,Y,Z,f,g,h) is a distinguished triangle. First observe that (X,Y,Z,f,g,h) is a special triangle (terminology from Remark 4.4) as a direct summand of the distinguished hence special triangle (X ⊕ X0,Y ⊕ Y 0,Z ⊕ Z0, f ⊕ f 0, g ⊕ g0, h ⊕ h0). Using TR1 let (X, Y, Q, f, g00, h00) be a distinguished triangle. By TR3 there exists a morphism of distinguished triangles (X ⊕ X0,Y ⊕ DERIVED CATEGORIES 9

Y 0,Z ⊕ Z0, f ⊕ f 0, g ⊕ g0, h ⊕ h0) → (X, Y, Q, f, g00, h00). Composing this with the inclusion map we get a morphism of triangles (1, 1, c):(X,Y,Z,f,g,h) −→ (X, Y, Q, f, g00, h00) By Remark 4.4 we ﬁnd that c is an isomorphism and we conclude that (2) holds. 05QT Lemma 4.11. Let D be a pre-triangulated category. Let (X,Y,Z,f,g,h) be a distinguished triangle. (1) If h = 0, then there exists a right inverse s : Z → Y to g. (2) For any right inverse s : Z → Y of g the map f ⊕ s : X ⊕ Z → Y is an isomorphism. (3) For any objects X0,Z0 of D the triangle (X0,X0 ⊕ Z0,Z0, (1, 0), (0, 1), 0) is distinguished.

Proof. To see (1) use that HomD(Z,Y ) → HomD(Z,Z) → HomD(Z,X[1]) is exact by Lemma 4.2. By the same token, if s is as in (2), then h = 0 and the sequence

0 → HomD(W, X) → HomD(W, Y ) → HomD(W, Z) → 0 is split exact (split by s : Z → Y ). Hence by Yoneda’s lemma we see that X⊕Z → Y is an isomorphism. The last assertion follows from TR1 and Lemma 4.10. 05QU Lemma 4.12. Let D be a pre-triangulated category. Let f : X → Y be a morphism of D. The following are equivalent (1) f has a kernel, (2) f has a cokernel, (3) f is the isomorphic to a composition K ⊕ Z → Z → Z ⊕ Q of a projection and coprojection for some objects K,Z,Q of D. Proof. Any morphism isomorphic to a map of the form X0 ⊕Z → Z ⊕Y 0 has both a kernel and a cokernel. Hence (3) ⇒ (1), (2). Next we prove (1) ⇒ (3). Suppose ﬁrst that f : X → Y is a monomorphism, i.e., its kernel is zero. By TR1 there exists a distinguished triangle (X,Y,Z,f,g,h). By Lemma 4.1 the composition f ◦ h[−1] = 0. As f is a monomorphism we see that h[−1] = 0 and hence h = 0. Then Lemma 4.11 implies that Y = X ⊕ Z, i.e., we see that (3) holds. Next, assume f has a kernel K. As K → X is a monomorphism we conclude X = K ⊕X0 0 0 0 and f|X0 : X → Y is a monomorphism. Hence Y = X ⊕ Y and we win. The implication (2) ⇒ (3) is dual to this. 0CRG Lemma 4.13. Let D be a pre-triangulated category. Let I be a set.

(1) Let Xi, i ∈ I be a family of objects of D. Q Q Q (a) If Xi exists, then ( Xi)[1] = Xi[1]. L L L (b) If Xi exists, then ( Xi)[1] = Xi[1]. (2) Let Xi → Yi → Zi → Xi[1] be a family of distinguished triangles of D. Q Q Q Q Q Q Q (a) If Xi, Yi, Zi exist, then Xi → Yi → Zi → Xi[1] is a distinguished triangle. L L L L L L L (b) If Xi, Yi, Zi exist, then Xi → Yi → Zi → Xi[1] is a distinguished triangle. Proof. Part (1) is true because [1] is an autoequivalence of D and because direct sums and products are deﬁned in terms of the category structure. Let us prove Q Q Q (2)(a). Choose a distinguished triangle Xi → Yi → Z → Xi[1]. For each j we can use TR3 to choose a morphism pj : Z → Zj ﬁtting into a morphism of DERIVED CATEGORIES 10

Q Q distinguished triangles with the projection maps Xi → Xj and Yi → Yj. Using Q Q the definition of products we obtain a map pi : Z → Zi fitting into a morphism of triangles from the distinguished triangle to the triangle made out of the products. Q Q Q Q Observe that the “product” triangle Xi → Yi → Zi → Xi[1] is special in the terminology of Remark 4.4 because products of exact sequences of abelian groups are exact. Hence Remark 4.4 shows that the morphism of triangles is an isomorphism and we conclude by TR1. The proof of (2)(b) is dual. 05QW Lemma 4.14. Let D be a pre-triangulated category. If D has countable products, then D is Karoubian. If D has countable coproducts, then D is Karoubian. Proof. Assume D has countable products. By Homology, Lemma 4.3 it suffices to check that morphisms which have a right inverse have kernels. Any morphism which has a right inverse is an epimorphism, hence has a kernel by Lemma 4.12. The second statement is dual to the first. The following lemma makes it slightly easier to prove that a pre-triangulated category is triangulated. 014C Lemma 4.15. Let D be a pre-triangulated category. In order to prove TR4 it suffices to show that given any pair of composable morphisms f : X → Y and g : Y → Z there exist (1) isomorphisms i : X0 → X, j : Y 0 → Y and k : Z0 → Z, and then setting f 0 = j−1fi : X0 → Y 0 and g0 = k−1gj : Y 0 → Z0 there exist 0 0 0 0 0 0 0 (2) distinguished triangles (X ,Y ,Q1, f , p1, d1), (X ,Z ,Q2, g ◦f , p2, d2) and 0 0 0 (Y ,Z ,Q3, g , p3, d3), such that the assertion of TR4 holds. Proof. The replacement of X,Y,Z by X0,Y 0,Z0 is harmless by our definition of distinguished triangles and their isomorphisms. The lemma follows from the fact 0 0 0 0 0 0 0 that the distinguished triangles (X ,Y ,Q1, f , p1, d1), (X ,Z ,Q2, g ◦f , p2, d2) and 0 0 0 (Y ,Z ,Q3, g , p3, d3) are unique up to isomorphism by Lemma 4.7. 05QX Lemma 4.16. Let D be a pre-triangulated category. Assume that D0 is an additive full subcategory of D. The following are equivalent (1) there exists a set of triangles T 0 such that (D0, T 0) is a pre-triangulated subcategory of D, (2) D0 is preserved under [1], [−1] and given any morphism f : X → Y in D0 there exists a distinguished triangle (X,Y,Z,f,g,h) in D such that Z is isomorphic to an object of D0. In this case T 0 as in (1) is the set of distinguished triangles (X,Y,Z,f,g,h) of D such that X,Y,Z ∈ Ob(D0). Finally, if D is a triangulated category, then (1) and (2) are also equivalent to (3) D0 is a triangulated subcategory.

Proof. Omitted. 05QY Lemma 4.17. An exact functor of pre-triangulated categories is additive. Proof. Let F : D → D0 be an exact functor of pre-triangulated categories. Since (0, 0, 0, 10, 10, 0) is a distinguished triangle of D the triangle

(F (0),F (0),F (0), 1F (0), 1F (0),F (0)) DERIVED CATEGORIES 11

0 is distinguished in D . This implies that 1F (0) ◦ 1F (0) is zero, see Lemma 4.1. Hence F (0) is the zero object of D0. This also implies that F applied to any zero morphism is zero (since a morphism in an additive category is zero if and only if it factors through the zero object). Next, using that (X,X ⊕ Y,Y, (1, 0), (0, 1), 0) is a distinguished triangle, we see that (F (X),F (X ⊕ Y ),F (Y ),F (1, 0),F (0, 1), 0) is one too. This implies that the map F (1, 0) ⊕ F (0, 1) : F (X) ⊕ F (Y ) → F (X ⊕ Y ) is an isomorphism, see Lemma 4.11. We omit the rest of the argument. 05SQ Lemma 4.18. Let F : D → D0 be a fully faithful exact functor of pre-triangulated categories. Then a triangle (X,Y,Z,f,g,h) of D is distinguished if and only if (F (X),F (Y ),F (Z),F (f),F (g),F (h)) is distinguished in D0. Proof. The “only if” part is clear. Assume (F (X),F (Y ),F (Z)) is distinguished in D0. Pick a distinguished triangle (X,Y,Z0, f, g0, h0) in D. By Lemma 4.7 there exists an isomorphism of triangles (1, 1, c0):(F (X),F (Y ),F (Z)) −→ (F (X),F (Y ),F (Z0)). Since F is fully faithful, there exists a morphism c : Z → Z0 such that F (c) = c0. Then (1, 1, c) is an isomorphism between (X,Y,Z) and (X,Y,Z0). Hence (X,Y,Z) is distinguished by TR1. 014Y Lemma 4.19. Let D, D0, D00 be pre-triangulated categories. Let F : D → D0 and F 0 : D0 → D00 be exact functors. Then F 0 ◦ F is an exact functor.

Proof. Omitted. 05QZ Lemma 4.20. Let D be a pre-triangulated category. Let A be an abelian category. Let H : D → A be a homological functor. (1) Let D0 be a pre-triangulated category. Let F : D0 → D be an exact functor. Then the composition H ◦ F is a homological functor as well. (2) Let A0 be an abelian category. Let G : A → A0 be an exact functor. Then G ◦ H is a homological functor as well.

Proof. Omitted. 0151 Lemma 4.21. Let D be a triangulated category. Let A be an abelian category. Let G : A → D be a δ-functor. (1) Let D0 be a triangulated category. Let F : D → D0 be an exact functor. Then the composition F ◦ G is a δ-functor as well. (2) Let A0 be an abelian category. Let H : A0 → A be an exact functor. Then G ◦ H is a δ-functor as well.

Proof. Omitted. 05SR Lemma 4.22. Let D be a triangulated category. Let A and B be abelian categories. Let G : A → D be a δ-functor. Let H : D → B be a homological functor. Assume that H−1(G(A)) = 0 for all A in A. Then the collection

n n {H ◦ G, H (δA→B→C )}n≥0 is a δ-functor from A → B, see Homology, Deﬁnition 12.1. DERIVED CATEGORIES 12

Proof. The notation signiﬁes the following. If 0 → A −→a B −→b C → 0 is a short exact sequence in A, then

δ = δA→B→C : G(C) → G(A)[1] is a morphism in D such that (G(A),G(B),G(C), a, b, δ) is a distinguished triangle, see Deﬁnition 3.6. Then Hn(δ): Hn(G(C)) → Hn(G(A)[1]) = Hn+1(G(A)) is clearly functorial in the short exact sequence. Finally, the long exact cohomology sequence (3.5.1) combined with the vanishing of H−1(G(C)) gives a long exact sequence H0(δ) 0 → H0(G(A)) → H0(G(B)) → H0(G(C)) −−−−→ H1(G(A)) → ... in B as desired. The proof of the following result uses TR4. 05R0 Proposition 4.23. Let D be a triangulated category. Any commutative diagram X / Y

X0 / Y 0 can be extended to a diagram X / Y / Z / X[1]

X0 / Y 0 / Z0 / X0[1]

X00 / Y 00 / Z00 / X00[1]

X[1] / Y [1] / Z[1] / X[2] where all the squares are commutative, except for the lower right square which is anticommutative. Moreover, each of the rows and columns are distinguished triangles. Finally, the morphisms on the bottom row (resp. right column) are obtained from the morphisms of the top row (resp. left column) by applying [1]. Proof. During this proof we avoid writing the arrows in order to make the proof leg- ible. Choose distinguished triangles (X,Y,Z), (X0,Y 0,Z0), (X,X0,X00), (Y,Y 0,Y 00), and (X,Y 0,A). Note that the morphism X → Y 0 is both equal to the composition X → Y → Y 0 and equal to the composition X → X0 → Y 0. Hence, we can ﬁnd morphisms (1) a : Z → A and b : A → Y 00, and (2) a0 : X00 → A and b0 : A → Z0 as in TR4. Denote c : Y 00 → Z[1] the composition Y 00 → Y [1] → Z[1] and denote c0 : Z0 → X00[1] the composition Z0 → X0[1] → X00[1]. The conclusion of our application TR4 are that (1) (Z, A, Y 00, a, b, c), (X00, A, Z0, a0, b0, c0) are distinguished triangles, DERIVED CATEGORIES 13

(2) (X,Y,Z) → (X,Y 0,A), (X,Y 0,A) → (Y,Y 0,Y 00), (X,X0,X00) → (X,Y 0,A), (X,Y 0,A) → (X0,Y 0,Z0) are morphisms of triangles. First using that (X,X0,X00) → (X,Y 0,A) and (X,Y 0,A) → (Y,Y 0,Y 00). are morphisms of triangles we see the ﬁrst of the diagrams

X0 / Y 0 Y / Z / X[1] 0 00 b◦a 00 X / Y and b0◦a Y 0 / Z0 / X0[1] X[1] / Y [1]

is commutative. The second is commutative too using that (X,Y,Z) → (X,Y 0,A) and (X,Y 0,A) → (X0,Y 0,Z0) are morphisms of triangles. At this point we choose a distinguished triangle (X00,Y 00,Z00) starting with the map b ◦ a0 : X00 → Y 00.

Next we apply TR4 one more time to the morphisms X00 → A → Y 00 and the triangles (X00, A, Z0, a0, b0, c0), (X00,Y 00,Z00), and (A, Y 00,Z[1], b, c, −a[1]) to get morphisms a00 : Z0 → Z00 and b00 : Z00 → Z[1]. Then (Z0,Z00,Z[1], a00, b00, −b0[1] ◦ a[1]) is a distinguished triangle, hence also (Z,Z0,Z00, −b0 ◦ a, a00, −b00) and hence also (Z,Z0,Z00, b0◦a, a00, b00). Moreover, (X00, A, Z0) → (X00,Y 00,Z00) and (X00,Y 00,Z00) → (A, Y 00,Z[1], b, c, −a[1]) are morphisms of triangles. At this point we have deﬁned all the distinguished triangles and all the morphisms, and all that’s left is to verify some commutativity relations.

To see that the middle square in the diagram commutes, note that the arrow Y 0 → Z0 factors as Y 0 → A → Z0 because (X,Y 0,A) → (X0,Y 0,Z0) is a morphism of triangles. Similarly, the morphism Y 0 → Y 00 factors as Y 0 → A → Y 00 because (X,Y 0,A) → (Y,Y 0,Y 00) is a morphism of triangles. Hence the middle square commutes because the square with sides (A, Z0,Z00,Y 00) commutes as (X00, A, Z0) → (X00,Y 00,Z00) is a morphism of triangles (by TR4). The square with sides (Y 00,Z00,Y [1],Z[1]) commutes because (X00,Y 00,Z00) → (A, Y 00,Z[1], b, c, −a[1]) is a morphism of triangles and c : Y 00 → Z[1] is the composition Y 00 → Y [1] → Z[1]. The square with sides (Z0,X0[1],X00[1],Z00) is commutative because (X00, A, Z0) → (X00,Y 00,Z00) is a morphism of triangles and c0 : Z0 → X00[1] is the composition Z0 → X0[1] → X00[1]. Finally, we have to show that the square with sides (Z00,X00[1],Z[1],X[2]) anticommutes. This holds because (X00,Y 00,Z00) → 00 (A, Y ,Z[1], b, c, −a[1]) is a morphism of triangles and we’re done.

5. Localization of triangulated categories 05R1 In order to construct the derived category starting from the homotopy category of complexes, we will use a localization process.

05R2Deﬁnition 5.1. Let D be a pre-triangulated category. We say a multiplicative system S is compatible with the triangulated structure if the following two conditions hold: MS5 For s ∈ S we have s[n] ∈ S for all n ∈ Z. DERIVED CATEGORIES 14

MS6 Given a solid commutative square X / Y / Z / X[1]

s s0 s[1] X0 / Y 0 / Z0 / X0[1] whose rows are distinguished triangles with s, s0 ∈ S there exists a morphism s00 : Z → Z0 in S such that (s, s0, s00) is a morphism of triangles. It turns out that these axioms are not independent of the axioms defining multiplicative systems. 05R3 Lemma 5.2. Let D be a pre-triangulated category. Let S be a set of morphisms of D and assume that axioms MS1, MS5, MS6 hold (see Categories, Definition 27.1 and Definition 5.1). Then MS2 holds. Proof. Suppose that f : X → Y is a morphism of D and t : X → X0 an element of S. Choose a distinguished triangle (X,Y,Z,f,g,h). Next, choose a distinguished triangle (X0,Y 0, Z, f 0, g0, t[1] ◦ h) (here we use TR1 and TR2). By MS5, MS6 (and TR2 to rotate) we can find the dotted arrow in the commutative diagram

X / Y / Z / X[1]

t s0 1 t[1] X0 / Y 0 / Z / X0[1] 0 with moreover s ∈ S. This proves LMS2. The proof of RMS2 is dual. 05R4 Lemma 5.3. Let F : D → D0 be an exact functor of pre-triangulated categories. Let S = {f ∈ Arrows(D) | F (f) is an isomorphism} Then S is a saturated (see Categories, Definition 27.20) multiplicative system compatible with the triangulated structure on D. Proof. We have to prove axioms MS1 – MS6, see Categories, Definitions 27.1 and 27.20 and Definition 5.1. MS1, MS4, and MS5 are direct from the definitions. MS6 follows from TR3 and Lemma 4.3. By Lemma 5.2 we conclude that MS2 holds. To finish the proof we have to show that MS3 holds. To do this let f, g : X → Y be morphisms of D, and let t : Z → X be an element of S such that f ◦ t = g ◦ t. As D is additive this simply means that a ◦ t = 0 with a = f − g. Choose a distinguished triangle (Z, X, Q, t, d, h) using TR1. Since a ◦ t = 0 we see by Lemma 4.2 there exists a morphism i : Q → Y such that i ◦ d = a. Finally, using TR1 again we can choose a triangle (Q, Y, W, i, j, k). Here is a picture

Z / X / Q / Z[1] t d 1 i X a / Y

j W DERIVED CATEGORIES 15

OK, and now we apply the functor F to this diagram. Since t ∈ S we see that F (Q) = 0, see Lemma 4.9. Hence F (j) is an isomorphism by the same lemma, i.e., j ∈ S. Finally, j ◦ a = j ◦ i ◦ d = 0 as j ◦ i = 0. Thus j ◦ f = j ◦ g and we see that LMS3 holds. The proof of RMS3 is dual. 05R5 Lemma 5.4. Let H : D → A be a homological functor between a pre-triangulated category and an abelian category. Let S = {f ∈ Arrows(D) | Hi(f) is an isomorphism for all i ∈ Z} Then S is a saturated (see Categories, Definition 27.20) multiplicative system compatible with the triangulated structure on D. Proof. We have to prove axioms MS1 – MS6, see Categories, Definitions 27.1 and 27.20 and Definition 5.1. MS1, MS4, and MS5 are direct from the definitions. MS6 follows from TR3 and the long exact cohomology sequence (3.5.1). By Lemma 5.2 we conclude that MS2 holds. To finish the proof we have to show that MS3 holds. To do this let f, g : X → Y be morphisms of D, and let t : Z → X be an element of S such that f ◦ t = g ◦ t. As D is additive this simply means that a ◦ t = 0 with a = f − g. Choose a distinguished triangle (Z, X, Q, t, g, h) using TR1 and TR2. Since a ◦ t = 0 we see by Lemma 4.2 there exists a morphism i : Q → Y such that i ◦ g = a. Finally, using TR1 again we can choose a triangle (Q, Y, W, i, j, k). Here is a picture Z X Q Z[1] t / g / /

1 i X a / Y

j W OK, and now we apply the functors Hi to this diagram. Since t ∈ S we see that Hi(Q) = 0 by the long exact cohomology sequence (3.5.1). Hence Hi(j) is an isomorphism for all i by the same argument, i.e., j ∈ S. Finally, j ◦ a = j ◦ i ◦ g = 0 as j ◦ i = 0. Thus j ◦ f = j ◦ g and we see that LMS3 holds. The proof of RMS3 is dual. 05R6 Proposition 5.5. Let D be a pre-triangulated category. Let S be a multiplicative system compatible with the triangulated structure. Then there exists a unique structure of a pre-triangulated category on S−1D such that the localization functor Q : D → S−1D is exact. Moreover, if D is a triangulated category, so is S−1D. Proof. We have seen that S−1D is an additive category and that the localization functor Q is additive in Homology, Lemma 8.2. It is clear that we may deﬁne Q(X)[n] = Q(X[n]) since S is preserved under the shift functors [n] by MS5. Finally, we say a triangle of S−1D is distinguished if it is isomorphic to the image of a distinguished triangle under the localization functor Q. Proof of TR1. The only thing to prove here is that if a : Q(X) → Q(Y ) is a morphism of S−1D, then a ﬁts into a distinguished triangle. Write a = Q(s)−1 ◦ Q(f) for some s : Y → Y 0 in S and f : X → Y 0. Choose a distinguished triangle (X,Y 0, Z, f, g, h) in D. Then we see that (Q(X),Q(Y ),Q(Z), a, Q(g) ◦ Q(s),Q(h)) is a distinguished triangle of S−1D. DERIVED CATEGORIES 16

Proof of TR2. This is immediate from the deﬁnitions. Proof of TR3. Note that the existence of the dotted arrow which is required to exist may be proven after replacing the two triangles by isomorphic triangles. Hence we may assume given distinguished triangles (X,Y,Z,f,g,h) and (X0,Y 0,Z0, f 0, g0, h0) of D and a commutative diagram Q(X) / Q(Y ) Q(f)

a b Q(f 0) Q(X0) / Q(Y 0) in S−1D. Now we apply Categories, Lemma 27.10 to ﬁnd a morphism f 00 : X00 → Y 00 in D and a commutative diagram X / X00 o X0 k s f f 00 f 0 l t Y / Y 00 o Y 0 in D with s, t ∈ S and a = s−1k, b = t−1l. At this point we can use TR3 for D and MS6 to ﬁnd a commutative diagram X / Y / Z / X[1]

k l m g[1] X00 / Y 00 / Z00 / X00[1] O O O O s t r s[1]

X0 / Y 0 / Z0 / X0[1] with r ∈ S. It follows that setting c = Q(r)−1Q(m) we obtain the desired morphism of triangles (Q(X),Q(Y ),Q(Z),Q(f),Q(g),Q(h))

(a,b,c) (Q(X0),Q(Y 0),Q(Z0),Q(f 0),Q(g0),Q(h0))

This proves the first statement of the lemma. If D is also a triangulated category, then we still have to prove TR4 in order to show that S−1D is triangulated as well. To do this we reduce by Lemma 4.15 to the following statement: Given composable morphisms a : Q(X) → Q(Y ) and b : Q(Y ) → Q(Z) we have to produce an octahedron after possibly replacing Q(X),Q(Y ),Q(Z) by isomorphic objects. To do this we may first replace Y by an object such that a = Q(f) for some morphism f : X → Y in D. (More precisely, write a = s−1f with s : Y → Y 0 in S and f : X → Y 0. Then replace Y by Y 0.) After this we similarly replace Z by an object such that b = Q(g) for some morphism g : Y → Z. Now we can find distinguished triangles (X,Y,Q1, f, p1, d1), (X,Z,Q2, g ◦ f, p2, d2), and (Y,Z,Q3, g, p3, d3) in D (by TR1), and morphisms a : Q1 → Q2 and b : Q2 → Q3 as in TR4. Then it is immediately verified that applying the functor Q to all these data gives a −1 corresponding structure in S D DERIVED CATEGORIES 17

The universal property of the localization of a triangulated category is as follows (we formulate this for pre-triangulated categories, hence it holds a fortiori for triangulated categories). 05R7 Lemma 5.6. Let D be a pre-triangulated category. Let S be a multiplicative system compatible with the triangulated category. Let Q : D → S−1D be the localization functor, see Proposition 5.5. (1) If H : D → A is a homological functor into an abelian category A such that H(s) is an isomorphism for all s ∈ S, then the unique factorization H0 : S−1D → A such that H = H0 ◦ Q (see Categories, Lemma 27.8) is a homological functor too. (2) If F : D → D0 is an exact functor into a pre-triangulated category D0 such that F (s) is an isomorphism for all s ∈ S, then the unique factorization F 0 : S−1D → D0 such that F = F 0 ◦ Q (see Categories, Lemma 27.8) is an exact functor too.

Proof. This lemma proves itself. Details omitted. The following lemma describes the kernel (see Deﬁnition 6.5) of the localization functor. 05R8 Lemma 5.7. Let D be a pre-triangulated category. Let S be a multiplicative system compatible with the triangulated structure. Let Z be an object of D. The following are equivalent (1) Q(Z) = 0 in S−1D, (2) there exists Z0 ∈ Ob(D) such that 0 : Z → Z0 is an element of S, (3) there exists Z0 ∈ Ob(D) such that 0 : Z0 → Z is an element of S, and (4) there exists an object Z0 and a distinguished triangle (X,Y,Z ⊕ Z0, f, g, h) such that f ∈ S. If S is saturated, then these are also equivalent to (5) the morphism 0 → Z is an element of S, (6) the morphism Z → 0 is an element of S, (7) there exists a distinguished triangle (X,Y,Z,f,g,h) such that f ∈ S. Proof. The equivalence of (1), (2), and (3) is Homology, Lemma 8.3. If (2) holds, then (Z0[−1],Z0[−1] ⊕ Z,Z, (1, 0), (0, 1), 0) is a distinguished triangle (see Lemma 4.11) with “0 ∈ S”. By rotating we conclude that (4) holds. If (X,Y,Z⊕Z0, f, g, h) is a distinguished triangle with f ∈ S then Q(f) is an isomorphism hence Q(Z ⊕Z0) = 0 hence Q(Z) = 0. Thus (1) – (4) are all equivalent. Next, assume that S is saturated. Note that each of (5), (6), (7) implies one of the equivalent conditions (1) – (4). Suppose that Q(Z) = 0. Then 0 → Z is a morphism of D which becomes an isomorphism in S−1D. According to Categories, Lemma 27.21 the fact that S is saturated implies that 0 → Z is in S. Hence (1) ⇒ (5). Dually (1) ⇒ (6). Finally, if 0 → Z is in S, then the triangle (0,Z,Z, 0, idZ , 0) is distinguished by TR1 and TR2 and is a triangle as in (4). 05R9 Lemma 5.8. Let D be a triangulated category. Let S be a saturated multiplicative system in D that is compatible with the triangulated structure. Let (X,Y,Z,f,g,h) be a distinguished triangle in D. Consider the category of morphisms of triangles I = {(s, s0, s00):(X,Y,Z,f,g,h) → (X0,Y 0,Z0, f 0, g0, h0) | s, s0, s00 ∈ S} DERIVED CATEGORIES 18

Then I is a filtered category and the functors I → X/S, I → Y/S, and I → Z/S are cofinal. Proof. We strongly suggest the reader skip the proof of this lemma and instead work it out on a napkin. The first remark is that using rotation of distinguished triangles (TR2) gives an equivalence of categories between I and the corresponding category for the distinguished triangle (Y,Z,X[1], g, h, −f[1]). Using this we see for example that if we prove the functor I → X/S is cofinal, then the same thing is true for the functors I → Y/S and I → Z/S. Note that if s : X → X0 is a morphism of S, then using MS2 we can find s0 : Y → Y 0 and f 0 : X0 → Y 0 such that f 0 ◦ s = s0 ◦ f, whereupon we can use MS6 to complete this into an object of I. Hence the functor I → X/S is surjective on objects. Using rotation as above this implies the same thing is true for the functors I → Y/S and I → Z/S.

Suppose given objects s1 : X → X1 and s2 : X → X2 in X/S and a morphism a : X1 → X2 in X/S. Since S is saturated, we see that a ∈ S, see Categories, Lemma 27.21. By the argument of the previous paragraph we can complete s1 : X → X1 0 00 to an object (s1, s1, s1 ):(X,Y,Z,f,g,h) → (X1,Y1,Z1, f1, g1, h1) in I. Then we can repeat and find (a, b, c):(X1,Y1,Z1, f1, g1, h1) → (X2,Y2,Z2, f2, g2, h2) with a, b, c ∈ S completing the given a : X1 → X2. But then (a, b, c) is a morphism in I. In this way we conclude that the functor I → X/S is also surjective on arrows. Using rotation as above, this implies the same thing is true for the functors I → Y/S and I → Z/S. The category I is nonempty as the identity provides an object. This proves the condition (1) of the definition of a filtered category, see Categories, Definition 19.1. We check condition (2) of Categories, Definition 19.1 for the category I. Suppose 0 00 0 00 given objects (s1, s1, s1 ):(X,Y,Z,f,g,h) → (X1,Y1,Z1, f1, g1, h1) and (s2, s2, s2 ): (X,Y,Z,f,g,h) → (X2,Y2,Z2, f2, g2, h2) in I. We want to find an object of I which is the target of an arrow from both (X1,Y1,Z1, f1, g1, h1) and (X2,Y2,Z2, f2, g2, h2). By Categories, Remark 27.7 the categories X/S, Y/S, Z/S are filtered. Thus we can find X → X3 in X/S and morphisms s : X2 → X3 and a : X1 → X3. By the above 0 00 we can find a morphism (s, s , s ):(X2,Y2,Z2, f2, g2, h2) → (X3,Y3,Z3, f3, g3, h3) 0 00 with s , s ∈ S. After replacing (X2,Y2,Z2) by (X3,Y3,Z3) we may assume that there exists a morphism a : X1 → X2 in X/S. Repeating the argument for Y and Z (by rotating as above) we may assume there is a morphism a : X1 → X2 in X/S, b : Y1 → Y2 in Y/S, and c : Z1 → Z2 in Z/S. However, these morphisms do not necessarily give rise to a morphism of distinguished triangles. On the other hand, the necessary diagrams do commute in S−1D. Hence we see (for example) 0 0 0 that there exists a morphism s2 : Y2 → Y3 in S such that s2 ◦ f2 ◦ a = s2 ◦ b ◦ f1. Another replacement of (X2,Y2,Z2) as above then gets us to the situation where f2 ◦ a = b ◦ f1. Rotating and applying the same argument two more times we see that we may assume (a, b, c) is a morphism of triangles. This proves condition (2). 0 00 Next we check condition (3) of Categories, Definition 19.1. Suppose (s1, s1, s1 ): 0 00 (X,Y,Z) → (X1,Y1,Z1) and (s2, s2, s2 ):(X,Y,Z) → (X2,Y2,Z2) are objects of 0 0 0 I, and suppose (a, b, c), (a , b , c ) are two morphisms between them. Since a ◦ s1 = 0 0 a ◦ s1 there exists a morphism s3 : X2 → X3 such that s3 ◦ a = s3 ◦ a . Using the DERIVED CATEGORIES 19

0 00 surjectivity statement we can complete this to a morphism of triangles (s3, s3, s3 ): 0 00 0 0 00 00 (X2,Y2,Z2) → (X3,Y3,Z3) with s3, s3, s3 ∈ S. Thus (s3 ◦ s2, s3 ◦ s2, s3 ◦ s2 ): (X,Y,Z) → (X3,Y3,Z3) is also an object of I and after composing the maps 0 0 0 0 00 0 (a, b, c), (a , b , c ) with (s3, s3, s3 ) we obtain a = a . By rotating we may do the same to get b = b0 and c = c0. Finally, we check that I → X/S is cofinal, see Categories, Definition 17.1. The first condition is true as the functor is surjective. Suppose that we have an object s : 0 0 00 X → X in X/S and two objects (s1, s1, s1 ):(X,Y,Z,f,g,h) → (X1,Y1,Z1, f1, g1, h1) 0 00 and (s2, s2, s2 ):(X,Y,Z,f,g,h) → (X2,Y2,Z2, f2, g2, h2) in I as well as morphisms 0 0 t1 : X → X1 and t2 : X → X2 in X/S. By property (2) of I proved above we 0 00 can find morphisms (s3, s3, s3 ):(X1,Y1,Z1, f1, g1, h1) → (X3,Y3,Z3, f3, g3, h3) 0 00 and (s4, s4, s4 ):(X2,Y2,Z2, f2, g2, h2) → (X3,Y3,Z3, f3, g3, h3) in I. We would 0 0 be done if the compositions X → X1 → X3 and X → X1 → X3 where equal (see displayed equation in Categories, Definition 17.1). If not, then, because X/S is filtered, we can choose a morphism X3 → X4 in S such that the compositions 0 0 X → X1 → X3 → X4 and X → X1 → X3 → X4 are equal. Then we finally complete X3 → X4 to a morphism (X3,Y3,Z3) → (X4,Y4,Z4) in I and compose with that morphism to see that the result is true.

6. Quotients of triangulated categories 05RA Given a triangulated category and a triangulated subcategory we can construct another triangulated category by taking the “quotient”. The construction uses a localization. This is similar to the quotient of an abelian category by a Serre subcategory, see Homology, Section 10. Before we do the actual construction we briefly discuss kernels of exact functors. 05RBDefinition 6.1. Let D be a pre-triangulated category. We say a full pre-triangulated subcategory D0 of D is saturated if whenever X ⊕ Y is isomorphic to an object of D0 then both X and Y are isomorphic to objects of D0. A saturated triangulated subcategory is sometimes called a thick triangulated subcategory. In some references, this is only used for strictly full triangulated subcategories (and sometimes the definition is written such that it implies strictness). There is another notion, that of an épaisse triangulated subcategory. The definition is that given a commutative diagram S ?

X / Y / T / X[1] where the second line is a distinguished triangle and S and T isomorphic to objects of D0, then also X and Y are isomorphic to objects of D0. It turns out that this is equivalent to being saturated (this is elementary and can be found in [Ric89]) and the notion of a saturated category is easier to work with. 05RC Lemma 6.2. Let F : D → D0 be an exact functor of pre-triangulated categories. Let D00 be the full subcategory of D with objects Ob(D00) = {X ∈ Ob(D) | F (X) = 0} DERIVED CATEGORIES 20

Then D00 is a strictly full saturated pre-triangulated subcategory of D. If D is a triangulated category, then D00 is a triangulated subcategory. Proof. It is clear that D00 is preserved under [1] and [−1]. If (X,Y,Z,f,g,h) is a distinguished triangle of D and F (X) = F (Y ) = 0, then also F (Z) = 0 as (F (X),F (Y ),F (Z),F (f),F (g),F (h)) is distinguished. Hence we may apply Lemma 4.16 to see that D00 is a pre-triangulated subcategory (respectively a triangulated subcategory if D is a triangulated category). The ﬁnal assertion of being saturated follows from F (X) ⊕ F (Y ) = 0 ⇒ F (X) = F (Y ) = 0. 05RD Lemma 6.3. Let H : D → A be a homological functor of a pre-triangulated category into an abelian category. Let D0 be the full subcategory of D with objects Ob(D0) = {X ∈ Ob(D) | H(X[n]) = 0 for all n ∈ Z} Then D0 is a strictly full saturated pre-triangulated subcategory of D. If D is a triangulated category, then D0 is a triangulated subcategory. Proof. It is clear that D0 is preserved under [1] and [−1]. If (X,Y,Z,f,g,h) is a distinguished triangle of D and H(X[n]) = H(Y [n]) = 0 for all n, then also H(Z[n]) = 0 for all n by the long exact sequence (3.5.1). Hence we may apply Lemma 4.16 to see that D0 is a pre-triangulated subcategory (respectively a triangulated subcategory if D is a triangulated category). The assertion of being saturated follows from H((X ⊕ Y )[n]) = 0 ⇒ H(X[n] ⊕ Y [n]) = 0 ⇒ H(X[n]) ⊕ H(Y [n]) = 0 ⇒ H(X[n]) = H(Y [n]) = 0

for all n ∈ Z. 05RE Lemma 6.4. Let H : D → A be a homological functor of a pre-triangulated + − b category into an abelian category. Let DH , DH , DH be the full subcategory of D with objects + Ob(DH ) = {X ∈ Ob(D) | H(X[n]) = 0 for all n 0} − Ob(DH ) = {X ∈ Ob(D) | H(X[n]) = 0 for all n 0} b Ob(DH ) = {X ∈ Ob(D) | H(X[n]) = 0 for all |n| 0} Each of these is a strictly full saturated pre-triangulated subcategory of D. If D is a triangulated category, then each is a triangulated subcategory. + Proof. Let us prove this for DH . It is clear that it is preserved under [1] and [−1]. If (X,Y,Z,f,g,h) is a distinguished triangle of D and H(X[n]) = H(Y [n]) = 0 for all n 0, then also H(Z[n]) = 0 for all n 0 by the long exact sequence (3.5.1). + Hence we may apply Lemma 4.16 to see that DH is a pre-triangulated subcategory (respectively a triangulated subcategory if D is a triangulated category). The assertion of being saturated follows from H((X ⊕ Y )[n]) = 0 ⇒ H(X[n] ⊕ Y [n]) = 0 ⇒ H(X[n]) ⊕ H(Y [n]) = 0 ⇒ H(X[n]) = H(Y [n]) = 0

for all n ∈ Z. DERIVED CATEGORIES 21

05RFDefinition 6.5. Let D be a (pre-)triangulated category. (1) Let F : D → D0 be an exact functor. The kernel of F is the strictly full saturated (pre-)triangulated subcategory described in Lemma 6.2. (2) Let H : D → A be a homological functor. The kernel of H is the strictly full saturated (pre-)triangulated subcategory described in Lemma 6.3. These are sometimes denoted Ker(F ) or Ker(H). The proof of the following lemma uses TR4. 05RG Lemma 6.6. Let D be a triangulated category. Let D0 ⊂ D be a full triangulated subcategory. Set f ∈ Arrows(D) such that there exists a distinguished triangle (6.6.1)05RH S = (X,Y,Z,f,g,h) of D with Z isomorphic to an object of D0 Then S is a multiplicative system compatible with the triangulated structure on D. In this situation the following are equivalent (1) S is a saturated multiplicative system, (2) D0 is a saturated triangulated subcategory. Proof. To prove the first assertion we have to prove that MS1, MS2, MS3 and MS5, MS6 hold. Proof of MS1. It is clear that identities are in S because (X,X, 0, 1, 0, 0) is distinguished for every object X of D and because 0 is an object of D0. Let f : X → Y and g : Y → Z be composable morphisms contained in S. Choose distinguished triangles (X,Y,Q1, f, p1, d1), (X,Z,Q2, g ◦ f, p2, d2), and (Y,Z,Q3, g, p3, d3). By 0 assumption we know that Q1 and Q3 are isomorphic to objects of D . By TR4 we 0 know there exists a distinguished triangle (Q1,Q2,Q3, a, b, c). Since D is a trian- 0 gulated subcategory we conclude that Q2 is isomorphic to an object of D . Hence g ◦ f ∈ S. Proof of MS3. Let a : X → Y be a morphism and let t : Z → X be an element of S such that a ◦ t = 0. To prove LMS3 it suffices to find an s ∈ S such that s ◦ a = 0, compare with the proof of Lemma 5.3. Choose a distinguished triangle (Z, X, Q, t, g, h) using TR1 and TR2. Since a ◦ t = 0 we see by Lemma 4.2 there exists a morphism i : Q → Y such that i ◦ g = a. Finally, using TR1 again we can choose a triangle (Q, Y, W, i, s, k). Here is a picture Z X Q Z[1] t / g / /

1 i X a / Y s W Since t ∈ S we see that Q is isomorphic to an object of D0. Hence s ∈ S. Finally, s ◦ a = s ◦ i ◦ g = 0 as s ◦ i = 0 by Lemma 4.1. We conclude that LMS3 holds. The proof of RMS3 is dual. Proof of MS5. Follows as distinguished triangles and D0 are stable under transla- tions DERIVED CATEGORIES 22

Proof of MS6. Suppose given a commutative diagram

X / Y

s s0 X0 / Y 0 with s, s0 ∈ S. By Proposition 4.23 we can extend this to a nine square diagram. As s, s0 are elements of S we see that X00,Y 00 are isomorphic to objects of D0. Since D0 is a full triangulated subcategory we see that Z00 is also isomorphic to an object of D0. Whence the morphism Z → Z0 is an element of S. This proves MS6. MS2 is a formal consequence of MS1, MS5, and MS6, see Lemma 5.2. This ﬁnishes the proof of the ﬁrst assertion of the lemma. Let’s assume that S is saturated. (In the following we will use rotation of distinguished triangles without further mention.) Let X⊕Y be an object isomorphic to an object of D0. Consider the morphism f : 0 → X. The composition 0 → X → X ⊕Y is an element of S as (0,X ⊕ Y,X ⊕ Y, 0, 1, 0) is a distinguished triangle. The composition Y [−1] → 0 → X is an element of S as (X,X ⊕ Y,Y, (1, 0), (0, 1), 0) is a distinguished triangle, see Lemma 4.11. Hence 0 → X is an element of S (as S is saturated). Thus X is isomorphic to an object of D0 as desired. Finally, assume D0 is a saturated triangulated subcategory. Let

g f W −→h X −→ Y −→ Z be composable morphisms of D such that fg, gh ∈ S. We will build up a picture of objects as in the diagram below.

Q12 Q23 = =

+1 ! +1 ! Q1 o Q2 o Q3 a a ` +1 +1 +1

~ } } W / X / Y / Z

First choose distinguished triangles (W, X, Q1), (X,Y,Q2), (Y,Z,Q3)(W, Y, Q12), and (X,Z,Q23). Denote s : Q2 → Q1[1] the composition Q2 → X[1] → Q1[1]. Denote t : Q3 → Q2[1] the composition Q3 → Y [1] → Q2[1]. By TR4 applied to the composition W → X → Y and the composition X → Y → Z there exist distinguished triangles (Q1,Q12,Q2) and (Q2,Q23,Q3) which use the morphisms s 0 and t. The objects Q12 and Q23 are isomorphic to objects of D as W → Y and X → Z are assumed in S. Hence also s[1]t is an element of S as S is closed under compositions and shifts. Note that s[1]t = 0 as Y [1] → Q2[1] → X[2] is zero, see 0 Lemma 4.1. Hence Q3[1] ⊕ Q1[2] is isomorphic to an object of D , see Lemma 4.11. 0 By assumption on D we conclude that Q3 and Q1 are isomorphic to objects of 0 D . Looking at the distinguished triangle (Q1,Q12,Q2) we conclude that Q2 is also 0 isomorphic to an object of D . Looking at the distinguished triangle (X,Y,Q2) we ﬁnally conclude that g ∈ S. (It is also follows that h, f ∈ S, but we don’t need this.) DERIVED CATEGORIES 23

05RIDefinition 6.7. Let D be a triangulated category. Let B be a full triangulated subcategory. We define the quotient category D/B by the formula D/B = S−1D, where S is the multiplicative system of D associated to B via Lemma 6.6. The localization functor Q : D → D/B is called the quotient functor in this case. Note that the quotient functor Q : D → D/B is an exact functor of triangulated categories, see Proposition 5.5. The universal property of this construction is the following. 05RJ Lemma 6.8. Let D be a triangulated category. Let B be a full triangulated subcategory of D. Let Q : D → D/B be the quotient functor. (1) If H : D → A is a homological functor into an abelian category A such that B ⊂ Ker(H) then there exists a unique factorization H0 : D/B → A such that H = H0 ◦ Q and H0 is a homological functor too. (2) If F : D → D0 is an exact functor into a pre-triangulated category D0 such that B ⊂ Ker(F ) then there exists a unique factorization F 0 : D/B → D0 such that F = F 0 ◦ Q and F 0 is an exact functor too. Proof. This lemma follows from Lemma 5.6. Namely, if f : X → Y is a morphism of D such that for some distinguished triangle (X,Y,Z,f,g,h) the object Z is isomorphic to an object of B, then H(f), resp. F (f) is an isomorphism under the assumptions of (1), resp. (2). Details omitted. The kernel of the quotient functor can be described as follows. 05RK Lemma 6.9. Let D be a triangulated category. Let B be a full triangulated subcategory. The kernel of the quotient functor Q : D → D/B is the strictly full subcategory of D whose objects are Z ∈ Ob(D) such that there exists a Z0 ∈ Ob(D) Ob(Ker(Q)) = such that Z ⊕ Z0 is isomorphic to an object of B In other words it is the smallest strictly full saturated triangulated subcategory of D containing B. Proof. First note that the kernel is automatically a strictly full triangulated subcategory containing summands of any of its objects, see Lemma 6.2. The description of its objects follows from the definitions and Lemma 5.7 part (4). Let D be a triangulated category. At this point we have constructions which induce order preserving maps between (1) the partially ordered set of multiplicative systems S in D compatible with the triangulated structure, and (2) the partially ordered set of full triangulated subcategories B ⊂ D. Namely, the constructions are given by S 7→ B(S) = Ker(Q : D → S−1D) and B 7→ S(B) where S(B) is the multiplicative set of (6.6.1), i.e., f ∈ Arrows(D) such that there exists a distinguished triangle S(B) = (X,Y,Z,f,g,h) of D with Z isomorphic to an object of B Note that it is not the case that these operations are mutually inverse. 05RL Lemma 6.10. Let D be a triangulated category. The operations described above have the following properties DERIVED CATEGORIES 24

(1) S(B(S)) is the “saturation” of S, i.e., it is the smallest saturated multiplicative system in D containing S, and (2) B(S(B)) is the “saturation” of B, i.e., it is the smallest strictly full saturated triangulated subcategory of D containing B. In particular, the constructions deﬁne mutually inverse maps between the (partially ordered) set of saturated multiplicative systems in D compatible with the triangulated structure on D and the (partially ordered) set of strictly full saturated triangulated subcategories of D.

Proof. First, let’s start with a full triangulated subcategory B. Then B(S(B)) = Ker(Q : D → D/B) and hence (2) is the content of Lemma 6.9. Next, suppose that S is multiplicative system in D compatible with the triangula- tion on D. Then B(S) = Ker(Q : D → S−1D). Hence (using Lemma 4.9 in the localized category) f ∈ Arrows(D) such that there exists a distinguished S(B(S)) = triangle (X,Y,Z,f,g,h) of D with Q(Z) = 0 = {f ∈ Arrows(D) | Q(f) is an isomorphism} = Sˆ = S0 in the notation of Categories, Lemma 27.21. The ﬁnal statement of that lemma ﬁnishes the proof.

05RM Lemma 6.11. Let H : D → A be a homological functor from a triangulated category D to an abelian category A, see Deﬁnition 3.5. The subcategory Ker(H) of D is a strictly full saturated triangulated subcategory of D whose corresponding saturated multiplicative system (see Lemma 6.10) is the set

S = {f ∈ Arrows(D) | Hi(f) is an isomorphism for all i ∈ Z}. The functor H factors through the quotient functor Q : D → D/ Ker(H).

Proof. The category Ker(H) is a strictly full saturated triangulated subcategory of D by Lemma 6.3. The set S is a saturated multiplicative system compatible with the triangulated structure by Lemma 5.4. Recall that the multiplicative system corresponding to Ker(H) is the set f ∈ Arrows(D) such that there exists a distinguished triangle (X,Y,Z,f,g,h) with Hi(Z) = 0 for all i

By the long exact cohomology sequence, see (3.5.1), it is clear that f is an element of this set if and only if f is an element of S. Finally, the factorization of H through Q is a consequence of Lemma 6.8.

7. Adjoints for exact functors 0A8C Results on adjoint functors between triangulated categories.

0A8D Lemma 7.1. Let F : D → D0 be an exact functor between triangulated categories. If F admits a right adjoint G : D0 → D, then G is also an exact functor. DERIVED CATEGORIES 25

Proof. Let X be an object of D and A an object of D0. Since F is an exact functor we see that

MorD(X,G(A[1]) = MorD0 (F (X),A[1])

= MorD0 (F (X)[−1],A)

= MorD0 (F (X[−1]),A)

= MorD(X[−1],G(A))

= MorD(X,G(A)[1]) By Yoneda’s lemma (Categories, Lemma 3.5) we obtain a canonical isomorphism G(A)[1] = G(A[1]). Let A → B → C → A[1] be a distinguished triangle in D0. Choose a distinguished triangle G(A) → G(B) → X → G(A)[1] in D. Then F (G(A)) → F (G(B)) → F (X) → F (G(A))[1] is a distinguished triangle in D0. By TR3 we can choose a morphism of distinguished triangles F (G(A)) / F (G(B)) / F (X) / F (G(A))[1]

A / B / C / A[1] Since G is the adjoint the new morphism determines a morphism X → G(C) such that the diagram G(A) / G(B) / X / G(A)[1]

G(A) / G(B) / G(C) / G(A)[1]

0 commutes. Applying the homological functor HomD0 (W, −) for an object W of D we deduce from the 5 lemma that

HomD0 (W, X) → HomD0 (W, G(C)) is a bijection and using the Yoneda lemma once more we conclude that X → G(C) is an isomorphism. Hence we conclude that G(A) → G(B) → G(C) → G(A)[1] is a distinguished triangle which is what we wanted to show. 09J1 Lemma 7.2. Let D, D0 be triangulated categories. Let F : D → D0 and G : D0 → D be functors. Assume that (1) F and G are exact functors, (2) F is fully faithful, (3) G is a right adjoint to F , and (4) the kernel of G is zero. Then F is an equivalence of categories. Proof. Since F is fully faithful the adjunction map id → G ◦ F is an isomorphism (Categories, Lemma 24.4). Let X be an object of D0. Choose a distinguished triangle F (G(X)) → X → Y → F (G(X))[1] DERIVED CATEGORIES 26

in D0. Applying G and using that G(F (G(X))) = G(X) we ﬁnd a distinguished triangle G(X) → G(X) → G(Y ) → G(X)[1]

Hence G(Y ) = 0. Thus Y = 0. Thus F (G(X)) → X is an isomorphism.

8. The homotopy category 05RN Let A be an additive category. The homotopy category K(A) of A is the category of complexes of A with morphisms given by morphisms of complexes up to homotopy. Here is the formal deﬁnition.

013HDeﬁnition 8.1. Let A be an additive category. (1) We set Comp(A) = CoCh(A) be the category of (cochain) complexes. (2) A complex K• is said to be bounded below if Kn = 0 for all n 0. (3) A complex K• is said to be bounded above if Kn = 0 for all n 0. (4) A complex K• is said to be bounded if Kn = 0 for all |n| 0. (5) We let Comp+(A), Comp−(A), resp. Compb(A) be the full subcategory of Comp(A) whose objects are the complexes which are bounded below, bounded above, resp. bounded. (6) We let K(A) be the category with the same objects as Comp(A) but as morphisms homotopy classes of maps of complexes (see Homology, Lemma 13.7). (7) We let K+(A), K−(A), resp. Kb(A) be the full subcategory of K(A) whose objects are bounded below, bounded above, resp. bounded complexes of A.

It will turn out that the categories K(A), K+(A), K−(A), and Kb(A) are triangulated categories. To prove this we ﬁrst develop some machinery related to cones and split exact sequences.

9. Cones and termwise split sequences 014D Let A be an additive category, and let K(A) denote the category of complexes of A with morphisms given by morphisms of complexes up to homotopy. Note that the shift functors [n] on complexes, see Homology, Deﬁnition 14.7, give rise to functors [n]: K(A) → K(A) such that [n] ◦ [m] = [n + m] and [0] = id.

014EDeﬁnition 9.1. Let A be an additive category. Let f : K• → L• be a morphism of complexes of A. The cone of f is the complex C(f)• given by C(f)n = Ln ⊕ Kn+1 and diﬀerential n n+1 n dL f dC(f) = n+1 0 −dK It comes equipped with canonical morphisms of complexes i : L• → C(f)• and p : C(f)• → K•[1] induced by the obvious maps Ln → C(f)n → Kn+1.

In other words (K, L, C(f), f, i, p) forms a triangle: K• → L• → C(f)• → K•[1] The formation of this triangle is functorial in the following sense. DERIVED CATEGORIES 27

014F Lemma 9.2. Suppose that • • K1 / L1 f1 a b

• f2 • K2 / L2 is a diagram of morphisms of complexes which is commutative up to homotopy. • • Then there exists a morphism c : C(f1) → C(f2) which gives rise to a mor- • • • • • • phism of triangles (a, b, c):(K1 ,L1,C(f1) , f1, i1, p1) → (K2 ,L2,C(f2) , f2, i2, p2) of K(A).

n n n−1 Proof. Let h : K1 → L2 be a family of morphisms such that b ◦ f1 − f2 ◦ a = d ◦ h + h ◦ d. Deﬁne cn by the matrix bn hn+1 cn = : Ln ⊕ Kn+1 → Ln ⊕ Kn+1 0 an+1 1 1 2 2 A matrix computation show that c is a morphism of complexes. It is trivial that c ◦ i1 = i2 ◦ b, and it is trivial also to check that p2 ◦ c = a ◦ p1. • • Note that the morphism c : C(f1) → C(f2) constructed in the proof of Lemma 9.2 in general depends on the chosen homotopy h between f2 ◦ a and b ◦ f1. 08RI Lemma 9.3. Suppose that f : K• → L• and g : L• → M • are morphisms of complexes such that g ◦ f is homotopic to zero. Then (1) g factors through a morphism C(f)• → M •, and (2) f factors through a morphism K• → C(g)•[−1]. Proof. The assumptions say that the diagram K• / L• f g 0 / M • commutes up to homotopy. Since the cone on 0 → M • is M • the map C(f)• → C(0 → M •) = M • of Lemma 9.2 is the map in (1). The cone on K• → 0 is K•[1] and applying Lemma 9.2 gives a map K•[1] → C(g)•. Applying [−1] we obtain the map in (2). Note that the morphisms C(f)• → M • and K• → C(g)•[−1] constructed in the proof of Lemma 9.3 in general depend on the chosen homotopy. 014GDeﬁnition 9.4. Let A be an additive category. A termwise split injection α : A• → B• is a morphism of complexes such that each An → Bn is isomorphic to the inclusion of a direct summand. A termwise split surjection β : B• → C• is a morphism of complexes such that each Bn → Cn is isomorphic to the projection onto a direct summand. 014H Lemma 9.5. Let A be an additive category. Let

A• / B• f a b g C• / D• DERIVED CATEGORIES 28

be a diagram of morphisms of complexes commuting up to homotopy. If f is a termwise split injection, then b is homotopic to a morphism which makes the diagram commute. If g is a split surjection, then a is homotopic to a morphism which makes the diagram commute. Proof. Let hn : An → Dn−1 be a collection of morphisms such that bf − ga = dh + hd. Suppose that πn : Bn → An are morphisms splitting the morphisms f n. Take b0 = b − dhπ − hπd. Suppose sn : Dn → Cn are morphisms splitting the n n n 0 morphisms g : C → D . Take a = a + dsh + shd. Computations omitted. The following lemma can be used to replace a morphism of complexes by a morphism where in each degree the map is the injection of a direct summand. 013N Lemma 9.6. Let A be an additive category. Let α : K• → L• be a morphism of complexes of A. There exists a factorization

K• α˜ L˜• π L• / /6 α such that (1) α˜ is a termwise split injection (see Definition 9.4), • • (2) there is a map of complexes s : L → L˜ such that π ◦ s = idL• and such that s ◦ π is homotopic to idL˜• . Moreover, if both K• and L• are in K+(A), K−(A), or Kb(A), then so is L˜•. Proof. We set Lñ = Ln ⊕ Kn ⊕ Kn+1 and we define  n  dL 0 0 n n n+1 dL˜ =  0 dK idK  n+1 0 0 −dK • • In other words, L˜ = L ⊕ C(1K• ). Moreover, we set  α  α˜ = idKn  0 which is clearly a split injection. It is also clear that it defines a morphism of complexes. We define π = idLn 0 0 so that clearly π ◦ α˜ = α. We set   idLn s =  0  0

n n n−1 so that π ◦ s = idL• . Finally, let h : L˜ → L˜ be the map which maps the summand Kn of L˜n via the identity morphism to the summand Kn of L˜n−1. Then it is a trivial matter (see computations in remark below) to prove that

idL˜• − s ◦ π = d ◦ h + h ◦ d which ﬁnishes the proof of the lemma. DERIVED CATEGORIES 29

013ORemark 9.7. To see the last displayed equality in the proof above we can argue with elements as follows. We have sπ(l, k, k+) = (l, 0, 0). Hence the morphism of the left hand side maps (l, k, k+) to (0, k, k+). On the other hand h(l, k, k+) = (0, 0, k) and d(l, k, k+) = (dl, dk + k+, −dk+). Hence (dh + hd)(l, k, k+) = d(0, 0, k) + h(dl, dk + k+, −dk+) = (0, k, −dk) + (0, 0, dk + k+) = (0, k, k+) as desired. 0642 Lemma 9.8. Let A be an additive category. Let α : K• → L• be a morphism of complexes of A. There exists a factorization

K• i K˜ • α˜ L• / /6 α such that (1) α˜ is a termwise split surjection (see Definition 9.4), • • (2) there is a map of complexes s : K˜ → K such that s ◦ i = idK• and such that i ◦ s is homotopic to idK˜ • . Moreover, if both K• and L• are in K+(A), K−(A), or Kb(A), then so is K˜ •. Proof. Dual to Lemma 9.6. Take K˜ n = Kn ⊕ Ln−1 ⊕ Ln and we define  n  dK 0 0 n n−1 n dK˜ =  0 −dL idL  n 0 0 dL ˜ • • in other words K = K ⊕ C(1L•[−1]). Moreover, we set α˜ = α 0 idLn which is clearly a split surjection. It is also clear that it defines a morphism of complexes. We define   idKn i =  0  0 so that clearly α˜ ◦ i = α. We set s = idKn 0 0

n n n−1 so that s ◦ i = idK• . Finally, let h : K˜ → K˜ be the map which maps the summand Ln−1 of K˜ n via the identity morphism to the summand Ln−1 of K˜ n−1. Then it is a trivial matter to prove that

idK˜ • − i ◦ s = d ◦ h + h ◦ d which ﬁnishes the proof of the lemma. 014IDeﬁnition 9.9. Let A be an additive category. A termwise split exact sequence of complexes of A is a complex of complexes β 0 → A• −→α B• −→ C• → 0 together with given direct sum decompositions Bn = An ⊕ Cn compatible with αn and βn. We often write sn : Cn → Bn and πn : Bn → An for the maps induced by DERIVED CATEGORIES 30

the direct sum decompositions. According to Homology, Lemma 14.10 we get an associated morphism of complexes δ : C• −→ A•[1]

n+1 n n • • • which in degree n is the map π ◦ dB ◦ s . In other words (A ,B ,C , α, β, δ) forms a triangle A• → B• → C• → A•[1] This will be the triangle associated to the termwise split sequence of complexes. 05SS Lemma 9.10. Let A be an additive category. Let 0 → A• → B• → C• → 0 be termwise split exact sequences as in Deﬁnition 9.9. Let (π0)n, (s0)n be a second collection of splittings. Denote δ0 : C• −→ A•[1] the morphism associated to this second set of splittings. Then (1, 1, 1) : (A•,B•,C•, α, β, δ) −→ (A•,B•,C•, α, β, δ0) is an isomorphism of triangles in K(A). Proof. The statement simply means that δ and δ0 are homotopic maps of complexes. This is Homology, Lemma 14.12.

• • • 014JRemark 9.11. Let A be an additive category. Let 0 → Ai → Bi → Ci → 0, • • • • i = 1, 2 be termwise split exact sequences. Suppose that a : A1 → A2, b : B1 → B2 , • • and c : C1 → C2 are morphisms of complexes such that

• • • A1 / B1 / C1

a b c

• • • A2 / B2 / C2

0 • • commutes in K(A). In general, there does not exist a morphism b : B1 → B2 which is homotopic to b such that the diagram above commutes in the category of complexes. Namely, consider Examples, Equation (62.0.1). If we could replace the middle map there by a homotopic one such that the diagram commutes, then we would have additivity of traces which we do not.

• • • 086L Lemma 9.12. Let A be an additive category. Let 0 → Ai → Bi → Ci → 0, • • i = 1, 2, 3 be termwise split exact sequences of complexes. Let b : B1 → B2 and 0 • • b : B2 → B3 be morphisms of complexes such that

• • • • • • A1 / B1 / C1 A2 / B2 / C2

0 b 0 and 0 b0 0

• • • • • • A2 / B2 / C2 A3 / B3 / C3 commute in K(A). Then b0 ◦ b = 0 in K(A). Proof. By Lemma 9.5 we can replace b and b0 by homotopic maps such that the right square of the left diagram commutes and the left square of the right diagram n n n 0 n commutes. In other words, we have Im(b ) ⊂ Im(A2 → B2 ) and Ker((b ) ) ⊃ n n 0 Im(A2 → B2 ). Then b ◦ b = 0 as a map of complexes. DERIVED CATEGORIES 31

• • • • 014K Lemma 9.13. Let A be an additive category. Let f1 : K1 → L1 and f2 : K2 → L2 be morphisms of complexes. Let

• • • • • • (a, b, c):(K1 ,L1,C(f1) , f1, i1, p1) −→ (K2 ,L2,C(f2) , f2, i2, p2) be any morphism of triangles of K(A). If a and b are homotopy equivalences then so is c.

−1 • • Proof. Let a : K2 → K1 be a morphism of complexes which is inverse to a −1 • • in K(A). Let b : L2 → L1 be a morphism of complexes which is inverse to b 0 • • in K(A). Let c : C(f2) → C(f1) be the morphism from Lemma 9.2 applied −1 −1 0 0 to f1 ◦ a = b ◦ f2. If we can show that c ◦ c and c ◦ c are isomorphisms in K(A) then we win. Hence it suﬃces to prove the following: Given a morphism of triangles (1, 1, c):(K•,L•,C(f)•, f, i, p) in K(A) the morphism c is an isomorphism in K(A). By assumption the two squares in the diagram

L• / C(f)• / K•[1]

1 c 1 L• / C(f)• / K•[1]

commute up to homotopy. By construction of C(f)• the rows form termwise split sequences of complexes. Thus we see that (c − 1)2 = 0 in K(A) by Lemma 9.12. Hence c is an isomorphism in K(A) with inverse 2 − c. Hence if a and b are homotopy equivalences then the resulting morphism of triangles is an isomorphism of triangles in K(A). It turns out that the collection of triangles of K(A) given by cones and the collection of triangles of K(A) given by termwise split sequences of complexes are the same up to isomorphisms, at least up to sign! 014L Lemma 9.14. Let A be an additive category. (1) Given a termwise split sequence of complexes (α : A• → B•, β : B• → C•, sn, πn) there exists a homotopy equivalence C(α)• → C• such that the diagram A• B• C(α)• A•[1] / / −p /

δ A• / B• / C• / A•[1] deﬁnes an isomorphism of triangles in K(A). (2) Given a morphism of complexes f : K• → L• there exists an isomorphism of triangles

K• / L˜• / M • / K•[1] δ

−p K• / L• / C(f)• / K•[1] where the upper triangle is the triangle associated to a termwise split exact sequence K• → L˜• → M •. DERIVED CATEGORIES 32

Proof. Proof of (1). We have C(α)n = Bn ⊕ An+1 and we simply define C(α)n → Cn via the projection onto Bn followed by βn. This defines a morphism of complexes because the compositions An+1 → Bn+1 → Cn+1 are zero. To get a homotopy inverse we take C• → C(α)• given by (sn, −δn) in degree n. This is a morphism of complexes because the morphism δn can be characterized as the unique morphism Cn → An+1 such that d ◦ sn − sn+1 ◦ d = α ◦ δn, see proof of Homology, Lemma 14.10. The composition C• → C(α)• → C• is the identity. The composition C(α)• → C• → C(α)• is equal to the morphism sn ◦ βn 0 −δn ◦ βn 0 To see that this is homotopic to the identity map use the homotopy hn : C(α)n → C(α)n−1 given by the matrix 0 0 : C(α)n = Bn ⊕ An+1 → Bn−1 ⊕ An = C(α)n−1 πn 0 It is trivial to verify that 1 0 sn d αn 0 0 0 0 d αn+1 − βn 0 = + 0 1 −δn 0 −d πn 0 πn+1 0 0 −d To finish the proof of (1) we have to show that the morphisms −p : C(α)• → A•[1] (see Definition 9.1) and C(α)• → C• → A•[1] agree up to homotopy. This is clear from the above. Namely, we can use the homotopy inverse (s, −δ): C• → C(α)• and check instead that the two maps C• → A•[1] agree. And note that p ◦ (s, −δ) = −δ as desired. Proof of (2). We let f˜ : K• → L˜•, s : L• → L˜• and π : L˜• → L• be as in Lemma 9.6. By Lemmas 9.2 and 9.13 the triangles (K•,L•,C(f), i, p) and (K•, L˜•,C(f˜),˜i, p˜) are isomorphic. Note that we can compose isomorphisms of triangles. Thus we may replace L• by L˜• and f by f˜. In other words we may assume that f is a termwise split injection. In this case the result follows from part (1). • • • 014M Lemma 9.15. Let A be an additive category. Let A1 → A2 → ... → An be a sequence of composable morphisms of complexes. There exists a commutative diagram • • • A1 / A2 / ... / An O O O

• • • B1 / B2 / ... / Bn • • • • such that each morphism Bi → Bi+1 is a split injection and each Bi → Ai is a • + − b homotopy equivalence. Moreover, if all Ai are in K (A), K (A), or K (A), then • so are the Bi . Proof. The case n = 1 is without content. Lemma 9.6 is the case n = 2. Suppose • we have constructed the diagram except for Bn. Apply Lemma 9.6 to the compo- • • • • • • sition Bn−1 → An−1 → An. The result is a factorization Bn−1 → Bn → An as desired. 014N Lemma 9.16. Let A be an additive category. Let (α : A• → B•, β : B• → C•, sn, πn) be a termwise split sequence of complexes. Let (A•,B•,C•, α, β, δ) be DERIVED CATEGORIES 33

the associated triangle. Then the triangle (C•[−1],A•,B•, δ[−1], α, β) is isomorphic to the triangle (C•[−1],A•,C(δ[−1])•, δ[−1], i, p). Proof. We write Bn = An ⊕ Cn and we identify αn and βn with the natural inclusion and projection maps. By construction of δ we have n n n dA δ dB = n 0 dC On the other hand the cone of δ[−1] : C•[−1] → A• is given as C(δ[−1])n = An⊕Cn with diﬀerential identical with the matrix above! Whence the lemma. 014O Lemma 9.17. Let A be an additive category. Let f : K• → L• be a morphism of complexes. The triangle (L•,C(f)•,K•[1], i, p, f[1]) is the triangle associated to the termwise split sequence 0 → L• → C(f)• → K•[1] → 0 coming from the deﬁnition of the cone of f.

Proof. Immediate from the definitions. 10. Distinguished triangles in the homotopy category 014P Since we want our boundary maps in long exact sequences of cohomology to be given by the maps in the snake lemma without signs we define distinguished triangles in the homotopy category as follows. 014QDefinition 10.1. Let A be an additive category. A triangle (X,Y,Z,f,g,h) of K(A) is called a distinguished triangle of K(A) if it is isomorphic to the triangle associated to a termwise split exact sequence of complexes, see Definition 9.9. Same definition for K+(A), K−(A), and Kb(A). Note that according to Lemma 9.14 a triangle of the form (K•,L•,C(f)•, f, i, −p) is a distinguished triangle. This does indeed lead to a triangulated category, see Proposition 10.3. Before we can prove the proposition we need one more lemma in order to be able to prove TR4. 014R Lemma 10.2. Let A be an additive category. Suppose that α : A• → B• and β : B• → C• are split injections of complexes. Then there exist distinguished • • • • • • • • • triangles (A ,B ,Q1, α, p1, d1), (A ,C ,Q2, β ◦α, p2, d2) and (B ,C ,Q3, β, p3, d3) for which TR4 holds.

n n n n n n Proof. Say π1 : B → A , and π3 : C → B are the splittings. Then also • • n n n • • A → C is a split injection with splittings π2 = π1 ◦ π3 . Let us write Q1, Q2 and • n n n n Q3 for the “quotient” complexes. In other words, Q1 = Ker(π1 ), Q3 = Ker(π3 ) n n n n n and Q2 = Ker(π2 ). Note that the kernels exist. Then B = A ⊕ Q1 and Cn = n n n n B ⊕ Q3 , where we think of A as a subobject of B and so on. This implies n n n n n n n n n C = A ⊕ Q1 ⊕ Q3 . Note that π2 = π1 ◦ π3 is zero on both Q1 and Q3 . Hence n n n Q2 = Q1 ⊕ Q3 . Consider the commutative diagram • • • 0 → A → B → Q1 → 0 ↓ ↓ ↓ • • • 0 → A → C → Q2 → 0 ↓ ↓ ↓ • • • 0 → B → C → Q3 → 0 DERIVED CATEGORIES 34

The rows of this diagram are termwise split exact sequences, and hence determine distinguished triangles by definition. Moreover downward arrows in the diagram above are compatible with the chosen splittings and hence define morphisms of triangles • • • • • • • • (A → B → Q1 → A [1]) −→ (A → C → Q2 → A [1]) and • • • • • • • • (A → C → Q2 → A [1]) −→ (B → C → Q3 → B [1]). n n Note that the splittings Q3 → C of the bottom split sequence in the diagram • • • provides a splitting for the split sequence 0 → Q1 → Q2 → Q3 → 0 upon composing n n • • with C → Q2 . It follows easily from this that the morphism δ : Q3 → Q1[1] in the corresponding distinguished triangle • • • • (Q1 → Q2 → Q3 → Q1[1]) • • • is equal to the composition Q3 → B [1] → Q1[1]. Hence we get a structure as in the conclusion of axiom TR4. 014S Proposition 10.3. Let A be an additive category. The category K(A) of complexes up to homotopy with its natural translation functors and distinguished triangles as defined above is a triangulated category. Proof. Proof of TR1. By definition every triangle isomorphic to a distinguished one is distinguished. Also, any triangle (A•,A•, 0, 1, 0, 0) is distinguished since 0 → A• → A• → 0 → 0 is a termwise split sequence of complexes. Finally, given any morphism of complexes f : K• → L• the triangle (K, L, C(f), f, i, −p) is distinguished by Lemma 9.14. Proof of TR2. Let (X,Y,Z,f,g,h) be a triangle. Assume (Y,Z,X[1], g, h, −f[1]) is distinguished. Then there exists a termwise split sequence of complexes A• → B• → C• such that the associated triangle (A•,B•,C•, α, β, δ) is isomorphic to (Y,Z,X[1], g, h, −f[1]). Rotating back we see that (X,Y,Z,f,g,h) is isomorphic to (C•[−1],A•,B•, −δ[−1], α, β). It follows from Lemma 9.16 that the triangle (C•[−1],A•,B•, δ[−1], α, β) is isomorphic to (C•[−1],A•,C(δ[−1])•, δ[−1], i, p). Pre- composing the previous isomorphism of triangles with −1 on Y it follows that (X,Y,Z,f,g,h) is isomorphic to (C•[−1],A•,C(δ[−1])•, δ[−1], i, −p). Hence it is distinguished by Lemma 9.14. On the other hand, suppose that (X,Y,Z,f,g,h) is distinguished. By Lemma 9.14 this means that it is isomorphic to a triangle of the form (K•,L•,C(f), f, i, −p) for some morphism of complexes f. Then the ro- tated triangle (Y,Z,X[1], g, h, −f[1]) is isomorphic to (L•,C(f),K•[1], i, −p, −f[1]) which is isomorphic to the triangle (L•,C(f),K•[1], i, p, f[1]). By Lemma 9.17 this triangle is distinguished. Hence (Y,Z,X[1], g, h, −f[1]) is distinguished as desired. Proof of TR3. Let (X,Y,Z,f,g,h) and (X0,Y 0,Z0, f 0, g0, h0) be distinguished triangles of K(A) and let a : X → X0 and b : Y → Y 0 be morphisms such that f 0 ◦ a = b ◦ f. By Lemma 9.14 we may assume that (X,Y,Z,f,g,h) = (X,Y,C(f), f, i, −p) and (X0,Y 0,Z0, f 0, g0, h0) = (X0,Y 0,C(f 0), f 0, i0, −p0). At this point we simply apply Lemma 9.2 to the commutative diagram given by f, f 0, a, b. Proof of TR4. At this point we know that K(A) is a pre-triangulated category. Hence we can use Lemma 4.15. Let A• → B• and B• → C• be composable morphisms of K(A). By Lemma 9.15 we may assume that A• → B• and B• → C• are split injective morphisms. In this case the result follows from Lemma 10.2. DERIVED CATEGORIES 35

05RPRemark 10.4. Let A be an additive category. Exactly the same proof as the proof of Proposition 10.3 shows that the categories K+(A), K−(A), and Kb(A) are triangulated categories. Namely, the cone of a morphism between bounded (above, below) is bounded (above, below). But we prove below that these are triangulated subcategories of K(A) which gives another proof. 05RQ Lemma 10.5. Let A be an additive category. The categories K+(A), K−(A), and Kb(A) are full triangulated subcategories of K(A). Proof. Each of the categories mentioned is a full additive subcategory. We use the criterion of Lemma 4.16 to show that they are triangulated subcategories. It is clear that each of the categories K+(A), K−(A), and Kb(A) is preserved under the shift functors [1], [−1]. Finally, suppose that f : A• → B• is a morphism in K+(A), K−(A), or Kb(A). Then (A•,B•,C(f)•, f, i, −p) is a distinguished triangle of K(A) with C(f)• ∈ K+(A), K−(A), or Kb(A) as is clear from the construction of the cone. Thus the lemma is proved. (Alternatively, K• → L• is isomorphic to an termwise split injection of complexes in K+(A), K−(A), or Kb(A), see Lemma 9.6 and then one can directly take the associated distinguished triangle.) 014X Lemma 10.6. Let A, B be additive categories. Let F : A → B be an additive functor. The induced functors F : K(A) −→ K(B) F : K+(A) −→ K+(B) F : K−(A) −→ K−(B) F : Kb(A) −→ Kb(B) are exact functors of triangulated categories. Proof. Suppose A• → B• → C• is a termwise split sequence of complexes of A with splittings (sn, πn) and associated morphism δ : C• → A•[1], see Deﬁnition 9.9. Then F (A•) → F (B•) → F (C•) is a termwise split sequence of complexes with splittings (F (sn),F (πn)) and associated morphism F (δ): F (C•) → F (A•)[1]. Thus F transforms distinguished triangles into distinguished triangles. 0G6C Lemma 10.7. Let A be an additive category. Let (A•,B•,C•, a, b, c) be a distinguished triangle in K(A). Then there exists an isomorphic distinguished triangle (A•, (B0)•,C•, a0, b0, c) such that 0 → An → (B0)n → Cn → 0 is a split short exact sequence for all n. Proof. We will use that K(A) is a triangulated category by Proposition 10.3. Let W • be the cone on c : C• → A•[1] with its maps i : A•[1] → W • and p : W • → C•[1]. Then (C•,A•[1],W •, c, i, −p) is a distinguished triangle by Lemma 9.14. Rotating backwards twice we see that (A•,W •[−1],C•, −i[−1], p[−1], c) is a distinguished triangle. By TR3 there is a morphism of distinguished triangles (id, β, id):(A•,B•,C•, a, b, c) → (A•,W •[−1],C•, −i[−1], p[−1], c) which must be an isomorphism by Lemma 4.3. This ﬁnishes the proof because 0 → A• → W •[−1] → C• → 0 is a termwise split short exact sequence of complexes by the very construction of cones in Section 9. 0G6DRemark 10.8. Let A be an additive category with countable direct sums. Let DoubleComp(A) denote the category of double complexes in A, see Homology, Section 18. We can use this category to construct two triangulated categories. DERIVED CATEGORIES 36

(1) We can consider an object A•,• of DoubleComp(A) as a complex of complexes as follows ... → A•,−1 → A•,0 → A•,1 → ...

and take the homotopy category Kfirst(DoubleComp(A)) with the corresponding triangulated structure given by Proposition 10.3. By Homology, Remark 18.6 the functor

Tot : Kfirst(DoubleComp(A)) −→ K(A) is an exact functor of triangulated categories. (2) We can consider an object A•,• of DoubleComp(A) as a complex of complexes as follows ... → A−1,• → A0,• → A1,• → ...

and take the homotopy category Ksecond(DoubleComp(A)) with the corresponding triangulated structure given by Proposition 10.3. By Homology, Remark 18.7 the functor

Tot : Ksecond(DoubleComp(A)) −→ K(A) is an exact functor of triangulated categories. 0G6ERemark 10.9. Let A, B, C be additive categories and assume C has countable direct sums. Suppose that ⊗ : A × B −→ C, (X,Y ) 7−→ X ⊗ Y is a functor which is bilinear on morphisms. This determines a functor Comp(A) × Comp(B) −→ DoubleComp(C), (X•,Y •) 7−→ X• ⊗ Y • See Homology, Example 18.2. (1) For a ﬁxed object X• of Comp(A) the functor K(B) −→ K(C),Y • 7−→ Tot(X• ⊗ Y •) is an exact functor of triangulated categories. (2) For a ﬁxed object Y • of Comp(B) the functor K(A) −→ K(C),X• 7−→ Tot(X• ⊗ Y •) is an exact functor of triangulated categories. This follows from Remark 10.8 since the functors Comp(A) → DoubleComp(C), Y • 7→ X• ⊗ Y • and Comp(B) → DoubleComp(C), X• 7→ X• ⊗ Y • are immediately seen to be compatible with homotopies and termwise split short exact sequences and hence induce exact functors of triangulated categories

K(B) → Kfirst(DoubleComp(C)) and K(A) → Ksecond(DoubleComp(C)) Observe that for the ﬁrst of the two the isomorphism Tot(X• ⊗ Y •[1]) =∼ Tot(X• ⊗ Y •)[1] involves signs (this goes back to the signs chosen in Homology, Remark 18.5). DERIVED CATEGORIES 37

11. Derived categories 05RR In this section we construct the derived category of an abelian category A by invert- ing the quasi-isomorphisms in K(A). Before we do this recall that the functors Hi : Comp(A) → A factor through K(A), see Homology, Lemma 13.11. Moreover, in Homology, Definition 14.8 we have defined identifications Hi(K•[n]) = Hi+n(K•). At this point it makes sense to redefine Hi(K•) = H0(K•[i]) in order to avoid confusion and possible sign errors. 05RS Lemma 11.1. Let A be an abelian category. The functor H0 : K(A) −→ A is homological. Proof. Because H0 is a functor, and by our definition of distinguished triangles it suffices to prove that given a termwise split short exact sequence of complexes 0 → A• → B• → C• → 0 the sequence H0(A•) → H0(B•) → H0(C•) is exact. This follows from Homology, Lemma 13.12. In particular, this lemma implies that a distinguished triangle (X,Y,Z,f,g,h) in K(A) gives rise to a long exact cohomology sequence

Hi(f) Hi(g) Hi(h) (11.1.1)05ST ... / Hi(X) / Hi(Y ) / Hi(Z) / Hi+1(X) / ... see (3.5.1). Moreover, there is a compatibility with the long exact sequence of cohomology associated to a short exact sequence of complexes (insert future reference here). For example, if (A•,B•,C•, α, β, δ) is the distinguished triangle associated to a termwise split exact sequence of complexes (see Definition 9.9), then the cohomology sequence above agrees with the one defined using the snake lemma, see Homology, Lemma 13.12 and for agreement of sequences, see Homology, Lemma 14.11. Recall that a complex K• is acyclic if Hi(K•) = 0 for all i ∈ Z. Moreover, recall that a morphism of complexes f : K• → L• is a quasi-isomorphism if and only if Hi(f) is an isomorphism for all i. See Homology, Definition 13.10. 05RT Lemma 11.2. Let A be an abelian category. The full subcategory Ac(A) of K(A) consisting of acyclic complexes is a strictly full saturated triangulated subcategory of K(A). The corresponding saturated multiplicative system (see Lemma 6.10) of K(A) is the set Qis(A) of quasi-isomorphisms. In particular, the kernel of the localization functor Q : K(A) → Qis(A)−1K(A) is Ac(A) and the functor H0 factors through Q. Proof. We know that H0 is a homological functor by Lemma 11.1. Thus this lemma is a special case of Lemma 6.11. 05RUDefinition 11.3. Let A be an abelian category. Let Ac(A) and Qis(A) be as in Lemma 11.2. The derived category of A is the triangulated category D(A) = K(A)/Ac(A) = Qis(A)−1K(A). We denote H0 : D(A) → A the unique functor whose composition with the quotient functor gives back the functor H0 defined above. Using Lemma 6.4 we introduce DERIVED CATEGORIES 38

the strictly full saturated triangulated subcategories D+(A),D−(A),Db(A) whose sets of objects are Ob(D+(A)) = {X ∈ Ob(D(A)) | Hn(X) = 0 for all n 0} Ob(D−(A)) = {X ∈ Ob(D(A)) | Hn(X) = 0 for all n 0} Ob(Db(A)) = {X ∈ Ob(D(A)) | Hn(X) = 0 for all |n| 0} The category Db(A) is called the bounded derived category of A. If K• and L• are complexes of A then we sometimes say “K• is quasi-isomorphic to L•” to indicate that K• and L• are isomorphic objects of D(A). 09PARemark 11.4. In this chapter, we consistently work with “small” abelian categories (as is the convention in the Stacks project). For a “big” abelian category A, it isn’t clear that the derived category D(A) exists, because it isn’t clear that morphisms in the derived category are sets. In fact, in general they aren’t, see Ex- amples, Lemma 60.1. However, if A is a Grothendieck abelian category, and given K•,L• in K(A), then by Injectives, Theorem 12.6 there exists a quasi-isomorphism L• → I• to a K-injective complex I• and Lemma 31.2 shows that • • • • HomD(A)(K ,L ) = HomK(A)(K ,I ) which is a set. Some examples of Grothendieck abelian categories are the category of modules over a ring, or more generally the category of sheaves of modules on a ringed site. Each of the variants D+(A),D−(A),Db(A) can be constructed as a localization of the corresponding homotopy category. This relies on the following simple lemma. 05RV Lemma 11.5. Let A be an abelian category. Let K• be a complex. (1) If Hn(K•) = 0 for all n 0, then there exists a quasi-isomorphism K• → L• with L• bounded below. (2) If Hn(K•) = 0 for all n 0, then there exists a quasi-isomorphism M • → K• with M • bounded above. (3) If Hn(K•) = 0 for all |n| 0, then there exists a commutative diagram of morphisms of complexes K• / L• O O

M • / N • where all the arrows are quasi-isomorphisms, L• bounded below, M • bounded above, and N • a bounded complex. • • • • • • Proof. Pick a 0 b and set M = τ≤bK , L = τ≥aK , and N = τ≤bL = • τ≥aM . See Homology, Section 15 for the truncation functors. To state the following lemma denote Ac+(A), Ac−(A), resp. Acb(A) the intersection of K+(A), K−(A), resp. Kb(A) with Ac(A). Denote Qis+(A), Qis−(A), resp. Qisb(A) the intersection of K+(A), K−(A), resp. Kb(A) with Qis(A). 05RW Lemma 11.6. Let A be an abelian category. The subcategories Ac+(A), Ac−(A), resp. Acb(A) are strictly full saturated triangulated subcategories of K+(A), K−(A), resp. Kb(A). The corresponding saturated multiplicative systems (see Lemma 6.10) are the sets Qis+(A), Qis−(A), resp. Qisb(A). DERIVED CATEGORIES 39

(1) The kernel of the functor K+(A) → D+(A) is Ac+(A) and this induces an equivalence of triangulated categories K+(A)/Ac+(A) = Qis+(A)−1K+(A) −→ D+(A) (2) The kernel of the functor K−(A) → D−(A) is Ac−(A) and this induces an equivalence of triangulated categories K−(A)/Ac−(A) = Qis−(A)−1K−(A) −→ D−(A)

(3) The kernel of the functor Kb(A) → Db(A) is Acb(A) and this induces an equivalence of triangulated categories Kb(A)/Acb(A) = Qisb(A)−1Kb(A) −→ Db(A) Proof. The initial statements follow from Lemma 6.11 by considering the restric- tion of the homological functor H0. The statement on kernels in (1), (2), (3) is a consequence of the definitions in each case. Each of the functors is essentially surjective by Lemma 11.5. To finish the proof we have to show the functors are fully faithful. We first do this for the bounded below version. Suppose that K•,L• are bounded above complexes. A morphism between these in D(A) is of the form s−1f for a pair f : K• → (L0)•, s : L• → (L0)• where s is a quasi-isomorphism. This implies that (L0)• has cohomology bounded below. Hence by Lemma 11.5 we can choose a quasi-isomorphism s0 :(L0)• → (L00)• with (L00)• bounded below. Then the pair (s0 ◦ f, s0 ◦ s) defines a morphism in Qis+(A)−1K+(A). Hence the functor is “full”. Finally, suppose that the pair f : K• → (L0)•, s : L• → (L0)• defines a morphism in Qis+(A)−1K+(A) which is zero in D(A). This means that there exists a quasi-isomorphism s0 :(L0)• → (L00)• such that s0 ◦ f = 0. Using Lemma 11.5 once more we obtain a quasi-isomorphism s00 :(L00)• → (L000)• with (L000)• bounded below. Thus we see that s00 ◦ s0 ◦ f = 0 which implies that s−1f is zero in Qis+(A)−1K+(A). This finishes the proof that the functor in (1) is an equivalence. The proof of (2) is dual to the proof of (1). To prove (3) we may use the result of (2). Hence it suffices to prove that the functor Qisb(A)−1Kb(A) → Qis−(A)−1K−(A) is fully faithful. The argument given in the previous paragraph applies directly to show this where we consistently work with complexes which are already bounded above.

12. The canonical delta-functor 014Z The derived category should be the receptacle for the universal cohomology functor. In order to state the result we use the notion of a δ-functor from an abelian category into a triangulated category, see Deﬁnition 3.6. Consider the functor Comp(A) → K(A). This functor is not a δ-functor in general. The easiest way to see this is to consider a nonsplit short exact sequence 0 → A → B → C → 0 of objects of A. Since HomK(A)(C[0],A[1]) = 0 we see that any distinguished triangle arising from this short exact sequence would look like (A[0],B[0],C[0], a, b, 0). But the existence of such a distinguished triangle in K(A) implies that the extension is split. A contradiction. DERIVED CATEGORIES 40

It turns out that the functor Comp(A) → D(A) is a δ-functor. In order to see this we have to define the morphisms δ associated to a short exact sequence 0 → A• −→a B• −→b C• → 0 of complexes in the abelian category A. Consider the cone C(a)• of the morphism a. We have C(a)n = Bn ⊕ An+1 and we define qn : C(a)n → Cn via the projection to Bn followed by bn. Hence a morphism of complexes q : C(a)• −→ C•. It is clear that q ◦ i = b where i is as in Definition 9.1. Note that, as a• is injective in each degree, the kernel of q is identified with the cone of idA• which is acyclic. Hence we see that q is a quasi-isomorphism. According to Lemma 9.14 the triangle (A, B, C(a), a, i, −p) is a distinguished triangle in K(A). As the localization functor K(A) → D(A) is exact we see that (A, B, C(a), a, i, −p) is a distinguished triangle in D(A). Since q is a quasi-isomorphism we see that q is an isomorphism in D(A). Hence we deduce that (A, B, C, a, b, −p ◦ q−1) is a distinguished triangle of D(A). This suggests the following lemma. 0152 Lemma 12.1. Let A be an abelian category. The functor Comp(A) → D(A) defined has the natural structure of a δ-functor, with −1 δA•→B•→C• = −p ◦ q with p and q as explained above. The same construction turns the functors Comp+(A) → D+(A), Comp−(A) → D−(A), and Compb(A) → Db(A) into δ-functors. Proof. We have already seen that this choice leads to a distinguished triangle whenever given a short exact sequence of complexes. We have to show that given a commutative diagram 0 / A• / B• / C• / 0 a b f g h a0 b0 0 / (A0)• / (B0)• / (C0)• / 0 we get the desired commutative diagram of Definition 3.6 (2). By Lemma 9.2 the pair (f, g) induces a canonical morphism c : C(a)• → C(a0)•. It is a simple computation to show that q0 ◦ c = h ◦ q and f[1] ◦ p = p0 ◦ c. From this the result follows directly. 0153 Lemma 12.2. Let A be an abelian category. Let 0 / A• / B• / C• / 0

0 / D• / E• / F • / 0 be a commutative diagram of morphisms of complexes such that the rows are short exact sequences of complexes, and the vertical arrows are quasi-isomorphisms. The δ-functor of Lemma 12.1 above maps the short exact sequences 0 → A• → B• → C• → 0 and 0 → D• → E• → F • → 0 to isomorphic distinguished triangles. DERIVED CATEGORIES 41

Proof. Trivial from the fact that K(A) → D(A) transforms quasi-isomorphisms into isomorphisms and that the associated distinguished triangles are functorial.

0154 Lemma 12.3. Let A be an abelian category. Let

0 / A• / B• / C• / 0 be a short exact sequences of complexes. Assume this short exact sequence is termwise split. Let (A•,B•,C•, α, β, δ) be the distinguished triangle of K(A) associated to the sequence. The δ-functor of Lemma 12.1 above maps the short exact sequences 0 → A• → B• → C• → 0 to a triangle isomorphic to the distinguished triangle (A•,B•,C•, α, β, δ).

Proof. Follows from Lemma 9.14.

08J5Remark 12.4. Let A be an abelian category. Let K• be a complex of A. Let a ∈ Z. We claim there is a canonical distinguished triangle

• • • • τ≤aK → K → τ≥a+1K → (τ≤aK )[1] in D(A). Here we have used the canonical truncation functors τ from Homology, Section 15. Namely, we ﬁrst take the distinguished triangle associated by our δ- functor (Lemma 12.1) to the short exact sequence of complexes

• • • • 0 → τ≤aK → K → K /τ≤aK → 0

• • Next, we use that the map K → τ≥a+1K factors through a quasi-isomorphism • • • K /τ≤aK → τ≥a+1K by the description of cohomology groups in Homology, Section 15. In a similar way we obtain canonical distinguished triangles

• • a+1 • • τ≤aK → τ≤a+1K → H (K )[−a − 1] → (τ≤aK )[1] and a • • • a • H (K )[−a] → τ≥aK → τ≥a+1K → H (K )[−a + 1] 08Q2 Lemma 12.5. Let A be an abelian category. Let

• • • K0 → K1 → ... → Kn be maps of complexes such that i • (1) H (K0 ) = 0 for i > 0, −j • −j • (2) H (Kj ) → H (Kj+1) is zero. • • • • Then the composition K0 → Kn factors through τ≤−nKn → Kn in D(A). Dually, given maps of complexes

• • • Kn → Kn−1 → ... → K0 such that i • (1) H (K0 ) = 0 for i < 0, j • j • (2) H (Kj+1) → H (Kj ) is zero, • • • • then the composition Kn → K0 factors through Kn → τ≥nKn in D(A). DERIVED CATEGORIES 42

• • • • Proof. The case n = 1. Since τ≤0K0 = K0 in D(A) we can replace K0 by τ≤0K0 • • and K1 by τ≤0K1 . Consider the distinguished triangle • • 0 • • τ≤−1K1 → K1 → H (K1 )[0] → (τ≤−1K1 )[1] • • 0 • (Remark 12.4). The composition K0 → K1 → H (K1 )[0] is zero as it is equal • 0 • 0 • to K0 → H (K0 )[0] → H (K1 )[0] which is zero by assumption. The fact that • HomD(A)(K0 , −) is a homological functor (Lemma 4.2), allows us to ﬁnd the desired • • factorization. For n = 2 we get a factorization K0 → τ≤−1K1 by the case n = 1 • • and we can apply the case n = 1 to the map of complexes τ≤−1K1 → τ≤−1K2 to • • get a factorization τ≤−1K1 → τ≤−2K2 . The general case is proved in exactly the same manner.

13. Filtered derived categories 05RX A reference for this section is [Ill72, I, Chapter V]. Let A be an abelian category. In this section we will define the filtered derived category DF (A) of A. In short, we will define it as the derived category of the exact category of objects of A endowed with a finite filtration. (Thus our construction is a special case of a more general construction of the derived category of an exact category, see for example [Büh10], [Kel90].) Illusie’s filtered derived category is the full subcategory of ours consisting of those objects whose filtration is finite. (In our category the filtration is still finite in each degree, but may not be uniformly bounded.) The rationale for our choice is that it is not harder and it allows us to apply the discussion to the spectral sequences of Lemma 21.3, see also Remark 21.4. We will use the notation regarding filtered objects introduced in Homology, Section 19. The category of filtered objects of A is denoted Fil(A). All filtrations will be decreasing by fiat. 05RYDefinition 13.1. Let A be an abelian category. The category of finite filtered objects of A is the category of filtered objects (A, F ) of A whose filtration F is finite. We denote it Filf (A). Thus Filf (A) is a full subcategory of Fil(A). For each p ∈ Z there is a functor grp : Filf (A) → A. There is a functor M gr = grp : Filf (A) → Gr(A) p∈Z where Gr(A) is the category of graded objects of A, see Homology, Definition 16.1. Finally, there is a functor (forget F ): Filf (A) −→ A which associates to the filtered object (A, F ) the underlying object of A. The category Filf (A) is an additive category, but not abelian in general, see Homology, Example 3.13. Because the functors grp, gr, (forget F ) are additive they induce exact functors of triangulated categories grp, (forget F ): K(Filf (A)) → K(A) and gr : K(Filf (A)) → K(Gr(A)) by Lemma 10.6. By analogy with the case of the homotopy category of an abelian category we make the following definitions. DERIVED CATEGORIES 43

05RZDefinition 13.2. Let A be an abelian category. (1) Let α : K• → L• be a morphism of K(Filf (A)). We say that α is a filtered quasi-isomorphism if the morphism gr(α) is a quasi-isomorphism. (2) Let K• be an object of K(Filf (A)). We say that K• is filtered acyclic if the complex gr(K•) is acyclic. Note that α : K• → L• is a filtered quasi-isomorphism if and only if each grp(α) is a quasi-isomorphism. Similarly a complex K• is filtered acyclic if and only if each grp(K•) is acyclic. 05S0 Lemma 13.3. Let A be an abelian category. (1) The functor K(Filf (A)) −→ Gr(A), K• 7−→ H0(gr(K•)) is homological. (2) The functor K(Filf (A)) → A, K• 7−→ H0(grp(K•)) is homological. (3) The functor K(Filf (A)) −→ A, K• 7−→ H0((forget F )K•) is homological. Proof. This follows from the fact that H0 : K(A) → A is homological, see Lemma 11.1 and the fact that the functors gr, grp, (forget F ) are exact functors of triangulated categories. See Lemma 4.20. 05S1 Lemma 13.4. Let A be an abelian category. The full subcategory FAc(A) of K(Filf (A)) consisting of filtered acyclic complexes is a strictly full saturated triangulated subcategory of K(Filf (A)). The corresponding saturated multiplicative system (see Lemma 6.10) of K(Filf (A)) is the set FQis(A) of filtered quasi-isomorphisms. In particular, the kernel of the localization functor Q : K(Filf (A)) −→ FQis(A)−1K(Filf (A)) is FAc(A) and the functor H0 ◦ gr factors through Q. Proof. We know that H0 ◦ gr is a homological functor by Lemma 13.3. Thus this lemma is a special case of Lemma 6.11. 05S2Definition 13.5. Let A be an abelian category. Let FAc(A) and FQis(A) be as in Lemma 13.4. The filtered derived category of A is the triangulated category DF (A) = K(Filf (A))/FAc(A) = FQis(A)−1K(Filf (A)). 05S3 Lemma 13.6. The functors grp, gr, (forget F ) induce canonical exact functors grp, gr, (forget F ): DF (A) −→ D(A) which commute with the localization functors. Proof. This follows from the universal property of localization, see Lemma 5.6, provided we can show that a filtered quasi-isomorphism is turned into a quasi- isomorphism by each of the functors grp, gr, (forget F ). This is true by definition for the first two. For the last one the statement we have to do a little bit of work. Let f : K• → L• be a filtered quasi-isomorphism in K(Filf (A)). Choose a distinguished triangle (K•,L•,M •, f, g, h) which contains f. Then M • is filtered acyclic, see Lemma 13.4. Hence by the corresponding lemma for K(A) it suffices to show that a filtered acyclic complex is an acyclic complex if we forget the filtration. This follows from Homology, Lemma 19.15. DERIVED CATEGORIES 44

05S4Definition 13.7. Let A be an abelian category. The bounded filtered derived category DF b(A) is the full subcategory of DF (A) with objects those X such that gr(X) ∈ Db(A). Similarly for the bounded below filtered derived category DF +(A) and the bounded above filtered derived category DF −(A). 05S5 Lemma 13.8. Let A be an abelian category. Let K• ∈ K(Filf (A)). (1) If Hn(gr(K•)) = 0 for all n < a, then there exists a filtered quasi-isomorphism K• → L• with Ln = 0 for all n < a. (2) If Hn(gr(K•)) = 0 for all n > b, then there exists a filtered quasi-isomorphism M • → K• with M n = 0 for all n > b. (3) If Hn(gr(K•)) = 0 for all |n| 0, then there exists a commutative diagram of morphisms of complexes K• / L• O O

M • / N • where all the arrows are ﬁltered quasi-isomorphisms, L• bounded below, M • bounded above, and N • a bounded complex. Proof. Suppose that Hn(gr(K•)) = 0 for all n < a. By Homology, Lemma 19.15 the sequence a−2 a−1 Ka−1 −−−→d Ka−1 −−−→d Ka is an exact sequence of objects of A and the morphisms da−2 and da−1 are strict. Hence Coim(da−1) = Im(da−1) in Filf (A) and the map gr(Im(da−1)) → gr(Ka) is injective with image equal to the image of gr(Ka−1) → gr(Ka), see Homology, • • Lemma 19.13. This means that the map K → τ≥aK into the truncation • a a−1 a+1 τ≥aK = (... → 0 → K / Im(d ) → K → ...) is a ﬁltered quasi-isomorphism. This proves (1). The proof of (2) is dual to the proof of (1). Part (3) follows formally from (1) and (2). To state the following lemma denote FAc+(A), FAc−(A), resp. FAcb(A) the intersection of K+(Filf A), K−(Filf A), resp. Kb(Filf A) with FAc(A). Denote FQis+(A), FQis−(A), resp. FQisb(A) the intersection of K+(Filf A), K−(Filf A), resp. Kb(Filf A) with FQis(A). 05S6 Lemma 13.9. Let A be an abelian category. The subcategories FAc+(A), FAc−(A), resp. FAcb(A) are strictly full saturated triangulated subcategories of K+(Filf A), K−(Filf A), resp. Kb(Filf A). The corresponding saturated multiplicative systems (see Lemma 6.10) are the sets FQis+(A), FQis−(A), resp. FQisb(A). (1) The kernel of the functor K+(Filf A) → DF +(A) is FAc+(A) and this induces an equivalence of triangulated categories K+(Filf A)/FAc+(A) = FQis+(A)−1K+(Filf A) −→ DF +(A) (2) The kernel of the functor K−(Filf A) → DF −(A) is FAc−(A) and this induces an equivalence of triangulated categories K−(Filf A)/FAc−(A) = FQis−(A)−1K−(Filf A) −→ DF −(A) DERIVED CATEGORIES 45

(3) The kernel of the functor Kb(Filf A) → DF b(A) is FAcb(A) and this induces an equivalence of triangulated categories Kb(Filf A)/FAcb(A) = FQisb(A)−1Kb(Filf A) −→ DF b(A) Proof. This follows from the results above, in particular Lemma 13.8, by exactly the same arguments as used in the proof of Lemma 11.6.

14. Derived functors in general 05S7 A reference for this section is Deligne’s exposé XVII in [AGV71]. A very general notion of right and left derived functors exists where we have an exact functor between triangulated categories, a multiplicative system in the source category and we want to find the “correct” extension of the exact functor to the localized category. 05S8Situation 14.1. Here F : D → D0 is an exact functor of triangulated categories and S is a saturated multiplicative system in D compatible with the structure of triangulated category on D. Let X ∈ Ob(D). Recall from Categories, Remark 27.7 the filtered category X/S of arrows s : X → X0 in S with source X. Dually, in Categories, Remark 27.15 we defined the cofiltered category S/X of arrows s : X0 → X in S with target X. 05S9Definition 14.2. Assumptions and notation as in Situation 14.1. Let X ∈ Ob(D). (1) we say the right derived functor RF is defined at X if the ind-object (X/S) −→ D0, (s : X → X0) 7−→ F (X0) is essentially constant4; in this case the value Y in D0 is called the value of RF at X. (2) we say the left derived functor LF is defined at X if the pro-object (S/X) −→ D0, (s : X0 → X) 7−→ F (X0) is essentially constant; in this case the value Y in D0 is called the value of LF at X. By abuse of notation we often denote the values simply RF (X) or LF (X). It will turn out that the full subcategory of D consisting of objects where RF is defined is a triangulated subcategory, and RF will define a functor on this subcategory which transforms morphisms of S into isomorphisms. 05SA Lemma 14.3. Assumptions and notation as in Situation 14.1. Let f : X → Y be a morphism of D. (1) If RF is defined at X and Y then there exists a unique morphism RF (f): RF (X) → RF (Y ) between the values such that for any commutative diagram 0 X s / X

f f 0 s0 Y / Y 0

4For a discussion of when an ind-object or pro-object of a category is essentially constant we refer to Categories, Section 22. DERIVED CATEGORIES 46

with s, s0 ∈ S the diagram

F (X) / F (X0) / RF (X)

F (Y ) / F (Y 0) / RF (Y ) commutes. (2) If LF is deﬁned at X and Y then there exists a unique morphism LF (f): LF (X) → LF (Y ) between the values such that for any commutative diagram 0 X s / X

f 0 f s0 Y 0 / Y with s, s0 in S the diagram

LF (X) / F (X0) / F (X)

LF (Y ) / F (Y 0) / F (Y ) commutes. Proof. Part (1) holds if we only assume that the colimits 0 0 RF (X) = colims:X→X0 F (X ) and RF (Y ) = colims0:Y →Y 0 F (Y ) exist. Namely, to give a morphism RF (X) → RF (Y ) between the colimits is the same thing as giving for each s : X → X0 in Ob(X/S) a morphism F (X0) → RF (Y ) compatible with morphisms in the category X/S. To get the morphism we choose a commutative diagram 0 X s / X

f f 0 s0 Y / Y 0 with s, s0 in S as is possible by MS2 and we set F (X0) → RF (Y ) equal to the composition F (X0) → F (Y 0) → RF (Y ). To see that this is independent of the choice of the diagram above use MS3. Details omitted. The proof of (2) is dual. 05SB Lemma 14.4. Assumptions and notation as in Situation 14.1. Let s : X → Y be an element of S. (1) RF is defined at X if and only if it is defined at Y . In this case the map RF (s): RF (X) → RF (Y ) between values is an isomorphism. (2) LF is defined at X if and only if it is defined at Y . In this case the map LF (s): LF (X) → LF (Y ) between values is an isomorphism.

Proof. Omitted. 05SU Lemma 14.5. Assumptions and notation as in Situation 14.1. Let X be an object of D and n ∈ Z. DERIVED CATEGORIES 47

(1) RF is defined at X if and only if it is defined at X[n]. In this case there is a canonical isomorphism RF (X)[n] = RF (X[n]) between values. (2) LF is defined at X if and only if it is defined at X[n]. In this case there is a canonical isomorphism LF (X)[n] → LF (X[n]) between values.

Proof. Omitted. 05SC Lemma 14.6. Assumptions and notation as in Situation 14.1. Let (X,Y,Z,f,g,h) be a distinguished triangle of D. If RF is defined at two out of three of X,Y,Z, then it is defined at the third. Moreover, in this case (RF (X),RF (Y ),RF (Z),RF (f),RF (g),RF (h)) is a distinguished triangle in D0. Similarly for LF . Proof. Say RF is defined at X,Y with values A, B. Let RF (f): A → B be the induced morphism, see Lemma 14.3. We may choose a distinguished triangle (A, B, C, RF (f), b, c) in D0. We claim that C is a value of RF at Z. To see this pick s : X → X0 in S such that there exists a morphism α : A → F (X0) as in Categories, Definition 22.1. We may choose a commutative diagram 0 X s / X

f f 0 s0 Y / Y 0 with s0 ∈ S by MS2. Using that Y/S is ﬁltered we can (after replacing s0 by some s00 : Y → Y 00 in S) assume that there exists a morphism β : B → F (Y 0) as in Categories, Deﬁnition 22.1. Picture 0 A α / F (X ) / A

RF (f) F (f 0) RF (f) β B / F (Y 0) / B It may not be true that the left square commutes, but the outer and right squares 0 commute. The assumption that the ind-object {F (Y )}s0:Y 0→Y is essentially constant means that there exists a s00 : Y → Y 00 in S and a morphism h : Y 0 → Y 00 such that s00 = h ◦ s0 and such that F (h) equal to F (Y 0) → B → F (Y 0) → F (Y 00). Hence after replacing Y 0 by Y 00 and β by F (h) ◦ β the diagram will commute (by direct computation with arrows). Using MS6 choose a morphism of triangles (s, s0, s00):(X,Y,Z,f,g,h) −→ (X0,Y 0,Z0, f 0, g0, h0) with s00 ∈ S. By TR3 choose a morphism of triangles (α, β, γ):(A, B, C, RF (f), b, c) −→ (F (X0),F (Y 0),F (Z0),F (f 0),F (g0),F (h0))

By Lemma 14.4 it suﬃces to prove that RF (Z0) is deﬁned and has value C. Consider the category I of Lemma 5.8 of triangles I = {(t, t0, t00):(X0,Y 0,Z0, f 0, g0, h0) → (X00,Y 00,Z00, f 00, g00, h00) | (t, t0, t00) ∈ S} To show that the system F (Z00) is essentially constant over the category Z0/S is equivalent to showing that the system of F (Z00) is essentially constant over I DERIVED CATEGORIES 48

because I → Z0/S is coﬁnal, see Categories, Lemma 22.11 (coﬁnality is proven in Lemma 5.8). For any object W in D0 we consider the diagram

00 colimI MorD0 (W, F (X ))o MorD0 (W, A) O O

00 colimI MorD0 (W, F (Y )) o MorD0 (W, B) O O

00 colimI MorD0 (W, F (Z )) o MorD0 (W, C) O O

00 colimI MorD0 (W, F (X [1])) o MorD0 (W, A[1]) O O

00 colimI MorD0 (W, F (Y [1])) o MorD0 (W, B[1]) where the horizontal arrows are given by composing with (α, β, γ). Since ﬁltered colimits are exact (Algebra, Lemma 8.8) the left column is an exact sequence. Thus the 5 lemma (Homology, Lemma 5.20) tells us the 00 colimI MorD0 (W, F (Z )) −→ MorD0 (W, C) is bijective. Choose an object (t, t0, t00):(X0,Y 0,Z0) → (X00,Y 00,Z00) of I. Applying 00 what we just showed to W = F (Z ) and the element idF (X00) of the colimit we ﬁnd 00 00 00 00 a unique morphism c(X00,Y 00,Z00) : F (Z ) → C such that for some (X ,Y ,Z ) → (X000,Y 000,Z00) in I

c(X00,Y 00,Z00) γ F (Z00) −−−−−−−−→ C −→ F (Z0) → F (Z00) → F (Z000) equals F (Z00) → F (Z000) 00 The family of morphisms c(X00,Y 00,Z00) form an element c of limI MorD0 (F (Z ),C) 00 by uniqueness (computation omitted). Finally, we show that colimI F (Z ) = C via the morphisms c(X00,Y 00,Z00) which will finish the proof by Categories, Lemma 22.9. 0 00 Namely, let W be an object of D and let d(X00,Y 00,Z00) : F (Z ) → W be a family of 00 maps corresponding to an element of limI MorD0 (F (Z ),W ). If d(X0,Y 0,Z0) ◦ γ = 0, 00 00 00 then for every object (X ,Y ,Z ) of I the morphism d(X00,Y 00,Z00) is zero by the 00 00 00 000 000 00 existence of c(X00,Y 00,Z00) and the morphism (X ,Y ,Z ) → (X ,Y ,Z ) in I satisfying the displayed equality above. Hence the map 00 limI MorD0 (F (Z ),W ) −→ MorD0 (C,W ) (coming from precomposing by γ) is injective. However, it is also surjective because the element c gives a left inverse. We conclude that C is the colimit by Categories, Remark 14.4. 05SD Lemma 14.7. Assumptions and notation as in Situation 14.1. Let X,Y be objects of D. (1) If RF is defined at X and Y , then RF is defined at X ⊕ Y . (2) If D0 is Karoubian and RF is defined at X ⊕ Y , then RF is defined at both X and Y . In either case we have RF (X ⊕ Y ) = RF (X) ⊕ RF (Y ). Similarly for LF . DERIVED CATEGORIES 49

Proof. If RF is defined at X and Y , then the distinguished triangle X → X ⊕Y → Y → X[1] (Lemma 4.11) and Lemma 14.6 shows that RF is defined at X ⊕ Y and that we have a distinguished triangle RF (X) → RF (X ⊕ Y ) → RF (Y ) → RF (X)[1]. Applying Lemma 4.11 to this once more we find that RF (X ⊕ Y ) = RF (X) ⊕ RF (Y ). This proves (1) and the final assertion. Conversely, assume that RF is defined at X ⊕ Y and that D0 is Karoubian. Since S is a saturated system S is the set of arrows which become invertible under the additive localization functor Q : D → S−1D, see Categories, Lemma 27.21. Thus for any s : X → X0 and s0 : Y → Y 0 in S the morphism s ⊕ s0 : X ⊕ Y → X0 ⊕ Y 0 is an element of S. In this way we obtain a functor X/S × Y/S −→ (X ⊕ Y )/S Recall that the categories X/S, Y/S, (X ⊕ Y )/S are filtered (Categories, Remark 0 27.7). By Categories, Lemma 22.12 X/S × Y/S is filtered and F |X/S : X/S → D 0 (resp. G|Y/S : Y/S → D ) is essentially constant if and only if F |X/S ◦ pr1 : X/S × 0 0 Y/S → D (resp. G|Y/S ◦ pr2 : X/S × Y/S → D ) is essentially constant. Below we will show that the displayed functor is cofinal, hence by Categories, Lemma 22.11. we see that F |(X⊕Y )/S is essentially constant implies that F |X/S ◦pr1 ⊕F |Y/S ◦pr2 : X/S × Y/S → D0 is essentially constant. By Homology, Lemma 30.3 (and this is 0 where we use that D is Karoubian) we see that F |X/S ◦ pr1 ⊕ F |Y/S ◦ pr2 being essentially constant implies F |X/S ◦ pr1 and F |Y/S ◦ pr2 are essentially constant proving that RF is defined at X and Y . Proof that the displayed functor is cofinal. To do this pick any t : X ⊕ Y → Z in S. Using MS2 we can find morphisms Z → X0, Z → Y 0 and s : X → X0, s0 : Y → Y 0 in S such that X o X ⊕ Y / Y

s s0 X0 o Z / Y 0 commutes. This proves there is a map Z → X0 ⊕ Y 0 in (X ⊕ Y )/S, i.e., we get part (1) of Categories, Definition 17.1. To prove part (2) it suffices to prove that 0 0 0 given t : X ⊕ Y → Z and morphisms si ⊕ si : Z → Xi ⊕ Yi , i = 1, 2 in (X ⊕ Y )/S 0 0 0 0 0 0 0 0 we can find morphisms a : X1 → X , b : X2 → X , c : Y1 → Y , d : Y2 → Y 0 0 in S such that a ◦ s1 = b ◦ s2 and c ◦ s1 = d ◦ s2. To do this we first choose any X0 and Y 0 and maps a, b, c, d in S; this is possible as X/S and Y/S are filtered. 0 −1 Then the two maps a ◦ s1, b ◦ s2 : Z → X become equal in S D. Hence we can find a morphism X0 → X00 in S equalizing them. Similarly we find Y 0 → Y 00 in S 0 0 0 00 0 00 equalizing c ◦ s1 and d ◦ s2. Replacing X by X and Y by Y we get a ◦ s1 = b ◦ s2 0 0 and c ◦ s1 = d ◦ s2. The proof of the corresponding statements for LF are dual. 05SE Proposition 14.8. Assumptions and notation as in Situation 14.1. (1) The full subcategory E of D consisting of objects at which RF is defined is a strictly full triangulated subcategory of D. (2) We obtain an exact functor RF : E −→ D0 of triangulated categories. (3) Elements of S with either source or target in E are morphisms of E. −1 −1 (4) The functor SE E → S D is a fully faithful exact functor of triangulated categories. DERIVED CATEGORIES 50

(5) Any element of SE = Arrows(E) ∩ S is mapped to an isomorphism by RF . (6) We obtain an exact functor −1 0 RF : SE E −→ D . (7) If D0 is Karoubian, then E is a saturated triangulated subcategory of D. A similar result holds for LF . Proof. Since S is saturated it contains all isomorphisms (see remark following Categories, Definition 27.20). Hence (1) follows from Lemmas 14.4, 14.6, and 14.5. We get (2) from Lemmas 14.3, 14.5, and 14.6. We get (3) from Lemma 14.4. The fully faithfulness in (4) follows from (3) and the definitions. The fact that −1 −1 −1 SE E → S D is exact follows from the fact that a triangle in SE E is distinguished if and only if it is isomorphic to the image of a distinguished triangle in E, see proof of Proposition 5.5. Part (5) follows from Lemma 14.4. The factorization of 0 −1 0 RF : E → D through an exact functor SE E → D follows from Lemma 5.6. Part (7) follows from Lemma 14.7. Proposition 14.8 tells us that RF lives on a maximal strictly full triangulated subcategory of S−1D and is an exact functor on this triangulated category. Picture: D / D0 F < Q RF fully faithful S−1D S−1E o exact E 05SVDefinition 14.9. In Situation 14.1. We say F is right derivable, or that RF everywhere defined if RF is defined at every object of D. We say F is left derivable, or that LF everywhere defined if LF is defined at every object of D. In this case we obtain a right (resp. left) derived functor (14.9.1)05SW RF : S−1D −→ D0, (resp. LF : S−1D −→ D0), see Proposition 14.8. In most interesting situations it is not the case that RF ◦ Q is equal to F . In fact, it might happen that the canonical map F (X) → RF (X) is never an isomorphism. In practice this does not happen, because in practice we only know how to prove F is right derivable by showing that RF can be computed by evaluating F at judiciously chosen objects of the triangulated category D. This warrants a definition. 05SXDefinition 14.10. In Situation 14.1. (1) An object X of D computes RF if RF is defined at X and the canonical map F (X) → RF (X) is an isomorphism. (2) An object X of D computes LF if LF is defined at X and the canonical map LF (X) → F (X) is an isomorphism. 05SY Lemma 14.11. Assumptions and notation as in Situation 14.1. Let X be an object of D and n ∈ Z. (1) X computes RF if and only if X[n] computes RF . (2) X computes LF if and only if X[n] computes LF .

Proof. Omitted. DERIVED CATEGORIES 51

05SZ Lemma 14.12. Assumptions and notation as in Situation 14.1. Let (X,Y,Z,f,g,h) be a distinguished triangle of D. If X,Y compute RF then so does Z. Similar for LF . Proof. By Lemma 14.6 we know that RF is deﬁned at Z and that RF applied to the triangle produces a distinguished triangle. Consider the morphism of distinguished triangles (F (X),F (Y ),F (Z),F (f),F (g),F (h))

(RF (X),RF (Y ),RF (Z),RF (f),RF (g),RF (h))

Two out of three maps are isomorphisms, hence so is the third. 05T0 Lemma 14.13. Assumptions and notation as in Situation 14.1. Let X,Y be objects of D. If X ⊕ Y computes RF , then X and Y compute RF . Similarly for LF . Proof. If X ⊕ Y computes RF , then RF (X ⊕ Y ) = F (X) ⊕ F (Y ). In the proof of Lemma 14.7 we have seen that the functor X/S ×Y/S → (X ⊕Y )/S, (s, s0) 7→ s⊕s0 is cofinal. We will use this without further mention. Let s : X → X0 be an element of S. Then F (X) → F (X0) has a section, namely, F (X0) → F (X0 ⊕ Y ) → RF (X0 ⊕ Y ) = RF (X ⊕ Y ) = F (X) ⊕ F (Y ) → F (X). where we have used Lemma 14.4. Hence F (X0) = F (X) ⊕ E for some object E of D0 such that E → F (X0 ⊕ Y ) → RF (X0 ⊕ Y ) = RF (X ⊕ Y ) is zero (Lemma 4.12). Because RF is defined at X0 ⊕ Y with value F (X) ⊕ F (Y ) we can find a morphism t : X0 ⊕ Y → Z of S such that F (t) annihilates E. We may assume Z = X00 ⊕ Y 00 and t = t0 ⊕ t00 with t0, t00 ∈ S. Then F (t0) annihilates E. It follows that F is essentially constant on X/S with value F (X) as desired. 05T1 Lemma 14.14. Assumptions and notation as in Situation 14.1. (1) If for every object X ∈ Ob(D) there exists an arrow s : X → X0 in S such that X0 computes RF , then RF is everywhere defined. (2) If for every object X ∈ Ob(D) there exists an arrow s : X0 → X in S such that X0 computes LF , then LF is everywhere defined.

Proof. This is clear from the definitions. 06XN Lemma 14.15. Assumptions and notation as in Situation 14.1. If there exists a subset I ⊂ Ob(D) such that (1) for all X ∈ Ob(D) there exists s : X → X0 in S with X0 ∈ I, and (2) for every arrow s : X → X0 in S with X,X0 ∈ I the map F (s): F (X) → F (X0) is an isomorphism, then RF is everywhere defined and every X ∈ I computes RF . Dually, if there exists a subset P ⊂ Ob(D) such that (1) for all X ∈ Ob(D) there exists s : X0 → X in S with X0 ∈ P, and (2) for every arrow s : X → X0 in S with X,X0 ∈ P the map F (s): F (X) → F (X0) is an isomorphism, then LF is everywhere defined and every X ∈ P computes LF . DERIVED CATEGORIES 52

Proof. Let X be an object of D. Assumption (1) implies that the arrows s : X → X0 in S with X0 ∈ I are cofinal in the category X/S. Assumption (2) implies that F is constant on this cofinal subcategory. Clearly this implies that F :(X/S) → D0 0 0 0 is essentially constant with value F (X ) for any s : X → X in S with X ∈ I. 05T2 Lemma 14.16. Let A, B, C be triangulated categories. Let S, resp. S0 be a saturated multiplicative system in A, resp. B compatible with the triangulated structure. Let F : A → B and G : B → C be exact functors. Denote F 0 : A → (S0)−1B the composition of F with the localization functor. (1) If RF 0, RG, R(G ◦ F ) are everywhere defined, then there is a canonical transformation of functors t : R(G ◦ F ) −→ RG ◦ RF 0. (2) If LF 0, LG, L(G ◦ F ) are everywhere defined, then there is a canonical transformation of functors t : LG ◦ LF 0 → L(G ◦ F ). Proof. In this proof we try to be careful. Hence let us think of the derived functors as the functors RF 0 : S−1A → (S0)−1B,R(G ◦ F ): S−1A → C,RG :(S0)−1B → C. −1 0 −1 Let us denote QA : A → S A and QB : B → (S ) B the localization functors. 0 Then F = QB ◦ F . Note that for every object Y of B there is a canonical map

G(Y ) −→ RG(QB(Y )) 0 in other words, there is a transformation of functors t : G → RG ◦ QB. Let X be an object of A. We have 0 R(G ◦ F )(QA(X)) = colims:X→X0∈S G(F (X )) 0 t 0 −→ colims:X→X0∈S RG(QB(F (X ))) 0 0 = colims:X→X0∈S RG(F (X )) 0 0 = RG(colims:X→X0∈S F (X )) = RG(RF 0(X)). The system F 0(X0) is essentially constant in the category (S0)−1B. Hence we may pull the colimit inside the functor RG in the third equality of the diagram above, see Categories, Lemma 22.8 and its proof. We omit the proof this deﬁnes a transformation of functors. The case of left derived functors is similar.

15. Derived functors on derived categories 05T3 In practice derived functors come about most often when given an additive functor between abelian categories. 05T4Situation 15.1. Here F : A → B is an additive functor between abelian categories. This induces exact functors F : K(A) → K(B),K+(A) → K+(B),K−(A) → K−(B). See Lemma 10.6. We also denote F the composition K(A) → D(B), K+(A) → D+(B), and K−(A) → D−(B) of F with the localization functor K(B) → D(B), etc. This situation leads to four derived functors we will consider in the following. (1) The right derived functor of F : K(A) → D(B) relative to the multiplicative system Qis(A). DERIVED CATEGORIES 53

(2) The right derived functor of F : K+(A) → D+(B) relative to the multiplicative system Qis+(A). (3) The left derived functor of F : K(A) → D(B) relative to the multiplicative system Qis(A). (4) The left derived functor of F : K−(A) → D−(B) relative to the multiplicative system Qis−(A). Each of these cases is an example of Situation 14.1. Some of the ambiguity that may arise is alleviated by the following. 05T5 Lemma 15.2. In Situation 15.1. (1) Let X be an object of K+(A). The right derived functor of K(A) → D(B) is defined at X if and only if the right derived functor of K+(A) → D+(B) is defined at X. Moreover, the values are canonically isomorphic. (2) Let X be an object of K+(A). Then X computes the right derived functor of K(A) → D(B) if and only if X computes the right derived functor of K+(A) → D+(B). (3) Let X be an object of K−(A). The left derived functor of K(A) → D(B) is defined at X if and only if the left derived functor of K−(A) → D−(B) is defined at X. Moreover, the values are canonically isomorphic. (4) Let X be an object of K−(A). Then X computes the left derived functor of K(A) → D(B) if and only if X computes the left derived functor of K−(A) → D−(B). Proof. Let X be an object of K+(A). Consider a quasi-isomorphism s : X → X0 in K(A). By Lemma 11.5 there exists quasi-isomorphism X0 → X00 with X00 bounded below. Hence we see that X/Qis+(A) is cofinal in X/Qis(A). Thus it is clear that (1) holds. Part (2) follows directly from part (1). Parts (3) and (4) are dual to parts (1) and (2). Given an object A of an abelian category A we get a complex A[0] = (... → 0 → A → 0 → ...) where A is placed in degree zero. Hence a functor A → K(A), A 7→ A[0]. Let us temporarily say that a partial functor is one that is defined on a subcategory. 0157Definition 15.3. In Situation 15.1. (1) The right derived functors of F are the partial functors RF associated to cases (1) and (2) of Situation 15.1. (2) The left derived functors of F are the partial functors LF associated to cases (3) and (4) of Situation 15.1. (3) An object A of A is said to be right acyclic for F , or acyclic for RF if A[0] computes RF . (4) An object A of A is said to be left acyclic for F , or acyclic for LF if A[0] computes LF . The following few lemmas give some criteria for the existence of enough acyclics. 05T7 Lemma 15.4. Let A be an abelian category. Let P ⊂ Ob(A) be a subset containing 0 such that every object of A is a quotient of an element of P. Let a ∈ Z. (1) Given K• with Kn = 0 for n > a there exists a quasi-isomorphism P • → K• with P n ∈ P and P n → Kn surjective for all n and P n = 0 for n > a. DERIVED CATEGORIES 54

(2) Given K• with Hn(K•) = 0 for n > a there exists a quasi-isomorphism P • → K• with P n ∈ P for all n and P n = 0 for n > a.

Proof. Proof of part (1). Consider the following induction hypothesis IHn: There are P j ∈ P, j ≥ n, with P j = 0 for j > a, maps dj : P j → P j+1 for j ≥ n, and surjective maps αj : P j → Kj for j ≥ n such that the diagram

P n / P n+1 / P n+2 / ...

α α α ... / Kn−1 / Kn / Kn+1 / Kn+2 / ... is commutative, such that dj+1 ◦dj = 0 for j ≥ n, such that α induces isomorphisms j • j j−1 n n H (K ) → Ker(d )/ Im(d ) for j > n, and such that α : Ker(d ) → Ker(dK ) is surjective. Then we choose a surjection

n−1 n−1 n n−1 n P −→ K ×Kn Ker(d ) = K × n Ker(d ) Ker(dK ) with P n−1 in P. This allows us to extend the diagram above to

P n−1 / P n / P n+1 / P n+2 / ...

α α α α ... / Kn−1 / Kn / Kn+1 / Kn+2 / ...

The reader easily checks that IHn−1 holds with this choice.

We ﬁnish the proof of (1) as follows. First we note that IHn is true for n = a + 1 since we can just take P j = 0 for j > a. Hence we see that proceeding by descending induction we produce a complex P • with P n = 0 for n > a consisting of objects from P, and a termwise surjective quasi-isomorphism α : P • → K• as desired.

• • Proof of part (2). The assumption implies that the morphism τ≤aK → K (Ho- • • mology, Section 15) is a quasi-isomorphism. Apply part (1) to ﬁnd P → τ≤aK . • • The composition P → K is the desired quasi-isomorphism.

05T6 Lemma 15.5. Let A be an abelian category. Let I ⊂ Ob(A) be a subset containing 0 such that every object of A is a subobject of an element of I. Let a ∈ Z. (1) Given K• with Kn = 0 for n < a there exists a quasi-isomorphism K• → I• with Kn → In injective and In ∈ I for all n and In = 0 for n < a, (2) Given K• with Hn(K•) = 0 for n < a there exists a quasi-isomorphism K• → I• with In ∈ I and In = 0 for n < a.

Proof. This lemma is dual to Lemma 15.4.

05T8 Lemma 15.6. In Situation 15.1. Let I ⊂ Ob(A) be a subset with the following properties: (1) every object of A is a subobject of an element of I, (2) for any short exact sequence 0 → P → Q → R → 0 of A with P,Q ∈ I, then R ∈ I, and 0 → F (P ) → F (Q) → F (R) → 0 is exact. Then every object of I is acyclic for RF . DERIVED CATEGORIES 55

Proof. We may add 0 to I if necessary. Pick A ∈ I. Let A[0] → K• be a quasi-isomorphism with K• bounded below. Then we can find a quasi-isomorphism K• → I• with I• bounded below and each In ∈ I, see Lemma 15.5. Hence we see that these resolutions are cofinal in the category A[0]/Qis+(A). To finish the proof it therefore suffices to show that for any quasi-isomorphism A[0] → I• with I• bounded above and In ∈ I we have F (A)[0] → F (I•) is a quasi-isomorphism. n To see this suppose that I = 0 for n < n0. Of course we may assume that n0 < 0. n−1 n −1 Starting with n = n0 we prove inductively that Im(d ) = Ker(d ) and Im(d ) are elements of I using property (2) and the exact sequences 0 → Ker(dn) → In → Im(dn) → 0. Moreover, property (2) also guarantees that the complex 0 → F (In0 ) → F (In0+1) → ... → F (I−1) → F (Im(d−1)) → 0 is exact. The exact sequence 0 → Im(d−1) → I0 → I0/ Im(d−1) → 0 implies that I0/ Im(d−1) is an element of I. The exact sequence 0 → A → I0/ Im(d−1) → Im(d0) → 0 then implies that Im(d0) = Ker(d1) is an elements of I and from then on one continues as before to show that Im(dn−1) = Ker(dn) is an element of I for all n > 0. Applying F to each of the short exact sequences mentioned above and • using (2) we observe that F (A)[0] → F (I ) is an isomorphism as desired. 05T9 Lemma 15.7. In Situation 15.1. Let P ⊂ Ob(A) be a subset with the following properties: (1) every object of A is a quotient of an element of P, (2) for any short exact sequence 0 → P → Q → R → 0 of A with Q, R ∈ P, then P ∈ P, and 0 → F (P ) → F (Q) → F (R) → 0 is exact. Then every object of P is acyclic for LF .

Proof. Dual to the proof of Lemma 15.6.

16. Higher derived functors 05TB The following simple lemma shows that right derived functors “move to the right”. 05TC Lemma 16.1. Let F : A → B be an additive functor between abelian categories. Let K• be a complex of A and a ∈ Z. (1) If Hi(K•) = 0 for all i < a and RF is defined at K•, then Hi(RF (K•)) = 0 for all i < a. • • i • i • (2) If RF is defined at K and τ≤aK , then H (RF (τ≤aK )) = H (RF (K )) for all i ≤ a. Proof. Assume K• satisfies the assumptions of (1). Let K• → L• be any quasi- • • isomorphism. Then it is also true that K → τ≥aL is a quasi-isomorphism by our assumption on K•. Hence in the category K•/Qis+(A) the quasi-isomorphisms s : K• → L• with Ln = 0 for n < a are cofinal. Thus RF is the value of the essentially constant ind-object F (L•) for these s it follows that Hi(RF (K•)) = 0 for i < a. To prove (2) we use the distinguished triangle • • • • τ≤aK → K → τ≥a+1K → (τ≤aK )[1] DERIVED CATEGORIES 56

• of Remark 12.4 to conclude via Lemma 14.6 that RF is defined at τ≥a+1K as well and that we have a distinguished triangle • • • • RF (τ≤aK ) → RF (K ) → RF (τ≥a+1K ) → RF (τ≤aK )[1] • in D(B). By part (1) we see that RF (τ≥a+1K ) has vanishing cohomology in degrees < a + 1. The long exact cohomology sequence of this distinguished triangle then shows what we want. 015ADefinition 16.2. Let F : A → B be an additive functor between abelian categories. Assume RF : D+(A) → D+(B) is everywhere defined. Let i ∈ Z. The ith right derived functor RiF of F is the functor RiF = Hi ◦ RF : A −→ B The following lemma shows that it really does not make a lot of sense to take the right derived functor unless the functor is left exact. 05TD Lemma 16.3. Let F : A → B be an additive functor between abelian categories and assume RF : D+(A) → D+(B) is everywhere defined. (1) We have RiF = 0 for i < 0, (2) R0F is left exact, (3) the map F → R0F is an isomorphism if and only if F is left exact. Proof. Let A be an object of A. Let A[0] → K• be any quasi-isomorphism. Then • it is also true that A[0] → τ≥0K is a quasi-isomorphism. Hence in the category A[0]/Qis+(A) the quasi-isomorphisms s : A[0] → K• with Kn = 0 for n < 0 are cofinal. Thus it is clear that Hi(RF (A[0])) = 0 for i < 0. Moreover, for such an s the sequence 0 → A → K0 → K1 is exact. Hence if F is left exact, then 0 → F (A) → F (K0) → F (K1) is exact as well, and we see that F (A) → H0(F (K•)) is an isomorphism for every s : A[0] → K• as above which implies that H0(RF (A[0])) = F (A). Let 0 → A → B → C → 0 be a short exact sequence of A. By Lemma 12.1 we obtain a distinguished triangle (A[0],B[0],C[0], a, b, c) in K+(A). From the long exact cohomology sequence (and the vanishing for i < 0 proved above) we deduce that 0 → R0F (A) → R0F (B) → R0F (C) is exact. Hence R0F is left exact. Of 0 course this also proves that if F → R F is an isomorphism, then F is left exact. 015C Lemma 16.4. Let F : A → B be an additive functor between abelian categories and assume RF : D+(A) → D+(B) is everywhere defined. Let A be an object of A. (1) A is right acyclic for F if and only if F (A) → R0F (A) is an isomorphism and RiF (A) = 0 for all i > 0, (2) if F is left exact, then A is right acyclic for F if and only if RiF (A) = 0 for all i > 0. Proof. If A is right acyclic for F , then RF (A[0]) = F (A)[0] and in particular F (A) → R0F (A) is an isomorphism and RiF (A) = 0 for i 6= 0. Conversely, if F (A) → R0F (A) is an isomorphism and RiF (A) = 0 for all i > 0 then F (A[0]) → RF (A[0]) is a quasi-isomorphism by Lemma 16.3 part (1) and hence A is acyclic. 0 If F is left exact then F = R F , see Lemma 16.3. DERIVED CATEGORIES 57

015D Lemma 16.5. Let F : A → B be a left exact functor between abelian categories and assume RF : D+(A) → D+(B) is everywhere defined. Let 0 → A → B → C → 0 be a short exact sequence of A. (1) If A and C are right acyclic for F then so is B. (2) If A and B are right acyclic for F then so is C. (3) If B and C are right acyclic for F and F (B) → F (C) is surjective then A is right acyclic for F . In each of the three cases 0 → F (A) → F (B) → F (C) → 0 is a short exact sequence of B. Proof. By Lemma 12.1 we obtain a distinguished triangle (A[0],B[0],C[0], a, b, c) in K+(A). As RF is an exact functor and since RiF = 0 for i < 0 and R0F = F (Lemma 16.3) we obtain an exact cohomology sequence 0 → F (A) → F (B) → F (C) → R1F (A) → ... in the abelian category B. Thus the lemma follows from the characterization of acyclic objects in Lemma 16.4. 05TE Lemma 16.6. Let F : A → B be an additive functor between abelian categories and assume RF : D+(A) → D+(B) is everywhere defined. (1) The functors RiF , i ≥ 0 come equipped with a canonical structure of a δ-functor from A → B, see Homology, Definition 12.1. (2) If every object of A is a subobject of a right acyclic object for F , then i {R F, δ}i≥0 is a universal δ-functor, see Homology, Definition 12.3. Proof. The functor A → Comp+(A), A 7→ A[0] is exact. The functor Comp+(A) → D+(A) is a δ-functor, see Lemma 12.1. The functor RF : D+(A) → D+(B) is exact. Finally, the functor H0 : D+(B) → B is a homological functor, see Definition 11.3. Hence we get the structure of a δ-functor from Lemma 4.22 and Lemma 4.21. Part (2) follows from Homology, Lemma 12.4 and the description of acyclics in Lemma 16.4. 015E Lemma 16.7 (Leray’s acyclicity lemma). Let F : A → B be an additive functor between abelian categories. Let A• be a bounded below complex of right F -acyclic objects such that RF is defined at A•5. The canonical map F (A•) −→ RF (A•) is an isomorphism in D+(B), i.e., A• computes RF . Proof. Let A• be a bounded complex of right F -acyclic objects. We claim that RF is defined at A• and that F (A•) → RF (A•) is an isomorphism in D+(B). Namely, it holds for complexes with at most one nonzero right F -acyclic object for example by Lemma 16.4. Next, suppose that An = 0 for n 6∈ [a, b]. Using the “stupid” truncations we obtain a termwise split short exact sequence of complexes • • • 0 → σ≥a+1A → A → σ≤aA → 0

5For example this holds if RF : D+(A) → D+(B) is everywhere deﬁned. DERIVED CATEGORIES 58

• • • see Homology, Section 15. Thus a distinguished triangle (σ≥a+1A ,A , σ≤aA ). By induction hypothesis RF is defined for the two outer complexes and these complexes compute RF . Then the same is true for the middle one by Lemma 14.12. Suppose that A• is a bounded below complex of acyclic objects such that RF is defined at A•. To show that F (A•) → RF (A•) is an isomorphism in D+(B) it suffices to show that Hi(F (A•)) → Hi(RF (A•)) is an isomorphism for all i. Pick i. Consider the termwise split short exact sequence of complexes • • • 0 → σ≥i+2A → A → σ≤i+1A → 0. Note that this induces a termwise split short exact sequence • • • 0 → σ≥i+2F (A ) → F (A ) → σ≤i+1F (A ) → 0. Hence we get distinguished triangles • • • • • • (σ≥i+2A ,A , σ≤i+1A ) and (σ≥i+2F (A ),F (A ), σ≤i+1F (A )) • • Since RF is defined at A (by assumption) and at σ≤i+1A (by the first paragraph) • we see that RF is defined at σ≥i+1A and we get a distinghuished triangle • • • (RF (σ≥i+2A ),RF (A ),RF (σ≤i+1A )) See Lemma 14.6. Using these distinguished triangles we obtain a map of exact sequences

i • i • i • i+1 • H (σ≥i+2F (A )) / H (F (A )) / H (σ≤i+1F (A )) / H (σ≥i+2F (A ))

α β i • i • i • i+1 • H (RF (σ≥i+2A )) / H (RF (A )) / H (RF (σ≤i+1A )) / H (RF (σ≥i+2A )) By the results of the first paragraph the map β is an isomorphism. By inspection the objects on the upper left and the upper right are zero. Hence to finish the proof i • i+1 • it suffices to show that H (RF (σ≥i+2A )) = 0 and H (RF (σ≥i+2A )) = 0. This follows immediately from Lemma 16.1. 05TA Proposition 16.8. Let F : A → B be an additive functor of abelian categories. (1) If every object of A injects into an object acyclic for RF , then RF is defined on all of K+(A) and we obtain an exact functor RF : D+(A) −→ D+(B) see (14.9.1). Moreover, any bounded below complex A• whose terms are acyclic for RF computes RF . (2) If every object of A is quotient of an object acyclic for LF , then LF is defined on all of K−(A) and we obtain an exact functor LF : D−(A) −→ D−(B) see (14.9.1). Moreover, any bounded above complex A• whose terms are acyclic for LF computes LF . Proof. Assume every object of A injects into an object acyclic for RF . Let I be the set of objects acyclic for RF . Let K• be a bounded below complex in A. By Lemma 15.5 there exists a quasi-isomorphism α : K• → I• with I• bounded below and In ∈ I. Hence in order to prove (1) it suffices to show that F (I•) → F ((I0)•) is a quasi-isomorphism when s : I• → (I0)• is a quasi-isomorphism of bounded below DERIVED CATEGORIES 59

complexes of objects from I, see Lemma 14.15. Note that the cone C(s)• is an acyclic bounded below complex all of whose terms are in I. Hence it suﬃces to show: given an acyclic bounded below complex I• all of whose terms are in I the complex F (I•) is acyclic.

n n n • Say I = 0 for n < n0. Setting J = Im(d ) we break I into short exact se- n n+1 n+1 quences 0 → J → I → J → 0 for n ≥ n0. These sequences induce distinguished triangles (J n,In+1,J n+1) in D+(A) by Lemma 12.1. For each k ∈ Z denote Hk the assertion: For all n ≤ k the right derived functor RF is deﬁned n i n at J and R F (J ) = 0 for i 6= 0. Then Hk holds trivially for k ≤ n0. If n+1 Hn holds, then, using Proposition 14.8, we see that RF is deﬁned at J and (RF (J n),RF (In+1),RF (J n+1)) is a distinguished triangle of D+(B). Thus the long exact cohomology sequence (11.1.1) associated to this triangle gives an exact sequence 0 → R−1F (J n+1) → R0F (J n) → F (In+1) → R0F (J n+1) → 0 and gives that RiF (J n+1) = 0 for i 6∈ {−1, 0}. By Lemma 16.1 we see that −1 n+1 R F (J ) = 0. This proves that Hn+1 is true hence Hk holds for all k. We also conclude that 0 → R0F (J n) → F (In+1) → R0F (J n+1) → 0 is short exact for all n. This in turn proves that F (I•) is exact.

The proof in the case of LF is dual. 015F Lemma 16.9. Let F : A → B be an exact functor of abelian categories. Then (1) every object of A is right acyclic for F , (2) RF : D+(A) → D+(B) is everywhere deﬁned, (3) RF : D(A) → D(B) is everywhere deﬁned, (4) every complex computes RF , in other words, the canonical map F (K•) → RF (K•) is an isomorphism for all complexes, and (5) RiF = 0 for i 6= 0. Proof. This is true because F transforms acyclic complexes into acyclic complexes and quasi-isomorphisms into quasi-isomorphisms. Details omitted.

17. Triangulated subcategories of the derived category 06UP Let A be an abelian category. In this section we look at certain strictly full saturated triangulated subcategories D0 ⊂ D(A). Let B ⊂ A be a weak Serre subcategory, see Homology, Deﬁnition 10.1 and Lemma 10.3. We let DB(A) the full subcategory of D(A) whose objects are n Ob(DB(A)) = {X ∈ Ob(D(A)) | H (X) is an object of B for all n} + + We also deﬁne DB (A) = D (A) ∩ DB(A) and similarly for the other bounded versions. 06UQ Lemma 17.1. Let A be an abelian category. Let B ⊂ A be a weak Serre subcategory. The category DB(A) is a strictly full saturated triangulated subcategory of D(A). Similarly for the bounded versions. DERIVED CATEGORIES 60

Proof. It is clear that DB(A) is an additive subcategory preserved under the trans- n n lation functors. If X ⊕ Y is in DB(A), then both H (X) and H (Y ) are kernels of maps between maps of objects of B as Hn(X ⊕ Y ) = Hn(X) ⊕ Hn(Y ). Hence both X and Y are in DB(A). By Lemma 4.16 it therefore suffices to show that given a distinguished triangle (X,Y,Z,f,g,h) such that X and Y are in DB(A) then Z is an object of DB(A). The long exact cohomology sequence (11.1.1) and the definition of a weak Serre subcategory (see Homology, Definition 10.1) show n that H (Z) is an object of B for all n. Thus Z is an object of DB(A). We continue to assume that B is a weak Serre subcategory of the abelian category A. Then B is an abelian category and the inclusion functor B → A is exact. Hence we obtain a derived functor D(B) → D(A), see Lemma 16.9. Clearly the functor D(B) → D(A) factors through a canonical exact functor

(17.1.1)06UR D(B) −→ DB(A)

After all a complex made from objects of B certainly gives rise to an object of DB(A) and as distinguished triangles in DB(A) are exactly the distinguished triangles of D(A) whose vertices are in DB(A) we see that the functor is exact since D(B) → + + − − D(A) is exact. Similarly we obtain functors D (B) → DB (A), D (B) → DB (A), b b and D (B) → DB(A) for the bounded versions. A key question in many cases is whether the displayed functor is an equivalence. Now, suppose that B is a Serre subcategory of A. In this case we have the quotient functor A → A/B, see Homology, Lemma 10.6. In this case DB(A) is the kernel of the functor D(A) → D(A/B). Thus we obtain a canonical functor

D(A)/DB(A) −→ D(A/B) by Lemma 6.8. Similarly for the bounded versions. 06XL Lemma 17.2. Let A be an abelian category. Let B ⊂ A be a Serre subcategory. Then D(A) → D(A/B) is essentially surjective. Proof. We will use the description of the category A/B in the proof of Homology, Lemma 10.6. Let (X•, d•) be a complex of A/B. This means that Xi is an object of A and di : Xi → Xi+1 is a morphism in A/B such that di ◦ di−1 = 0 in A/B. For i ≥ 0 we may write di = (si, f i) where si : Y i → Xi is a morphism of A whose kernel and cokernel are in B (equivalently si becomes an isomorphism in the quotient category) and f i : Y i → Xi+1 is a morphism of A. By induction we will construct a commutative diagram

(X0)1 / (X0)2 / ... < O O

X0 X1 X2 ... O < O : O < s0 s1 s2 f 0 f 1 f 2 Y 0 Y 1 Y 2 ... where the vertical arrows Xi → (X0)i become isomorphisms in the quotient category. Namely, we ﬁrst let (X0)1 = Coker(Y 0 → X0 ⊕ X1) (or rather the pushout of the diagram with arrows s0 and f 0) which gives the ﬁrst commutative diagram. DERIVED CATEGORIES 61

Next, we take (X0)2 = Coker(Y 1 → (X0)1 ⊕ X2). And so on. Setting additionally (X0)n = Xn for n ≤ 0 we see that the map (X•, d•) → ((X0)•, (d0)•) is an isomorphism of complexes in A/B. Hence we may assume dn : Xn → Xn+1 is given by a map Xn → Xn+1 in A for n ≥ 0. Dually, for i < 0 we may write di = (gi, ti+1) where ti+1 : Xi+1 → Zi+1 is an isomorphism in the quotient category and gi : Xi → Zi+1 is a morphism. By induction we will construct a commutative diagram ...Z−2 Z−1 Z0 O 9 O ; O

t−2 t−1 t0 g−2 g−1 ...X−2 X−1 X0 O O ;

... (X0)−2 / (X0)−1 where the vertical arrows (X0)i → Xi become isomorphisms in the quotient cate- 0 −1 −1 0 0 −2 −2 gory. Namely, we take (X ) = X ×Z0 X . Then we take (X ) = X ×Z−1 (X0)−1. And so on. Setting additionally (X0)n = Xn for n ≥ 0 we see that the map ((X0)•, (d0)•) → (X•, d•) is an isomorphism of complexes in A/B. Hence we may assume dn : Xn → Xn+1 is given by a map dn : Xn → Xn+1 in A for all n ∈ Z. In this case we know the compositions dn ◦ dn−1 are zero in A/B. If for n > 0 we replace Xn by X (X0)n = Xn/ Im(Im(Xk−2 → Xk) → Xn) 0

HomD(A)/DB(A)(X,Y ) −→ HomD(A/B)(vX, vY ) DERIVED CATEGORIES 62

is bijective because u gives an inverse (by the remarks above). For certain Serre subcategories B ⊂ A we can prove that the functor D(B) → DB(A) is fully faithful. 0FCL Lemma 17.4. Let A be an abelian category. Let B ⊂ A be a Serre subcategory. Assume that for every surjection X → Y with X ∈ Ob(A) and Y ∈ Ob(B) there exists X0 ⊂ X, X0 ∈ Ob(B) which surjects onto Y . Then the functor D−(B) → − DB (A) of (17.1.1) is an equivalence. Proof. Let X• be a bounded above complex of A such that Hi(X•) ∈ Ob(B) for all i ∈ Z. Moreover, suppose we are given Bi ⊂ Xi, Bi ∈ Ob(B) for all i ∈ Z. Claim: there exists a subcomplex Y • ⊂ X• such that (1) Y • → X• is a quasi-isomorphism, (2) Y i ∈ Ob(B) for all i ∈ Z, and (3) Bi ⊂ Y i for all i ∈ Z. To prove the claim, using the assumption of the lemma we ﬁrst choose Ci ⊂ Ker(di : Xi → Xi+1), Ci ∈ Ob(B) surjecting onto Hi(X•). Setting Di = Ci +di−1(Bi−1)+ Bi we ﬁnd a subcomplex D• satisfying (2) and (3) such that Hi(D•) → Hi(X•) is surjective for all i ∈ Z. For any choice of Ei ⊂ Xi with Ei ∈ Ob(B) and di(Ei) ⊂ Di+1 + Ei+1 we see that setting Y i = Di + Ei gives a subcomplex whose terms are in B and whose cohomology surjects onto the cohomology of X•. Clearly, if di(Ei) = (Di+1 + Ei+1) ∩ Im(di) then we see that the map on cohomology is also injective. For n 0 we can take En equal to 0. By descending induction we can choose Ei for all i with the desired property. Namely, given Ei+1,Ei+2,... we choose Ei ⊂ Xi such that di(Ei) = (Di+1 + Ei+1) ∩ Im(di). This is possible by our assumption in the lemma combined with the fact that (Di+1 + Ei+1) ∩ Im(di) is in B as B is a Serre subcategory of A. The claim above implies the lemma. Essential surjectivity is immediate from the claim. Let us prove faithfulness. Namely, suppose we have a morphism f : U • → V • of bounded above complexes of B whose image in D(A) is zero. Then there exists a quasi-isomorphism s : V • → X• into a bounded above complex of A such that s ◦ f is homotopic to zero. Choose a homotopy hi : U i → Xi−1 between 0 and s ◦ f. Apply the claim with Bi = hi+1(U i+1) + si(V i). The resulting map s0 : V • → Y • is a quasi-isomorphism as well and s0 ◦ f is homotopic to zero as is clear from the fact that hi factors through Y i−1. This proves faithfulness. Fully faithfulness is proved in the exact same manner.

18. Injective resolutions 013G In this section we prove some lemmas regarding the existence of injective resolutions in abelian categories having enough injectives. 013IDeﬁnition 18.1. Let A be an abelian category. Let A ∈ Ob(A). An injective resolution of A is a complex I• together with a map A → I0 such that: (1) We have In = 0 for n < 0. (2) Each In is an injective object of A. (3) The map A → I0 is an isomorphism onto Ker(d0). (4) We have Hi(I•) = 0 for i > 0. DERIVED CATEGORIES 63

Hence A[0] → I• is a quasi-isomorphism. In other words the complex ... → 0 → A → I0 → I1 → ... is acyclic. Let K• be a complex in A. An injective resolution of K• is a complex I• together with a map α : K• → I• of complexes such that (1) We have In = 0 for n 0, i.e., I• is bounded below. (2) Each In is an injective object of A. (3) The map α : K• → I• is a quasi-isomorphism. In other words an injective resolution K• → I• gives rise to a diagram

... / Kn−1 / Kn / Kn+1 / ...

... / In−1 / In / In+1 / ... which induces an isomorphism on cohomology objects in each degree. An injective resolution of an object A of A is almost the same thing as an injective resolution of the complex A[0]. 013J Lemma 18.2. Let A be an abelian category. Let K• be a complex of A. (1) If K• has an injective resolution then Hn(K•) = 0 for n 0. (2) If Hn(K•) = 0 for all n 0 then there exists a quasi-isomorphism K• → L• with L• bounded below.

• • Proof. Omitted. For the second statement use L = τ≥nK for some n 0. See Homology, Section 15 for the deﬁnition of the truncation τ≥n. 013K Lemma 18.3. Let A be an abelian category. Assume A has enough injectives. (1) Any object of A has an injective resolution. (2) If Hn(K•) = 0 for all n 0 then K• has an injective resolution. (3) If K• is a complex with Kn = 0 for n < a, then there exists an injective resolution α : K• → I• with In = 0 for n < a such that each αn : Kn → In is injective. Proof. Proof of (1). First choose an injection A → I0 of A into an injective object 1 of A. Next, choose an injection I0/A → I into an injective object of A. Denote d0 the induced map I0 → I1. Next, choose an injection I1/ Im(d0) → I2 into an injective object of A. Denote d1 the induced map I1 → I2. And so on. By Lemma 18.2 part (2) follows from part (3). Part (3) is a special case of Lemma 15.5. 013R Lemma 18.4. Let A be an abelian category. Let K• be an acyclic complex. Let I• be bounded below and consisting of injective objects. Any morphism K• → I• is homotopic to zero. Proof. Let α : K• → I• be a morphism of complexes. Assume that αj = 0 for j < n. We will show that there exists a morphism h : Kn+1 → In such that αn = h ◦ d. Thus α will be homotopic to the morphism of complexes β deﬁned by   0 if j ≤ n βj = αn+1 − d ◦ h if j = n + 1  αj if j > n + 1 DERIVED CATEGORIES 64

This will clearly prove the lemma (by induction). To prove the existence of h note n n−1 • n−1 n−1 that α |dn−1(Kn−1) = 0 since α = 0. Since K is acyclic we have d (K ) = Ker(Kn → Kn+1). Hence we can think of αn as a map into In deﬁned on the subobject Im(Kn → Kn+1) of Kn+1. By injectivity of the object In we can extend n+1 n this to a map h : K → I as desired. 05TFRemark 18.5. Let A be an abelian category. Using the fact that K(A) is a triangulated category we may use Lemma 18.4 to obtain proofs of some of the lemmas below which are usually proved by chasing through diagrams. Namely, suppose that α : K• → L• is a quasi-isomorphism of complexes. Then (K•,L•,C(α)•, α, i, −p) is a distinguished triangle in K(A) (Lemma 9.14) and C(α)• is an acyclic complex (Lemma 11.2). Next, let I• be a bounded below complex of injective objects. Then

• • • • • • HomK(A)(C(α) ,I ) / HomK(A)(L ,I ) / HomK(A)(K ,I )

• r • HomK(A)(C(α) [−1],I )

is an exact sequence of abelian groups, see Lemma 4.2. At this point Lemma • • 18.4 guarantees that the outer two groups are zero and hence HomK(A)(L ,I ) = • • HomK(A)(K ,I ). 013P Lemma 18.6. Let A be an abelian category. Consider a solid diagram

• • K α / L γ β } I• where I• is bounded below and consists of injective objects, and α is a quasi- isomorphism. (1) There exists a map of complexes β making the diagram commute up to homotopy. (2) If α is injective in every degree then we can find a β which makes the diagram commute. Proof. The “correct” proof of part (1) is explained in Remark 18.5. We also give a direct proof here. We first show that (2) implies (1). Namely, let α˜ : K → L˜•, π, s be as in Lemma 9.6. Since α˜ is injective by (2) there exists a morphism β˜ : L˜• → I• such that γ = β˜ ◦ α˜. Set β = β˜ ◦ s. Then we have β ◦ α = β˜ ◦ s ◦ π ◦ α˜ ∼ β˜ ◦ α˜ = γ as desired. Assume that α : K• → L• is injective. Suppose we have already defined β in all degrees ≤ n − 1 compatible with differentials and such that γj = βj ◦ αj for all DERIVED CATEGORIES 65

j ≤ n − 1. Consider the commutative solid diagram

Kn−1 / Kn

α α γ Ln−1 / Ln γ

β In−1 / In Thus we see that the dotted arrow is prescribed on the subobjects α(Kn) and dn−1(Ln−1). Moreover, these two arrows agree on α(dn−1(Kn−1)). Hence if (18.6.1)013Q α(dn−1(Kn−1)) = α(Kn) ∩ dn−1(Ln−1)

then these morphisms glue to a morphism α(Kn) + dn−1(Ln−1) → In and, using the injectivity of In, we can extend this to a morphism from all of Ln into In. After this by induction we get the morphism β for all n simultaneously (note that we can set βn = 0 for all n 0 since I• is bounded below – in this way starting the induction). It remains to prove the equality (18.6.1). The reader is encouraged to argue this for themselves with a suitable diagram chase. Nonetheless here is our argument. Note that the inclusion α(dn−1(Kn−1)) ⊂ α(Kn) ∩ dn−1(Ln−1) is obvious. Take an object T of A and a morphism x : T → Ln whose image is contained in the subobject α(Kn) ∩ dn−1(Ln−1). Since α is injective we see that x = α ◦ x0 for some x0 : T → Kn. Moreover, since x lies in dn−1(Ln−1) we see that dn ◦ x = 0. Hence using injectivity of α again we see that dn ◦ x0 = 0. Thus x0 gives a morphism [x0]: T → Hn(K•). On the other hand the corresponding map [x]: T → Hn(L•) induced by x is zero by assumption. Since α is a quasi-isomorphism we conclude that [x0] = 0. This of course means exactly that the image of x0 is contained in n−1 n−1 d (K ) and we win. 013S Lemma 18.7. Let A be an abelian category. Consider a solid diagram

• • K α / L γ β } i I• where I• is bounded below and consists of injective objects, and α is a quasi- isomorphism. Any two morphisms β1, β2 making the diagram commute up to homotopy are homotopic. Proof. This follows from Remark 18.5. We also give a direct argument here.

• Let α˜ : K → L˜ , π, s be as in Lemma 9.6. If we can show that β1 ◦ π is homotopic to β2 ◦ π, then we deduce that β1 ∼ β2 because π ◦ s is the identity. Hence we may assume αn : Kn → Ln is the inclusion of a direct summand for all n. Thus we get a short exact sequence of complexes 0 → K• → L• → M • → 0 DERIVED CATEGORIES 66

which is termwise split and such that M • is acyclic. We choose splittings Ln = n n n n n n n n n K ⊕ M , so we have βi : K ⊕ M → I and γ : K → I . In this case the n n n−1 condition on βi is that there are morphisms hi : K → I such that n n n n+1 γ − βi |Kn = d ◦ hi + hi ◦ d Thus we see that n n n n n+1 n+1 β1 |Kn − β2 |Kn = d ◦ (h1 − h2 ) + (h1 − h2 ) ◦ d n n n n−1 n n Consider the map h : K ⊕M → I which equals h1 −h2 on the ﬁrst summand and zero on the second. Then we see that n n n n+1 β1 − β2 − (d ◦ h + h ) ◦ d is a morphism of complexes L• → I• which is identically zero on the subcomplex K•. Hence it factors as L• → M • → I•. Thus the result of the lemma follows from Lemma 18.4. 05TG Lemma 18.8. Let A be an abelian category. Let I• be bounded below complex consisting of injective objects. Let L• ∈ K(A). Then • • • • MorK(A)(L ,I ) = MorD(A)(L ,I ). Proof. Let a be an element of the right hand side. We may represent a = γα−1 where α : K• → L• is a quasi-isomorphism and γ : K• → I• is a map of complexes. By Lemma 18.6 we can ﬁnd a morphism β : L• → I• such that β ◦ α is homotopic to γ. This proves that the map is surjective. Let b be an element of the left hand side which maps to zero in the right hand side. Then b is the homotopy class of a morphism β : L• → I• such that there exists a quasi-isomorphism α : K• → L• with β ◦ α homotopic to zero. Then Lemma 18.7 shows that β is homotopic to zero also, i.e., b = 0. 013T Lemma 18.9. Let A be an abelian category. Assume A has enough injectives. For any short exact sequence 0 → A• → B• → C• → 0 of Comp+(A) there exists a commutative diagram in Comp+(A)

0 / A• / B• / C• / 0

• • • 0 / I1 / I2 / I3 / 0 where the vertical arrows are injective resolutions and the rows are short exact sequences of complexes. In fact, given any injective resolution A• → I• we may • • assume I1 = I . Proof. Step 1. Choose an injective resolution A• → I• (see Lemma 18.3) or use the given one. Recall that Comp+(A) is an abelian category, see Homology, Lemma 13.9. Hence we may form the pushout along the injective map A• → I• to get

0 / A• / B• / C• / 0

0 / I• / E• / C• / 0 DERIVED CATEGORIES 67

Note that the lower short exact sequence is termwise split, see Homology, Lemma 27.2. Hence it suffices to prove the lemma when 0 → A• → B• → C• → 0 is termwise split. Step 2. Choose splittings. In other words, write Bn = An ⊕ Cn. Denote δ : C• → A•[1] the morphism as in Homology, Lemma 14.10. Choose injective resolutions • • • • • f1 : A → I1 and f3 : C → I3 . (If A is a complex of injectives, then use • • I1 = A .) We may assume f3 is injective in every degree. By Lemma 18.6 we may 0 • • 0 find a morphism δ : I3 → I1 [1] such that δ ◦ f3 = f1[1] ◦ δ (equality of morphisms n n n of complexes). Set I2 = I1 ⊕ I3 . Define dn (δ0)n dn = I1 I2 0 dn I3 n n n n and define the maps B → I2 to be given as the sum of the maps A → I1 and n n C → I3 . Everything is clear.

19. Projective resolutions 0643 This section is dual to Section 18. We give deﬁnitions and state results, but we do not reprove the lemmas. 0644Deﬁnition 19.1. Let A be an abelian category. Let A ∈ Ob(A). An projective resolution of A is a complex P • together with a map P 0 → A such that: (1) We have P n = 0 for n > 0. (2) Each P n is an projective object of A. (3) The map P 0 → A induces an isomorphism Coker(d−1) → A. (4) We have Hi(P •) = 0 for i < 0. Hence P • → A[0] is a quasi-isomorphism. In other words the complex ... → P −1 → P 0 → A → 0 → ... is acyclic. Let K• be a complex in A. An projective resolution of K• is a complex P • together with a map α : P • → K• of complexes such that (1) We have P n = 0 for n 0, i.e., P • is bounded above. (2) Each P n is an projective object of A. (3) The map α : P • → K• is a quasi-isomorphism. 0645 Lemma 19.2. Let A be an abelian category. Let K• be a complex of A. (1) If K• has a projective resolution then Hn(K•) = 0 for n 0. (2) If Hn(K•) = 0 for n 0 then there exists a quasi-isomorphism L• → K• with L• bounded above.

Proof. Dual to Lemma 18.2. 0646 Lemma 19.3. Let A be an abelian category. Assume A has enough projectives. (1) Any object of A has a projective resolution. (2) If Hn(K•) = 0 for all n 0 then K• has a projective resolution. (3) If K• is a complex with Kn = 0 for n > a, then there exists a projective resolution α : P • → K• with P n = 0 for n > a such that each αn : P n → Kn is surjective.

Proof. Dual to Lemma 18.3. DERIVED CATEGORIES 68

0647 Lemma 19.4. Let A be an abelian category. Let K• be an acyclic complex. Let P • be bounded above and consisting of projective objects. Any morphism P • → K• is homotopic to zero.

Proof. Dual to Lemma 18.4. 0648Remark 19.5. Let A be an abelian category. Suppose that α : K• → L• is a quasi- isomorphism of complexes. Let P • be a bounded above complex of projectives. Then • • • • HomK(A)(P ,K ) −→ HomK(A)(P ,L ) is an isomorphism. This is dual to Remark 18.5. 0649 Lemma 19.6. Let A be an abelian category. Consider a solid diagram K• o L• O α =

β P • where P • is bounded above and consists of projective objects, and α is a quasi- isomorphism. (1) There exists a map of complexes β making the diagram commute up to homotopy. (2) If α is surjective in every degree then we can ﬁnd a β which makes the diagram commute.

Proof. Dual to Lemma 18.6. 064A Lemma 19.7. Let A be an abelian category. Consider a solid diagram K• o L• O α =

βi P • where P • is bounded above and consists of projective objects, and α is a quasi- isomorphism. Any two morphisms β1, β2 making the diagram commute up to homotopy are homotopic.

Proof. Dual to Lemma 18.7. 064B Lemma 19.8. Let A be an abelian category. Let P • be bounded above complex consisting of projective objects. Let L• ∈ K(A). Then • • • • MorK(A)(P ,L ) = MorD(A)(P ,L ). Proof. Dual to Lemma 18.8. 064C Lemma 19.9. Let A be an abelian category. Assume A has enough projectives. For any short exact sequence 0 → A• → B• → C• → 0 of Comp+(A) there exists a commutative diagram in Comp+(A) • • • 0 / P1 / P2 / P3 / 0

0 / A• / B• / C• / 0 DERIVED CATEGORIES 69

where the vertical arrows are projective resolutions and the rows are short exact sequences of complexes. In fact, given any projective resolution P • → C• we may • • assume P3 = P . Proof. Dual to Lemma 18.9. 064D Lemma 19.10. Let A be an abelian category. Let P •, K• be complexes. Let n ∈ Z. Assume that (1) P • is a bounded complex consisting of projective objects, (2) P i = 0 for i < n, and (3) Hi(K•) = 0 for i ≥ n. • • • • Then HomK(A)(P ,K ) = HomD(A)(P ,K ) = 0. Proof. The first equality follows from Lemma 19.8. Note that there is a distinguished triangle • • • (τ≤n−1K ,K , τ≥nK , f, g, h) • • by Remark 12.4. Hence, by Lemma 4.2 it suffices to prove HomK(A)(P , τ≤n−1K ) = • • 0 and HomK(A)(P , τ≥nK ) = 0. The first vanishing is trivial and the second is Lemma 19.4. 064E Lemma 19.11. Let A be an abelian category. Let β : P • → L• and α : E• → L• be maps of complexes. Let n ∈ Z. Assume (1) P • is a bounded complex of projectives and P i = 0 for i < n, (2) Hi(α) is an isomorphism for i > n and surjective for i = n. Then there exists a map of complexes γ : P • → E• such that α ◦ γ and β are homotopic. Proof. Consider the cone C• = C(α)• with map i : L• → C•. Note that i ◦ β is • zero by Lemma 19.10. Hence we can lift β to E by Lemma 4.2. 20. Right derived functors and injective resolutions 0156 At this point we can use the material above to define the right derived functors of an additive functor between an abelian category having enough injectives and a general abelian category. 05TH Lemma 20.1. Let A be an abelian category. Let I ∈ Ob(A) be an injective object. Let I• be a bounded below complex of injectives in A. (1) I• computes RF relative to Qis+(A) for any exact functor F : K+(A) → D into any triangulated category D. (2) I is right acyclic for any additive functor F : A → B into any abelian category B. Proof. Part (2) is a direct consequences of part (1) and Definition 15.3. To prove (1) let α : I• → K• be a quasi-isomorphism into a complex. By Lemma 18.6 we see that α has a left inverse. Hence the category I•/Qis+(A) is essentially constant with value id : I• → I•. Thus also the ind-object I•/Qis+(A) −→ D, (I• → K•) 7−→ F (K•) is essentially constant with value F (I•). This proves (1), see Definitions 14.2 and 14.10. 05TI Lemma 20.2. Let A be an abelian category with enough injectives. DERIVED CATEGORIES 70

(1) For any exact functor F : K+(A) → D into a triangulated category D the right derived functor RF : D+(A) −→ D is everywhere defined. (2) For any additive functor F : A → B into an abelian category B the right derived functor RF : D+(A) −→ D+(B) is everywhere defined. Proof. Combine Lemma 20.1 and Proposition 16.8 for the second assertion. To see the first assertion combine Lemma 18.3, Lemma 20.1, Lemma 14.14, and Equation (14.9.1). 0159 Lemma 20.3. Let A be an abelian category with enough injectives. Let F : A → B be an additive functor. (1) The functor RF is an exact functor D+(A) → D+(B). (2) The functor RF induces an exact functor K+(A) → D+(B). (3) The functor RF induces a δ-functor Comp+(A) → D+(B). (4) The functor RF induces a δ-functor A → D+(B). Proof. This lemma simply reviews some of the results obtained so far. Note that by Lemma 20.2 RF is everywhere defined. Here are some references: (1) The derived functor is exact: This boils down to Lemma 14.6. (2) This is true because K+(A) → D+(A) is exact and compositions of exact functors are exact. (3) This is true because Comp+(A) → D+(A) is a δ-functor, see Lemma 12.1. (4) This is true because A → Comp+(A) is exact and precomposing a δ-functor by an exact functor gives a δ-functor. 015B Lemma 20.4. Let A be an abelian category with enough injectives. Let F : A → B be a left exact functor. (1) For any short exact sequence 0 → A• → B• → C• → 0 of complexes in Comp+(A) there is an associated long exact sequence ... → Hi(RF (A•)) → Hi(RF (B•)) → Hi(RF (C•)) → Hi+1(RF (A•)) → ... (2) The functors RiF : A → B are zero for i < 0. Also R0F = F : A → B. (3) We have RiF (I) = 0 for i > 0 and I injective. (4) The sequence (RiF, δ) forms a universal δ-functor (see Homology, Defini- tion 12.3) from A to B. Proof. This lemma simply reviews some of the results obtained so far. Note that by Lemma 20.2 RF is everywhere defined. Here are some references: (1) This follows from Lemma 20.3 part (3) combined with the long exact cohomology sequence (11.1.1) for D+(B). (2) This is Lemma 16.3. (3) This is the fact that injective objects are acyclic. (4) This is Lemma 16.6. DERIVED CATEGORIES 71

21. Cartan-Eilenberg resolutions 015G This section can be expanded. The material can be generalized and applied in more cases. Resolutions need not use injectives and the method also works in the unbounded case in some situations. 015HDefinition 21.1. Let A be an abelian category. Let K• be a bounded below complex. A Cartan-Eilenberg resolution of K• is given by a double complex I•,• and a morphism of complexes : K• → I•,0 with the following properties: (1) There exists a i 0 such that Ip,q = 0 for all p < i and all q. (2) We have Ip,q = 0 if q < 0. (3) The complex Ip,• is an injective resolution of Kp. p,• p (4) The complex Ker(d1 ) is an injective resolution of Ker(dK ). p,• p (5) The complex Im(d1 ) is an injective resolution of Im(dK ). p •,• p • (6) The complex HI (I ) is an injective resolution of H (K ). 015I Lemma 21.2. Let A be an abelian category with enough injectives. Let K• be a bounded below complex. There exists a Cartan-Eilenberg resolution of K•. Proof. Suppose that Kp = 0 for p < n. Decompose K• into short exact sequences as follows: Set Zp = Ker(dp), Bp = Im(dp−1), Hp = Zp/Bp, and consider 0 → Zn → Kn → Bn+1 → 0 0 → Bn+1 → Zn+1 → Hn+1 → 0 0 → Zn+1 → Kn+1 → Bn+2 → 0 0 → Bn+2 → Zn+2 → Hn+2 → 0 ... Set Ip,q = 0 for p < n. Inductively we choose injective resolutions as follows: n n,• (1) Choose an injective resolution Z → JZ . n n,• n+1 n+1,• (2) Using Lemma 18.9 choose injective resolutions K → I , B → JB , n,• n,• n+1,• and an exact sequence of complexes 0 → JZ → I → JB → 0 compatible with the short exact sequence 0 → Zn → Kn → Bn+1 → 0. n+1 n+1,• n+1 (3) Using Lemma 18.9 choose injective resolutions Z → JZ , H → n+1,• n+1,• n+1,• JH , and an exact sequence of complexes 0 → JB → JZ → n+1,• n+1 n+1 JH → 0 compatible with the short exact sequence 0 → B → Z → Hn+1 → 0. (4) Etc. • p,• p+1,• p,• p+1,• p+1,• Taking as maps d1 : I → I the compositions I → JB → JZ → p+1,• I everything is clear. 015J Lemma 21.3. Let F : A → B be a left exact functor of abelian categories. Let K• be a bounded below complex of A. Let I•,• be a Cartan-Eilenberg resolution for • 0 0 00 00 K . The spectral sequences ( Er, dr)r≥0 and ( Er, dr)r≥0 associated to the double complex F (I•,•) satisfy the relations 0 p,q q p 00 p,q p q • E1 = R F (K ) and E2 = R F (H (K )) Moreover, these spectral sequences are bounded, converge to H∗(RF (K•)), and the associated induced filtrations on Hn(RF (K•)) are finite. Proof. We will use the following remarks without further mention: DERIVED CATEGORIES 72

(1) As Ip,• is an injective resolution of Kp we see that RF is defined at Kp[0] with value F (Ip,•). p •,• p • (2) As HI (I ) is an injective resolution of H (K ) the derived functor RF is p • p •,• defined at H (K )[0] with value F (HI (I )). (3) By Homology, Lemma 25.4 the total complex Tot(I•,•) is an injective resolution of K•. Hence RF is defined at K• with value F (Tot(I•,•)). Consider the two spectral sequences associated to the double complex L•,• = F (I•,•), see Homology, Lemma 25.1. These are both bounded, converge to H∗(Tot(L•,•)), and induce finite filtrations on Hn(Tot(L•,•)), see Homology, Lemma 25.3. Since Tot(L•,•) = Tot(F (I•,•)) = F (Tot(I•,•)) computes Hn(RF (K•)) we find the final assertion of the lemma holds true. 0 p,q q p,• Computation of the first spectral sequence. We have E1 = H (L ) in other words 0 p,q q p,• q p E1 = H (F (I )) = R F (K ) 0 p,q 0 p,q 0 p+1,q as desired. Observe for later use that the maps d1 : E1 → E1 are the maps RqF (Kp) → RqF (Kp+1) induced by Kp → Kp+1 and the fact that RqF is a functor. 00 p,q q •,p Computation of the second spectral sequence. We have E1 = H (L ) = Hq(F (I•,p)). Note that the complex I•,p is bounded below, consists of injectives, and moreover each kernel, image, and cohomology group of the differentials is an injective object of A. Hence we can split the differentials, i.e., each differential is a split surjection onto a direct summand. It follows that the same is true after 00 p,q q •,p q •,p applying F . Hence E1 = F (H (I )) = F (HI (I )). The differentials on this q p •,• are (−1) times F applied to the differential of the complex HI (I ) which is an p • injective resolution of H (K ). Hence the description of the E2 terms. 015KRemark 21.4. The spectral sequences of Lemma 21.3 are functorial in the complex K•. This follows from functoriality properties of Cartan-Eilenberg resolutions. On the other hand, they are both examples of a more general spectral sequence which may be associated to a filtered complex of A. The functoriality will follow from its construction. We will return to this in the section on the filtered derived category, see Remark 26.15.

22. Composition of right derived functors 015L Sometimes we can compute the right derived functor of a composition. Suppose that A, B, C be abelian categories. Let F : A → B and G : B → C be left exact functors. Assume that the right derived functors RF : D+(A) → D+(B), RG : D+(B) → D+(C), and R(G ◦ F ): D+(A) → D+(C) are everywhere deﬁned. Then there exists a canonical transformation t : R(G ◦ F ) −→ RG ◦ RF of functors from D+(A) to D+(C), see Lemma 14.16. This transformation need not always be an isomorphism. 015M Lemma 22.1. Let A, B, C be abelian categories. Let F : A → B and G : B → C be left exact functors. Assume A, B have enough injectives. The following are equivalent (1) F (I) is right acyclic for G for each injective object I of A, and DERIVED CATEGORIES 73

(2) the canonical map t : R(G ◦ F ) −→ RG ◦ RF. is isomorphism of functors from D+(A) to D+(C). Proof. If (2) holds, then (1) follows by evaluating the isomorphism t on RF (I) = F (I). Conversely, assume (1) holds. Let A• be a bounded below complex of A. Choose an injective resolution A• → I•. The map t is given (see proof of Lemma 14.16) by the maps R(G ◦ F )(A•) = (G ◦ F )(I•) = G(F (I•))) → RG(F (I•)) = RG(RF (A•)) where the arrow is an isomorphism by Lemma 16.7. 015N Lemma 22.2 (Grothendieck spectral sequence). With assumptions as in Lemma 22.1 and assuming the equivalent conditions (1) and (2) hold. Let X be an object of + D (A). There exists a spectral sequence (Er, dr)r≥0 consisting of bigraded objects Er of C with dr of bidegree (r, −r + 1) and with p,q p q E2 = R G(H (RF (X))) Moreover, this spectral sequence is bounded, converges to H∗(R(G ◦ F )(X)), and induces a finite filtration on each Hn(R(G ◦ F )(X)). p,q p q p+q For an object A of A we get E2 = R G(R F (A)) converging to R (G ◦ F )(A). Proof. We may represent X by a bounded below complex A•. Choose an injective resolution A• → I•. Choose a Cartan-Eilenberg resolution F (I•) → I•,• using Lemma 21.2. Apply the second spectral sequence of Lemma 21.3. 23. Resolution functors 013U Let A be an abelian category with enough injectives. Denote I the full additive subcategory of A whose objects are the injective objects of A. It turns out that K+(I) and D+(A) are equivalent in this case (see Proposition 23.1). For many purposes it therefore makes sense to think of D+(A) as the (easier to grok) category K+(I) in this case. 013V Proposition 23.1. Let A be an abelian category. Assume A has enough injectives. Denote I ⊂ A the strictly full additive subcategory whose objects are the injective objects of A. The functor K+(I) −→ D+(A) is exact, fully faithful and essentially surjective, i.e., an equivalence of triangulated categories. Proof. It is clear that the functor is exact. It is essentially surjective by Lemma 18.3. Fully faithfulness is a consequence of Lemma 18.8. Proposition 23.1 implies that we can find resolution functors. It turns out that we can prove resolution functors exist even in some cases where the abelian category A is a “big” category, i.e., has a class of objects. 013WDefinition 23.2. Let A be an abelian category with enough injectives. A resolution functor6 for A is given by the following data:

6This is likely nonstandard terminology. DERIVED CATEGORIES 74

(1) for all K• ∈ Ob(K+(A)) a bounded below complex of injectives j(K•), and • + • • (2) for all K ∈ Ob(K (A)) a quasi-isomorphism iK• : K → j(K ). 05TJ Lemma 23.3. Let A be an abelian category with enough injectives. Given a resolution functor (j, i) there is a unique way to turn j into a functor and i into a 2-isomorphism producing a 2-commutative diagram

K+(A) / K+(I) j

$ z D+(A) where I is the full additive subcategory of A consisting of injective objects. Proof. For every morphism α : K• → L• of K+(A) there is a unique morphism j(α): j(K•) → j(L•) in K+(I) such that

• • K α / L

iK• iL• j(α) j(K•) / j(L•) is commutative in K+(A). To see this either use Lemmas 18.6 and 18.7 or the equivalent Lemma 18.8. The uniqueness implies that j is a functor, and the commutativity of the diagram implies that i gives a 2-morphism which witnesses the 2-commutativity of the diagram of categories in the statement of the lemma. 013X Lemma 23.4. Let A be an abelian category. Assume A has enough injectives. Then a resolution functor j exists and is unique up to unique isomorphism of functors. Proof. Consider the set of all objects K• of K+(A). (Recall that by our conventions any category has a set of objects unless mentioned otherwise.) By Lemma 18.3 every object has an injective resolution. By the axiom of choice we can choose • • • for each K an injective resolution iK• : K → j(K ). 014W Lemma 23.5. Let A be an abelian category with enough injectives. Any resolution functor j : K+(A) → K+(I) is exact.

• • Proof. Denote iK• : K → j(K ) the canonical maps of Deﬁnition 23.2. First we discuss the existence of the functorial isomorphism j(K•[1]) → j(K•)[1]. Consider the diagram K•[1] K•[1]

iK•[1] iK• [1]

ξ • j(K•[1]) K / j(K•)[1] + By Lemmas 18.6 and 18.7 there exists a unique dotted arrow ξK• in K (I) making the diagram commute in K+(A). We omit the veriﬁcation that this gives a functorial isomorphism. (Hint: use Lemma 18.7 again.) Let (K•,L•,M •, f, g, h) be a distinguished triangle of K+(A). We have to show • • • + that (j(K ), j(L ), j(M ), j(f), j(g), ξK• ◦j(h)) is a distinguished triangle of K (I). DERIVED CATEGORIES 75

Note that we have a commutative diagram K• / L• / M • / K•[1] f g h

j(f) j(g) ξK• ◦j(h) j(K•) / j(L•) / j(M •) / j(K•)[1]

+ in K (A) whose vertical arrows are the quasi-isomorphisms iK , iL, iM . Hence we • • • + see that the image of (j(K ), j(L ), j(M ), j(f), j(g), ξK• ◦ j(h)) in D (A) is isomorphic to a distinguished triangle and hence a distinguished triangle by TR1. • • • Thus we see from Lemma 4.18 that (j(K ), j(L ), j(M ), j(f), j(g), ξK• ◦ j(h)) is + a distinguished triangle in K (I). 05TK Lemma 23.6. Let A be an abelian category which has enough injectives. Let j be a resolution functor. Write Q : K+(A) → D+(A) for the natural functor. Then j = j0 ◦ Q for a unique functor j0 : D+(A) → K+(I) which is quasi-inverse to the canonical functor K+(I) → D+(A). Proof. By Lemma 11.6 Q is a localization functor. To prove the existence of j0 it suffices to show that any element of Qis+(A) is mapped to an isomorphism under the functor j, see Lemma 5.6. This is true by the remarks following Definition 23.2. 013YRemark 23.7. Suppose that A is a “big” abelian category with enough injectives such as the category of abelian groups. In this case we have to be slightly more careful in constructing our resolution functor since we cannot use the axiom of choice with a quantifier ranging over a class. But note that the proof of the lemma does show that any two localization functors are canonically isomorphic. Namely, given quasi-isomorphisms i : K• → I• and i0 : K• → J • of a bounded below complex K• into bounded below complexes of injectives there exists a unique(!) morphism a : I• → J • in K+(I) such that i0 = i ◦ a as morphisms in K+(I). Hence the only issue is existence, and we will see how to deal with this in the next section.

24. Functorial injective embeddings and resolution functors 0140 In this section we redo the construction of a resolution functor K+(A) → K+(I) in case the category A has functorial injective embeddings. There are two reasons for this: (1) the proof is easier and (2) the construction also works if A is a “big” abelian category. See Remark 24.3 below. Let A be an abelian category. As before denote I the additive full subcategory of A consisting of injective objects. Consider the category InjRes(A) of arrows α : K• → I• where K• is a bounded below complex of A, I• is a bounded below complex of injectives of A and α is a quasi-isomorphism. In other words, α is an injective resolution and K• is bounded below. There is an obvious functor s : InjRes(A) −→ Comp+(A) deﬁned by (α : K• → I•) 7→ K•. There is also a functor t : InjRes(A) −→ K+(I) deﬁned by (α : K• → I•) 7→ I•. DERIVED CATEGORIES 76

0141 Lemma 24.1. Let A be an abelian category. Assume A has functorial injective embeddings, see Homology, Definition 27.5. (1) There exists a functor inj : Comp+(A) → InjRes(A) such that s ◦ inj = id. (2) For any functor inj : Comp+(A) → InjRes(A) such that s ◦ inj = id we obtain a resolution functor, see Definition 23.2. Proof. Let A 7→ (A → J(A)) be a functorial injective embedding, see Homology, Definition 27.5. We first note that we may assume J(0) = 0. Namely, if not then for any object A we have 0 → A → 0 which gives a direct sum decomposition J(A) = J(0) ⊕ Ker(J(A) → J(0)). Note that the functorial morphism A → J(A) has to map into the second summand. Hence we can replace our functor by J 0(A) = Ker(J(A) → J(0)) if needed. Let K• be a bounded below complex of A. Say Kp = 0 if p < B. We are going to construct a double complex I•,• of injectives, together with a map α : K• → I•,0 such that α induces a quasi-isomorphism of K• with the associated total complex of I•,•. First we set Ip,q = 0 whenever q < 0. Next, we set Ip,0 = J(Kp) and αp : Kp → Ip,0 the functorial embedding. Since J is a functor we see that I•,0 is a complex and that α is a morphism of complexes. Each αp is injective. And Ip,0 = 0 for p < B because J(0) = 0. Next, we set Ip,1 = J(Coker(Kp → Ip,0)). Again by functoriality we see that I•,1 is a complex. And again we get that Ip,1 = 0 for p < B. It is also clear that Kp maps isomorphically onto Ker(Ip,0 → Ip,1). As our third step we take Ip,2 = J(Coker(Ip,0 → Ip,1)). And so on and so forth. At this point we can apply Homology, Lemma 25.4 to get that the map α : K• −→ Tot(I•,•) is a quasi-isomorphism. To prove we get a functor inj it rests to show that the construction above is functorial. This verification is omitted. Suppose we have a functor inj such that s ◦ inj = id. For every object K• of Comp+(A) we can write

• • • inj(K ) = (iK• : K → j(K ))

This provides us with a resolution functor as in Deﬁnition 23.2.

05TLRemark 24.2. Suppose inj is a functor such that s ◦ inj = id as in part (2) of • • • Lemma 24.1. Write inj(K ) = (iK• : K → j(K )) as in the proof of that lemma. Suppose α : K• → L• is a map of bounded below complexes. Consider the map inj(α) in the category InjRes(A). It induces a commutative diagram

α K• / L•

iK iL inj(α) j(K)• / j(L)•

of morphisms of complexes. Hence, looking at the proof of Lemma 23.3 we see that the functor j : K+(A) → K+(I) is given by the rule

• • j(α up to homotopy) = inj(α) up to homotopy ∈ HomK+(I)(j(K ), j(L )) DERIVED CATEGORIES 77

Hence we see that j matches t ◦ inj in this case, i.e., the diagram

Comp+(A) / K+(I) t◦inj :

j & K+(A) is commutative.

0142Remark 24.3. Let Mod(OX ) be the category of OX -modules on a ringed space (X, OX ) (or more generally on a ringed site). We will see later that Mod(OX ) has enough injectives and in fact functorial injective embeddings, see Injectives, Theorem 8.4. Note that the proof of Lemma 23.4 does not apply to Mod(OX ). But the proof of Lemma 24.1 does apply to Mod(OX ). Thus we obtain + + j : K (Mod(OX )) −→ K (I)

which is a resolution functor where I is the additive category of injective OX - modules. This argument also works in the following cases:

(1) The category ModR of R-modules over a ring R. (2) The category PMod(O) of presheaves of O-modules on a site endowed with a presheaf of rings. (3) The category Mod(O) of sheaves of O-modules on a ringed site. (4) Add more here as needed.

25. Right derived functors via resolution functors 05TM The content of the following lemma is that we can simply define RF (K•) = F (j(K•)) if we are given a resolution functor j. 05TN Lemma 25.1. Let A be an abelian category with enough injectives Let F : A → B be an additive functor into an abelian category. Let (i, j) be a resolution functor, see Definition 23.2. The right derived functor RF of F fits into the following 2- commutative diagram

j0 D+(A) / K+(I)

RF F $ z D+(B) where j0 is the functor from Lemma 23.6.

• • Proof. By Lemma 20.1 we have RF (K ) = F (j(K )). 0158Remark 25.2. In the situation of Lemma 25.1 we see that we have actually lifted the right derived functor to an exact functor F ◦ j0 : D+(A) → K+(B). It is occasionally useful to use such a factorization.

26. Filtered derived category and injective resolutions 015O Let A be an abelian category. In this section we will show that if A has enough injectives, then so does the category Filf (A) in some sense. One can use this observation to compute in the ﬁltered derived category of A. DERIVED CATEGORIES 78

The category Filf (A) is an example of an exact category, see Injectives, Remark 9.6. A special role is played by the strict morphisms, see Homology, Definition 19.3, i.e., the morphisms f such that Coim(f) = Im(f). We will say that a complex A → B → C in Filf (A) is exact if the sequence gr(A) → gr(B) → gr(C) is exact in A. This implies that A → B and B → C are strict morphisms, see Homology, Lemma 19.15. 015PDefinition 26.1. Let A be an abelian category. We say an object I of Filf (A) is filtered injective if each grp(I) is an injective object of A.

05TP Lemma 26.2. Let A be an abelian category. An object I of Filf (A) is filtered injective if and only if there exist a ≤ b, injective objects In, a ≤ n ≤ b of A and ∼ L p L an isomorphism I = a≤n≤b In such that F I = n≥p In. Proof. Follows from the fact that any injection J → M of A is split if J is an injective object. Details omitted. 05TQ Lemma 26.3. Let A be an abelian category. Any strict monomorphism u : I → A of Filf (A) where I is a filtered injective object is a split injection. Proof. Let p be the largest integer such that F pI 6= 0. In particular grp(I) = F pI. Let I0 be the object of Filf (A) whose underlying object of A is F pI and with filtration given by F nI0 = 0 for n > p and F nI0 = I0 = F pI for n ≤ p. Note that I0 → I is a strict monomorphism too. The fact that u is a strict monomorphism implies that F pI → A/F p+1(A) is injective, see Homology, Lemma 19.13. Choose a splitting s : A/F p+1A → F pI in A. The induced morphism s0 : A → I0 is a strict morphism of filtered objects splitting the composition I0 → I → A. Hence we can 0 0 0 0 0 0 write A = I ⊕ Ker(s ) and I = I ⊕ Ker(s |I ). Note that Ker(s |I ) → Ker(s ) is a 0 strict monomorphism and that Ker(s |I ) is a filtered injective object. By induction 0 0 on the length of the filtration on I the map Ker(s |I ) → Ker(s ) is a split injection. Thus we win. 05TR Lemma 26.4. Let A be an abelian category. Let u : A → B be a strict monomorphism of Filf (A) and f : A → I a morphism from A into a filtered injective object in Filf (A). Then there exists a morphism g : B → I such that f = g ◦ u.

0 Proof. The pushout f : I → I qA B of f by u is a strict monomorphism, see Homology, Lemma 19.10. Hence the result follows formally from Lemma 26.3. 05TS Lemma 26.5. Let A be an abelian category with enough injectives. For any object A of Filf (A) there exists a strict monomorphism A → I where I is a filtered injective object. Proof. Pick a ≤ b such that grp(A) = 0 unless p ∈ {a, a + 1, . . . , b}. For each n+1 n ∈ {a, a + 1, . . . , b} choose an injection un : A/F A → In with In an injective L p L object. Set I = a≤n≤b In with filtration F I = n≥p In and set u : A → I equal to the direct sum of the maps un. 05TT Lemma 26.6. Let A be an abelian category with enough injectives. For any object A of Filf (A) there exists a filtered quasi-isomorphism A[0] → I• where I• is a complex of filtered injective objects with In = 0 for n < 0. DERIVED CATEGORIES 79

0 Proof. First choose a strict monomorphism u0 : A → I of A into a filtered injective object, see Lemma 26.5. Next, choose a strict monomorphism u1 : Coker(u0) → I1 into a filtered injective object of A. Denote d0 the induced map I0 → I1. Next, 2 choose a strict monomorphism u2 : Coker(u1) → I into a filtered injective object of A. Denote d1 the induced map I1 → I2. And so on. This works because each of the sequences n+1 0 → Coker(un) → I → Coker(un+1) → 0 is short exact, i.e., induces a short exact sequence on applying gr. To see this use Homology, Lemma 19.13. 05TU Lemma 26.7. Let A be an abelian category with enough injectives. Let f : A → B be a morphism of Filf (A). Given filtered quasi-isomorphisms A[0] → I• and B[0] → J • where I•,J • are complexes of filtered injective objects with In = J n = 0 for n < 0, then there exists a commutative diagram

A[0] / B[0]

I• / J • Proof. As A[0] → I• and C[0] → J • are ﬁltered quasi-isomorphisms we conclude 0 0 n n that a : A → I , b : B → J and all the morphisms dI , dJ are strict, see Ho- mology, Lemma 19.15. We will inductively construct the maps f n in the following commutative diagram

0 1 2 A a / I / I / I / ...

f f 0 f 1 f 2 b B / J 0 / J 1 / J 2 / ... Because A → I0 is a strict monomorphism and because J 0 is filtered injective, we can find a morphism f 0 : I0 → J 0 such that f 0 ◦ a = b ◦ f, see Lemma 26.4. The 0 0 0 0 0 composition dJ ◦ b ◦ f is zero, hence dJ ◦ f ◦ a = 0, hence dJ ◦ f factors through a unique morphism 0 0 1 Coker(a) = Coim(dI ) = Im(dI ) −→ J . 0 1 As Im(dI ) → I is a strict monomorphism we can extend the displayed arrow to a 1 1 1 morphism f : I → J by Lemma 26.4 again. And so on. 05TV Lemma 26.8. Let A be an abelian category with enough injectives. Let 0 → A → B → C → 0 be a short exact sequence in Filf (A). Given filtered quasi-isomorphisms A[0] → I• and C[0] → J • where I•,J • are complexes of filtered injective objects with In = J n = 0 for n < 0, then there exists a commutative diagram

0 / A[0] / B[0] / C[0] / 0

0 / I• / M • / J • / 0 where the lower row is a termwise split sequence of complexes. DERIVED CATEGORIES 80

Proof. As A[0] → I• and C[0] → J • are filtered quasi-isomorphisms we conclude 0 0 n n that a : A → I , c : C → J and all the morphisms dI , dJ are strict, see Homology, Lemma 13.4. We are going to step by step construct the south-east and the south arrows in the following commutative diagram B / C / J 0 / J 1 / ... O β c b b α δ0 δ1 a A / I0 / I1 / I2 / ... As A → B is a strict monomorphism, we can find a morphism b : B → I0 such that b ◦ α = a, see Lemma 26.4. As A is the kernel of the strict morphism I0 → I1 and β = Coker(α) we obtain a unique morphism b : C → I1 fitting into the diagram. As c is a strict monomorphism and I1 is filtered injective we can find δ0 : J 0 → I1, see Lemma 26.4. Because B → C is a strict epimorphism and 0 1 2 1 2 1 0 because B → I → I → I is zero, we see that C → I → I is zero. Hence dI ◦ δ ∼ 1 0 is zero on C = Im(c). Hence dI ◦ δ factors through a unique morphism 0 0 2 Coker(c) = Coim(dJ ) = Im(dJ ) −→ I . 2 0 1 As I is filtered injective and Im(dJ ) → J is a strict monomorphism we can extend the displayed morphism to a morphism δ1 : J 1 → I2, see Lemma 26.4. And so on. We set M • = I• ⊕ J • with differential n n+1 n n dI (−1) δ dM = n 0 dJ • 0 0 Finally, the map B[0] → M is given by b ⊕ c ◦ β : M → I ⊕ J . 05TW Lemma 26.9. Let A be an abelian category with enough injectives. For every K• ∈ K+(Filf (A)) there exists a filtered quasi-isomorphism K• → I• with I• bounded below, each In a filtered injective object, and each Kn → In a strict monomorphism. Proof. After replacing K• by a shift (which is harmless for the proof) we may assume that Kn = 0 for n < 0. Consider the short exact sequences 0 0 0 0 → Ker(dK ) → K → Coim(dK ) → 0 1 1 1 0 → Ker(dK ) → K → Coim(dK ) → 0 2 2 2 0 → Ker(dK ) → K → Coim(dK ) → 0 ...

f i i+1 of the exact category Fil (A) and the maps ui : Coim(dK ) → Ker(dK ). For each i ≥ 0 we may choose ﬁltered quasi-isomorphisms i • Ker(dK )[0] → Iker,i i • Coim(dK )[0] → Icoim,i n n with Iker,i,Icoim,i ﬁltered injective and zero for n < 0, see Lemma 26.6. By Lemma • • • 26.7 we may lift ui to a morphism of complexes ui : Icoim,i → Iker,i+1. Finally, for each i ≥ 0 we may complete the diagrams

i i i 0 / Ker(dK )[0] / K [0] / Coim(dK )[0] / 0

• αi • βi • 0 / Iker,i / Ii / Icoim,i / 0 DERIVED CATEGORIES 81

with the lower sequence a termwise split exact sequence, see Lemma 26.8. For i ≥ 0 • • • set di : Ii → Ii+1 equal to di = αi+1 ◦ ui ◦ βi. Note that di ◦ di−1 = 0 because βi ◦ αi = 0. Hence we have constructed a commutative diagram • • • I0 / I1 / I2 / ... O O O

K0[0] / K1[0] / K2[0] / ... Here the vertical arrows are filtered quasi-isomorphisms. The upper row is a complex of complexes and each complex consists of filtered injective objects with no a,b b nonzero objects in degree < 0. Thus we obtain a double complex by setting I = Ia and using a,b a,b b b a+1,b d1 : I = Ia → Ia+1 = I b the map da and using for a,b a,b b b+1 a,b+1 d2 : I = Ia → Ia = I the map db . Denote Tot(I•,•) the total complex associated to this double complex, Ia n • see Homology, Definition 18.3. Observe that the maps K [0] → In come from maps Kn → In,0 which give rise to a map of complexes K• −→ Tot(I•,•) We claim this is a filtered quasi-isomorphism. As gr(−) is an additive functor, we see that gr(Tot(I•,•)) = Tot(gr(I•,•)). Thus we can use Homology, Lemma 25.4 to • •,• conclude that gr(K ) → gr(Tot(I )) is a quasi-isomorphism as desired. 05TX Lemma 26.10. Let A be an abelian category. Let K•,I• ∈ K(Filf (A)). Assume K• is filtered acyclic and I• bounded below and consisting of filtered injective objects. • • • • Any morphism K → I is homotopic to zero: HomK(Filf (A))(K ,I ) = 0. Proof. Let α : K• → I• be a morphism of complexes. Assume that αj = 0 for j < n. We will show that there exists a morphism h : Kn+1 → In such that αn = h ◦ d. Thus α will be homotopic to the morphism of complexes β defined by   0 if j ≤ n βj = αn+1 − d ◦ h if j = n + 1  αj if j > n + 1 This will clearly prove the lemma (by induction). To prove the existence of h note n n−1 n−1 • n−1 that α ◦ dK = 0 since α = 0. Since K is filtered acyclic we see that dK n and dK are strict and that n−1 n n 0 → Im(dK ) → K → Im(dK ) → 0 is an exact sequence of the exact category Filf (A), see Homology, Lemma 19.15. n n n n Hence we can think of α as a map into I defined on Im(dK ). Using that Im(dK ) → Kn+1 is a strict monomorphism and that In is filtered injective we may lift this n+1 n map to a map h : K → I as desired, see Lemma 26.4. 05TY Lemma 26.11. Let A be an abelian category. Let I• ∈ K(Filf (A)) be a bounded below complex consisting of filtered injective objects. DERIVED CATEGORIES 82

(1) Let α : K• → L• in K(Filf (A)) be a ﬁltered quasi-isomorphism. Then the map • • • • HomK(Filf (A))(L ,I ) → HomK(Filf (A))(K ,I ) is bijective. (2) Let L• ∈ K(Filf (A)). Then • • • • HomK(Filf (A))(L ,I ) = HomDF (A)(L ,I ). Proof. Proof of (1). Note that (K•,L•,C(α)•, α, i, −p) is a distinguished triangle in K(Filf (A)) (Lemma 9.14) and C(α)• is a ﬁltered acyclic complex (Lemma 13.4). Then

• • • • • • HomK(Filf (A))(C(α) ,I ) / HomK(Filf (A))(L ,I ) / HomK(Filf (A))(K ,I )

• q • HomK(Filf (A))(C(α) [−1],I ) is an exact sequence of abelian groups, see Lemma 4.2. At this point Lemma • • 26.10 guarantees that the outer two groups are zero and hence HomK(A)(L ,I ) = • • HomK(A)(K ,I ). Proof of (2). Let a be an element of the right hand side. We may represent a = γα−1 where α : K• → L• is a filtered quasi-isomorphism and γ : K• → I• is a map of complexes. By part (1) we can find a morphism β : L• → I• such that β ◦ α is homotopic to γ. This proves that the map is surjective. Let b be an element of the left hand side which maps to zero in the right hand side. Then b is the homotopy class of a morphism β : L• → I• such that there exists a filtered quasi-isomorphism α : K• → L• with β ◦α homotopic to zero. Then part (1) shows that β is homotopic to zero also, i.e., b = 0. 015Q Lemma 26.12. Let A be an abelian category with enough injectives. Let If ⊂ Filf (A) denote the strictly full additive subcategory whose objects are the filtered injective objects. The canonical functor K+(If ) −→ DF +(A) is exact, fully faithful and essentially surjective, i.e., an equivalence of triangulated categories. Furthermore the diagrams

K+(If ) / DF +(A) K+(If ) / DF +(A)

grp grp forget F forget F K+(I) / D+(A) K+(I) / D+(A) are commutative, where I ⊂ A is the strictly full additive subcategory whose objects are the injective objects. Proof. The functor K+(If ) → DF +(A) is essentially surjective by Lemma 26.9. It is fully faithful by Lemma 26.11. It is an exact functor by our deﬁnitions regarding distinguished triangles. The commutativity of the squares is immediate. DERIVED CATEGORIES 83

015RRemark 26.13. We can invert the arrow of the lemma only if A is a category in our sense, namely if it has a set of objects. However, suppose given a big abelian category A with enough injectives, such as Mod(OX ) for example. Then for any 0 given set of objects {Ai}i∈I there is an abelian subcategory A ⊂ A containing all of them and having enough injectives, see Sets, Lemma 12.1. Thus we may use the lemma above for A0. This essentially means that if we use a set worth of diagrams, etc then we will never run into trouble using the lemma. Let A, B be abelian categories. Let T : A → B be a left exact functor. (We cannot use the letter F for the functor since this would conflict too much with our use of the letter F to indicate filtrations.) Note that T induces an additive functor T : Filf (A) → Filf (B) by the rule T (A, F ) = (T (A),F ) where F pT (A) = T (F pA) which makes sense as T is left exact. (Warning: It may not be the case that gr(T (A)) = T (gr(A)).) This induces functors of triangulated categories (26.13.1)05TZ T : K+(Filf (A)) −→ K+(Filf (B)) The filtered right derived functor of T is the right derived functor of Definition 14.2 for this exact functor composed with the exact functor K+(Filf (B)) → DF +(B) and the multiplicative set FQis+(A). Assume A has enough injectives. At this point we can redo the discussion of Section 20 to define the filtered right derived functors (26.13.2)015S RT : DF +(A) −→ DF +(B) of our functor T . However, instead we will proceed as in Section 25, and it will turn out that we can define RT even if T is just additive. Namely, we first choose a quasi-inverse j0 : DF +(A) → K+(If ) of the equivalence of Lemma 26.12. By Lemma 4.18 we see that j0 is an exact functor of triangulated categories. Next, we note that for a filtered injective object I we have a (noncanonical) decomposition ∼ M p M (26.13.3)015T I = Ip, with F I = Iq p∈Z q≥p by Lemma 26.2. Hence if T is any additive functor T : A → B then we get an additive functor f f (26.13.4)05U0 Text : I → Fil (B) L p L by setting Text(I) = T (Ip) with F Text(I) = q≥p T (Iq). Note that we have the property gr(Text(I)) = T (gr(I)) by construction. Hence we obtain a functor + f + f (26.13.5)05U1 Text : K (I ) → K (Fil (B)) which commutes with gr. Then we define (26.13.2) by the composition 0 (26.13.6)05U2 RT = Text ◦ j . Since RT : D+(A) → D+(B) is computed by injective resolutions as well, see Lemmas 20.1, the commutation of T with gr, and the commutative diagrams of Lemma 26.12 imply that (26.13.7)015U grp ◦ RT =∼ RT ◦ grp DERIVED CATEGORIES 84

and (26.13.8)015V (forget F ) ◦ RT =∼ RT ◦ (forget F ) as functors DF +(A) → D+(B). The ﬁltered derived functor RT (26.13.2) induces functors RT : Filf (A) → DF +(B), RT : Comp+(Filf (A)) → DF +(B), RT : KF +(A) → DF +(B).

Note that since Filf (A), and Comp+(Filf (A)) are no longer abelian it does not make sense to say that RT restricts to a δ-functor on them. (This can be repaired by thinking of these categories as exact categories and formulating the notion of a δ-functor from an exact category into a triangulated category.) But it does make sense, and it is true by construction, that RT is an exact functor on the triangulated category KF +(A). 015W Lemma 26.14. Let A, B be abelian categories. Let T : A → B be a left exact functor. Assume A has enough injectives. Let (K•,F ) be an object of Comp+(Filf (A)). There exists a spectral sequence (Er, dr)r≥0 consisting of bigraded objects Er of B and dr of bidegree (r, −r + 1) and with p,q p+q p • E1 = R T (gr (K )) Moreover, this spectral sequence is bounded, converges to R∗T (K•), and induces a finite filtration on each RnT (K•). The construction of this spectral sequence is • + f functorial in the object K of Comp (Fil (A)) and the terms (Er, dr) for r ≥ 1 do not depend on any choices. Proof. Choose a filtered quasi-isomorphism K• → I• with I• a bounded below complex of filtered injective objects, see Lemma 26.9. Consider the complex • • RT (K ) = Text(I ), see (26.13.6). Thus we can consider the spectral sequence (Er, dr)r≥0 associated to this as a filtered complex in B, see Homology, Section p,q p+q p • 24. By Homology, Lemma 24.2 we have E1 = H (gr (T (I ))). By Equa- p,q p+q p • tion (26.13.3) we have E1 = H (T (gr (I ))), and by definition of a filtered injective resolution the map grp(K•) → grp(I•) is an injective resolution. Hence p,q p+q p • E1 = R T (gr (K )). On the other hand, each In has a finite filtration and hence each T (In) has a finite filtration. Thus we may apply Homology, Lemma 24.11 to conclude that the spectral sequence is bounded, converges to Hn(T (I•)) = RnT (K•) moreover inducing finite filtrations on each of the terms. Suppose that K• → L• is a morphism of Comp+(Filf (A)). Choose a filtered quasi-isomorphism L• → J • with J • a bounded below complex of filtered injective objects, see Lemma 26.9. By our results above, for example Lemma 26.11, there exists a diagram K• / L•

I• / J • DERIVED CATEGORIES 85

which commutes up to homotopy. Hence we get a morphism of filtered complexes T (I•) → T (J •) which gives rise to the morphism of spectral sequences, see Homol- ogy, Lemma 24.4. The last statement follows from this. 015XRemark 26.15. As promised in Remark 21.4 we discuss the connection of the lemma above with the constructions using Cartan-Eilenberg resolutions. Namely, let T : A → B be a left exact functor of abelian categories, assume A has enough injectives, and let K• be a bounded below complex of A. We give an alternative construction of the spectral sequences 0E and 00E of Lemma 21.3. First spectral sequence. Consider the “stupid” filtration on K• obtained by set- p • • ting F (K ) = σ≥p(K ), see Homology, Section 15. Note that this stupid in the sense that d(F p(K•)) ⊂ F p+1(K•), compare Homology, Lemma 24.3. Note that grp(K•) = Kp[−p] with this filtration. According to Lemma 26.14 there is a spectral sequence with E1 term p,q p+q p q p E1 = R T (K [−p]) = R T (K ) 0 p,q as in the spectral sequence Er. Observe moreover that the differentials E1 → p+1,q 0 E1 agree with the differentials in E1, see Homology, Lemma 24.3 part (2) and 0 the description of d1 in the proof of Lemma 21.3. Second spectral sequence. Consider the filtration on the complex K• obtained by p • • setting F (K ) = τ≤−p(K ), see Homology, Section 15. The minus sign is necessary to get a decreasing filtration. Note that grp(K•) is quasi-isomorphic to H−p(K•)[p] with this filtration. According to Lemma 26.14 there is a spectral sequence with E1 term p,q p+q −p • 2p+q −p • 00 i,j E1 = R T (H (K )[p]) = R T (H (K )) = E2 with i = 2p+q and j = −p. (This looks unnatural, but note that we could just have well developed the whole theory of filtered complexes using increasing filtrations, with the end result that this then looks natural, but the other one doesn’t.) We leave it to the reader to see that the differentials match up. Actually, given a Cartan-Eilenberg resolution K• → I•,• the induced morphism K• → Tot(I•,•) into the associated total complex will be a filtered injective resolution for either filtration using suitable filtrations on Tot(I•,•). This can be used to match up the spectral sequences exactly.

27. Ext groups 06XP In this section we start describing the Ext groups of objects of an abelian category. First we have the following very general deﬁnition. 06XQDeﬁnition 27.1. Let A be an abelian category. Let i ∈ Z. Let X,Y be objects of D(A). The ith extension group of X by Y is the group i ExtA(X,Y ) = HomD(A)(X,Y [i]) = HomD(A)(X[−i],Y ). i i If A, B ∈ Ob(A) we set ExtA(A, B) = ExtA(A[0],B[0]).

Since HomD(A)(X, −), resp. HomD(A)(−,Y ) is a homological, resp. cohomological functor, see Lemma 4.2, we see that a distinguished triangle (Y,Y 0,Y 00), resp. (X,X0,X00) leads to a long exact sequence i i 0 i 00 i+1 ... → ExtA(X,Y ) → ExtA(X,Y ) → ExtA(X,Y ) → ExtA (X,Y ) → ... DERIVED CATEGORIES 86

respectively i 00 i 0 i i+1 00 ... → ExtA(X ,Y ) → ExtA(X ,Y ) → ExtA(X,Y ) → ExtA (X ,Y ) → ... Note that since D+(A), D−(A), Db(A) are full subcategories we may compute the Ext groups by Hom groups in these categories provided X, Y are contained in them. In case the category A has enough injectives or enough projectives we can compute the Ext groups using injective or projective resolutions. To avoid confusion, recall that having an injective (resp. projective) resolution implies vanishing of homology in all low (resp. high) degrees, see Lemmas 18.2 and 19.2. 06XR Lemma 27.2. Let A be an abelian category. Let X•,Y • ∈ Ob(K(A)). (1) Let Y • → I• be an injective resolution (Definition 18.1). Then i • • • • ExtA(X ,Y ) = HomK(A)(X ,I [i]). (2) Let P • → X• be a projective resolution (Definition 19.1). Then i • • • • ExtA(X ,Y ) = HomK(A)(P [−i],Y ). Proof. Follows immediately from Lemma 18.8 and Lemma 19.8. In the rest of this section we discuss extensions of objects of the abelian category itself. First we observe the following. 06XS Lemma 27.3. Let A be an abelian category. (1) Let X, Y be objects of D(A). Given a, b ∈ Z such that Hi(X) = 0 for i > a j n and H (Y ) = 0 for j < b, we have ExtA(X,Y ) = 0 for n < b − a and b−a a b ExtA (X,Y ) = HomA(H (X),H (Y )) i (2) Let A, B ∈ Ob(A). For i < 0 we have ExtA(B,A) = 0. We have 0 ExtA(B,A) = HomA(B,A). • • • • Proof. Choose complexes X and Y representing X and Y . Since Y → τ≥bY is a quasi-isomorphism, we may assume that Y j = 0 for j < b. Let L• → X• • • be any quasi-isomorphism. Then τ≤aL → X is a quasi-isomorphism. Hence a morphism X → Y [n] in D(A) can be represented as fs−1 where s : L• → X• is a quasi-isomorphism, f : L• → Y •[n] a morphism, and Li = 0 for i < a. Note that f maps Li to Y i+n. Thus f = 0 if n < b − a because always either Li or Y i+n is zero. If n = b − a, then f corresponds exactly to a morphism Ha(X) → Hb(Y ). Part (2) is a special case of (1). Let A be an abelian category. Suppose that 0 → A → A0 → A00 → 0 is a short exact sequence of objects of A. Then 0 → A[0] → A0[0] → A00[0] → 0 leads to a distinguished triangle in D(A) (see Lemma 12.1) hence a long exact sequence of Ext groups 0 0 0 0 00 1 0 → ExtA(B,A) → ExtA(B,A ) → ExtA(B,A ) → ExtA(B,A) → ... Similarly, given a short exact sequence 0 → B → B0 → B00 → 0 we obtain a long exact sequence of Ext groups 0 00 0 0 0 1 00 0 → ExtA(B ,A) → ExtA(B ,A) → ExtA(B,A) → ExtA(B ,A) → ... We may view these Ext groups as an application of the construction of the derived category. It shows one can define Ext groups and construct the long exact sequence DERIVED CATEGORIES 87

of Ext groups without needing the existence of enough injectives or projectives. There is an alternative construction of the Ext groups due to Yoneda which avoids the use of the derived category, see [Yon60]. 06XTDeﬁnition 27.4. Let A be an abelian category. Let A, B ∈ Ob(A). A degree i Yoneda extension of B by A is an exact sequence

E : 0 → A → Zi−1 → Zi−2 → ... → Z0 → B → 0 in A. We say two Yoneda extensions E and E0 of the same degree are equivalent if there exists a commutative diagram

0 / A / Zi−1 / ... / Z0 / B / 0 O O O O id id

00 00 0 / A / Zi−1 / ... / Z0 / B / 0

id id

0 0 0 / A / Zi−1 / ... / Z0 / B / 0 where the middle row is a Yoneda extension as well. It is not immediately clear that the equivalence of the deﬁnition is an equivalence relation. Although it is instructive to prove this directly this will also follow from Lemma 27.5 below. Let A be an abelian category with objects A, B. Given a Yoneda extension E : 0 → A → Zi−1 → Zi−2 → ... → Z0 → B → 0 we deﬁne an associated element δ(E) ∈ Exti(B,A) as the morphism δ(E) = fs−1 : B[0] → A[i] where s is the quasi-isomorphism

(... → 0 → A → Zi−1 → ... → Z0 → 0 → ...) −→ B[0] and f is the morphism of complexes

(... → 0 → A → Zi−1 → ... → Z0 → 0 → ...) −→ A[i] We call δ(E) = fs−1 the class of the Yoneda extension. It turns out that this class characterizes the equivalence class of the Yoneda extension. 06XU Lemma 27.5. Let A be an abelian category with objects A, B. Any element in i ExtA(B,A) is δ(E) for some degree i Yoneda extension of B by A. Given two Yoneda extensions E, E0 of the same degree then E is equivalent to E0 if and only if δ(E) = δ(E0).

i −1 Proof. Let ξ : B[0] → A[i] be an element of ExtA(B,A). We may write ξ = fs for some quasi-isomorphism s : L• → B[0] and map f : L• → A[i]. After replacing • • j L by τ≤0L we may assume that L = 0 for j > 0. Picture L−i−1 / L−i / ... / L0 / B / 0

A −i+1 −i −j Then setting Zi−1 = (L ⊕ A)/L and Zj = L for j = i − 2,..., 0 we see that we obtain a degree i extension E of B by A whose class δ(E) equals ξ. DERIVED CATEGORIES 88

It is immediate from the definitions that equivalent Yoneda extensions have the same class. Suppose that E : 0 → A → Zi−1 → Zi−2 → ... → Z0 → B → 0 and 0 0 0 0 E : 0 → A → Zi−1 → Zi−2 → ... → Z0 → B → 0 are Yoneda extensions with the same class. By construction of D(A) as the localization of K(A) at the set of quasi-isomorphisms, this means there exists a complex L• and quasi-isomorphisms • t : L → (... → 0 → A → Zi−1 → ... → Z0 → 0 → ...) and 0 • 0 0 t : L → (... → 0 → A → Zi−1 → ... → Z0 → 0 → ...) such that s ◦ t = s0 ◦ t0 and f ◦ t = f 0 ◦ t0, see Categories, Section 27. Let E00 be the degree i extension of B by A constructed from the pair L• → B[0] and L• → A[i] in the first paragraph of the proof. Then the reader sees readily that there exists “morphisms” of degree i Yoneda extensions E00 → E and E00 → E0 as in the definition of equivalent Yoneda extensions (details omitted). This finishes the proof. 06XV Lemma 27.6. Let A be an abelian category. Let A, B be objects of A. Then 1 ExtA(B,A) is the group ExtA(B,A) constructed in Homology, Definition 6.2. Proof. This is the case i = 1 of Lemma 27.5. p 0EWW Lemma 27.7. Let A be an abelian category and let p ≥ 0. If ExtA(B,A) = 0 i for any pair of objects A, B of A, then ExtA(B,A) = 0 for i ≥ p and any pair of objects A, B of A. Proof. For i > p write any class ξ as δ(E) where E is a Yoneda extension

E : 0 → A → Zi−1 → Zi−2 → ... → Z0 → B → 0

This is possible by Lemma 27.5. Set C = Ker(Zp−1 → Zp) = Im(Zp → Zp−1). Then δ(E) is the composition of δ(E0) and δ(E00) where 0 E : 0 → C → Zp−1 → ... → Z0 → B → 0 and 00 E : 0 → A → Zi−1 → Zi−2 → ... → Zp → C → 0 0 p Since δ(E ) ∈ ExtA(B,C) = 0 we conclude. 0GM4 Lemma 27.8. Let A be an abelian category. Let K be an object of Db(A) such p i j that ExtA(H (K),H (K)) = 0 for all p ≥ 2 and i > j. Then K is isomorphic to the direct sum of its cohomologies: K =∼ L Hi(K)[−i]. Proof. Choose a, b such that Hi(K) = 0 for i 6∈ [a, b]. We will prove the lemma by induction on b − a. If b − a ≤ 0, then the result is clear. If b − a > 0, then we look at the distinguished triangle of truncations b τ≤b−1K → K → H (K)[−b] → (τ≤b−1K)[1] ∼ see Remark 12.4. By Lemma 4.11 if the last arrow is zero, then K = τ≤b−1K ⊕ Hb(K)[−b] and we win by induction. Again using induction we see that

b M b−i+1 b i HomD(A)(H (K)[−b], (τ≤b−1K)[1]) = Ext (H (K),H (K)) i

2 0EWX Lemma 27.9. Let A be an abelian category. Assume ExtA(B,A) = 0 for any pair of objects A, B of A. Then any object K of Db(A) is isomorphic to the direct sum of its cohomologies: K =∼ L Hi(K)[−i].

i Proof. The assumption implies that ExtA(B,A) = 0 for i ≥ 2 and any pair of objects A, B of A by Lemma 27.7. Hence this lemma is a special case of Lemma 27.8.

28. K-groups

0FCM A tiny bit about K0 of a triangulated category.

0FCNDeﬁnition 28.1. Let D be a triangulated category. We denote K0(D) the zeroth K-group of D. It is the abelian group constructed as follows. Take the free abelian group on the objects on D and for every distinguished triangle X → Y → Z impose the relation [Y ] − [X] − [Z] = 0. Observe that this implies that [X[n]] = (−1)n[X] because we have the distinguished triangle (X, 0,X[1], 0, 0, −id[1]). 0FCP Lemma 28.2. Let A be an abelian category. Then there is a canonical identiﬁca- b tion K0(D (A)) = K0(A) of zeroth K-groups. Proof. Given an object A of A denote A[0] the object A viewed as a complex sitting in degree 0. If 0 → A → A0 → A00 → 0 is a short exact sequence, then we get a distinguished triangle A[0] → A0[0] → A00[0] → A[1], see Section 12. This b shows that we obtain a map K0(A) → K0(D (A)) by sending [A] to [A[0]] with apologies for the horrendous notation. On the other hand, given an object X of Db(A) we can consider the element

X i i c(X) = (−1) [H (X)] ∈ K0(A) Given a distinguished triangle X → Y → Z the long exact sequence of cohomology (11.1.1) and the relations in K0(A) show that c(Y ) = c(X) + c(Z). Thus c factors b through a map c : K0(D (A)) → K0(A). We want to show that the two maps above are mutually inverse. It is clear that b • the composition K0(A) → K0(D (A)) → K0(A) is the identity. Suppose that X is a bounded complex of A. The existence of the distinguished triangles of “stupid truncations” (see Homology, Section 15) • • n−1 • σ≥nX → σ≥n−1X → X [−n + 1] → (σ≥nX )[1] and induction show that X [X•] = (−1)i[Xi[0]]

b in K0(D (A)) (with again apologies for the notation). It follows that the composi- b tion K0(A) → K0(D (A)) is surjective which ﬁnishes the proof. 0FCQ Lemma 28.3. Let F : D → D0 be an exact functor of triangulated categories. 0 Then F induces a group homomorphism K0(D) → K0(D ).

Proof. Omitted. DERIVED CATEGORIES 90

0FCR Lemma 28.4. Let H : D → A be a homological functor from a triangulated category to an abelian category. Assume that for any X in D only a finite number of the objects H(X[i]) are nonzero in A. Then H induces a group homomorphism P i K0(D) → K0(A) sending [X] to (−1) [H(X[i])]. Proof. Omitted. 0FCS Lemma 28.5. Let B be a weak Serre subcategory of the abelian category A. Then there are canonical maps b K0(B) −→ K0(DB(A)) −→ K0(B) whose composition is zero. The second arrow sends the class [X] of the object X to P i i the element (−1) [H (X)] of K0(B). Proof. Omitted. 0FCT Lemma 28.6. Let D, D0, D00 be triangulated categories. Let ⊗ : D × D0 −→ D00 be a functor such that for fixed X in D the functor X ⊗ − : D0 → D00 is an exact functor and for fixed X0 in D0 the functor − ⊗ X0 : D → D00 is an exact functor. 0 00 0 Then ⊗ induces a bilinear map K0(D) × K0(D ) → K0(D ) which sends ([X], [X ]) to [X ⊗ X0].

Proof. Omitted. 29. Unbounded complexes 06XW A reference for the material in this section is [Spa88]. The following lemma is useful to ﬁnd “good” left resolutions of unbounded complexes. 06XX Lemma 29.1. Let A be an abelian category. Let P ⊂ Ob(A) be a subset. Assume P contains 0, is closed under (ﬁnite) direct sums, and every object of A is a quotient of an element of P. Let K• be a complex. There exists a commutative diagram • • P1 / P2 / ...

• • τ≤1K / τ≤2K / ... in the category of complexes such that (1) the vertical arrows are quasi-isomorphisms and termwise surjective, • (2) Pn is a bounded above complex with terms in P, • • (3) the arrows Pn → Pn+1 are termwise split injections and each cokernel i i Pn+1/Pn is an element of P. Proof. We are going to use that the homotopy category K(A) is a triangulated category, see Proposition 10.3. By Lemma 15.4 we can ﬁnd a termwise surjective • • map of complexes P1 → τ≤1K which is a quasi-isomorphism such that the terms • • • • of P1 are in P. By induction it suﬃces, given P1 ,...,Pn to construct Pn+1 and • • • • the maps Pn → Pn+1 and Pn+1 → τ≤n+1K . • • • • Choose a distinguished triangle Pn → τ≤n+1K → C → Pn [1] in K(A). Applying Lemma 15.4 we choose a map of complexes Q• → C• which is a quasi-isomorphism such that the terms of Q• are in P. By the axioms of triangulated categories we DERIVED CATEGORIES 91

• • • • may ﬁt the composition Q → C → Pn [1] into a distinguished triangle Pn → • • • • Pn+1 → Q → Pn [1] in K(A). By Lemma 10.7 we may and do assume 0 → Pn → • • Pn+1 → Q → 0 is a termwise split short exact sequence. This implies that the • • • terms of Pn+1 are in P and that Pn → Pn+1 is a termwise split injection whose cokernels are in P. By the axioms of triangulated categories we obtain a map of distinguished triangles

• • • • Pn / Pn+1 / Q / Pn [1]

• • • • Pn / τ≤n+1K / C / Pn [1] in the triangulated category K(A). Choose an actual morphism of complexes f : • • Pn+1 → τ≤n+1K . The left square of the diagram above commutes up to homotopy, • • but as Pn → Pn+1 is a termwise split injection we can lift the homotopy and modify our choice of f to make it commute. Finally, f is a quasi-isomorphism, because • • • • both Pn → Pn and Q → C are. At this point we have all the properties we want, except we don’t know that the • • map f : Pn+1 → τ≤n+1K is termwise surjective. Since we have the commutative diagram • • Pn / Pn+1

• • τ≤nK / τ≤n+1K of complexes, by induction hypothesis we see that f is surjective on terms in all degrees except possibly n and n + 1. Choose an object P ∈ P and a surjection q : P → Kn. Consider the map

• 1 • g : P = (... → 0 → P −→ P → 0 → ...) −→ τ≤n+1K

with first copy of P in degree n and maps given by q in degree n and dK ◦ q in degree n + 1. This is a surjection in degree n and the cokernel in degree n + 1 is n+1 • • n+1 H (τ≤n+1K ); to see this recall that τ≤n+1K has Ker(dK ) in degree n + 1. However, since f is a quasi-isomorphism we know that Hn+1(f) is surjective. Hence • • • • • after replacing f : Pn+1 → τ≤n+1K by f ⊕ g : Pn+1 ⊕ P → τ≤n+1K we win. In some cases we can use the lemma above to show that a left derived functor is everywhere defined. 0794 Proposition 29.2. Let F : A → B be a right exact functor of abelian categories. Let P ⊂ Ob(A) be a subset. Assume (1) P contains 0, is closed under (finite) direct sums, and every object of A is a quotient of an element of P, (2) for any bounded above acyclic complex P • of A with P n ∈ P for all n the complex F (P •) is exact, (3) A and B have colimits of systems over N, (4) colimits over N are exact in both A and B, and (5) F commutes with colimits over N. Then LF is defined on all of D(A). DERIVED CATEGORIES 92

Proof. By (1) and Lemma 15.4 for any bounded above complex K• there exists a quasi-isomorphism P • → K• with P • bounded above and P n ∈ P for all n. Suppose that s : P • → (P 0)• is a quasi-isomorphism of bounded above complexes consisting of objects of P. Then F (P •) → F ((P 0)•) is a quasi-isomorphism because F (C(s)•) is acyclic by assumption (2). This already shows that LF is deﬁned on D−(A) and that a bounded above complex consisting of objects of P computes LF , see Lemma 14.15.

Next, let K• be an arbitrary complex of A. Choose a diagram

• • P1 / P2 / ...

 • • τ≤1K / τ≤2K / ...

• • as in Lemma 29.1. Note that the map colim Pn → K is a quasi-isomorphism i • i • because colimits over N in A are exact and H (Pn ) = H (K ) for n > i. We claim that • • F (colim Pn ) = colim F (Pn )

• • (termwise colimits) is LF (K ), i.e., that colim Pn computes LF . To see this, by Lemma 14.15, it suﬃces to prove the following claim. Suppose that

• • α • • colim Qn = Q −−→ P = colim Pn

• • is a quasi-isomorphism of complexes, such that each Pn , Qn is a bounded above • • • • complex whose terms are in P and the maps Pn → τ≤nP and Qn → τ≤nQ are quasi-isomorphisms. Claim: F (α) is a quasi-isomorphism.

The problem is that we do not assume that α is given as a colimit of maps between • • the complexes Pn and Qn. However, for each n we know that the solid arrows in the diagram R•

• • • Pn o L / Qn

τ α • ≤n • τ≤nP / τ≤nQ are quasi-isomorphisms. Because quasi-isomorphisms form a multiplicative system • • in K(A) (see Lemma 11.2) we can find a quasi-isomorphism L → Pn and map • • of complexes L → Qn such that the diagram above commutes up to homotopy. • • Then τ≤nL → L is a quasi-isomorphism. Hence (by the first part of the proof) we can find a bounded above complex R• whose terms are in P and a quasi- isomorphism R• → L• (as indicated in the diagram). Using the result of the first • • • • paragraph of the proof we see that F (R ) → F (Pn ) and F (R ) → F (Qn) are quasi- i • i • isomorphisms. Thus we obtain a isomorphisms H (F (Pn )) → H (F (Qn)) fitting DERIVED CATEGORIES 93

into the commutative diagram

i • i • H (F (Pn )) / H (F (Qn))

Hi(F (P •)) / Hi(F (Q•)) The exact same argument shows that these maps are also compatible as n varies. Since by (4) and (5) we have i • i • i • i • H (F (P )) = H (F (colim Pn )) = H (colim F (Pn )) = colim H (F (Pn )) and similarly for Q• we conclude that Hi(α): Hi(F (P •) → Hi(F (Q•) is an isomorphism and the claim follows. 070F Lemma 29.3. Let A be an abelian category. Let I ⊂ Ob(A) be a subset. Assume I contains 0, is closed under (ﬁnite) products, and every object of A is a subobject of an element of I. Let K• be a complex. There exists a commutative diagram • • ... / τ≥−2K / τ≥−1K

• • ... / I2 / I1 in the category of complexes such that (1) the vertical arrows are quasi-isomorphisms and termwise injective, • (2) In is a bounded below complex with terms in I, • • i i (3) the arrows In+1 → In are termwise split surjections and Ker(In+1 → In) is an element of I.

Proof. This lemma is dual to Lemma 29.1.

30. Deriving adjoints 0FNC Let F : D → D0 and G : D0 → D be exact functors of triangulated categories. Let S, resp. S0 be a multiplicative system for D, resp. D0 compatible with the triangulated structure. Denote Q : D → S−1D and Q0 : D0 → (S0)−1D0 the localization functors. In this situation, by abuse of notation, one often denotes RF the partially deﬁned right derived functor corresponding to Q0 ◦ F : D → (S0)−1D0 and the multiplicative system S. Similarly one denotes LG the partially deﬁned left derived functor corresponding to Q ◦ G : D0 → S−1D and the multiplicative system S0. Picture D / D0 D0 / D F G Q Q0 and Q0 Q RF LG S−1D / (S0)−1D0 (S0)−1D0 / S−1D

0FND Lemma 30.1. In the situation above assume F is right adjoint to G. Let K ∈ Ob(D) and M ∈ Ob(D0). If RF is deﬁned at K and LG is deﬁned at M, then there is a canonical isomorphism

Hom(S0)−1D0 (M,RF (K)) = HomS−1D(LG(M),K) DERIVED CATEGORIES 94

This isomorphism is functorial in both variables on the triangulated subcategories of S−1D and (S0)−1D0 where RF and LG are deﬁned. Proof. Since RF is deﬁned at K, we see that the rule which assigns to a s : K → I in S the object F (I) is essentially constant as an ind-object of (S0)−1D0 with value RF (K). Similarly, the rule which assigns to a t : P → M in S0 the object G(P ) is essentially constant as a pro-object of S−1D with value LG(M). Thus we have

Hom(S0)−1D0 (M,RF (K)) = colims:K→I Hom(S0)−1D0 (M,F (I))

= colims:K→I colimt:P →M HomD0 (P,F (I))

= colimt:P →M colims:K→I HomD0 (P,F (I))

= colimt:P →M colims:K→I HomD(G(P ),I)

= colimt:P →M HomS−1D(G(P ),K)

= HomS−1D(LG(M),K) The first equality holds by Categories, Lemma 22.9. The second equality holds by the definition of morphisms in D(B), see Categories, Remark 27.15. The third equality holds by Categories, Lemma 14.10. The fourth equality holds because F and G are adjoint. The fifth equality holds by definition of morphism in D(A), see Categories, Remark 27.7. The sixth equality holds by Categories, Lemma 22.10. We omit the proof of functoriality. 0DVC Lemma 30.2. Let F : A → B and G : B → A be functors of abelian categories such that F is a right adjoint to G. Let K• be a complex of A and let M • be a complex of B. If RF is defined at K• and LG is defined at M •, then there is a canonical isomorphism • • • • HomD(B)(M ,RF (K )) = HomD(A)(LG(M ),K ) This isomorphism is functorial in both variables on the triangulated subcategories of D(A) and D(B) where RF and LG are defined.

Proof. This is a special case of the very general Lemma 30.1. The following lemma is an example of why it is easier to work with unbounded derived categories. Namely, without having the unbounded derived functors, the lemma could not even be stated. 09T5 Lemma 30.3. Let F : A → B and G : B → A be functors of abelian categories such that F is a right adjoint to G. If the derived functors RF : D(A) → D(B) and LG : D(B) → D(A) exist, then RF is a right adjoint to LG.

Proof. Immediate from Lemma 30.2. 31. K-injective complexes 070G The following types of complexes can be used to compute right derived functors on the unbounded derived category. 070HDeﬁnition 31.1. Let A be an abelian category. A complex I• is K-injective if for • • • every acyclic complex M we have HomK(A)(M ,I ) = 0. • • In the situation of the deﬁnition we have in fact HomK(A)(M [i],I ) = 0 for all i as the translate of an acyclic complex is acyclic. DERIVED CATEGORIES 95

070I Lemma 31.2. Let A be an abelian category. Let I• be a complex. The following are equivalent (1) I• is K-injective, (2) for every quasi-isomorphism M • → N • the map • • • • HomK(A)(N ,I ) → HomK(A)(M ,I ) is bijective, and (3) for every complex N • the map • • • • HomK(A)(N ,I ) → HomD(A)(N ,I ) is an isomorphism.

• Proof. Assume (1). Then (2) holds because the functor HomK(A)(−,I ) is cohomological and the cone on a quasi-isomorphism is acyclic. Assume (2). A morphism N • → I• in D(A) is of the form fs−1 : N • → I• where s : M • → N • is a quasi-isomorphism and f : M • → I• is a map. By (2) this corresponds to a unique morphism N • → I• in K(A), i.e., (3) holds. Assume (3). If M • is acyclic then M • is isomorphic to the zero complex in D(A) • • • • hence HomD(A)(M ,I ) = 0, whence HomK(A)(M ,I ) = 0 by (3), i.e., (1) holds. 090X Lemma 31.3. Let A be an abelian category. Let (K, L, M, f, g, h) be a distinguished triangle of K(A). If two out of K, L, M are K-injective complexes, then the third is too. Proof. Follows from the deﬁnition, Lemma 4.2, and the fact that K(A) is a triangulated category (Proposition 10.3). 070J Lemma 31.4. Let A be an abelian category. A bounded below complex of injectives is K-injective.

Proof. Follows from Lemmas 31.2 and 18.8. 0BK6 Lemma 31.5. Let A be an abelian category. Let T be a set and for each t ∈ T let • n Q n • It be a K-injective complex. If I = t It exists for all n, then I is a K-injective • • complex. Moreover, I represents the product of the objects It in D(A). Proof. Let K• be an complex. Observe that the complex Y Y Y C : Hom(K−b,Ib−1) → Hom(K−b,Ib) → Hom(K−b,Ib+1) b b b • • has cohomology HomK(A)(K ,I ) in the middle. Similarly, the complex

Y −b b−1 Y −b b Y −b b+1 Ct : Hom(K ,It ) → Hom(K ,It ) → Hom(K ,It ) b b b • • computes HomK(A)(K ,It ). Next, observe that we have Y C = Ct t∈T as complexes of abelian groups by our choice of I. Taking products is an exact func- • • • tor on the category of abelian groups. Hence if K is acyclic, then HomK(A)(K ,It ) = • • 0, hence Ct is acyclic, hence C is acyclic, hence we get HomK(A)(K ,I ) = 0. Thus DERIVED CATEGORIES 96

we find that I• is K-injective. Having said this, we can use Lemma 31.2 to conclude that • • Y • • HomD(A)(K ,I ) = HomD(A)(K ,It ) t∈T • and indeed I represents the product in the derived category. 070Y Lemma 31.6. Let A be an abelian category. Let F : K(A) → D0 be an exact functor of triangulated categories. Then RF is defined at every complex in K(A) which is quasi-isomorphic to a K-injective complex. In fact, every K-injective complex computes RF . Proof. By Lemma 14.4 it suffices to show that RF is defined at a K-injective complex, i.e., it suffices to show a K-injective complex I• computes RF . Any quasi-isomorphism I• → N • is a homotopy equivalence as it has an inverse by • • • Lemma 31.2. Thus I → I is a final object of I /Qis(A) and we win. 070K Lemma 31.7. Let A be an abelian category. Assume every complex has a quasi- isomorphism towards a K-injective complex. Then any exact functor F : K(A) → D0 of triangulated categories has a right derived functor RF : D(A) −→ D0 and RF (I•) = F (I•) for K-injective complexes I•. Proof. To see this we apply Lemma 14.15 with I the collection of K-injective complexes. Since (1) holds by assumption, it suffices to prove that if I• → J • is a quasi-isomorphism of K-injective complexes, then F (I•) → F (J •) is an isomorphism. This is clear because I• → J • is a homotopy equivalence, i.e., an isomorphism in K(A), by Lemma 31.2. The following lemma can be generalized to limits over bigger ordinals. 070L Lemma 31.8. Let A be an abelian category. Let • • • ... → I3 → I2 → I1 be an inverse system of complexes. Assume • (1) each In is K-injective, m m (2) each map In+1 → In is a split surjection, m m (3) the limits I = lim In exist. Then the complex I• is K-injective. Proof. We urge the reader to skip the proof of this lemma. Let M • be an a b acyclic complex. Let us abbreviate Hn(a, b) = HomA(M ,In). With this nota- • • tion HomK(A)(M ,I ) is the cohomology of the complex Y Y Y Y lim Hn(m, m−2) → lim Hn(m, m−1) → lim Hn(m, m) → lim Hn(m, m+1) n n n n m m m m in the third spot from the left. We may exchange the order of Q and lim and each of the complexes Y Y Y Y Hn(m, m − 2) → Hn(m, m − 1) → Hn(m, m) → Hn(m, m + 1) m m m m is exact by assumption (1). By assumption (2) the maps in the systems Y Y Y ... → H3(m, m − 2) → H2(m, m − 2) → H1(m, m − 2) m m m DERIVED CATEGORIES 97

are surjective. Thus the lemma follows from Homology, Lemma 31.4. It appears that a combination of Lemmas 29.3, 31.4, and 31.8 produces “enough K- injectives” for any abelian category with enough injectives and countable products. Actually, this may not work! See Lemma 34.4 for an explanation. 08BJ Lemma 31.9. Let A and B be abelian categories. Let u : A → B and v : B → A be additive functors. Assume (1) u is right adjoint to v, and (2) v is exact. Then u transforms K-injective complexes into K-injective complexes. Proof. Let I• be a K-injective complex of A. Let M • be a acyclic complex of B. As v is exact we see that v(M •) is an acyclic complex. By adjointness we get • • • • 0 = HomK(A)(v(M ),I ) = HomK(B)(M , u(I )) hence the lemma follows. 32. Bounded cohomological dimension 07K5 There is another case where the unbounded derived functor exists. Namely, when the functor has bounded cohomological dimension. 07K6 Lemma 32.1. Let A be an abelian category. Let d : Ob(A) → {0, 1, 2,..., ∞} be a function. Assume that (1) every object of A is a subobject of an object A with d(A) = 0, (2) d(A ⊕ B) ≤ max{d(A), d(B)} for A, B ∈ A, and (3) if 0 → A → B → C → 0 is short exact, then d(C) ≤ max{d(A) − 1, d(B)}. Let K• be a complex such that n + d(Kn) tends to −∞ as n → −∞. Then there exists a quasi-isomorphism K• → L• with d(Ln) = 0 for all n ∈ Z.

• • n Proof. By Lemma 15.5 we can ﬁnd a quasi-isomorphism σ≥0K → M with M = 0 for n < 0 and d(M n) = 0 for n ≥ 0. Then K• is quasi-isomorphic to the complex ... → K−2 → K−1 → M 0 → M 1 → ... Hence we may assume that d(Kn) = 0 for n 0. Note that the condition n + d(Kn) → −∞ as n → −∞ is not violated by this replacement. We are going to improve K• by an (inﬁnite) sequence of elementary replacements. An elementary replacement is the following. Choose an index n such that d(Kn) > 0. Choose an injection Kn → M where d(M) = 0. Set M 0 = Coker(Kn → M ⊕ Kn+1). Consider the map of complexes

K• : Kn−1 / Kn / Kn+1 / Kn+2

(K0)• : Kn−1 / M / M 0 / Kn+2 It is clear that K• → (K0)• is a quasi-isomorphism. Moreover, it is clear that d((K0)n) = 0 and d((K0)n+1) ≤ max{d(Kn) − 1, d(M ⊕ Kn+1)} ≤ max{d(Kn) − 1, d(Kn+1)} and the other values are unchanged. DERIVED CATEGORIES 98

To finish the proof we carefuly choose the order in which to do the elementary • replacements so that for every integer m the complex σ≥mK is changed only a finite number of times. To do this set ξ(K•) = max{n + d(Kn) | d(Kn) > 0} and I = {n ∈ Z | ξ(K•) = n + d(Kn) and d(Kn) > 0} Our assumption that n + d(Kn) tends to −∞ as n → −∞ and the fact that d(Kn) = 0 for n >> 0 implies ξ(K•) < +∞ and that I is a finite set. It is clear that ξ((K0)•) ≤ ξ(K•) for an elementary transformation as above. An elementary transformation changes the complex in degrees ≤ ξ(K•) + 1. Hence if we can find finite sequence of elementary transformations which decrease ξ(K•), then we win. However, note that if we do an elementary transformation starting with the smallest element n ∈ I, then we either decrease the size of I, or we increase min I. Since • every element of I is ≤ ξ(K ) we see that we win after a finite number of steps. 07K7 Lemma 32.2. Let F : A → B be a left exact functor of abelian categories. Assume (1) every object of A is a subobject of an object which is right acyclic for F , (2) there exists an integer n ≥ 0 such that RnF = 0, Then (1) RF : D(A) → D(B) exists, (2) any complex consisting of right acyclic objects for F computes RF , (3) any complex is the source of a quasi-isomorphism into a complex consisting of right acyclic objects for F , (4) for E ∈ D(A) i i (a) H (RF (τ≤aE) → H (RF (E)) is an isomorphism for i ≤ a, i i (b) H (RF (E)) → H (RF (τ≥b−n+1E)) is an isomorphism for i ≥ b, (c) if Hi(E) = 0 for i 6∈ [a, b] for some −∞ ≤ a ≤ b ≤ ∞, then Hi(RF (E)) = 0 for i 6∈ [a, b + n − 1]. Proof. Note that the first assumption implies that RF : D+(A) → D+(B) exists, see Proposition 16.8. Let A be an object of A. Choose an injection A → A0 with A0 acyclic. Then we see that Rn+1F (A) = RnF (A0/A) = 0 by the long exact cohomology sequence. Hence we conclude that Rn+1F = 0. Continuing like this using induction we find that RmF = 0 for all m ≥ n. We are going to use Lemma 32.1 with the function d : Ob(A) → {0, 1, 2,...} given by d(A) = max{0} ∪ {i | RiF (A) 6= 0}. The first assumption of Lemma 32.1 is our assumption (1). The second assumption of Lemma 32.1 follows from the fact that RF (A ⊕ B) = RF (A) ⊕ RF (B). The third assumption of Lemma 32.1 follows from the long exact cohomology sequence. Hence for every complex K• there exists a quasi-isomorphism K• → L• into a complex of objects right acyclic for F . This proves statement (3). We claim that if L• → M • is a quasi-isomorphism of complexes of right acyclic objects for F , then F (L•) → F (M •) is a quasi-isomorphism. If we prove this claim then we get statements (1) and (2) of the lemma by Lemma 14.15. To prove the claim pick an integer i ∈ Z. Consider the distinguished triangle • • • σ≥i−n−1L → σ≥i−n−1M → Q , DERIVED CATEGORIES 99

i.e., let Q• be the cone of the ﬁrst map. Note that Q• is bounded below and that Hj(Q•) is zero except possibly for j = i−n−1 or j = i−n−2. We may apply RF to Q•. Using the second spectral sequence of Lemma 21.3 and the assumed vanishing of cohomology (2) we conclude that Hj(RF (Q•)) is zero except possibly for j ∈ • • {i−n−2, . . . , i−1}. Hence we see that RF (σ≥i−n−1L ) → RF (σ≥i−n−1M ) induces an isomorphism of cohomology objects in degrees ≥ i. By Proposition 16.8 we know • • • • that RF (σ≥i−n−1L ) = σ≥i−n−1F (L ) and RF (σ≥i−n−1M ) = σ≥i−n−1F (M ). We conclude that F (L•) → F (M •) is an isomorphism in degree i as desired. Part (4)(a) follows from Lemma 16.1. For part (4)(b) let E be represented by the complex L• of objects right acyclic for • • F . By part (2) RF (E) is represented by the complex F (L ) and RF (σ≥cL ) is • represented by σ≥cF (L ). Consider the distinguished triangle b−n • • • H (L )[n − b] → τ≥b−nL → τ≥b−n+1L i • of Remark 12.4. The vanishing established above gives that H (RF (τ≥b−nL )) i • agrees with H (RF (τ≥b−n+1L )) for i ≥ b. Consider the short exact sequence of complexes b−n−1 b−n • • 0 → Im(L → L )[n − b] → σ≥b−nL → τ≥b−nL → 0 Using the distinguished triangle associated to this (see Section 12) and the vanishing i • i • as before we conclude that H (RF (τ≥b−nL )) agrees with H (RF (σ≥b−nL )) for • • • i ≥ b. Since the map RF (σ≥b−nL ) → RF (L ) is represented by σ≥b−nF (L ) → F (L•) we conclude that this in turn agrees with Hi(RF (L•)) for i ≥ b as desired.

Proof of (4)(c). Under the assumption on E we have τ≤a−1E = 0 and we get the vanishing of Hi(RF (E)) for i ≤ a − 1 from part (4)(a). Similarly, we have i τ≥b+1E = 0 and hence we get the vanishing of H (RF (E)) for i ≥ b + n from part (4)(b). 07K8 Lemma 32.3. Let F : A → B be a right exact functor of abelian categories. If (1) every object of A is a quotient of an object which is left acyclic for F , (2) there exists an integer n ≥ 0 such that LnF = 0, Then (1) LF : D(A) → D(B) exists, (2) any complex consisting of left acyclic objects for F computes LF , (3) any complex is the target of a quasi-isomorphism from a complex consisting of left acyclic objects for F , (4) for E ∈ D(A) i i (a) H (LF (τ≤a+n−1E) → H (LF (E)) is an isomorphism for i ≤ a, i i (b) H (LF (E)) → H (LF (τ≥bE)) is an isomorphism for i ≥ b, (c) if Hi(E) = 0 for i 6∈ [a, b] for some −∞ ≤ a ≤ b ≤ ∞, then Hi(LF (E)) = 0 for i 6∈ [a − n + 1, b].

Proof. This is dual to Lemma 32.2.

33. Derived colimits 0A5K In a triangulated category there is a notion of derived colimit. DERIVED CATEGORIES 100

090ZDeﬁnition 33.1. Let D be a triangulated category. Let (Kn, fn) be a system of objects of D. We say an object K is a derived colimit, or a homotopy colimit of the L system (Kn) if the direct sum Kn exists and there is a distinguished triangle M M M Kn → Kn → K → Kn[1]

L L where the map Kn → Kn is given by 1 − fn in degree n. If this is the case, then we sometimes indicate this by the notation K = hocolimKn.

By TR3 a derived colimit, if it exists, is unique up to (non-unique) isomorphism. L Moreover, by TR1 a derived colimit of Kn exists as soon as Kn exists. The derived category D(Ab) of the category of abelian groups is an example of a triangulated category where all homotopy colimits exist.

The nonuniqueness makes it hard to pin down the derived colimit. In More on Algebra, Lemma 85.4 the reader ﬁnds an exact sequence

1 0 → R lim Hom(Kn,L[−1]) → Hom(hocolimKn,L) → lim Hom(Kn,L) → 0

describing the Homs out of a homotopy colimit in terms of the usual Homs.

0CRHRemark 33.2. Let D be a triangulated category. Let (Kn, fn) be a system of objects of D. We may think of a derived colimit as an object K of D endowed with morphisms in : Kn → K such that in+1 ◦ fn = in and such that there exists a L morphism c : K → Kn with the property that

M 1−fn M in c M Kn −−−→ Kn −→ K −→ Kn[1]

0 0 0 is a distinguished triangle. If (K , in, c ) is a second derived colimit, then there 0 0 0 exists an isomorphism ϕ : K → K such that ϕ ◦ in = in and c ◦ ϕ = c. The existence of ϕ is TR3 and the fact that ϕ is an isomorphism is Lemma 4.3.

0CRIRemark 33.3. Let D be a triangulated category. Let (an):(Kn, fn) → (Ln, gn) be a morphism of systems of objects of D. Let (K, in, c) be a derived colimit of the ﬁrst system and let (L, jn, d) be a derived colimit of the second system with notation as in Remark 33.2. Then there exists a morphism a : K → L such that a ◦ in = jn and d ◦ a = (an[1]) ◦ c. This follows from TR3 applied to the deﬁning distinguished triangles.

0CRJ Lemma 33.4. Let D be a triangulated category. Let (Kn, fn) be a system of objects L L of D. Let n1 < n2 < n3 < . . . be a sequence of integers. Assume Kn and Kni

exist. Then there exists an isomorphism hocolimKni → hocolimKn such that

Kni / hocolimKni

id Kni / hocolimKn

commutes for all i. DERIVED CATEGORIES 101

Proof. Let gi : Kni → Kni+1 be the composition fni+1−1 ◦ ... ◦ fni . We construct commutative diagrams L L L L i Kni / i Kni n Kn / n Kn 1−gi 1−fn b a and d c L 1−fn L L 1−gi L n Kn / n Kn i Kni / i Kni as follows. Let a = a| be the inclusion of K into the direct sum. In other i Kni ni words, a is the natural inclusion. Let b = b| be the map i Kni

1, fni , fni+1◦fni , ..., fni+1−2◦...◦fni Kni −−−−−−−−−−−−−−−−−−−−−−−−−→ Kni ⊕ Kni+1 ⊕ ... ⊕ Kni+1−1

If ni−1 < j ≤ ni, then we let cj = c|Kj be the map

fni−1◦...◦fj Kj −−−−−−−−→ Kni

We let dj = d|Kj be zero if j 6= ni for any i and we let dni be the natural inclusion

of Kni into the direct sum. In other words, d is the natural projection. By TR3 these diagrams deﬁne morphisms

ϕ : hocolimKni → hocolimKn and ψ : hocolimKn → hocolimKni Since c ◦ a and d ◦ b are the identity maps we see that ϕ ◦ ψ is an isomorphism by Lemma 4.3. The other way around we get the morphisms a ◦ c and b ◦ d. Consider L L the morphism h = (hj): Kn → Kn given by the rule: for ni−1 < j < ni we set 1, fj , fj+1◦fj , ..., fni−1◦...◦fj hj : Kj −−−−−−−−−−−−−−−−−−−−→ Kj ⊕ ... ⊕ Kni Then the reader verifies that (1 − f) ◦ h = id − a ◦ c and h ◦ (1 − f) = id − b ◦ d. This means that id − ψ ◦ ϕ has square zero by Lemma 4.5 (small argument omitted). In other words, ψ ◦ ϕ differs from the identity by a nilpotent endomorphism, hence is an isomorphism. Thus ϕ and ψ are isomorphisms as desired. 0A5L Lemma 33.5. Let A be an abelian category. If A has exact countable direct sums, • then D(A) has countable direct sums. In fact given a collection of complexes Ki L • indexed by a countable index set I the termwise direct sum Ki is the direct sum • of Ki in D(A). • • • Proof. Let L be a complex. Suppose given maps αi : Ki → L in D(A). This • • means there exist quasi-isomorphisms si : Mi → Ki of complexes and maps of • • −1 complexes fi : Mi → L such that αi = fisi . By assumption the map of complexes M • M • s : Mi −→ Ki L −1 is a quasi-isomorphism. Hence setting f = fi we see that α = fs is a map in • L • D(A) whose composition with the coprojection Ki → Ki is αi. We omit the verification that α is unique. 093W Lemma 33.6. Let A be an abelian category. Assume colimits over N exist and are exact. Then countable direct sums exists and are exact. Moreover, if (An, fn) is a system over N, then there is a short exact sequence M M 0 → An → An → colim An → 0 DERIVED CATEGORIES 102

where the ﬁrst map in degree n is given by 1 − fn. L Proof. The ﬁrst statement follows from An = colim(A1 ⊕ ... ⊕ An). For the second, note that for each n we have the short exact sequence

0 → A1 ⊕ ... ⊕ An−1 → A1 ⊕ ... ⊕ An → An → 0

where the ﬁrst map is given by the maps 1 − fi and the second map is the sum of the transition maps. Take the colimit to get the sequence of the lemma. • 0949 Lemma 33.7. Let A be an abelian category. Let Ln be a system of complexes of A. Assume colimits over N exist and are exact in A. Then the termwise colimit • • L = colim Ln is a homotopy colimit of the system in D(A). Proof. We have an exact sequence of complexes

M • M • • 0 → Ln → Ln → L → 0 by Lemma 33.6. The direct sums are direct sums in D(A) by Lemma 33.5. Thus the result follows from the deﬁnition of derived colimits in Deﬁnition 33.1 and the fact that a short exact sequence of complexes gives a distinguished triangle (Lemma 12.1). 0CRK Lemma 33.8. Let D be a triangulated category having countable direct sums. Let A be an abelian category with exact colimits over N. Let H : D → A be a homological functor commuting with countable direct sums. Then H(hocolimKn) = colim H(Kn) for any system of objects of D.

Proof. Write K = hocolimKn. Apply H to the deﬁning distinguished triangle to get M M M M H(Kn) → H(Kn) → H(K) → H(Kn[1]) → H(Kn[1])

where the ﬁrst map is given by 1−H(fn) and the last map is given by 1−H(fn[1]). Apply Lemma 33.6 to see that this proves the lemma. The following lemma tells us that taking maps out of a compact object (to be deﬁned later) commutes with derived colimits. 094A Lemma 33.9. Let D be a triangulated category with countable direct sums. Let K ∈ D be an object such that for every countable set of objects En ∈ D the canonical map M M HomD(K,En) −→ HomD(K, En)

is a bijection. Then, given any system Ln of D over N whose derived colimit L = hocolimLn exists we have that

colim HomD(K,Ln) −→ HomD(K,L) is a bijection. Proof. Consider the deﬁning distinguished triangle M M M Ln → Ln → L → Ln[1]

Apply the cohomological functor HomD(K, −) (see Lemma 4.2). By elementary considerations concerning colimits of abelian groups we get the result. DERIVED CATEGORIES 103

34. Derived limits 08TB In a triangulated category there is a notion of derived limit.

08TCDefinition 34.1. Let D be a triangulated category. Let (Kn, fn) be an inverse system of objects of D. We say an object K is a derived limit, or a homotopy limit Q of the system (Kn) if the product Kn exists and there is a distinguished triangle Y Y K → Kn → Kn → K[1] Q Q where the map Kn → Kn is given by (kn) 7→ (kn − fn+1(kn+1)). If this is the case, then we sometimes indicate this by the notation K = R lim Kn. By TR3 a derived limit, if it exists, is unique up to (non-unique) isomorphism. Q Moreover, by TR1 a derived limit R lim Kn exists as soon as Kn exists. The derived category D(Ab) of the category of abelian groups is an example of a triangulated category where all derived limits exist. The nonuniqueness makes it hard to pin down the derived limit. In More on Algebra, Lemma 85.3 the reader finds an exact sequence 1 0 → R lim Hom(L, Kn[−1]) → Hom(L, R lim Kn) → lim Hom(L, Kn) → 0 describing the Homs into a derived limit in terms of the usual Homs. 07KC Lemma 34.2. Let A be an abelian category with exact countable products. Then (1) D(A) has countable products, Q (2) countable products Ki in D(A) are obtained by taking termwise products of any complexes representing the Ki, and p Q Q p (3) H ( Ki) = H (Ki). • • Proof. Let Ki be a complex representing Ki in D(A). Let L be a complex. Sup- • • pose given maps αi : L → Ki in D(A). This means there exist quasi-isomorphisms • • • • si : Ki → Mi of complexes and maps of complexes fi : L → Mi such that −1 αi = si fi. By assumption the map of complexes Y • Y • s : Ki −→ Mi Q −1 is a quasi-isomorphism. Hence setting f = fi we see that α = s f is a map Q • • in D(A) whose composition with the projection Ki → Ki is αi. We omit the verification that α is unique. The duals of Lemmas 33.6, 33.7, and 33.9 should be stated here and proved. How- ever, we do not know any applications of these lemmas for now. 0BK7 Lemma 34.3. Let A be an abelian category with countable products and enough + injectives. Let (Kn) be an inverse system of D (A). Then R lim Kn exists. Q Proof. It suffices to show that Kn exists in D(A). For every n we can represent • Q Kn by a bounded below complex In of injectives (Lemma 18.3). Then Kn is Q • represented by In, see Lemma 31.5. 070M Lemma 34.4. Let A be an abelian category with countable products and enough • • injectives. Let K be a complex. Let In be the inverse system of bounded below • • complexes of injectives produced by Lemma 29.3. Then I = lim In exists, is K- injective, and the following are equivalent (1) the map K• → I• is a quasi-isomorphism, DERIVED CATEGORIES 104

• • (2) the canonical map K → R lim τ≥−nK is an isomorphism in D(A). • Proof. The statement of the lemma makes sense as R lim τ≥−nK exists by Lemma • 34.3. Each complex In is K-injective by Lemma 31.4. Choose direct sum decom- p p p p p • • positions In+1 = Cn+1 ⊕ In for all n ≥ 1. Set C1 = I1 . The complex I = lim In p Q p exists because we can take I = n≥1 Cn. Fix p ∈ Z. We claim there is a split short exact sequence p Y p Y p 0 → I → In → In → 0 p p of objects of A. Here the ﬁrst map is given by the projection maps I → In and the p p p p second map by (xn) 7→ (xn − fn+1(xn+1)) where fn : In → In−1 are the transition Q p Q p p maps. The splitting comes from the map In → Cn = I . We obtain a termwise split short exact sequence of complexes • Y • Y • 0 → I → In → In → 0 Hence a corresponding distinguished triangle in K(A) and D(A). By Lemma 31.5 the products are K-injective and represent the corresponding products in D(A). It • • • follows that I represents R lim In (Deﬁnition 34.1). Moreover, it follows that I is K-injective by Lemma 31.3. By the commutative diagram of Lemma 29.3 we obtain a corresponding commutative diagram • • K / R lim τ≥−nK

• • I / R lim In in D(A). Since the right vertical arrow is an isomorphism (as derived limits are • • deﬁned on the level of the derived category and since τ≥−nK → In is a quasi- isomorphism), the lemma follows. 090Y Lemma 34.5. Let A be an abelian category having enough injectives and exact countable products. Then for every complex there is a quasi-isomorphism to a K- injective complex.

Proof. By Lemma 34.4 it suﬃces to show that K → R lim τ≥−nK is an isomorphism for all K in D(A). Consider the deﬁning distinguished triangle Y Y R lim τ≥−nK → τ≥−nK → τ≥−nK → (R lim τ≥−nK)[1] By Lemma 34.2 we have p Y Y p H ( τ≥−nK) = H (K) p≥−n It follows in a straightforward manner from the long exact cohomology sequence of p p the displayed distinguished triangle that H (R lim τ≥−nK) = H (K). 35. Operations on full subcategories 0FX0 Let T be a triangulated category. We will identify full subcategories of T with subsets of Ob(T ). Given full subcategories A, B,... we let (1) A[a, b] for −∞ ≤ a ≤ b ≤ ∞ be the full subcategory of T consisting of all objects A[−i] with i ∈ [a, b] ∩ Z with A ∈ Ob(A) (note the minus sign!), (2) smd(A) be the full subcategory of T consisting of all objects which are isomorphic to direct summands of objects of A, DERIVED CATEGORIES 105

(3) add(A) be the full subcategory of T consisting of all objects which are isomorphic to ﬁnite direct sums of objects of A, (4) A ? B be the full subcategory of T consisting of all objects X of T which ﬁt into a distinguished triangle A → X → B with A ∈ Ob(A) and B ∈ Ob(B), (5) A?n = A ?...? A with n ≥ 1 factors (we will see ? is associative below), (6) smd(add(A)?n) = smd(add(A) ? . . . ? add(A)) with n ≥ 1 factors. If E is an object of T , then we think of E sometimes also as the full subcategory of T whose single object is E. Then we can consider things like add(E[−1, 2]) and so on and so forth. We warn the reader that this notation is not universally accepted. 0FX1 Lemma 35.1. Let T be a triangulated category. Given full subcategories A, B, C we have (A ? B) ? C = A ? (B ? C). Proof. If we have distinguished triangles A → X → B and X → Y → C then by Axiom TR4 we have distinguished triangles A → Y → Z and B → Z → C. 0FX2 Lemma 35.2. Let T be a triangulated category. Given full subcategories A, B we have smd(A) ? smd(B) ⊂ smd(A ? B) and smd(smd(A) ? smd(B)) = smd(A ? B).

Proof. Suppose we have a distinguished triangle A1 → X → B1 where A1 ⊕ A2 ∈ Ob(A) and B1 ⊕ B2 ∈ Ob(B). Then we obtain a distinguished triangle A1 ⊕ A2 → A2 ⊕ X ⊕ B2 → B1 ⊕ B2 which proves that X is in smd(A ? B). This proves the inclusion. The equality follows trivially from this. 0FX3 Lemma 35.3. Let T be a triangulated category. Given full subcategories A, B the full subcategories add(A) ? add(B) and smd(add(A)) are closed under direct sums. Proof. Namely, if A → X → B and A0 → X0 → B0 are distinguished triangles and A, A0 ∈ add(A) and B,B0 ∈ add(B) then A ⊕ A0 → X ⊕ X0 → B ⊕ B0 is a distinguished triangle with A ⊕ A0 ∈ add(A) and B ⊕ B0 ∈ add(B). The result for smd(add(A)) is trivial. 0FX4 Lemma 35.4. Let T be a triangulated category. Given a full subcategory A for n ≥ 1 the subcategory ?n Cn = smd(add(A) ) = smd(add(A) ? . . . ? add(A)) deﬁned above is a strictly full subcategory of T closed under direct sums and direct summands and Cn+m = smd(Cn ? Cm) for all n, m ≥ 1.

Proof. Immediate from Lemmas 35.1, 35.2, and 35.3. 0FX5Remark 35.5. Let F : T → T 0 be an exact functor of triangulated categories. Given a full subcategory A of T we denote F (A) the full subcategory of T 0 whose objects consists of all objects F (A) with A ∈ Ob(A). We have F (A[a, b]) = F (A)[a, b] F (smd(A)) ⊂ smd(F (A)), F (add(A)) ⊂ add(F (A)), F (A ? B) ⊂ F (A) ?F (B), F (A?n) ⊂ F (A)?n. We omit the trivial veriﬁcations. DERIVED CATEGORIES 106

0FX6Remark 35.6. Let T be a triangulated category. Given full subcategories A1 ⊂ A2 ⊂ A3 ⊂ ... and B of T we have [ [ Ai [a, b] = Ai[a, b]

[ [ smd Ai = smd(Ai), [ [ add Ai = add(Ai), [ [ Ai ? B = Ai ? B, [ [ B ? Ai = B ? Ai,

?n [ [ ?n Ai = Ai . We omit the trivial veriﬁcations. 0FX7 Lemma 35.7. Let A be an abelian category. Let D = D(A). Let E ⊂ Ob(A) be a subset which we view as a subset of Ob(D) also. Let K be an object of D. (1) Let b ≥ a and assume Hi(K) is zero for i 6∈ [a, b] and Hi(K) ∈ E if i ∈ [a, b]. Then K is in smd(add(E[a, b])?(b−a+1)). (2) Let b ≥ a and assume Hi(K) is zero for i 6∈ [a, b] and Hi(K) ∈ smd(add(E)) if i ∈ [a, b]. Then K is in smd(add(E[a, b])?(b−a+1)). (3) Let b ≥ a and assume K can be represented by a complex K• with Ki = 0 for i 6∈ [a, b] and Ki ∈ E for i ∈ [a, b]. Then K is in smd(add(E[a, b])?(b−a+1)). (4) Let b ≥ a and assume K can be represented by a complex K• with Ki = 0 for i 6∈ [a, b] and Ki ∈ smd(add(E)) for i ∈ [a, b]. Then K is in smd(add(E[a, b])?(b−a+1)). Proof. We will use Lemma 35.4 without further mention. We will prove (2) which trivially implies (1). We use induction on b − a. If b − a = 0, then K is isomorphic to Hi(K)[−a] in D and the result is immediate. If b − a > 0, then we consider the distinguished triangle • • b τ≤b−1K → K → K [−b] and we conclude by induction on b − a. We omit the proof of (3) and (4). 0FX8 Lemma 35.8. Let T be a triangulated category. Let H : T → A be a homological functor to an abelian category A. Let a ≤ b and E ⊂ Ob(T ) be a subset such that Hi(E) = 0 for E ∈ E and i 6∈ [a, b]. Then for X ∈ smd(add(E[−m, m])?n) we have Hi(X) = 0 for i 6∈ [−m + na, m + nb].

Proof. Omitted. Pleasant exercise in the deﬁnitions.

36. Generators of triangulated categories 09SI In this section we briefly introduce a few of the different notions of a generator for a triangulated category. Our terminology is taken from [BV03] (except that we use “saturated” for what they call “épaisse”, see Definition 6.1, and our definition of add(A) is different). DERIVED CATEGORIES 107

Let D be a triangulated category. Let E be an object of D. Denote hEi1 the strictly full subcategory of D consisting of objects in D isomorphic to direct summands of ﬁnite direct sums M E[ni] i=1,...,r of shifts of E. It is clear that in the notation of Section 35 we have

hEi1 = smd(add(E[−∞, ∞]))

For n > 1 let hEin denote the full subcategory of D consisting of objects of D isomorphic to direct summands of objects X which ﬁt into a distinguished triangle A → X → B → A[1]

where A is an object of hEi1 and B an object of hEin−1. In the notation of Section 35 we have hEin = smd(hEi1 ? hEin−1)

Each of the categories hEin is a strictly full additive (by Lemma 35.3) subcategory of D preserved under shifts and under taking summands. But, hEin is not necessarily closed under “taking cones” or “extensions”, hence not necessarily a triangulated subcategory. This will be true for the subcategory [ hEi = hEin n as will be shown in the lemmas below. 0FX9 Lemma 36.1. Let T be a triangulated category. Let E be an object of T . For n ≥ 1 we have ?n [ ?n hEin = smd(hEi1 ?...? hEi1) = smd(hEi1 ) = smd(add(E[−m, m]) ) m≥1 0 For n, n ≥ 1 we have hEin+n0 = smd(hEin ? hEin0 ). Proof. The left equality in the displayed formula follows from Lemmas 35.1 and 35.2 and induction. The middle equality is a matter of notation. Since hEi1 = S smd(add(E[−∞, ∞])]) and since E[−∞, ∞] = m≥1 E[−m, m] we see from Re- mark 35.6 and Lemma 35.2 that we get the equality on the right. Then the ﬁnal statement follows from the remark and the corresponding statement of Lemma 35.4. 0ATG Lemma 36.2. Let D be a triangulated category. Let E be an object of D. The subcategory [ [ ?n hEi = hEin = smd(add(E[−m, m]) ) n n,m≥1 is a strictly full, saturated, triangulated subcategory of D and it is the smallest such subcategory of D containing the object E. Proof. The equality on the right follows from Lemma 36.1. It is clear that hEi = S hEin contains E, is preserved under shifts, direct sums, direct summands. If A ∈ hEia and B ∈ hEib and if A → X → B → A[1] is a distinguished triangle, S then X ∈ hEia+b by Lemma 36.1. Hence hEin is also preserved under extensions and it follows that it is a triangulated subcategory. Finally, let D0 ⊂ D be a strictly full, saturated, triangulated subcategory of D containing E. Then D0[−∞, ∞] ⊂ D0, add(D) ⊂ D0, smd(D0) ⊂ D0, and D0 ? D0 ⊂ D0. In other words, all the operations we used to construct hEi out of E preserve 0 0 D . Hence hEi ⊂ D and this ﬁnishes the proof. DERIVED CATEGORIES 108

09SJDeﬁnition 36.3. Let D be a triangulated category. Let E be an object of D. (1) We say E is a classical generator of D if the smallest strictly full, saturated, triangulated subcategory of D containing E is equal to D, in other words, if hEi = D. (2) We say E is a strong generator of D if hEin = D for some n ≥ 1. (3) We say E is a weak generator or a generator of D if for any nonzero object K of D there exists an integer n and a nonzero map E → K[n]. This deﬁnition can be generalized to the case of a family of objects. 09SK Lemma 36.4. Let D be a triangulated category. Let E,K be objects of D. The following are equivalent (1) Hom(E,K[i]) = 0 for all i ∈ Z, (2) Hom(E0,K) = 0 for all E0 ∈ hEi. Proof. The implication (2) ⇒ (1) is immediate. Conversely, assume (1). Then Hom(X,K) = 0 for all X in hEi1. Arguing by induction on n and using Lemma 4.2 we see that Hom(X,K) = 0 for all X in hEin. 09SL Lemma 36.5. Let D be a triangulated category. Let E be an object of D. If E is a classical generator of D, then E is a generator. Proof. Assume E is a classical generator. Let K be an object of D such that Hom(E,K[i]) = 0 for all i ∈ Z. By Lemma 36.4 Hom(E0,K) = 0 for all E0 in hEi. However, since D = hEi we conclude that idK = 0, i.e., K = 0. 0FXA Lemma 36.6. Let D be a triangulated category which has a strong generator. Let E be an object of D. If E is a classical generator of D, then E is a strong generator.

0 0 Proof. Let E be an object of D such that D = hE in. Since D = hEi we see that 0 0 E ∈ hEim for some m ≥ 1 by Lemma 36.2. Then hE i1 ⊂ hEim hence 0 0 0 D = hE in = smd(hE i1 ?...? hE i1) ⊂ smd(hEim ?...? hEim) = hEinm as desired. Here we used Lemma 36.1. 0ATHRemark 36.7. Let D be a triangulated category. Let E be an object of D. Let T be a property of objects of D. Suppose that L (1) if Ki ∈ D(A), i = 1, . . . , r with T (Ki) for i = 1, . . . , r, then T ( Ki), (2) if K → L → M → K[1] is a distinguished triangle and T holds for two, then T holds for the third object, (3) if T (K ⊕ L) then T (K) and T (L), and (4) T (E[n]) holds for all n. Then T holds for all objects of hEi.

37. Compact objects 09SM Here is the deﬁnition. 07LSDeﬁnition 37.1. Let D be an additive category with arbitrary direct sums. A compact object of D is an object K such that the map M M HomD(K,Ei) −→ HomD(K, Ei) i∈I i∈I

is bijective for any set I and objects Ei ∈ Ob(D) parametrized by i ∈ I. DERIVED CATEGORIES 109

This notion turns out to be very useful in algebraic geometry. It is an intrinsic condition on objects that forces the objects to be, well, compact.

09QH Lemma 37.2. Let D be a (pre-)triangulated category with direct sums. Then the compact objects of D form the objects of a Karoubian, saturated, strictly full, (pre-)triangulated subcategory Dc of D. Proof. Let (X,Y,Z,f,g,h) be a distinguished triangle of D with X and Y compact. Then it follows from Lemma 4.2 and the ﬁve lemma (Homology, Lemma 5.20) that Z is a compact object too. It is clear that if X ⊕ Y is compact, then X, Y are compact objects too. Hence Dc is a saturated triangulated subcategory. Since D is Karoubian by Lemma 4.14 we conclude that the same is true for Dc.

09SN Lemma 37.3. Let D be a triangulated category with direct sums. Let Ei, i ∈ I be L a family of compact objects of D such that Ei generates D. Then every object X of D can be written as

X = hocolimXn

where X1 is a direct sum of shifts of the Ei and each transition morphism ﬁts into a distinguished triangle Yn → Xn → Xn+1 → Yn[1] where Yn is a direct sum of shifts of the Ei. L Proof. Set X1 = (i,m,ϕ) Ei[m] where the direct sum is over all triples (i, m, ϕ) such that i ∈ I, m ∈ Z and ϕ : Ei[m] → X. Then X1 comes equipped with L a canonical morphism X1 → X. Given Xn → X we set Yn = (i,m,ϕ) Ei[m] where the direct sum is over all triples (i, m, ϕ) such that i ∈ I, m ∈ Z, and ϕ : Ei[m] → Xn is a morphism such that Ei[m] → Xn → X is zero. Choose a distinguished triangle Yn → Xn → Xn+1 → Yn[1] and let Xn+1 → X be any morphism such that Xn → Xn+1 → X is the given one; such a morphism exists by our choice of Yn. We obtain a morphism hocolimXn → X by the construction of our maps Xn → X. Choose a distinguished triangle

C → hocolimXn → X → C[1]

Let Ei[m] → C be a morphism. Since Ei is compact, the composition Ei[m] → C → hocolimXn factors through Xn for some n, say by Ei[m] → Xn. Then the construction of Yn shows that the composition Ei[m] → Xn → Xn+1 is zero. In other words, the composition Ei[m] → C → hocolimXn is zero. This means that our morphism Ei[m] → C comes from a morphism Ei[m] → X[−1]. The construction of X1 then shows that such morphism lifts to hocolimXn and we conclude that L our morphism Ei[m] → C is zero. The assumption that Ei generates D implies that C is zero and the proof is done.

09SP Lemma 37.4. With assumptions and notation as in Lemma 37.3. If C is a compact object and C → Xn is a morphism, then there is a factorization C → E →

Xn where E is an object of hEi1 ⊕ ... ⊕ Eit i for some i1, . . . , it ∈ I. Proof. We prove this by induction on n. The base case n = 1 is clear. If n > 1 consider the composition C → Xn → Yn−1[1]. This can be factored through some 0 0 0 E [1] → Yn−1[1] where E is a ﬁnite direct sum of shifts of the Ei. Let I ⊂ I be DERIVED CATEGORIES 110

the ﬁnite set of indices that occur in this direct sum. Thus we obtain E0 / C0 / C / E0[1]

 Yn−1 / Xn−1 / Xn / Yn−1[1]

0 00 00 By induction the morphism C → Xn−1 factors through E → Xn−1 with E an L 00 object of h i∈I00 Eii for some finite subset I ⊂ I. Choose a distinguished triangle E0 → E00 → E → E0[1] L then E is an object of h i∈I0∪I00 Eii. By construction and the axioms of a triangulated category we can choose morphisms C → E and a morphism E → Xn fitting into morphisms of triangles (E0,C0,C) → (E0,E00,E) and (E0,E00,E) → (Yn−1,Xn−1,Xn). The composition C → E → Xn may not equal the given morphism C → Xn, but the compositions into Yn−1 are equal. Let C → Xn−1 be a morphism that lifts the difference. By induction assumption we can factor this 000 00 L through a morphism E → Xn−1 with E an object of h i∈I000 Eii for some finite 0 000 subset I ⊂ I. Thus we see that we get a solution on considering E ⊕ E → Xn 000 L because E ⊕ E is an object of h i∈I0∪I00∪I000 Eii. 09SQDefinition 37.5. Let D be a triangulated category with arbitrary direct sums. We say D is compactly generated if there exists a set Ei, i ∈ I of compact objects L such that Ei generates D. The following proposition clarifies the relationship between classical generators and weak generators. 09SR Proposition 37.6. Let D be a triangulated category with direct sums. Let E be a compact object of D. The following are equivalent

(1) E is a classical generator for Dc and D is compactly generated, and (2) E is a generator for D.

Proof. If E is a classical generator for Dc, then Dc = hEi. It follows formally from the assumption that D is compactly generated and Lemma 36.4 that E is a generator for D. The converse is more interesting. Assume that E is a generator for D. Let X be a compact object of D. Apply Lemma 37.3 with I = {1} and E1 = E to write

X = hocolimXn

as in the lemma. Since X is compact we ﬁnd that X → hocolimXn factors through Xn for some n (Lemma 33.9). Thus X is a direct summand of Xn. By Lemma 37.4 we see that X is an object of hEi and the lemma is proven.

38. Brown representability 0A8E A reference for the material in this section is [Nee96]. 0A8F Lemma 38.1. Let D be a triangulated category with direct sums which is com- [Nee96, Theorem pactly generated. Let H : D → Ab be a contravariant cohomological functor which 3.1]. transforms direct sums into products. Then H is representable. DERIVED CATEGORIES 111

L Proof. Let Ei, i ∈ I be a set of compact objects such that i∈I Ei generates D. We may and do assume that the set of objects {Ei} is preserved under shifts. Consider pairs (i, a) where i ∈ I and a ∈ H(Ei) and set M X1 = Ei (i,a) Q Since H(X1) = (i,a) H(Ei) we see that (a)(i,a) deﬁnes an element a1 ∈ H(X1). Set H1 = HomD(−,X1). By Yoneda’s lemma (Categories, Lemma 3.5) the element a1 deﬁnes a natural transformation H1 → H.

We are going to inductively construct Xn and transformations an : Hn → H where Hn = HomD(−,Xn). Namely, we apply the procedure above to the functor Ker(Hn → H) to get an object M Kn+1 = Ei (i,k), k∈Ker(Hn(Ei)→H(Ei)) and a transformation HomD(−,Kn+1) → Ker(Hn → H). By Yoneda’s lemma the composition HomD(−,Kn+1) → Hn gives a morphism Kn+1 → Xn. We choose a distinguished triangle

Kn+1 → Xn → Xn+1 → Kn+1[1] in D. The element an ∈ H(Xn) maps to zero in H(Kn+1) by construction. Since H is cohomological we can lift it to an element an+1 ∈ H(Xn+1).

We claim that X = hocolimXn represents H. Applying H to the deﬁning distinguished triangle M M M Xn → Xn → X → Xn[1] we obtain an exact sequence Y Y H(Xn) ← H(Xn) ← H(X) Q Thus there exists an element a ∈ H(X) mapping to (an) in H(Xn). Hence a natural transformation HomD(−,X) → H such that

HomD(−,X1) → HomD(−,X2) → HomD(−,X3) → ... → HomD(−,X) → H commutes. For each i the map HomD(Ei,X) → H(Ei) is surjective, by construction of X1. On the other hand, by construction of Xn → Xn+1 the kernel of HomD(Ei,Xn) → H(Ei) is killed by the map HomD(Ei,Xn) → HomD(Ei,Xn+1). Since HomD(Ei,X) = colim HomD(Ei,Xn) by Lemma 33.9 we see that HomD(Ei,X) → H(Ei) is injective. To ﬁnish the proof, consider the subcategory 0 D = {Y ∈ Ob(D) | HomD(Y [n],X) → H(Y [n]) is an isomorphism for all n}

As HomD(−,X) → H is a transformation between cohomological functors, the subcategory D0 is a strictly full, saturated, triangulated subcategory of D (details omitted; see proof of Lemma 6.3). Moreover, as both H and HomD(−,X) transform direct sums into products, we see that direct sums of objects of D0 are in D0. Thus 0 0 derived colimits of objects of D are in D . Since {Ei} is preserved under shifts, we 0 0 see that Ei is an object of D for all i. It follows from Lemma 37.3 that D = D and the proof is complete. DERIVED CATEGORIES 112

0A8G Proposition 38.2. Let D be a triangulated category with direct sums which is [Nee96, Theorem compactly generated. Let F : D → D0 be an exact functor of triangulated categories 4.1]. which transforms direct sums into direct sums. Then F has an exact right adjoint. Proof. For an object Y of D0 consider the contravariant functor

D → Ab,W 7→ HomD0 (F (W ),Y ) This is a cohomological functor as F is exact and transforms direct sums into products as F transforms direct sums into direct sums. Thus by Lemma 38.1 we find an object X of D such that HomD(W, X) = HomD0 (F (W ),Y ). The existence of the adjoint follows from Categories, Lemma 24.2. Exactness follows from Lemma 7.1. 39. Admissible subcategories 0CQP A reference for this section is [BK89, Section 1]. 0FXBDefinition 39.1. Let D be an additive category. Let A ⊂ D be a full subcategory. The right orthogonal A⊥ of A is the full subcategory consisting of the objects X of D such that Hom(A, X) = 0 for all A ∈ Ob(A). The left orthogonal ⊥A of A is the full subcategory consisting of the objects X of D such that Hom(X,A) = 0 for all A ∈ Ob(A). 0CQQ Lemma 39.2. Let D be a triangulated category. Let A ⊂ D be a full subcategory invariant under all shifts. Consider a distinguished triangle X → Y → Z → X[1] of D. The following are equivalent (1) Z is in A⊥, and (2) Hom(A, X) = Hom(A, Y ) for all A ∈ Ob(A). Proof. By Lemma 4.2 the functor Hom(A, −) is homological and hence we get a long exact sequence as in (3.5.1). Assume (1) and let A ∈ Ob(A). Then we consider the exact sequence Hom(A[1],Z) → Hom(A, X) → Hom(A, Y ) → Hom(A, Z) Since A[1] ∈ Ob(A) we see that the first and last groups are zero. Thus we get (2). Assume (2) and let A ∈ Ob(A). Then we consider the exact sequence Hom(A, X) → Hom(A, Y ) → Hom(A, Z) → Hom(A[−1],X) → Hom(A[−1],Y ) and we conclude that Hom(A, Z) = 0 as desired. 0FXC Lemma 39.3. Let D be a triangulated category. Let A ⊂ D be a full subcategory invariant under all shifts. Then both the right orthogonal A⊥ and the left orthogonal ⊥A of A are strictly full, saturated7, triangulated subcagories of D. Proof. It is immediate from the definitions that the orthogonals are preserved under taking shifts, direct sums, and direct summands. Consider a distinguished triangle X → Y → Z → X[1] of D. By Lemma 4.16 it suffices to show that if X and Y are in A⊥, then Z is in ⊥ A . This is immediate from Lemma 39.2. 7Definition 6.1. DERIVED CATEGORIES 113

0CQR Lemma 39.4. Let D be a triangulated category. Let A be a full triangulated subcategory of D. For an object X of D consider the property P (X): there exists a distinguished triangle A → X → B → A[1] in D with A in A and B in A⊥.

(1) If X1 → X2 → X3 → X1[1] is a distinguished triangle and P holds for two out of three, then it holds for the third. (2) If P holds for X1 and X2, then it holds for X1 ⊕ X2.

Proof. Let X1 → X2 → X3 → X1[1] be a distinguished triangle and assume P holds for X1 and X2. Choose distinguished triangles

A1 → X1 → B1 → A1[1] and A2 → X2 → B2 → A2[1]

as in condition P . Since Hom(A1,A2) = Hom(A1,X2) by Lemma 39.2 there is a unique morphism A1 → A2 such that the diagram

A1 / X1

 A2 / X2 commutes. Choose an extension of this to a diagram

A1 / X1 / Q1 / A1[1]

 A2 / X2 / Q2 / A2[1]

 A3 / X3 / Q3 / A3[1]

A1[1] / X1[1] / Q1[1] / A1[2] ∼ ∼ as in Proposition 4.23. By TR3 we see that Q1 = B1 and Q2 = B2 and hence ⊥ Q1,Q2 ∈ Ob(A ). As Q1 → Q2 → Q3 → Q1[1] is a distinguished triangle we see ⊥ that Q3 ∈ Ob(A ) by Lemma 39.3. Since A is a full triangulated subcategory, we see that A3 is isomorphic to an object of A. Thus X3 satisﬁes P . The other cases of (1) follow from this case by translation. Part (2) is a special case of (1) via Lemma 4.11. 0CQS Lemma 39.5. Let D be a triangulated category. Let A ⊂ D be a full triangulated subcategory. The following are equivalent (1) the inclusion functor A → D has a right adjoint, and (2) for every X in D there exists a distinguished triangle A → X → B → A[1] in D with A ∈ Ob(A) and B ∈ Ob(A⊥). If this holds, then A is saturated (Deﬁnition 6.1) and if A is strictly full in D, then A = ⊥(A⊥). DERIVED CATEGORIES 114

Proof. Assume (1) and denote v : D → A the right adjoint. Let X ∈ Ob(D). Set A = v(X). We may extend the adjunction mapping A → X to a distinguished triangle A → X → B → A[1]. Since 0 0 0 HomA(A ,A) = HomA(A , v(X)) = HomD(A ,X) for A0 ∈ Ob(A), we conclude that B ∈ Ob(A⊥) by Lemma 39.2. Assume (2). We will contruct the adjoint v explictly. Let X ∈ Ob(D). Choose A → X → B → A[1] as in (2). Set v(X) = A. Let f : X → Y be a morphism in D. Choose A0 → Y → B0 → A0[1] as in (2). Since Hom(A, A0) = Hom(A, Y ) by Lemma 39.2 there is a unique morphism f 0 : A → A0 such that the diagram A / X

f 0 f A0 / Y commutes. Hence we can set v(f) = f 0 to get a functor. To see that v is adjoint to the inclusion morphism use Lemma 39.2 again. Proof of the final statement. In other to prove that A is saturated we may replace A by the strictly full subcategory having the same isomorphism classes as A; details omitted. Assume A is strictly full. If we show that A = ⊥(A⊥), then A will be saturated by Lemma 39.3. Since the incusion A ⊂ ⊥(A⊥) is clear it suffices to prove the other inclusion. Let X be an object of ⊥(A⊥). Choose a distinguished triangle A → X → B → A[1] as in (2). As Hom(X,B) = 0 by assumption we see that A =∼ X ⊕ B[−1] by Lemma 4.11. Since Hom(A, B[−1]) = 0 as B ∈ A⊥ this ∼ implies B[−1] = 0 and A = X as desired. 0CQT Lemma 39.6. Let D be a triangulated category. Let A ⊂ D be a full triangulated subcategory. The following are equivalent (1) the inclusion functor A → D has a left adjoint, and (2) for every X in D there exists a distinguished triangle B → X → A → B[1] in D with A ∈ Ob(A) and B ∈ Ob(⊥A). If this holds, then A is saturated (Definition 6.1) and if A is strictly full in D, then A = (⊥A)⊥.

Proof. Omitted. Dual to Lemma 39.5. 0FXDDeﬁnition 39.7. Let D be a triangulated category. A right admissible subcategory of D is a strictly full triangulated subcategory satisfying the equivalent conditions of Lemma 39.5. A left admissible subcategory of D is a strictly full triangulated subcategory satisfying the equivalent conditions of Lemma 39.6. A two-sided admissible subcategory is one which is both right and left admissible.

40. Postnikov systems 0D7Y A reference for this section is [Orl97]. Let D be a triangulated category. Let

Xn → Xn−1 → ... → X0 be a complex in D. In this section we consider the problem of constructing a “totalization” of this complex. DERIVED CATEGORIES 115

0D7ZDeﬁnition 40.1. Let D be a triangulated category. Let

Xn → Xn−1 → ... → X0 be a complex in D.A Postnikov system is deﬁned inductively as follows.

(1) If n = 0, then it is an isomorphism Y0 → X0. (2) If n = 1, then it is a choice of a distinguished triangle

Y1 → X1 → Y0 → Y1[1]

where X1 → Y0 composed with Y0 → X0 is the given morphism X1 → X0. (3) If n > 1, then it is a choice of a Postnikov system for Xn−1 → ... → X0 and a choice of a distinguished triangle

Yn → Xn → Yn−1 → Yn[1]

where the morphism Xn → Yn−1 composed with Yn−1 → Xn−1 is the given morphism Xn → Xn−1. Given a morphism

Xn / Xn−1 / ... / X0 (40.1.1)0D80

 0 0 0 Xn / Xn−1 / ... / X0 between complexes of the same length in D there is an obvious notion of a morphism of Postnikov systems. Here is a key example.

0D8ZExample 40.2. Let A be an abelian category. Let ... → A2 → A1 → A0 be a chain complex in A. Then we can consider the objects

Xn = An and Yn = (An → An−1 → ... → A0)[−n]

of D(A). With the evident canonical maps Yn → Xn and Y0 → Y1[1] → Y2[2] → ... the distinguished triangles Yn → Xn → Yn−1 → Yn[1] define a Postnikov system as in Definition 40.1 for ... → X2 → X1 → X0. Here we are using the obvious extension of Postnikov systems for an infinite complex of D(A). Finally, if colimits over N exist and are exact in A then

hocolimYn[n] = (... → A2 → A1 → A0 → 0 → ...) in D(A). This follows immediately from Lemma 33.7.

Given a complex Xn → Xn−1 → ... → X0 and a Postnikov system as in Deﬁnition 40.1 we can consider the maps

Y0 → Y1[1] → ... → Yn[n] These maps ﬁt together in certain distinguished triangles and ﬁt with the given maps between the Xi. Here is a picture for n = 3:

Y0 / Y1[1] / Y2[2] / Y3[3] a c c +1 +1 +1

{ +1 { +1 { X1[1] o X2[2] o X3[3] DERIVED CATEGORIES 116

We encourage the reader to think of Yn[n] as obtained from X0,X1[1],...,Xn[n]; for L example if the maps Xi → Xi−1 are zero, then we can take Yn[n] = i=0,...,n Xi[i]. Postnikov systems do not always exist. Here is a simple lemma for low n. 0D81 Lemma 40.3. Let D be a triangulated category. Consider Postnikov systems for complexes of length n. (1) For n = 0 Postnikov systems always exist and any morphism (40.1.1) of complexes extends to a unique morphism of Postnikov systems. (2) For n = 1 Postnikov systems always exist and any morphism (40.1.1) of complexes extends to a (nonunique) morphism of Postnikov systems. (3) For n = 2 Postnikov systems always exist but morphisms (40.1.1) of complexes in general do not extend to morphisms of Postnikov systems. (4) For n > 2 Postnikov systems do not always exist. Proof. The case n = 0 is immediate as isomorphisms are invertible. The case n = 1 follows immediately from TR1 (existence of triangles) and TR3 (extending morphisms to triangles). For the case n = 2 we argue as follows. Set Y0 = X0. By the case n = 1 we can choose a Postnikov system

Y1 → X1 → Y0 → Y1[1]

Since the composition X2 → X1 → X0 is zero, we can factor X2 → X1 (nonuniquely) as X2 → Y1 → X1 by Lemma 4.2. Then we simply ﬁt the morphism X2 → Y1 into a distinguished triangle

Y2 → X2 → Y1 → Y2[1] to get the Postnikov system for n = 2. For n > 2 we cannot argue similarly, as we do not know whether the composition Xn → Xn−1 → Yn−1 is zero in D. 0D82 Lemma 40.4. Let D be a triangulated category. Given a map (40.1.1) consider the condition 0 (40.4.1)0DW1 Hom(Xi[i − j − 1],Xj) = 0 for i > j + 1 Then 0 0 0 (1) If we have a Postnikov system for Xn → Xn−1 → ... → X0 then property (40.4.1) implies that 0 Hom(Xi[i − j − 1],Yj ) = 0 for i > j + 1 (2) If we are given Postnikov systems for both complexes and we have (40.4.1), then the map extends to a (nonunique) map of Postnikov systems. Proof. We ﬁrst prove (1) by induction on j. For the base case j = 0 there is 0 0 nothing to prove as Y0 → X0 is an isomorphism. Say the result holds for j − 1. We consider the distinguished triangle 0 0 0 0 Yj → Xj → Yj−1 → Yj [1] The long exact sequence of Lemma 4.2 gives an exact sequence 0 0 0 Hom(Xi[i − j − 1],Yj−1[−1]) → Hom(Xi[i − j − 1],Yj ) → Hom(Xi[i − j − 1],Xj) From the induction hypothesis and (40.4.1) we conclude the outer groups are zero and we win. DERIVED CATEGORIES 117

Proof of (2). For n = 1 the existence of morphisms has been established in Lemma 40.3. For n > 1 by induction, we may assume given the map of Postnikov systems of length n − 1. The problem is that we do not know whether the diagram

Xn / Yn−1

0 0 Xn / Yn−1 0 is commutative. Denote α : Xn → Yn−1 the difference. Then we do know that 0 0 the composition of α with Yn−1 → Xn−1 is zero (because of what it means to be a 0 map of Postnikov systems of length n − 1). By the distinguished triangle Yn−1 → 0 0 0 0 0 Xn−1 → Yn−2 → Yn−1[1], this means that α is the composition of Yn−2[−1] → Yn−1 0 0 0 with a map α : Xn → Yn−2[−1]. Then (40.4.1) guarantees α is zero by part (1) of the lemma. Thus α is zero. To finish the proof of existence, the commutativity guarantees we can choose the dotted arrow fitting into the diagram

Yn−1[−1] / Yn / Xn / Yn−1

0 0 0 0 Yn−1[−1] / Yn / Xn / Yn−1 by TR3. 0FXE Lemma 40.5. Let D be a triangulated category. Given a map (40.1.1) assume we are given Postnikov systems for both complexes. If 0 (1) Hom(Xi[i],Yn[n]) = 0 for i = 1, . . . , n, or 0 (2) Hom(Yn[n],Xn−i[n − i]) = 0 for i = 1, . . . , n, or 0 0 (3) Hom(Xj−i[−i + 1],Xj) = 0 and Hom(Xj,Xj−i[−i]) = 0 for j ≥ i > 0, then there exists at most one morphism between these Postnikov systems. Proof. Proof of (1). Look at the following diagram

Y0 / Y1[1] / Y2[2] / ... / Yn[n]

{ 0 ru Yn[n] 0 The arrows are the composition of the morphism Yn[n] → Yn[n] and the morphism 0 Yi[i] → Yn[n]. The arrow Y0 → Yn[n] is determined as it is the composition 0 0 0 Y0 = X0 → X0 = Y0 → Yn[n]. Since we have the distinguished triangle Y0 → 0 Y1[1] → X1[1] we see that Hom(X1[1],Yn[n]) = 0 guarantees that the second vertical arrow is unique. Since we have the distinguished triangle Y1[1] → Y2[2] → X2[2] we 0 see that Hom(X2[2],Yn[n]) = 0 guarantees that the third vertical arrow is unique. And so on. 0 Proof of (2). The composition Yn[n] → Yn[n] → Xn[n] is the same as the com- 0 position Yn[n] → Xn[n] → Xn[n] and hence is unique. Then using the distin- 0 0 0 guished triangle Yn−1[n − 1] → Yn[n] → Xn[n] we see that it suﬃces to show 0 Hom(Yn[n],Yn−1[n − 1]) = 0. Using the distinguished triangles 0 0 0 Yn−i−1[n − i − 1] → Yn−i[n − i] → Xn−i[n − i] DERIVED CATEGORIES 118

we get this vanishing from our assumption. Small details omitted.

Proof of (3). Looking at the proof of Lemma 40.4 and arguing by induction on n it suﬃces to show that the dotted arrow in the morphism of triangles

Yn−1[−1] / Yn / Xn / Yn−1

0 0 0 0 Yn−1[−1] / Yn / Xn / Yn−1

0 is unique. By Lemma 4.8 part (5) it suﬃces to show that Hom(Yn−1,Xn) = 0 0 and Hom(Xn,Yn−1[−1]) = 0. To prove the ﬁrst vanishing we use the distinguished triangles Yn−i−1[−i] → Yn−i[−(i − 1)] → Xn−i[−(i − 1)] for i > 0 and induction on 0 i to see that the assumed vanishing of Hom(Xn−i[−i + 1],Xn) is enough. For the 0 0 second we similarly use the distinguished triangles Yn−i−1[−i − 1] → Yn−i[−i] → 0 0 Xn−i[−i] to see that the assumed vanishing of Hom(Xn,Xn−i[−i]) is enough as well.

0D83 Lemma 40.6. Let D be a triangulated category. Let Xn → Xn−1 → ... → X0 be a complex in D. If

Hom(Xi[i − j − 2],Xj) = 0 for i > j + 2 then there exists a Postnikov system. If we have

Hom(Xi[i − j − 1],Xj) = 0 for i > j + 1 then any two Postnikov systems are isomorphic.

Proof. We argue by induction on n. The cases n = 0, 1, 2 follow from Lemma 40.3. Assume n > 2. Suppose given a Postnikov system for the complex Xn−1 → Xn−2 → ... → X0. The only obstruction to extending this to a Postnikov system of length n is that we have to ﬁnd a morphism Xn → Yn−1 such that the composition Xn → Yn−1 → Xn−1 is equal to the given map Xn → Xn−1. Considering the distinguished triangle

Yn−1 → Xn−1 → Yn−2 → Yn−1[1]

and the associated long exact sequence coming from this and the functor Hom(Xn, −) (see Lemma 4.2) we ﬁnd that it suﬃces to show that the composition Xn → Xn−1 → Yn−2 is zero. Since we know that Xn → Xn−1 → Xn−2 is zero we can apply the distinguished triangle

Yn−2 → Xn−2 → Yn−3 → Yn−2[1]

to see that it suﬃces if Hom(Xn,Yn−3[−1]) = 0. Arguing exactly as in the proof of Lemma 40.4 part (1) the reader easily sees this follows from the condition stated in the lemma.

The statement on isomorphisms follows from the existence of a map between the Postnikov systems extending the identity on the complex proven in Lemma 40.4 part (2) and Lemma 4.3 to show all the maps are isomorphisms. DERIVED CATEGORIES 119

41. Essentially constant systems 0G38 Some preliminary lemmas on essentially constant systems in triangulated categories.

0G39 Lemma 41.1. Let D be a triangulated category. Let (Ai) be an inverse system in D. Then (Ai) is essentially constant (see Categories, Deﬁnition 22.1) if and only if there exists an i and for all j ≥ i a direct sum decomposition Aj = A ⊕ Zj such that (a) the maps Aj0 → Aj are compatible with the direct sum decompositions and 0 identity on A, (b) for all j ≥ i there exists some j ≥ j such that Zj0 → Zj is zero.

Proof. Assume (Ai) is essentially constant with value A. Then A = lim Ai and there exists an i and a morphism Ai → A such that (1) the composition A → Ai → 0 A is the identity on A and (2) for all j ≥ i there exists a j ≥ j such that Aj0 → Aj factors as Aj0 → Ai → A → Aj. From (1) we conclude that for j ≥ i the maps A → Aj and Aj → Ai → A compose to the identity on A. It follows that Aj → A has a kernel Zj and that the map A ⊕ Zj → Aj is an isomorphism, see Lemmas 4.12 and 4.11. These direct sum decompositions clearly satisfy (a). From (2) we 0 conclude that for all j there is a j ≥ j such that Zj0 → Zj is zero, so (b) holds. Proof of the converse is omitted. 0G3A Lemma 41.2. Let D be a triangulated category. Let

An → Bn → Cn → An[1]

be an inverse system of distinguished triangles in D. If (An) and (Cn) are essentially constant, then (Bn) is essentially constant and their values ﬁt into a distinguished triangle A → B → C → A[1] such that for some n ≥ 1 there is a map

An / Bn / Cn / An[1]

 A / B / C / A[1]

of distinguished triangles which induces an isomorphism limn0≥n An0 → A and similarly for B and C. 0 0 Proof. After renumbering we may assume that An = A ⊕ An and Cn = C ⊕ Cn 0 0 for inverse systems (An) and (Cn) which are essentially zero, see Lemma 41.1. In particular, the morphism 0 0 C ⊕ Cn → (A ⊕ An)[1] maps the summand C into the summand A[1] for all n by a map δ : C → A[1] which is independent of n. Choose a distinguished triangle A → B → C −→δ A[1] Next, choose a morphism of distingished triangles

(A1 → B1 → C1 → A1[1]) → (A → B → C → A[1]) which is possible by TR3. For any object D of D this induces a commutative diagram

... / HomD(C,D) / HomD(B,D) / HomD(A, D) / ...

 ... / colim HomD(Cn,D) / colim HomD(Bn,D) / colim HomD(An,D) / ... DERIVED CATEGORIES 120

The left and right vertical arrows are isomorphisms and so are the ones to the left and right of those. Thus by the 5-lemma we conclude that the middle arrow is an isomorphism. It follows that (Bn) is isomorphic to the constant inverse system with value B by the discussion in Categories, Remark 22.7. Since this is equivalent to (Bn) being essentially constant with value B by Categories, Remark 22.5 the proof is complete.

0G3B Lemma 41.3. Let A be an abelian category. Let An be an inverse system of objects of D(A). Assume i (1) there exist integers a ≤ b such that H (An) = 0 for i 6∈ [a, b], and i (2) the inverse systems H (An) of A are essentially constant for all i ∈ Z.

Then An is an essentially constant system of D(A) whose value A satisﬁes that i i H (A) is the value of the constant system H (An) for each i ∈ Z. Proof. By Remark 12.4 we obtain an inverse system of distinguished triangles

τ≤aAn → An → τ≥a+1An → (τ≤aAn)[1]

a Of course we have τ≤aAn = H (An)[−a] in D(A). Thus by assumption these form an essentially constant system. By induction on b − a we find that the inverse 0 system τ≥a+1An is essentially constant, say with value A . By Lemma 41.2 we find that An is an essentially constant system. We omit the proof of the statement on cohomologies (hint: use the final part of Lemma 41.2). 0G3C Lemma 41.4. Let D be a triangulated category. Let

An → Bn → Cn → An[1]

be an inverse system of distinguished triangles. If the system Cn is pro-zero (essentially constant with value 0), then the maps An → Bn determine a pro-isomorphism between the pro-object (An) and the pro-object (Bn). Proof. For any object X of D consider the exact sequence

colim Hom(Cn,X) → colim Hom(Bn,X) → colim Hom(An,X) → colim Hom(Cn[−1],X) → Exactness follows from Lemma 4.2 combined with Algebra, Lemma 8.8. By assumption the ﬁrst and last term are zero. Hence the map colim Hom(Bn,X) → colim Hom(An,X) is an isomorphism for all X. The lemma follows from this and Categories, Remark 22.7. 0G3D Lemma 41.5. Let A be an abelian category.

An → Bn be an inverse system of maps of D(A). Assume i i (1) there exist integers a ≤ b such that H (An) = 0 and H (Bn) = 0 for i 6∈ [a, b], and i i (2) the inverse system of maps H (An) → H (Bn) of A deﬁne an isomorphism of pro-objects of A for all i ∈ Z.

Then the maps An → Bn determine a pro-isomorphism between the pro-object (An) and the pro-object (Bn). DERIVED CATEGORIES 121

Proof. We can inductively extend the maps An → Bn to an inverse system of distinguished triangles An → Bn → Cn → An[1] by axiom TR3. By Lemma 41.4 it suﬃces to prove that Cn is pro-zero. By Lemma 41.3 it suﬃces to show that p H (Cn) is pro-zero for each p. This follows from assumption (2) and the long exact sequences

p αn p βn p δn p+1 n p+1 H (An) −−→ H (Bn) −−→ H (Cn) −→ H (An) −→ H (Bn)

Namely, for every n we can ﬁnd an m > n such that Im(βm) maps to zero in p p p H (Cn) because we may choose m such that H (Bm) → H (Bn) factors through p p αn : H (An) → H (Bn). For a similar reason we may then choose k > m such p+1 p p that Im(δk) maps to zero in H (Am). Then H (Ck) → H (Cn) is zero be- p p p p cause H (Ck) → H (Cm) maps into Ker(δm) and H (Cm) → H (Cn) annihilates Ker(δm) = Im(βm). 42. Other chapters

Preliminaries (31) Divisors (1) Introduction (32) Limits of Schemes (2) Conventions (33) Varieties (3) Set Theory (34) Topologies on Schemes (4) Categories (35) Descent (5) Topology (36) Derived Categories of Schemes (6) Sheaves on Spaces (37) More on Morphisms (7) Sites and Sheaves (38) More on Flatness (8) Stacks (39) Groupoid Schemes (9) Fields (40) More on Groupoid Schemes (10) Commutative Algebra (41) Étale Morphisms of Schemes (11) Brauer Groups Topics in Scheme Theory (12) Homological Algebra (13) Derived Categories (42) Chow Homology (14) Simplicial Methods (43) Intersection Theory (15) More on Algebra (44) Picard Schemes of Curves (16) Smoothing Ring Maps (45) Weil Cohomology Theories (17) Sheaves of Modules (46) Adequate Modules (18) Modules on Sites (47) Dualizing Complexes (19) Injectives (48) Duality for Schemes (20) Cohomology of Sheaves (49) Discriminants and Differents (21) Cohomology on Sites (50) de Rham Cohomology (22) Differential Graded Algebra (51) Local Cohomology (23) Divided Power Algebra (52) Algebraic and Formal Geometry (24) Differential Graded Sheaves (53) Algebraic Curves (25) Hypercoverings (54) Resolution of Surfaces (55) Semistable Reduction Schemes (56) Derived Categories of Varieties (26) Schemes (57) Fundamental Groups of Schemes (27) Constructions of Schemes (58) Étale Cohomology (28) Properties of Schemes (59) Crystalline Cohomology (29) Morphisms of Schemes (60) Pro-étale Cohomology (30) Cohomology of Schemes (61) More Étale Cohomology DERIVED CATEGORIES 122

(62) The Trace Formula (90) The Cotangent Complex Algebraic Spaces (91) Deformation Problems (63) Algebraic Spaces Algebraic Stacks (64) Properties of Algebraic Spaces (92) Algebraic Stacks (65) Morphisms of Algebraic Spaces (93) Examples of Stacks (66) Decent Algebraic Spaces (94) Sheaves on Algebraic Stacks (67) Cohomology of Algebraic Spaces (95) Criteria for Representability (68) Limits of Algebraic Spaces (96) Artin’s Axioms (69) Divisors on Algebraic Spaces (97) Quot and Hilbert Spaces (70) Algebraic Spaces over Fields (98) Properties of Algebraic Stacks (71) Topologies on Algebraic Spaces (99) Morphisms of Algebraic Stacks (72) Descent and Algebraic Spaces (100) Limits of Algebraic Stacks (73) Derived Categories of Spaces (101) Cohomology of Algebraic Stacks (74) More on Morphisms of Spaces (102) Derived Categories of Stacks (75) Flatness on Algebraic Spaces (103) Introducing Algebraic Stacks (76) Groupoids in Algebraic Spaces (104) More on Morphisms of Stacks (77) More on Groupoids in Spaces (105) The Geometry of Stacks (78) Bootstrap (79) Pushouts of Algebraic Spaces Topics in Moduli Theory Topics in Geometry (106) Moduli Stacks (80) Chow Groups of Spaces (107) Moduli of Curves (81) Quotients of Groupoids Miscellany (82) More on Cohomology of Spaces (108) Examples (83) Simplicial Spaces (109) Exercises (84) Duality for Spaces (110) Guide to Literature (85) Formal Algebraic Spaces (111) Desirables (86) Algebraization of Formal Spaces (112) Coding Style (87) Resolution of Surfaces Revisited (113) Obsolete Deformation Theory (114) GNU Free Documentation Li- (88) Formal Deformation Theory cense (89) Deformation Theory (115) Auto Generated Index

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[Nee01] , Triangulated categories, Annals of Mathematics Studies, vol. 148, Princeton University Press, Princeton, NJ, 2001. [Orl97] D. O. Orlov, Equivalences of derived categories and K3 surfaces, J. Math. Sci. (New York) 84 (1997), no. 5, 1361–1381, Algebraic geometry, 7. [Ric89] Jeremy Rickard, Derived categories and stable equivalence, J. Pure Appl. Algebra 61 (1989), no. 3, 303–317. [Spa88] Nicolas Spaltenstein, Resolutions of unbounded complexes, Compositio Math. 65 (1988), no. 2, 121–154. [Ver96] Jean-Louis Verdier, Des catégories dérivées des catégories abéliennes, Astérisque (1996), no. 239, xii+253, With a preface by Luc Illusie, Edited and with a note by Georges Maltsiniotis. [Yon60] Nobuo Yoneda, On Ext and exact sequences, J. Fac. Sci. Univ. Tokyo Sect. I 8 (1960), 507–576.