DERIVED CATEGORIES

05QI

Contents

1. Introduction 22. Triangulated categories 23. The definition of a triangulated category 24. Elementary results on triangulated categories 55. Localization of triangulated categories 136. Quotients of triangulated categories 197. Adjoints for exact functors 248. The homotopy category 269. Cones and termwise split sequences 2610. Distinguished triangles in the homotopy category 3311. Derived categories 3712. The canonical delta-functor 3913. Filtered derived categories 4214. Derived functors in general 4515. Derived functors on derived categories 5216. Higher derived functors 5517. Triangulated subcategories of the derived category 5918. Injective resolutions 6219. Projective resolutions 6720. Right derived functors and injective resolutions 6921. Cartan-Eilenberg resolutions 7022. Composition of right derived functors 7223. Resolution functors 7324. Functorial injective embeddings and resolution functors 7525. Right derived functors via resolution functors 7726. Filtered derived category and injective resolutions 7727. Ext groups 8528. K-groups 8929. Unbounded complexes 9030. Deriving adjoints 9331. K-injective complexes 9432. Bounded cohomological dimension 9733. Derived colimits 9934. Derived limits 10235. Operations on full subcategories 10436. Generators of triangulated categories 10637. Compact objects 10838. Brown representability 110

This is a chapter of the Stacks Project, version 41964eac, compiled on Dec 16, 2020.1

DERIVED CATEGORIES 2

39. Admissible subcategories 11240. Postnikov systems 11441. Essentially constant systems 11842. Other chapters 121References 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 categoryis a triangulated category. Once this is done we define the derived category ofan abelian category as the localization of the homotopy category with respect toquasi-isomorphisms. A good reference is Verdier’s thesis [Ver96].

2. Triangulated categories

0143 Triangulated categories are a convenient tool to describe the type of structureinherent 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.

Definition 3.1.0144 Let D be an additive category. Let [n] : D → D, E 7→ E[n] bea 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 asextuple (X,Y, Z, f, g, h) where X,Y, Z ∈ Ob(D) and f : X → Y , g : Y → Z andh : Z → X[1] are morphisms of D. A morphism of triangles (X,Y, Z, f, g, h) →(X ′, Y ′, Z ′, f ′, g′, h′) is given by morphisms a : X → X ′, b : Y → Y ′ and c : Z → Z ′

of D such that b f = f ′ a, c g = g′ b and a[1] h = h′ c.

A morphism of triangles is visualized by the following commutative diagram

X //

a

Y //

b

Z //

c

X[1]

a[1]

X ′ // Y ′ // Z ′ // X ′[1]

Here is the definition of a triangulated category as given in Verdier’s thesis.

Definition 3.2.0145 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

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TR1 Any triangle isomorphic to a distinguished triangle is a distinguished tri-angle. Any triangle of the form (X,X, 0, id, 0, 0) is distinguished. For anymorphism 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

Xf //

a

Yg //

b

Zh //

X[1]

a[1]

X ′f ′ // Y ′

g′ // Z ′h′ // X ′[1]

whose rows are distinguished triangles and which satisfies b f = f ′ a,there exists a morphism c : Z → Z ′ such that (a, b, c) is a morphism oftriangles.

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 → Q3such 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, gf, 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 asZ/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] etcwith morphisms of triangles. For a more precise reformulation of this idea see theproof of Lemma 10.2.The sign in TR2 means that if (X,Y, Z, f, g, h) is a distinguished triangle then inthe long sequence

(3.2.1)05QL . . .→ Z[−1] −h[−1]−−−−→ Xf−→ Y

g−→ Zh−→ X[1] −f [1]−−−→ Y [1] −g[1]−−−→ Z[1]→ . . .

each four term sequence gives a distinguished triangle.As usual we abuse notation and we simply speak of a (pre-)triangulated categoryD without explicitly introducing notation for the additional data. The notion of apre-triangulated category is useful in finding statements equivalent to TR4.We have the following definition of a triangulated functor.

Definition 3.3.014V Let D, D′ be pre-triangulated categories. An exact functor, or atriangulated functor from D to D′ is a functor F : D → D′ together with given func-torial isomorphisms ξX : F (X[1])→ F (X)[1] such that for every distinguished tri-angle (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 D′.

1We use [ ] as an abbreviation for the family [n]n∈Z.

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An exact functor is additive, see Lemma 4.17. When we say two triangulatedcategories are equivalent we mean that they are equivalent in the 2-category oftriangulated categories. A 2-morphism a : (F, ξ) → (F ′, ξ′) in this 2-category issimply a transformation of functors a : F → F ′ which is compatible with ξ and ξ′,i.e.,

F [1]ξ//

a?1

[1] F

1?a

F ′ [1] ξ′ // [1] F ′

commutes.

Definition 3.4.05QM Let (D, [ ], T ) be a pre-triangulated category. A pre-triangulatedsubcategory2 is a pair (D′, T ′) such that

(1) D′ is an additive subcategory of D which is preserved under [1] and [−1],(2) T ′ ⊂ T is a subset such that for every (X,Y, Z, f, g, h) ∈ T ′ we have

X,Y, Z ∈ Ob(D′) and f, g, h ∈ Arrows(D′), and(3) (D′, [ ], T ′) is a pre-triangulated category.

If D is a triangulated category, then we say (D′, T ′) is a triangulated subcategory ifit is a pre-triangulated subcategory and (D′, [ ], T ′) is a triangulated category.

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 along exact sequence.

Definition 3.5.0147 Let D be a pre-triangulated category. Let A be an abelian cate-gory. An additive functor H : D → A is called homological if for every distinguishedtriangle (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 calledcohomological 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 thatH(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)

0148 H−1(Z)H(h[−1]) // H0(X)

H(f) // H0(Y )H(g) // H0(Z)

H(h) // H1(X)

This will be called the long exact sequence associated to the distinguished triangleand the homological functor. As indicated we will not use any signs for the mor-phisms in the long exact sequence. This has the side effect that maps in the longexact sequence associated to the rotation (TR2) of a distinguished triangle differfrom the maps in the sequence above by some signs.

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

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Definition 3.6.0150 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 assignsto every short exact sequence

0→ Aa−→ 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 tri-

angle of D for any short exact sequence as above, and(2) for every morphism (A → B → C) → (A′ → B′ → C ′) of short exact

sequences the diagram

G(C)

δA→B→C

// G(A)[1]

G(C ′)

δA′→B′→C′ // G(A′)[1]

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

Note how a δ-functor comes equipped with additional structure. Strictly speakingit does not make sense to say that a given functor A → D is a δ-functor, but wewill 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 afortiori hold in any triangulated category.

Lemma 4.1.0146 Let D be a pre-triangulated category. Let (X,Y, Z, f, g, h) be a dis-tinguished 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 TR3to

X //

1

X //

f

0 //

X[1]

1[1]

Xf // Y

g // Zh // X[1]

Of course the dotted arrow is the zero map. Hence the commutativity of thediagram implies that g f = 0. For the other cases rotate the triangle, i.e., applyTR2.

Lemma 4.2.0149 Let D be a pre-triangulated category. For any object W of D thefunctor HomD(W,−) is homological, and the functor HomD(−,W ) is cohomologi-cal.

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

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g a = 0. Then we get a commutative diagram

W1//

b

W //

a

0 //

0

W [1]

b[1]

X // Y // Z // X[1]

Both rows are distinguished triangles (use TR1 for the top row). Hence we can fillthe dotted arrow b (first rotate using TR2, then apply TR3, and then rotate back).This proves the lemma.

Lemma 4.3.014A Let D be a pre-triangulated category. Let

(a, b, c) : (X,Y, Z, f, g, h)→ (X ′, Y ′, Z ′, f ′, g′, h′)

be a morphism of distinguished triangles. If two among a, b, c are isomorphisms sois the third.

Proof. Assume that a and c are isomorphisms. For any object W of D writeHW (−) = HomD(W,−). Then we get a commutative diagram of abelian groups

HW (Z[−1]) //

HW (X) //

HW (Y ) //

HW (Z) //

HW (X[1])

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 ishomological by Lemma 4.2 and the five lemma (Homology, Lemma 5.20) it followsthat the middle vertical arrow is an isomorphism. Hence by Yoneda’s lemma, seeCategories, Lemma 3.5 we see that b is an isomorphism. This implies the othercases by rotating (using TR2).

Remark 4.4.09WA Let D be an additive category with translation functors [n] as inDefinition 3.1. Let us call a triangle (X,Y, Z, f, g, h) special3 if for every object Wof 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)→ (X ′, Y ′, Z ′, f ′, g′, h′)

is a morphism of special triangles and if two among a, b, c are isomorphisms so is thethird. There is a dual statement for co-special triangles, i.e., triangles which turninto long exact sequences on applying the functor HomD(−,W ). Thus distinguishedtriangles are special and co-special, but in general there are many more (co-)specialtriangles, than there are distinguished triangles.

Lemma 4.5.05QP Let D be a pre-triangulated category. Let

(0, b, 0), (0, b′, 0) : (X,Y, Z, f, g, h)→ (X,Y, Z, f, g, h)

be endomorphisms of a distinguished triangle. Then bb′ = 0.

3This is nonstandard notation.

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Proof. PictureX //

0

Y //

b,b′

α

Z //

0β

X[1]

0

X // Y // Z // X[1]Applying Lemma 4.2 we find dotted arrows α and β such that b′ = f α andb = β g. Then bb′ = β g f α = 0 as g f = 0 by Lemma 4.1.

Lemma 4.6.05QQ Let D be a pre-triangulated category. Let (X,Y, Z, f, g, h) be a dis-tinguished triangle. If

Zh//

c

X[1]

a[1]

Zh // X[1]

is commutative and a2 = a, c2 = c, then there exists a morphism b : Y → Y withb2 = 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 b′ such that (a, b′, c) is an endomorphismof (X,Y, Z, f, g, h). Then (0, (b′)2 − b′, 0) is also an endomorphism. By Lemma4.5 we see that (b′)2 − b′ has square zero. Set b = b′ − (2b′ − 1)((b′)2 − b′) =3(b′)2 − 2(b′)3. A computation shows that (a, b, c) is an endomorphism and thatb2 − b = (4(b′)2 − 4b′ − 3)((b′)2 − b′)2 = 0.

Lemma 4.7.014B Let D be a pre-triangulated category. Let f : X → Y be a mor-phism of D. There exists a distinguished triangle (X,Y, Z, f, g, h) which is uniqueup to (nonunique) isomorphism of triangles. More precisely, given a second suchdistinguished triangle (X,Y, Z ′, f, g′, h′) there exists an isomorphism

(1, 1, c) : (X,Y, Z, f, g, h) −→ (X,Y, Z ′, f, g′, h′)

Proof. Existence by TR1. Uniqueness up to isomorphism by TR3 and Lemma4.3.

Lemma 4.8.0FWZ Let D be a pre-triangulated category. Let(a, b, c) : (X,Y, Z, f, g, h)→ (X ′, Y ′, Z ′, f ′, g′, h′)

be a morphism of distinguished triangles. If one of the following conditions holds(1) Hom(Y,X ′) = 0,(2) Hom(Z, Y ′) = 0,(3) Hom(X,X ′) = Hom(Z,X ′) = 0,(4) Hom(Z,X ′) = Hom(Z,Z ′) = 0, or(5) Hom(X[1], Z ′) = Hom(Z,X ′) = 0

then b is the unique morphism from Y → Y ′ such that (a, b, c) is a morphism oftriangles.

Proof. If we have a second morphism of triangles (a, b′, c) then (0, b − b′, 0) is amorphism of triangles. Hence we have to show: the only morphism b : Y → Y ′ suchthat the X → Y → Y ′ and Y → Y ′ → Z ′ are zero is 0. We will use Lemma 4.2without further mention. In particular, condition (3) implies (1). Given condition(1) if the composition b : Y → Y ′ → Z ′ is zero, then b lifts to a morphism Y → X ′

which has to be zero. This proves (1).

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The proof of (2) and (4) are dual to this argument.

Assume (5). Consider the diagram

Xf//

0

Yg//

b

Zh//

0

ε

X[1]

0

X ′f ′ // Y ′

g′ // Z ′h′ // X ′[1]

We may choose ε such that b = g ε. Then g′ ε g = 0 which implies thatg′ ε = δ h for some δ ∈ Hom(X[1], Z ′). Since Hom(X[1], Z ′) = 0 we concludethat g′ ε = 0. Hence ε = f ′ γ for some γ ∈ Hom(Z,X ′). Since Hom(Z,X ′) = 0we conclude that ε = 0 and hence b = 0 as desired.

Lemma 4.9.05QR Let D be a pre-triangulated category. Let f : X → Y be a morphismof 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 isomor-phism of triangles by Lemma 4.3 and Z = 0. Conversely, if Z = 0, then (1, f, 0) isan isomorphism of triangles as well, hence f is an isomorphism.

Lemma 4.10.05QS Let D be a pre-triangulated category. Let (X,Y, Z, f, g, h) and(X ′, Y ′, Z ′, f ′, g′, h′) be triangles. The following are equivalent

(1) (X ⊕X ′, Y ⊕ Y ′, Z ⊕ Z ′, f ⊕ f ′, g ⊕ g′, h⊕ h′) is a distinguished triangle,(2) both (X,Y, Z, f, g, h) and (X ′, Y ′, Z ′, f ′, g′, h′) are distinguished triangles.

Proof. Assume (2). By TR1 we may choose a distinguished triangle (X ⊕X ′, Y ⊕Y ′, Q, f ⊕ f ′, g′′, h′′). By TR3 we can find morphisms of distinguished triangles(X,Y, Z, f, g, h) → (X ⊕X ′, Y ⊕ Y ′, Q, f ⊕ f ′, g′′, h′′) and (X ′, Y ′, Z ′, f ′, g′, h′) →(X ⊕X ′, Y ⊕ Y ′, Q, f ⊕ f ′, g′′, h′′). Taking the direct sum of these morphisms weobtain a morphism of triangles

(X ⊕X ′, Y ⊕ Y ′, Z ⊕ Z ′, f ⊕ f ′, g ⊕ g′, h⊕ h′)

(1,1,c)

(X ⊕X ′, Y ⊕ Y ′, Q, f ⊕ f ′, g′′, h′′).

In the terminology of Remark 4.4 this is a map of special triangles (because a directsum 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. Firstobserve that (X,Y, Z, f, g, h) is a special triangle (terminology from Remark 4.4) asa direct summand of the distinguished hence special triangle (X ⊕X ′, Y ⊕ Y ′, Z ⊕Z ′, f ⊕ f ′, g ⊕ g′, h ⊕ h′). Using TR1 let (X,Y,Q, f, g′′, h′′) be a distinguishedtriangle. By TR3 there exists a morphism of distinguished triangles (X ⊕X ′, Y ⊕

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Y ′, Z ⊕ Z ′, f ⊕ f ′, g ⊕ g′, h ⊕ h′) → (X,Y,Q, f, g′′, h′′). Composing this with theinclusion map we get a morphism of triangles

(1, 1, c) : (X,Y, Z, f, g, h) −→ (X,Y,Q, f, g′′, h′′)By Remark 4.4 we find that c is an isomorphism and we conclude that (2) holds.

Lemma 4.11.05QT Let D be a pre-triangulated category. Let (X,Y, Z, f, g, h) be adistinguished 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 X ′, Z ′ of D the triangle (X ′, X ′ ⊕ Z ′, Z ′, (1, 0), (0, 1), 0) is

distinguished.

Proof. To see (1) use that HomD(Z, Y )→ HomD(Z,Z)→ HomD(Z,X[1]) is exactby 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)→ 0is split exact (split by s : Z → Y ). Hence by Yoneda’s lemma we see thatX⊕Z → Yis an isomorphism. The last assertion follows from TR1 and Lemma 4.10.

Lemma 4.12.05QU Let D be a pre-triangulated category. Let f : X → Y be a morphismof 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 X ′⊕Z → Z⊕Y ′ has botha kernel and a cokernel. Hence (3) ⇒ (1), (2). Next we prove (1) ⇒ (3). Supposefirst that f : X → Y is a monomorphism, i.e., its kernel is zero. By TR1 thereexists a distinguished triangle (X,Y, Z, f, g, h). By Lemma 4.1 the compositionf 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⊕X ′and f |X′ : X ′ → Y is a monomorphism. Hence Y = X ′ ⊕ Y ′ and we win. Theimplication (2) ⇒ (3) is dual to this.

Lemma 4.13.0CRG Let D be a pre-triangulated category. Let I be a set.(1) Let Xi, i ∈ I be a family of objects of D.

(a) If∏Xi exists, then (

∏Xi)[1] =

∏Xi[1].

(b) If⊕Xi exists, then (

⊕Xi)[1] =

⊕Xi[1].

(2) Let Xi → Yi → Zi → Xi[1] be a family of distinguished triangles of D.(a) If

∏Xi,

∏Yi,∏Zi exist, then

∏Xi →

∏Yi →

∏Zi →

∏Xi[1] is a

distinguished triangle.(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 directsums and products are defined in terms of the category structure. Let us prove(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

distinguished triangles with the projection maps∏Xi → Xj and

∏Yi → Yj . Using

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.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 abeliangroups are exact. Hence Remark 4.4 shows that the morphism of triangles is anisomorphism and we conclude by TR1. The proof of (2)(b) is dual.

Lemma 4.14.05QW 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 sufficesto check that morphisms which have a right inverse have kernels. Any morphismwhich 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 cate-gory is triangulated.

Lemma 4.15.014C Let D be a pre-triangulated category. In order to prove TR4 itsuffices to show that given any pair of composable morphisms f : X → Y andg : Y → Z there exist

(1) isomorphisms i : X ′ → X, j : Y ′ → Y and k : Z ′ → Z, and then settingf ′ = j−1fi : X ′ → Y ′ and g′ = k−1gj : Y ′ → Z ′ there exist

(2) distinguished triangles (X ′, Y ′, Q1, f′, p1, d1), (X ′, Z ′, Q2, g

′ f ′, p2, d2) and(Y ′, Z ′, Q3, g

′, p3, d3), such that the assertion of TR4 holds.

Proof. The replacement of X,Y, Z by X ′, Y ′, Z ′ is harmless by our definition ofdistinguished triangles and their isomorphisms. The lemma follows from the factthat the distinguished triangles (X ′, Y ′, Q1, f

′, p1, d1), (X ′, Z ′, Q2, g′f ′, p2, d2) and

(Y ′, Z ′, Q3, g′, p3, d3) are unique up to isomorphism by Lemma 4.7.

Lemma 4.16.05QX Let D be a pre-triangulated category. Assume that D′ is an additivefull subcategory of D. The following are equivalent

(1) there exists a set of triangles T ′ such that (D′, T ′) is a pre-triangulatedsubcategory of D,

(2) D′ is preserved under [1], [−1] and given any morphism f : X → Y in D′there exists a distinguished triangle (X,Y, Z, f, g, h) in D such that Z isisomorphic to an object of D′.

In this case T ′ as in (1) is the set of distinguished triangles (X,Y, Z, f, g, h) of Dsuch that X,Y, Z ∈ Ob(D′). Finally, if D is a triangulated category, then (1) and(2) are also equivalent to

(3) D′ is a triangulated subcategory.

Proof. Omitted.

Lemma 4.17.05QY An exact functor of pre-triangulated categories is additive.

Proof. Let F : D → D′ 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))

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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 D′. This also implies that F applied to any zeromorphism is zero (since a morphism in an additive category is zero if and only ifit factors through the zero object). Next, using that (X,X ⊕Y, Y, (1, 0), (0, 1), 0) isa distinguished triangle, we see that (F (X), F (X ⊕ Y ), F (Y ), F (1, 0), F (0, 1), 0) isone 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.

Lemma 4.18.05SQ Let F : D → D′ be a fully faithful exact functor of pre-triangulatedcategories. 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 D′.

Proof. The “only if” part is clear. Assume (F (X), F (Y ), F (Z)) is distinguishedin D′. Pick a distinguished triangle (X,Y, Z ′, f, g′, h′) in D. By Lemma 4.7 thereexists an isomorphism of triangles

(1, 1, c′) : (F (X), F (Y ), F (Z)) −→ (F (X), F (Y ), F (Z ′)).

Since F is fully faithful, there exists a morphism c : Z → Z ′ such that F (c) = c′.Then (1, 1, c) is an isomorphism between (X,Y, Z) and (X,Y, Z ′). Hence (X,Y, Z)is distinguished by TR1.

Lemma 4.19.014Y Let D,D′,D′′ be pre-triangulated categories. Let F : D → D′ andF ′ : D′ → D′′ be exact functors. Then F ′ F is an exact functor.

Proof. Omitted.

Lemma 4.20.05QZ Let D be a pre-triangulated category. Let A be an abelian category.Let H : D → A be a homological functor.

(1) Let D′ be a pre-triangulated category. Let F : D′ → D be an exact functor.Then the composition H F is a homological functor as well.

(2) Let A′ be an abelian category. Let G : A → A′ be an exact functor. ThenG H is a homological functor as well.

Proof. Omitted.

Lemma 4.21.0151 Let D be a triangulated category. Let A be an abelian category.Let G : A → D be a δ-functor.

(1) Let D′ be a triangulated category. Let F : D → D′ be an exact functor.Then the composition F G is a δ-functor as well.

(2) Let A′ be an abelian category. Let H : A′ → A be an exact functor. ThenG H is a δ-functor as well.

Proof. Omitted.

Lemma 4.22.05SR 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. Assumethat H−1(G(A)) = 0 for all A in A. Then the collection

Hn G,Hn(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 → Aa−→ B

b−→ C → 0 is a shortexact 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)) isclearly functorial in the short exact sequence. Finally, the long exact cohomologysequence (3.5.1) combined with the vanishing of H−1(G(C)) gives a long exactsequence

0→ H0(G(A))→ H0(G(B))→ H0(G(C)) H0(δ)−−−−→ H1(G(A))→ . . .

in B as desired.

The proof of the following result uses TR4.

Proposition 4.23.05R0 Let D be a triangulated category. Any commutative diagram

X //

Y

X ′ // Y ′

can be extended to a diagram

X //

Y //

Z //

X[1]

X ′ //

Y ′ //

Z ′ //

X ′[1]

X ′′ //

Y ′′ //

Z ′′ //

X ′′[1]

X[1] // Y [1] // Z[1] // X[2]

where all the squares are commutative, except for the lower right square which isanticommutative. Moreover, each of the rows and columns are distinguished trian-gles. Finally, the morphisms on the bottom row (resp. right column) are obtainedfrom 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), (X ′, Y ′, Z ′), (X,X ′, X ′′), (Y, Y ′, Y ′′),and (X,Y ′, A). Note that the morphism X → Y ′ is both equal to the compositionX → Y → Y ′ and equal to the composition X → X ′ → Y ′. Hence, we can findmorphisms

(1) a : Z → A and b : A→ Y ′′, and(2) a′ : X ′′ → A and b′ : A→ Z ′

as in TR4. Denote c : Y ′′ → Z[1] the composition Y ′′ → Y [1] → Z[1] and denotec′ : Z ′ → X ′′[1] the composition Z ′ → X ′[1] → X ′′[1]. The conclusion of ourapplication TR4 are that

(1) (Z,A, Y ′′, a, b, c), (X ′′, A, Z ′, a′, b′, c′) are distinguished triangles,

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(2) (X,Y, Z)→ (X,Y ′, A), (X,Y ′, A)→ (Y, Y ′, Y ′′), (X,X ′, X ′′)→ (X,Y ′, A),(X,Y ′, A)→ (X ′, Y ′, Z ′) are morphisms of triangles.

First using that (X,X ′, X ′′) → (X,Y ′, A) and (X,Y ′, A) → (Y, Y ′, Y ′′). are mor-phisms of triangles we see the first of the diagrams

X ′ //

Y ′

X ′′

ba′ //

Y ′′

X[1] // Y [1]

and

Y //

Z

b′a

// X[1]

Y ′ // Z ′ // X ′[1]

is commutative. The second is commutative too using that (X,Y, Z)→ (X,Y ′, A)and (X,Y ′, A)→ (X ′, Y ′, Z ′) are morphisms of triangles. At this point we choosea distinguished triangle (X ′′, Y ′′, Z ′′) starting with the map b a′ : X ′′ → Y ′′.

Next we apply TR4 one more time to the morphisms X ′′ → A → Y ′′ and the tri-angles (X ′′, A, Z ′, a′, b′, c′), (X ′′, Y ′′, Z ′′), and (A, Y ′′, Z[1], b, c,−a[1]) to get mor-phisms a′′ : Z ′ → Z ′′ and b′′ : Z ′′ → Z[1]. Then (Z ′, Z ′′, Z[1], a′′, b′′,−b′[1] a[1])is a distinguished triangle, hence also (Z,Z ′, Z ′′,−b′ a, a′′,−b′′) and hence also(Z,Z ′, Z ′′, b′a, a′′, b′′). Moreover, (X ′′, A, Z ′)→ (X ′′, Y ′′, Z ′′) and (X ′′, Y ′′, Z ′′)→(A, Y ′′, Z[1], b, c,−a[1]) are morphisms of triangles. At this point we have definedall the distinguished triangles and all the morphisms, and all that’s left is to verifysome commutativity relations.

To see that the middle square in the diagram commutes, note that the arrowY ′ → Z ′ factors as Y ′ → A → Z ′ because (X,Y ′, A) → (X ′, Y ′, Z ′) is a mor-phism of triangles. Similarly, the morphism Y ′ → Y ′′ factors as Y ′ → A → Y ′′

because (X,Y ′, A) → (Y, Y ′, Y ′′) is a morphism of triangles. Hence the mid-dle square commutes because the square with sides (A,Z ′, Z ′′, Y ′′) commutes as(X ′′, A, Z ′)→ (X ′′, Y ′′, Z ′′) is a morphism of triangles (by TR4). The square withsides (Y ′′, Z ′′, Y [1], Z[1]) commutes because (X ′′, Y ′′, Z ′′)→ (A, Y ′′, Z[1], b, c,−a[1])is a morphism of triangles and c : Y ′′ → Z[1] is the composition Y ′′ → Y [1]→ Z[1].The square with sides (Z ′, X ′[1], X ′′[1], Z ′′) is commutative because (X ′′, A, Z ′)→(X ′′, Y ′′, Z ′′) is a morphism of triangles and c′ : Z ′ → X ′′[1] is the compo-sition Z ′ → X ′[1] → X ′′[1]. Finally, we have to show that the square withsides (Z ′′, X ′′[1], Z[1], X[2]) anticommutes. This holds because (X ′′, Y ′′, Z ′′) →(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 ofcomplexes, we will use a localization process.

Definition 5.1.05R2 Let D be a pre-triangulated category. We say a multiplicativesystem S is compatible with the triangulated structure if the following two conditionshold:

MS5 For s ∈ S we have s[n] ∈ S for all n ∈ Z.

DERIVED CATEGORIES 14

MS6 Given a solid commutative square

X //

s

Y //

s′

Z //

X[1]

s[1]

X ′ // Y ′ // Z ′ // X ′[1]

whose rows are distinguished triangles with s, s′ ∈ S there exists a mor-phism s′′ : Z → Z ′ in S such that (s, s′, s′′) is a morphism of triangles.

It turns out that these axioms are not independent of the axioms defining multi-plicative systems.

Lemma 5.2.05R3 Let D be a pre-triangulated category. Let S be a set of morphisms ofD and assume that axioms MS1, MS5, MS6 hold (see Categories, Definition 27.1and Definition 5.1). Then MS2 holds.

Proof. Suppose that f : X → Y is a morphism of D and t : X → X ′ an element ofS. Choose a distinguished triangle (X,Y, Z, f, g, h). Next, choose a distinguishedtriangle (X ′, Y ′, Z, f ′, g′, t[1] h) (here we use TR1 and TR2). By MS5, MS6 (andTR2 to rotate) we can find the dotted arrow in the commutative diagram

X //

t

Y //

s′

Z //

1

X[1]

t[1]

X ′ // Y ′ // Z // X ′[1]

with moreover s′ ∈ S. This proves LMS2. The proof of RMS2 is dual.

Lemma 5.3.05R4 Let F : D → D′ be an exact functor of pre-triangulated categories.Let

S = f ∈ Arrows(D) | F (f) is an isomorphismThen S is a saturated (see Categories, Definition 27.20) multiplicative system com-patible with the triangulated structure on D.

Proof. We have to prove axioms MS1 – MS6, see Categories, Definitions 27.1 and27.20 and Definition 5.1. MS1, MS4, and MS5 are direct from the definitions. MS6follows from TR3 and Lemma 4.3. By Lemma 5.2 we conclude that MS2 holds. Tofinish the proof we have to show that MS3 holds. To do this let f, g : X → Y bemorphisms of D, and let t : Z → X be an element of S such that f t = g t. As Dis additive this simply means that a t = 0 with a = f − g. Choose a distinguishedtriangle (Z,X,Q, t, d, h) using TR1. Since a t = 0 we see by Lemma 4.2 thereexists a morphism i : Q→ Y such that i d = a. Finally, using TR1 again we canchoose a triangle (Q,Y,W, i, j, k). Here is a picture

Zt// X

d//

1

Q //

i

Z[1]

Xa// Y

j

W

DERIVED CATEGORIES 15

OK, and now we apply the functor F to this diagram. Since t ∈ S we see thatF (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 thatLMS3 holds. The proof of RMS3 is dual.

Lemma 5.4.05R5 Let H : D → A be a homological functor between a pre-triangulatedcategory and an abelian category. Let

S = f ∈ Arrows(D) | Hi(f) is an isomorphism for all i ∈ ZThen S is a saturated (see Categories, Definition 27.20) multiplicative system com-patible with the triangulated structure on D.

Proof. We have to prove axioms MS1 – MS6, see Categories, Definitions 27.1 and27.20 and Definition 5.1. MS1, MS4, and MS5 are direct from the definitions. MS6follows from TR3 and the long exact cohomology sequence (3.5.1). By Lemma 5.2we 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 elementof S such that f t = g t. As D is additive this simply means that a t = 0 witha = 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 thati g = a. Finally, using TR1 again we can choose a triangle (Q,Y,W, i, j, k). Hereis a picture

Zt// X

g//

1

Q //

i

Z[1]

Xa// Y

j

W

OK, and now we apply the functors Hi to this diagram. Since t ∈ S we see thatHi(Q) = 0 by the long exact cohomology sequence (3.5.1). Hence Hi(j) is anisomorphism for all i by the same argument, i.e., j ∈ S. Finally, j a = j i g = 0as j i = 0. Thus j f = j g and we see that LMS3 holds. The proof of RMS3 isdual.

Proposition 5.5.05R6 Let D be a pre-triangulated category. Let S be a multiplica-tive system compatible with the triangulated structure. Then there exists a uniquestructure of a pre-triangulated category on S−1D such that the localization functorQ : 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 localizationfunctor Q is additive in Homology, Lemma 8.2. It is clear that we may defineQ(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 imageof 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 amorphism of S−1D, then a fits into a distinguished triangle. Write a = Q(s)−1 Q(f) for some s : Y → Y ′ in S and f : X → Y ′. Choose a distinguished triangle(X,Y ′, 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.

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Proof of TR2. This is immediate from the definitions.Proof of TR3. Note that the existence of the dotted arrow which is required to existmay be proven after replacing the two triangles by isomorphic triangles. Hence wemay assume given distinguished triangles (X,Y, Z, f, g, h) and (X ′, Y ′, Z ′, f ′, g′, h′)of D and a commutative diagram

Q(X)Q(f)

//

a

Q(Y )

b

Q(X ′)

Q(f ′) // Q(Y ′)

in S−1D. Now we apply Categories, Lemma 27.10 to find a morphism f ′′ : X ′′ →Y ′′ in D and a commutative diagram

X

f

k// X ′′

f ′′

X ′

f ′

soo

Yl // Y ′′ Y ′

too

in D with s, t ∈ S and a = s−1k, b = t−1l. At this point we can use TR3 for D andMS6 to find a commutative diagram

X //

k

Y //

l

Z //

m

X[1]

g[1]

X ′′ // Y ′′ // Z ′′ // X ′′[1]

X ′ //

s

OO

Y ′ //

t

OO

Z ′ //

r

OO

X ′[1]

s[1]

OO

with r ∈ S. It follows that setting c = Q(r)−1Q(m) we obtain the desired morphismof triangles

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

(a,b,c)

(Q(X ′), Q(Y ′), Q(Z ′), Q(f ′), Q(g′), Q(h′))

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 composablemorphisms a : Q(X) → Q(Y ) and b : Q(Y ) → Q(Z) we have to produce anoctahedron after possibly replacing Q(X), Q(Y ), Q(Z) by isomorphic objects. Todo this we may first replace Y by an object such that a = Q(f) for some morphismf : X → Y in D. (More precisely, write a = s−1f with s : Y → Y ′ in S andf : X → Y ′. Then replace Y by Y ′.) After this we similarly replace Z by an objectsuch that b = Q(g) for some morphism g : Y → Z. Now we can find distinguishedtriangles (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. Thenit is immediately verified that applying the functor Q to all these data gives acorresponding structure in S−1D

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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 tri-angulated categories).

Lemma 5.6.05R7 Let D be a pre-triangulated category. Let S be a multiplicative systemcompatible with the triangulated category. Let Q : D → S−1D be the localizationfunctor, see Proposition 5.5.

(1) If H : D → A is a homological functor into an abelian category A suchthat H(s) is an isomorphism for all s ∈ S, then the unique factorizationH ′ : S−1D → A such that H = H ′ Q (see Categories, Lemma 27.8) is ahomological functor too.

(2) If F : D → D′ is an exact functor into a pre-triangulated category D′ suchthat F (s) is an isomorphism for all s ∈ S, then the unique factorizationF ′ : S−1D → D′ such that F = F ′ Q (see Categories, Lemma 27.8) is anexact functor too.

Proof. This lemma proves itself. Details omitted.

The following lemma describes the kernel (see Definition 6.5) of the localizationfunctor.

Lemma 5.7.05R8 Let D be a pre-triangulated category. Let S be a multiplicative systemcompatible with the triangulated structure. Let Z be an object of D. The followingare equivalent

(1) Q(Z) = 0 in S−1D,(2) there exists Z ′ ∈ Ob(D) such that 0 : Z → Z ′ is an element of S,(3) there exists Z ′ ∈ Ob(D) such that 0 : Z ′ → Z is an element of S, and(4) there exists an object Z ′ and a distinguished triangle (X,Y, Z ⊕ Z ′, 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 (Z ′[−1], Z ′[−1] ⊕ Z,Z, (1, 0), (0, 1), 0) is a distinguished triangle (see Lemma4.11) with “0 ∈ S”. By rotating we conclude that (4) holds. If (X,Y, Z⊕Z ′, f, g, h) isa distinguished triangle with f ∈ S then Q(f) is an isomorphism hence Q(Z⊕Z ′) =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 ofthe equivalent conditions (1) – (4). Suppose that Q(Z) = 0. Then 0 → Z is amorphism 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).

Lemma 5.8.05R9 Let D be a triangulated category. Let S be a saturated multiplicativesystem 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, s′, s′′) : (X,Y, Z, f, g, h)→ (X ′, Y ′, Z ′, f ′, g′, h′) | s, s′, s′′ ∈ S

DERIVED CATEGORIES 18

Then I is a filtered category and the functors I → X/S, I → Y/S, and I → Z/Sare cofinal.

Proof. We strongly suggest the reader skip the proof of this lemma and insteadwork it out on a napkin.The first remark is that using rotation of distinguished triangles (TR2) gives anequivalence of categories between I and the corresponding category for the distin-guished triangle (Y,Z,X[1], g, h,−f [1]). Using this we see for example that if weprove the functor I → X/S is cofinal, then the same thing is true for the functorsI → Y/S and I → Z/S.Note that if s : X → X ′ is a morphism of S, then using MS2 we can find s′ : Y → Y ′

and f ′ : X ′ → Y ′ such that f ′ s = s′ f , whereupon we can use MS6 to completethis into an object of I. Hence the functor I → X/S is surjective on objects. Usingrotation as above this implies the same thing is true for the functors I → Y/S andI → 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, Lemma27.21. By the argument of the previous paragraph we can complete s1 : X → X1to an object (s1, s

′1, s′′1) : (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) witha, b, c ∈ S completing the given a : X1 → X2. But then (a, b, c) is a morphismin I. In this way we conclude that the functor I → X/S is also surjective onarrows. Using rotation as above, this implies the same thing is true for the functorsI → Y/S and I → Z/S.The category I is nonempty as the identity provides an object. This proves thecondition (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. Supposegiven objects (s1, s

′1, s′′1) : (X,Y, Z, f, g, h)→ (X1, Y1, Z1, f1, g1, h1) and (s2, s

′2, s′′2) :

(X,Y, Z, f, g, h)→ (X2, Y2, Z2, f2, g2, h2) in I. We want to find an object of I whichis 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 categoriesX/S, Y/S, Z/S are filtered. Thus we canfind X → X3 in X/S and morphisms s : X2 → X3 and a : X1 → X3. By the abovewe can find a morphism (s, s′, s′′) : (X2, Y2, Z2, f2, g2, h2) → (X3, Y3, Z3, f3, g3, h3)with s′, s′′ ∈ S. After replacing (X2, Y2, Z2) by (X3, Y3, Z3) we may assume thatthere exists a morphism a : X1 → X2 in X/S. Repeating the argument for Y andZ (by rotating as above) we may assume there is a morphism a : X1 → X2 inX/S, b : Y1 → Y2 in Y/S, and c : Z1 → Z2 in Z/S. However, these morphismsdo not necessarily give rise to a morphism of distinguished triangles. On the otherhand, the necessary diagrams do commute in S−1D. Hence we see (for example)that there exists a morphism s′2 : Y2 → Y3 in S such that s′2 f2 a = s′2 b f1.Another replacement of (X2, Y2, Z2) as above then gets us to the situation wheref2 a = b f1. Rotating and applying the same argument two more times we seethat we may assume (a, b, c) is a morphism of triangles. This proves condition (2).Next we check condition (3) of Categories, Definition 19.1. Suppose (s1, s

′1, s′′1) :

(X,Y, Z) → (X1, Y1, Z1) and (s2, s′2, s′′2) : (X,Y, Z) → (X2, Y2, Z2) are objects of

I, and suppose (a, b, c), (a′, b′, c′) are two morphisms between them. Since a s1 =a′ s1 there exists a morphism s3 : X2 → X3 such that s3 a = s3 a′. Using the

DERIVED CATEGORIES 19

surjectivity statement we can complete this to a morphism of triangles (s3, s′3, s′′3) :

(X2, Y2, Z2) → (X3, Y3, Z3) with s3, s′3, s′′3 ∈ S. Thus (s3 s2, s

′3 s′2, s′′3 s′′2) :

(X,Y, Z) → (X3, Y3, Z3) is also an object of I and after composing the maps(a, b, c), (a′, b′, c′) with (s3, s

′3, s′′3) we obtain a = a′. By rotating we may do the

same to get b = b′ and c = c′.

Finally, we check that I → X/S is cofinal, see Categories, Definition 17.1. The firstcondition is true as the functor is surjective. Suppose that we have an object s :X → X ′ inX/S and two objects (s1, s

′1, s′′1) : (X,Y, Z, f, g, h)→ (X1, Y1, Z1, f1, g1, h1)

and (s2, s′2, s′′2) : (X,Y, Z, f, g, h)→ (X2, Y2, Z2, f2, g2, h2) in I as well as morphisms

t1 : X ′ → X1 and t2 : X ′ → X2 in X/S. By property (2) of I proved above wecan find morphisms (s3, s

′3, s′′3) : (X1, Y1, Z1, f1, g1, h1) → (X3, Y3, Z3, f3, g3, h3)

and (s4, s′4, s′′4) : (X2, Y2, Z2, f2, g2, h2) → (X3, Y3, Z3, f3, g3, h3) in I. We would

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/Sis filtered, we can choose a morphism X3 → X4 in S such that the compositionsX ′ → X1 → X3 → X4 and X ′ → X1 → X3 → X4 are equal. Then we finallycomplete X3 → X4 to a morphism (X3, Y3, Z3) → (X4, Y4, Z4) in I and composewith 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 constructanother triangulated category by taking the “quotient”. The construction uses alocalization. This is similar to the quotient of an abelian category by a Serresubcategory, see Homology, Section 10. Before we do the actual construction webriefly discuss kernels of exact functors.

Definition 6.1.05RB LetD be a pre-triangulated category. We say a full pre-triangulatedsubcategory D′ of D is saturated if whenever X ⊕ Y is isomorphic to an object ofD′ then both X and Y are isomorphic to objects of D′.

A saturated triangulated subcategory is sometimes called a thick triangulated sub-category. In some references, this is only used for strictly full triangulated sub-categories (and sometimes the definition is written such that it implies strictness).There is another notion, that of an épaisse triangulated subcategory. The definitionis 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 objectsof D′, then also X and Y are isomorphic to objects of D′. It turns out that this isequivalent to being saturated (this is elementary and can be found in [Ric89]) andthe notion of a saturated category is easier to work with.

Lemma 6.2.05RC Let F : D → D′ be an exact functor of pre-triangulated categories.Let D′′ be the full subcategory of D with objects

Ob(D′′) = X ∈ Ob(D) | F (X) = 0

DERIVED CATEGORIES 20

Then D′′ is a strictly full saturated pre-triangulated subcategory of D. If D is atriangulated category, then D′′ is a triangulated subcategory.

Proof. It is clear that D′′ 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) = 0as (F (X), F (Y ), F (Z), F (f), F (g), F (h)) is distinguished. Hence we may applyLemma 4.16 to see that D′′ is a pre-triangulated subcategory (respectively a trian-gulated subcategory if D is a triangulated category). The final assertion of beingsaturated follows from F (X)⊕ F (Y ) = 0⇒ F (X) = F (Y ) = 0.

Lemma 6.3.05RD Let H : D → A be a homological functor of a pre-triangulatedcategory into an abelian category. Let D′ be the full subcategory of D with objects

Ob(D′) = X ∈ Ob(D) | H(X[n]) = 0 for all n ∈ Z

Then D′ is a strictly full saturated pre-triangulated subcategory of D. If D is atriangulated category, then D′ is a triangulated subcategory.

Proof. It is clear that D′ is preserved under [1] and [−1]. If (X,Y, Z, f, g, h) isa distinguished triangle of D and H(X[n]) = H(Y [n]) = 0 for all n, then alsoH(Z[n]) = 0 for all n by the long exact sequence (3.5.1). Hence we may applyLemma 4.16 to see that D′ is a pre-triangulated subcategory (respectively a tri-angulated subcategory if D is a triangulated category). The assertion of beingsaturated 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.

Lemma 6.4.05RE Let H : D → A be a homological functor of a pre-triangulatedcategory into an abelian category. Let D+

H ,D−H ,DbH be the full subcategory of D

with objectsOb(D+

H) = X ∈ Ob(D) | H(X[n]) = 0 for all n 0Ob(D−H) = X ∈ Ob(D) | H(X[n]) = 0 for all n 0Ob(DbH) = 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 isa triangulated category, then each is a triangulated subcategory.

Proof. Let us prove this for D+H . 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 forall 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 D+

H is a pre-triangulated subcategory(respectively a triangulated subcategory if D is a triangulated category). Theassertion 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

Definition 6.5.05RF Let D be a (pre-)triangulated category.(1) Let F : D → D′ 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.

Lemma 6.6.05RG Let D be a triangulated category. Let D′ ⊂ D be a full triangulatedsubcategory. Set

(6.6.1)05RH S =f ∈ Arrows(D) such that there exists a distinguished triangle

(X,Y, Z, f, g, h) of D with Z isomorphic to an object of D′

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) D′ is a saturated triangulated subcategory.

Proof. To prove the first assertion we have to prove that MS1, MS2, MS3 andMS5, MS6 hold.

Proof of MS1. It is clear that identities are in S because (X,X, 0, 1, 0, 0) is distin-guished for every object X of D and because 0 is an object of D′. Let f : X → Yand g : Y → Z be composable morphisms contained in S. Choose distinguishedtriangles (X,Y,Q1, f, p1, d1), (X,Z,Q2, g f, p2, d2), and (Y, Z,Q3, g, p3, d3). Byassumption we know that Q1 and Q3 are isomorphic to objects of D′. By TR4 weknow there exists a distinguished triangle (Q1, Q2, Q3, a, b, c). Since D′ is a trian-gulated subcategory we conclude that Q2 is isomorphic to an object of D′. Henceg f ∈ S.

Proof of MS3. Let a : X → Y be a morphism and let t : Z → X be an elementof S such that a t = 0. To prove LMS3 it suffices to find an s ∈ S such thats 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 thereexists a morphism i : Q→ Y such that i g = a. Finally, using TR1 again we canchoose a triangle (Q,Y,W, i, s, k). Here is a picture

Zt// X

g//

1

Q //

i

Z[1]

Xa// Y

s

W

Since t ∈ S we see that Q is isomorphic to an object of D′. Hence s ∈ S. Finally,s a = s i g = 0 as s i = 0 by Lemma 4.1. We conclude that LMS3 holds. Theproof of RMS3 is dual.

Proof of MS5. Follows as distinguished triangles and D′ are stable under transla-tions

DERIVED CATEGORIES 22

Proof of MS6. Suppose given a commutative diagram

X //

s

Y

s′

X ′ // Y ′

with s, s′ ∈ S. By Proposition 4.23 we can extend this to a nine square diagram.As s, s′ are elements of S we see that X ′′, Y ′′ are isomorphic to objects of D′. SinceD′ is a full triangulated subcategory we see that Z ′′ is also isomorphic to an objectof D′. Whence the morphism Z → Z ′ is an element of S. This proves MS6.

MS2 is a formal consequence of MS1, MS5, and MS6, see Lemma 5.2. This finishesthe proof of the first assertion of the lemma.

Let’s assume that S is saturated. (In the following we will use rotation of distin-guished triangles without further mention.) LetX⊕Y be an object isomorphic to anobject of D′. Consider the morphism f : 0→ X. The composition 0→ X → X⊕Yis an element of S as (0, X ⊕ Y,X ⊕ Y, 0, 1, 0) is a distinguished triangle. The com-position Y [−1] → 0 → X is an element of S as (X,X ⊕ Y, Y, (1, 0), (0, 1), 0) is adistinguished triangle, see Lemma 4.11. Hence 0 → X is an element of S (as S issaturated). Thus X is isomorphic to an object of D′ as desired.

Finally, assume D′ is a saturated triangulated subcategory. Let

Wh−→ X

g−→ Yf−→ Z

be composable morphisms of D such that fg, gh ∈ S. We will build up a pictureof objects as in the diagram below.

Q12

!!

Q23

!!Q1

+1

~~

==

Q2+1

+1oo

==

Q3+1

+1oo

W // X

aa

// Y

aa

// 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 appliedto the composition W → X → Y and the composition X → Y → Z there existdistinguished triangles (Q1, Q12, Q2) and (Q2, Q23, Q3) which use the morphisms sand t. The objects Q12 and Q23 are isomorphic to objects of D′ as W → Y andX → Z are assumed in S. Hence also s[1]t is an element of S as S is closed undercompositions and shifts. Note that s[1]t = 0 as Y [1] → Q2[1] → X[2] is zero, seeLemma 4.1. Hence Q3[1]⊕Q1[2] is isomorphic to an object of D′, see Lemma 4.11.By assumption on D′ we conclude that Q3 and Q1 are isomorphic to objects ofD′. Looking at the distinguished triangle (Q1, Q12, Q2) we conclude that Q2 is alsoisomorphic to an object of D′. Looking at the distinguished triangle (X,Y,Q2) wefinally conclude that g ∈ S. (It is also follows that h, f ∈ S, but we don’t needthis.)

DERIVED CATEGORIES 23

Definition 6.7.05RI Let D be a triangulated category. Let B be a full triangulatedsubcategory. 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. Thelocalization 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 triangulatedcategories, see Proposition 5.5. The universal property of this construction is thefollowing.

Lemma 6.8.05RJ Let D be a triangulated category. Let B be a full triangulated sub-category 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 thatB ⊂ Ker(H) then there exists a unique factorization H ′ : D/B → A suchthat H = H ′ Q and H ′ is a homological functor too.

(2) If F : D → D′ is an exact functor into a pre-triangulated category D′ suchthat B ⊂ Ker(F ) then there exists a unique factorization F ′ : D/B → D′such that F = F ′ Q and F ′ is an exact functor too.

Proof. This lemma follows from Lemma 5.6. Namely, if f : X → Y is a morphismof D such that for some distinguished triangle (X,Y, Z, f, g, h) the object Z isisomorphic to an object of B, then H(f), resp. F (f) is an isomorphism under theassumptions of (1), resp. (2). Details omitted.

The kernel of the quotient functor can be described as follows.

Lemma 6.9.05RK Let D be a triangulated category. Let B be a full triangulated sub-category. The kernel of the quotient functor Q : D → D/B is the strictly fullsubcategory of D whose objects are

Ob(Ker(Q)) =Z ∈ Ob(D) such that there exists a Z ′ ∈ Ob(D)such that Z ⊕ Z ′ is isomorphic to an object of B

In other words it is the smallest strictly full saturated triangulated subcategory ofD containing B.

Proof. First note that the kernel is automatically a strictly full triangulated sub-category containing summands of any of its objects, see Lemma 6.2. The descriptionof 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 induceorder preserving maps between

(1) the partially ordered set of multiplicative systems S in D compatible withthe 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) andB 7→ S(B) where S(B) is the multiplicative set of (6.6.1), i.e.,

S(B) =f ∈ Arrows(D) such that there exists a distinguished triangle

(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.

Lemma 6.10.05RL Let D be a triangulated category. The operations described abovehave the following properties

DERIVED CATEGORIES 24

(1) S(B(S)) is the “saturation” of S, i.e., it is the smallest saturated multi-plicative system in D containing S, and

(2) B(S(B)) is the “saturation” of B, i.e., it is the smallest strictly full saturatedtriangulated subcategory of D containing B.

In particular, the constructions define mutually inverse maps between the (partiallyordered) set of saturated multiplicative systems in D compatible with the triangulatedstructure on D and the (partially ordered) set of strictly full saturated triangulatedsubcategories 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 thelocalized category)

S(B(S)) =f ∈ Arrows(D) such that there exists a distinguished

triangle (X,Y, Z, f, g, h) of D with Q(Z) = 0

= f ∈ Arrows(D) | Q(f) is an isomorphism

= S = S′

in the notation of Categories, Lemma 27.21. The final statement of that lemmafinishes the proof.

Lemma 6.11.05RM Let H : D → A be a homological functor from a triangulatedcategory 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 correspondingsaturated 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 ofD by Lemma 6.3. The set S is a saturated multiplicative system compatible withthe triangulated structure by Lemma 5.4. Recall that the multiplicative systemcorresponding 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 elementof this set if and only if f is an element of S. Finally, the factorization of H throughQ is a consequence of Lemma 6.8.

7. Adjoints for exact functors

0A8C Results on adjoint functors between triangulated categories.

Lemma 7.1.0A8D Let F : D → D′ be an exact functor between triangulated categories.If F admits a right adjoint G : D′ → D, then G is also an exact functor.

DERIVED CATEGORIES 25

Proof. Let X be an object of D and A an object of D′. Since F is an exact functorwe see that

MorD(X,G(A[1]) = MorD′(F (X), A[1])= MorD′(F (X)[−1], A)= MorD′(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 isomorphismG(A)[1] = G(A[1]). Let A → B → C → A[1] be a distinguished triangle in D′.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 distinguishedtriangle in D′. 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) suchthat the diagram

G(A) //

G(B) //

X //

G(A)[1]

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

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

HomD′(W,X)→ HomD′(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] isa distinguished triangle which is what we wanted to show.

Lemma 7.2.09J1 Let D, D′ be triangulated categories. Let F : D → D′ and G : D′ →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 D′. Choose a distinguishedtriangle

F (G(X))→ X → Y → F (G(X))[1]

DERIVED CATEGORIES 26

in D′. Applying G and using that G(F (G(X))) = G(X) we find a distinguishedtriangle

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 categoryK(A) of A is the category ofcomplexes of A with morphisms given by morphisms of complexes up to homotopy.Here is the formal definition.

Definition 8.1.013H 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 asmorphisms homotopy classes of maps of complexes (see Homology, Lemma13.7).

(7) We let K+(A), K−(A), resp. Kb(A) be the full subcategory of K(A) whoseobjects 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 trian-gulated categories. To prove this we first develop some machinery related to conesand 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 Awith morphisms given by morphisms of complexes up to homotopy. Note that theshift 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.

Definition 9.1.014E Let A be an additive category. Let f : K• → L• be a morphism ofcomplexes of A. The cone of f is the complex C(f)• given by C(f)n = Ln⊕Kn+1

and differential

dnC(f) =(dnL fn+1

0 −dn+1K

)It comes equipped with canonical morphisms of complexes i : L• → C(f)• andp : 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

Lemma 9.2.014F Suppose thatK•1 f1

//

a

L•1

b

K•2

f2 // L•2is 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) : (K•1 , L•1, C(f1)•, f1, i1, p1)→ (K•2 , L•2, C(f2)•, f2, i2, p2)of K(A).

Proof. Let hn : Kn1 → Ln−1

2 be a family of morphisms such that b f1 − f2 a =d h+ h d. Define cn by the matrix

cn =(bn hn+1

0 an+1

): Ln1 ⊕Kn+1

1 → Ln2 ⊕Kn+12

A matrix computation show that c is a morphism of complexes. It is trivial thatc 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 Lemma9.2 in general depends on the chosen homotopy h between f2 a and b f1.

Lemma 9.3.08RI Suppose that f : K• → L• and g : L• → M• are morphisms ofcomplexes 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•f//

L•

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 themap in (2).

Note that the morphisms C(f)• → M• and K• → C(g)•[−1] constructed in theproof of Lemma 9.3 in general depend on the chosen homotopy.

Definition 9.4.014G 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 tothe inclusion of a direct summand. A termwise split surjection β : B• → C• is amorphism of complexes such that each Bn → Cn is isomorphic to the projectiononto a direct summand.

Lemma 9.5.014H Let A be an additive category. Let

A•f//

a

B•

b

C•g // D•

DERIVED CATEGORIES 28

be a diagram of morphisms of complexes commuting up to homotopy. If f is atermwise split injection, then b is homotopic to a morphism which makes the dia-gram commute. If g is a split surjection, then a is homotopic to a morphism whichmakes 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 fn.Take b′ = b − dhπ − hπd. Suppose sn : Dn → Cn are morphisms splitting themorphisms gn : Cn → Dn. Take a′ = a+ dsh+ shd. Computations omitted.

The following lemma can be used to replace a morphism of complexes by a mor-phism where in each degree the map is the injection of a direct summand.

Lemma 9.6.013N Let A be an additive category. Let α : K• → L• be a morphism ofcomplexes of A. There exists a factorization

K•α //

α

66L•π // L•

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 setLn = Ln ⊕Kn ⊕Kn+1

and we define

dnL

=

dnL 0 00 dnK idKn+1

0 0 −dn+1K

In other words, L• = L• ⊕ C(1K•). Moreover, we set

α =

αidKn

0

which is clearly a split injection. It is also clear that it defines a morphism ofcomplexes. We define

π =(idLn 0 0

)so that clearly π α = α. We set

s =

idLn

00

so that π s = idL• . Finally, let hn : Ln → Ln−1 be the map which maps thesummand Kn of Ln via the identity morphism to the summand Kn of Ln−1. Thenit 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

Remark 9.7.013O To see the last displayed equality in the proof above we can arguewith elements as follows. We have sπ(l, k, k+) = (l, 0, 0). Hence the morphism of theleft 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.

Lemma 9.8.0642 Let A be an additive category. Let α : K• → L• be a morphism ofcomplexes of A. There exists a factorization

K•i //

α

66K•α // L•

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

Kn = Kn ⊕ Ln−1 ⊕ Ln

and we define

dnK

=

dnK 0 00 −dn−1

L idLn

0 0 dnL

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 ofcomplexes. We define

i =

idKn

00

so that clearly α i = α. We set

s =(idKn 0 0

)so that s i = idK• . Finally, let hn : Kn → Kn−1 be the map which maps thesummand Ln−1 of Kn via the identity morphism to the summand Ln−1 of Kn−1.Then it is a trivial matter to prove that

idK• − i s = d h+ h d

which finishes the proof of the lemma.

Definition 9.9.014I Let A be an additive category. A termwise split exact sequenceof complexes of A is a complex of complexes

0→ A•α−→ B•

β−→ C• → 0

together with given direct sum decompositions Bn = An ⊕Cn compatible with αnand 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 anassociated morphism of complexes

δ : C• −→ A•[1]

which in degree n is the map πn+1 dnB sn. 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.

Lemma 9.10.05SS Let A be an additive category. Let 0 → A• → B• → C• → 0 betermwise split exact sequences as in Definition 9.9. Let (π′)n, (s′)n be a secondcollection of splittings. Denote δ′ : C• −→ A•[1] the morphism associated to thissecond set of splittings. Then

(1, 1, 1) : (A•, B•, C•, α, β, δ) −→ (A•, B•, C•, α, β, δ′)

is an isomorphism of triangles in K(A).

Proof. The statement simply means that δ and δ′ are homotopic maps of com-plexes. This is Homology, Lemma 14.12.

Remark 9.11.014J Let A be an additive category. Let 0 → A•i → B•i → C•i → 0,i = 1, 2 be termwise split exact sequences. Suppose that a : A•1 → A•2, b : B•1 → B•2 ,and c : C•1 → C•2 are morphisms of complexes such that

A•1

a

// B•1 //

b

C•1

c

A•2 // B•2 // C•2

commutes in K(A). In general, there does not exist a morphism b′ : B•1 → B•2which is homotopic to b such that the diagram above commutes in the category ofcomplexes. Namely, consider Examples, Equation (59.0.1). If we could replace themiddle map there by a homotopic one such that the diagram commutes, then wewould have additivity of traces which we do not.

Lemma 9.12.086L Let A be an additive category. Let 0 → A•i → B•i → C•i → 0,i = 1, 2, 3 be termwise split exact sequences of complexes. Let b : B•1 → B•2 andb′ : B•2 → B•3 be morphisms of complexes such that

A•1

0

// B•1 //

b

C•1

0

A•2 // B•2 // C•2

and

A•2

0

// B•2 //

b′

C•2

0

A•3 // B•3 // C•3

commute in K(A). Then b′ b = 0 in K(A).

Proof. By Lemma 9.5 we can replace b and b′ by homotopic maps such that theright square of the left diagram commutes and the left square of the right diagramcommutes. In other words, we have Im(bn) ⊂ Im(An2 → Bn2 ) and Ker((b′)n) ⊃Im(An2 → Bn2 ). Then b b′ = 0 as a map of complexes.

DERIVED CATEGORIES 31

Lemma 9.13.014K Let A be an additive category. Let f1 : K•1 → L•1 and f2 : K•2 → L•2be morphisms of complexes. Let

(a, b, c) : (K•1 , L•1, C(f1)•, f1, i1, p1) −→ (K•2 , L•2, C(f2)•, f2, i2, p2)

be any morphism of triangles of K(A). If a and b are homotopy equivalences thenso is c.

Proof. Let a−1 : K•2 → K•1 be a morphism of complexes which is inverse to ain K(A). Let b−1 : L•2 → L•1 be a morphism of complexes which is inverse to bin K(A). Let c′ : C(f2)• → C(f1)• be the morphism from Lemma 9.2 appliedto f1 a−1 = b−1 f2. If we can show that c c′ and c′ c are isomorphisms inK(A) then we win. Hence it suffices to prove the following: Given a morphism oftriangles (1, 1, c) : (K•, L•, C(f)•, f, i, p) inK(A) the morphism c is an isomorphismin K(A). By assumption the two squares in the diagram

L• //

1

C(f)• //

c

K•[1]

1

L• // C(f)• // K•[1]

commute up to homotopy. By construction of C(f)• the rows form termwise splitsequences 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 trianglesis an isomorphism of triangles in K(A). It turns out that the collection of trianglesof K(A) given by cones and the collection of triangles of K(A) given by termwisesplit sequences of complexes are the same up to isomorphisms, at least up to sign!

Lemma 9.14.014L 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 thediagram

A• //

B•

// C(α)•−p//

A•[1]

A• // B• // C•

δ // 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•δ//

K•[1]

K• // L• // C(f)• −p // K•[1]

where the upper triangle is the triangle associated to a termwise split exactsequence 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 ontoBn followed by βn. This defines a morphism of complexesbecause the compositions An+1 → Bn+1 → Cn+1 are zero. To get a homotopyinverse we take C• → C(α)• given by (sn,−δn) in degree n. This is a morphism ofcomplexes because the morphism δn can be characterized as the unique morphismCn → An+1 such that d sn − sn+1 d = α δn, see proof of Homology, Lemma14.10. The composition C• → C(α)• → C• is the identity. The compositionC(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 0πn 0

): C(α)n = Bn ⊕An+1 → Bn−1 ⊕An = C(α)n−1

It is trivial to verify that(1 00 1

)−(sn

−δn)(

βn 0)

=(d αn

0 −d

)(0 0πn 0

)+(

0 0πn+1 0

)(d αn+1

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 clearfrom the above. Namely, we can use the homotopy inverse (s,−δ) : C• → C(α)• andcheck instead that the two maps C• → A•[1] agree. And note that p (s,−δ) = −δas 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 mayreplace L• by L• and f by f . In other words we may assume that f is a termwisesplit injection. In this case the result follows from part (1).

Lemma 9.15.014M Let A be an additive category. Let A•1 → A•2 → . . . → A•n bea sequence of composable morphisms of complexes. There exists a commutativediagram

A•1 // A•2 // . . . // A•n

B•1 //

OO

B•2 //

OO

. . . // B•n

OO

such that each morphism B•i → B•i+1 is a split injection and each B•i → A•i is ahomotopy equivalence. Moreover, if all A•i are in K+(A), K−(A), or Kb(A), thenso are the B•i .

Proof. The case n = 1 is without content. Lemma 9.6 is the case n = 2. Supposewe have constructed the diagram except for B•n. Apply Lemma 9.6 to the compo-sition B•n−1 → A•n−1 → A•n. The result is a factorization B•n−1 → B•n → A•n asdesired.

Lemma 9.16.014N 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 isomorphicto 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 naturalinclusion and projection maps. By construction of δ we have

dnB =(dnA δn

0 dnC

)On the other hand the cone of δ[−1] : C•[−1]→ A• is given as C(δ[−1])n = An⊕Cnwith differential identical with the matrix above! Whence the lemma.

Lemma 9.17.014O Let A be an additive category. Let f : K• → L• be a morphismof complexes. The triangle (L•, C(f)•,K•[1], i, p, f [1]) is the triangle associated tothe termwise split sequence

0→ L• → C(f)• → K•[1]→ 0coming 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 givenby the maps in the snake lemma without signs we define distinguished triangles inthe homotopy category as follows.

Definition 10.1.014Q Let A be an additive category. A triangle (X,Y, Z, f, g, h) ofK(A) is called a distinguished triangle of K(A) if it is isomorphic to the triangleassociated to a termwise split exact sequence of complexes, see Definition 9.9. Samedefinition 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, seeProposition 10.3. Before we can prove the proposition we need one more lemma inorder to be able to prove TR4.

Lemma 10.2.014R Let A be an additive category. Suppose that α : A• → B• andβ : B• → C• are split injections of complexes. Then there exist distinguishedtriangles (A•, B•, Q•1, α, p1, d1), (A•, C•, Q•2, βα, p2, d2) and (B•, C•, Q•3, β, p3, d3)for which TR4 holds.

Proof. Say πn1 : Bn → An, and πn3 : Cn → Bn are the splittings. Then alsoA• → C• is a split injection with splittings πn2 = πn1 πn3 . Let us write Q•1, Q•2 andQ•3 for the “quotient” complexes. In other words, Qn1 = Ker(πn1 ), Qn3 = Ker(πn3 )and Qn2 = Ker(πn2 ). Note that the kernels exist. Then Bn = An ⊕ Qn1 and Cn =Bn ⊕ Qn3 , where we think of An as a subobject of Bn and so on. This impliesCn = An ⊕Qn1 ⊕Qn3 . Note that πn2 = πn1 πn3 is zero on both Qn1 and Qn3 . HenceQn2 = Qn1 ⊕Qn3 . Consider the commutative diagram

0 → A• → B• → Q•1 → 0↓ ↓ ↓

0 → A• → C• → Q•2 → 0↓ ↓ ↓

0 → B• → C• → Q•3 → 0

DERIVED CATEGORIES 34

The rows of this diagram are termwise split exact sequences, and hence determinedistinguished triangles by definition. Moreover downward arrows in the diagramabove are compatible with the chosen splittings and hence define morphisms oftriangles

(A• → B• → Q•1 → A•[1]) −→ (A• → C• → Q•2 → A•[1])and

(A• → C• → Q•2 → A•[1]) −→ (B• → C• → Q•3 → B•[1]).Note that the splittings Qn3 → Cn of the bottom split sequence in the diagramprovides a splitting for the split sequence 0→ Q•1 → Q•2 → Q•3 → 0 upon composingwith Cn → Qn2 . It follows easily from this that the morphism δ : Q•3 → Q•1[1] inthe corresponding distinguished triangle

(Q•1 → Q•2 → Q•3 → Q•1[1])is equal to the composition Q•3 → B•[1] → Q•1[1]. Hence we get a structure as inthe conclusion of axiom TR4.

Proposition 10.3.014S Let A be an additive category. The category K(A) of com-plexes up to homotopy with its natural translation functors and distinguished trian-gles as defined above is a triangulated category.

Proof. Proof of TR1. By definition every triangle isomorphic to a distinguishedone is distinguished. Also, any triangle (A•, A•, 0, 1, 0, 0) is distinguished since0 → 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) isdistinguished 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 isomorphicto (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 isdistinguished 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 ofthe 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 thistriangle 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 (X ′, Y ′, Z ′, f ′, g′, h′) be distinguished trian-gles of K(A) and let a : X → X ′ and b : Y → Y ′ be morphisms such that f ′ 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 (X ′, Y ′, Z ′, f ′, g′, h′) = (X ′, Y ′, C(f ′), f ′, i′,−p′). At this point we simply ap-ply Lemma 9.2 to the commutative diagram given by f, f ′, 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 composablemorphisms 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

Remark 10.4.05RP Let A be an additive category. Exactly the same proof as theproof 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 aretriangulated subcategories of K(A) which gives another proof.

Lemma 10.5.05RQ 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 usethe criterion of Lemma 4.16 to show that they are triangulated subcategories. Itis clear that each of the categories K+(A), K−(A), and Kb(A) is preserved underthe shift functors [1], [−1]. Finally, suppose that f : A• → B• is a morphism inK+(A),K−(A), orKb(A). Then (A•, B•, C(f)•, f, i,−p) is a distinguished triangleof K(A) with C(f)• ∈ K+(A), K−(A), or Kb(A) as is clear from the constructionof the cone. Thus the lemma is proved. (Alternatively, K• → L• is isomorphic toan termwise split injection of complexes in K+(A), K−(A), or Kb(A), see Lemma9.6 and then one can directly take the associated distinguished triangle.)

Lemma 10.6.014X Let A, B be additive categories. Let F : A → B be an additivefunctor. 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 Awith splittings (sn, πn) and associated morphism δ : C• → A•[1], see Definition9.9. Then F (A•) → F (B•) → F (C•) is a termwise split sequence of complexeswith splittings (F (sn), F (πn)) and associated morphism F (δ) : F (C•)→ F (A•)[1].Thus F transforms distinguished triangles into distinguished triangles.

Lemma 10.7.0G6C Let A be an additive category. Let (A•, B•, C•, a, b, c) be a distin-guished triangle in K(A). Then there exists an isomorphic distinguished triangle(A•, (B′)•, C•, a′, b′, c) such that 0→ An → (B′)n → Cn → 0 is a split short exactsequence 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 • andp : W • → C•[1]. Then (C•, A•[1],W •, c, i,−p) is a distinguished triangle by Lemma9.14. Rotating backwards twice we see that (A•,W •[−1], C•,−i[−1], p[−1], c) isa 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 mustbe 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 thevery construction of cones in Section 9.

Remark 10.8.0G6D Let A be an additive category with countable direct sums. LetDoubleComp(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 com-plexes as follows

. . .→ A•,−1 → A•,0 → A•,1 → . . .

and take the homotopy category Kfirst(DoubleComp(A)) with the corre-sponding 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 com-

plexes as follows

. . .→ A−1,• → A0,• → A1,• → . . .

and take the homotopy category Ksecond(DoubleComp(A)) with the corre-sponding 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.

Remark 10.9.0G6E Let A, B, C be additive categories and assume C has countabledirect 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 immediatelyseen to be compatible with homotopies and termwise split short exact sequencesand 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, inHomology, 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.

Lemma 11.1.05RS Let A be an abelian category. The functor

H0 : K(A) −→ Ais homological.

Proof. Because H0 is a functor, and by our definition of distinguished trianglesit suffices to prove that given a termwise split short exact sequence of complexes0 → 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) inK(A) gives rise to a long exact cohomology sequence

(11.1.1)05ST . . . // Hi(X)Hi(f) // Hi(Y )

Hi(g) // Hi(Z)Hi(h)// Hi+1(X) // . . .

see (3.5.1). Moreover, there is a compatibility with the long exact sequence of co-homology associated to a short exact sequence of complexes (insert future referencehere). For example, if (A•, B•, C•, α, β, δ) is the distinguished triangle associatedto a termwise split exact sequence of complexes (see Definition 9.9), then the co-homology sequence above agrees with the one defined using the snake lemma, seeHomology, Lemma 13.12 and for agreement of sequences, see Homology, Lemma14.11.Recall that a complex K• is acyclic if Hi(K•) = 0 for all i ∈ Z. Moreover, recallthat a morphism of complexes f : K• → L• is a quasi-isomorphism if and only ifHi(f) is an isomorphism for all i. See Homology, Definition 13.10.

Lemma 11.2.05RT Let A be an abelian category. The full subcategory Ac(A) of K(A)consisting of acyclic complexes is a strictly full saturated triangulated subcategoryof K(A). The corresponding saturated multiplicative system (see Lemma 6.10) ofK(A) is the set Qis(A) of quasi-isomorphisms. In particular, the kernel of thelocalization 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 thislemma is a special case of Lemma 6.11.

Definition 11.3.05RU Let A be an abelian category. Let Ac(A) and Qis(A) be as inLemma 11.2. The derived category of A is the triangulated category

D(A) = K(A)/Ac(A) = Qis(A)−1K(A).We denoteH0 : D(A)→ A the unique functor whose composition with the quotientfunctor 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) whosesets of objects are

Ob(D+(A)) = X ∈ Ob(D(A)) | Hn(X) = 0 for all n 0Ob(D−(A)) = X ∈ Ob(D(A)) | Hn(X) = 0 for all n 0Ob(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-isomorphicto L•” to indicate that K• and L• are isomorphic objects of D(A).Remark 11.4.09PA In this chapter, we consistently work with “small” abelian cate-gories (as is the convention in the Stacks project). For a “big” abelian categoryA, it isn’t clear that the derived category D(A) exists, because it isn’t clear thatmorphisms in the derived category are sets. In fact, in general they aren’t, see Ex-amples, Lemma 57.1. However, if A is a Grothendieck abelian category, and givenK•, L• in K(A), then by Injectives, Theorem 12.6 there exists a quasi-isomorphismL• → 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 categoryof modules over a ring, or more generally the category of sheaves of modules on aringed site.Each of the variants D+(A), D−(A), Db(A) can be constructed as a localization ofthe corresponding homotopy category. This relies on the following simple lemma.Lemma 11.5.05RV 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 ofmorphisms of complexes

K• // L•

M•

OO

// N•

OO

where all the arrows are quasi-isomorphisms, L• bounded below,M• boundedabove, 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 intersectionof 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).

Lemma 11.6.05RW 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 anequivalence 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 anequivalence 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 anequivalence 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) isa consequence of the definitions in each case. Each of the functors is essentiallysurjective by Lemma 11.5. To finish the proof we have to show the functors arefully faithful. We first do this for the bounded below version.

Suppose that K•, L• are bounded above complexes. A morphism between thesein D(A) is of the form s−1f for a pair f : K• → (L′)•, s : L• → (L′)• where sis a quasi-isomorphism. This implies that (L′)• has cohomology bounded below.Hence by Lemma 11.5 we can choose a quasi-isomorphism s′ : (L′)• → (L′′)•with (L′′)• bounded below. Then the pair (s′ f, s′ s) defines a morphism inQis+(A)−1K+(A). Hence the functor is “full”. Finally, suppose that the pairf : K• → (L′)•, s : L• → (L′)• defines a morphism in Qis+(A)−1K+(A) which iszero in D(A). This means that there exists a quasi-isomorphism s′ : (L′)• → (L′′)•such that s′ f = 0. Using Lemma 11.5 once more we obtain a quasi-isomorphisms′′ : (L′′)• → (L′′′)• with (L′′′)• bounded below. Thus we see that s′′ s′ f = 0which implies that s−1f is zero in Qis+(A)−1K+(A). This finishes the proof thatthe 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 toshow this where we consistently work with complexes which are already boundedabove.

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 categoryinto 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 thatany 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 thiswe have to define the morphisms δ associated to a short exact sequence

0→ A•a−→ B•

b−→ C• → 0of complexes in the abelian category A. Consider the cone C(a)• of the morphisma. We have C(a)n = Bn ⊕An+1 and we define qn : C(a)n → Cn via the projectionto 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 injectivein 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) isexact we see that (A,B,C(a), a, i,−p) is a distinguished triangle in D(A). Since qis a quasi-isomorphism we see that q is an isomorphism in D(A). Hence we deducethat

(A,B,C, a, b,−p q−1)is a distinguished triangle of D(A). This suggests the following lemma.

Lemma 12.1.0152 Let A be an abelian category. The functor Comp(A) → D(A)defined has the natural structure of a δ-functor, with

δA•→B•→C• = −p q−1

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 trianglewhenever given a short exact sequence of complexes. We have to show that givena commutative diagram

0 // A•a//

f

B•b//

g

C• //

h

0

0 // (A′)• a′ // (B′)• b′ // (C ′)• // 0

we get the desired commutative diagram of Definition 3.6 (2). By Lemma 9.2the pair (f, g) induces a canonical morphism c : C(a)• → C(a′)•. It is a simplecomputation to show that q′ c = h q and f [1] p = p′ c. From this the resultfollows directly.

Lemma 12.2.0153 Let A be an abelian category. Let

0 // A• //

B• //

C• //

0

0 // D• // E• // F • // 0be a commutative diagram of morphisms of complexes such that the rows are shortexact sequences of complexes, and the vertical arrows are quasi-isomorphisms. Theδ-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-isomorphismsinto isomorphisms and that the associated distinguished triangles are functorial.

Lemma 12.3.0154 Let A be an abelian category. Let

0 // A• // B• // C• // 0

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

(A•, B•, C•, α, β, δ).

Proof. Follows from Lemma 9.14.

Remark 12.4.08J5 Let A be an abelian category. Let K• be a complex of A. Leta ∈ 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

τ≤aK• → τ≤a+1K

• → Ha+1(K•)[−a− 1]→ (τ≤aK•)[1]

andHa(K•)[−a]→ τ≥aK

• → τ≥a+1K• → Ha(K•)[−a+ 1]

Lemma 12.5.08Q2 Let A be an abelian category. Let

K•0 → K•1 → . . .→ K•n

be maps of complexes such that(1) Hi(K•0 ) = 0 for i > 0,(2) H−j(K•j )→ H−j(K•j+1) is zero.

Then the composition K•0 → K•n factors through τ≤−nK•n → K•n in D(A). Dually,given maps of complexes

K•n → K•n−1 → . . .→ K•0

such that(1) Hi(K•0 ) = 0 for i < 0,(2) Hj(K•j+1)→ Hj(K•j ) is zero,

then the composition K•n → K•0 factors through K•n → τ≥nK•n in D(A).

DERIVED CATEGORIES 42

Proof. The case n = 1. Since τ≤0K•0 = K•0 in D(A) we can replace K•0 by τ≤0K

•0

and K•1 by τ≤0K•1 . Consider the distinguished triangle

τ≤−1K•1 → K•1 → H0(K•1 )[0]→ (τ≤−1K

•1 )[1]

(Remark 12.4). The composition K•0 → K•1 → H0(K•1 )[0] is zero as it is equalto K•0 → H0(K•0 )[0] → H0(K•1 )[0] which is zero by assumption. The fact thatHomD(A)(K•0 ,−) is a homological functor (Lemma 4.2), allows us to find the desiredfactorization. For n = 2 we get a factorization K•0 → τ≤−1K

•1 by the case n = 1

and we can apply the case n = 1 to the map of complexes τ≤−1K•1 → τ≤−1K

•2 to

get a factorization τ≤−1K•1 → τ≤−2K

•2 . 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. Inthis section we will define the filtered derived category DF (A) of A. In short, wewill define it as the derived category of the exact category of objects of A endowedwith a finite filtration. (Thus our construction is a special case of a more generalconstruction of the derived category of an exact category, see for example [Büh10],[Kel90].) Illusie’s filtered derived category is the full subcategory of ours consistingof those objects whose filtration is finite. (In our category the filtration is still finitein each degree, but may not be uniformly bounded.) The rationale for our choiceis that it is not harder and it allows us to apply the discussion to the spectralsequences of Lemma 21.3, see also Remark 21.4.

We will use the notation regarding filtered objects introduced in Homology, Section19. The category of filtered objects of A is denoted Fil(A). All filtrations will bedecreasing by fiat.

Definition 13.1.05RY Let A be an abelian category. The category of finite filteredobjects of A is the category of filtered objects (A,F ) of A whose filtration F isfinite. We denote it Filf (A).

Thus Filf (A) is a full subcategory of Fil(A). For each p ∈ Z there is a functorgrp : Filf (A)→ A. There is a functor

gr =⊕

p∈Zgrp : Filf (A)→ Gr(A)

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. Thecategory 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 oftriangulated 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 abeliancategory we make the following definitions.

DERIVED CATEGORIES 43

Definition 13.2.05RZ Let A be an abelian category.

(1) Let α : K• → L• be a morphism of K(Filf (A)). We say that α is a filteredquasi-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 ifthe complex gr(K•) is acyclic.

Note that α : K• → L• is a filtered quasi-isomorphism if and only if each grp(α) isa quasi-isomorphism. Similarly a complex K• is filtered acyclic if and only if eachgrp(K•) is acyclic.

Lemma 13.3.05S0 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 Lemma11.1 and the fact that the functors gr, grp, (forget F ) are exact functors of triangu-lated categories. See Lemma 4.20.

Lemma 13.4.05S1 Let A be an abelian category. The full subcategory FAc(A) ofK(Filf (A)) consisting of filtered acyclic complexes is a strictly full saturated trian-gulated subcategory of K(Filf (A)). The corresponding saturated multiplicative sys-tem (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 thislemma is a special case of Lemma 6.11.

Definition 13.5.05S2 Let A be an abelian category. Let FAc(A) and FQis(A) be asin 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)).

Lemma 13.6.05S3 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 definitionfor the first two. For the last one the statement we have to do a little bit ofwork. Let f : K• → L• be a filtered quasi-isomorphism in K(Filf (A)). Choose adistinguished triangle (K•, L•,M•, f, g, h) which contains f . Then M• is filteredacyclic, see Lemma 13.4. Hence by the corresponding lemma for K(A) it suffices toshow that a filtered acyclic complex is an acyclic complex if we forget the filtration.This follows from Homology, Lemma 19.15.

DERIVED CATEGORIES 44

Definition 13.7.05S4 Let A be an abelian category. The bounded filtered derivedcategory DF b(A) is the full subcategory of DF (A) with objects those X such thatgr(X) ∈ Db(A). Similarly for the bounded below filtered derived category DF+(A)and the bounded above filtered derived category DF−(A).

Lemma 13.8.05S5 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 Mn = 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•

M•

OO

// N•

OO

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 19.15the sequence

Ka−1 da−2

−−−→ Ka−1 da−1

−−−→ 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

τ≥aK• = (. . .→ 0→ Ka/ Im(da−1)→ Ka+1 → . . .)

is a filtered quasi-isomorphism. This proves (1). The proof of (2) is dual to theproof of (1). Part (3) follows formally from (1) and (2).

To state the following lemma denote FAc+(A), FAc−(A), resp. FAcb(A) the inter-section ofK+(FilfA),K−(FilfA), resp.Kb(FilfA) with FAc(A). Denote FQis+(A),FQis−(A), resp. FQisb(A) the intersection ofK+(FilfA),K−(FilfA), resp.Kb(FilfA)with FQis(A).

Lemma 13.9.05S6 Let A be an abelian category. The subcategories FAc+(A), FAc−(A),resp. FAcb(A) are strictly full saturated triangulated subcategories of K+(FilfA),K−(FilfA), resp. Kb(FilfA). 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+(FilfA) → DF+(A) is FAc+(A) and thisinduces an equivalence of triangulated categories

K+(FilfA)/FAc+(A) = FQis+(A)−1K+(FilfA) −→ DF+(A)

(2) The kernel of the functor K−(FilfA) → DF−(A) is FAc−(A) and thisinduces an equivalence of triangulated categories

K−(FilfA)/FAc−(A) = FQis−(A)−1K−(FilfA) −→ DF−(A)

DERIVED CATEGORIES 45

(3) The kernel of the functor Kb(FilfA) → DF b(A) is FAcb(A) and this in-duces an equivalence of triangulated categories

Kb(FilfA)/FAcb(A) = FQisb(A)−1Kb(FilfA) −→ DF b(A)

Proof. This follows from the results above, in particular Lemma 13.8, by exactlythe 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 generalnotion of right and left derived functors exists where we have an exact functorbetween triangulated categories, a multiplicative system in the source category andwe want to find the “correct” extension of the exact functor to the localized category.

Situation 14.1.05S8 Here F : D → D′ is an exact functor of triangulated categoriesand S is a saturated multiplicative system in D compatible with the structure oftriangulated category on D.

Let X ∈ Ob(D). Recall from Categories, Remark 27.7 the filtered category X/S ofarrows s : X → X ′ in S with source X. Dually, in Categories, Remark 27.15 wedefined the cofiltered category S/X of arrows s : X ′ → X in S with target X.

Definition 14.2.05S9 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) −→ D′, (s : X → X ′) 7−→ F (X ′)

is essentially constant4; in this case the value Y in D′ is called the value ofRF at X.

(2) we say the left derived functor LF is defined at X if the pro-object

(S/X) −→ D′, (s : X ′ → X) 7−→ F (X ′)

is essentially constant; in this case the value Y in D′ is called the value ofLF 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 de-fined is a triangulated subcategory, and RF will define a functor on this subcategorywhich transforms morphisms of S into isomorphisms.

Lemma 14.3.05SA Assumptions and notation as in Situation 14.1. Let f : X → Y bea 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 dia-gram

X

f

s// X ′

f ′

Y

s′ // Y ′

4For a discussion of when an ind-object or pro-object of a category is essentially constant werefer to Categories, Section 22.

DERIVED CATEGORIES 46

with s, s′ ∈ S the diagram

F (X)

// F (X ′)

// RF (X)

F (Y ) // F (Y ′) // 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 dia-gram

X ′

f ′

s// X

f

Y ′

s′ // Y

with s, s′ in S the diagram

LF (X)

// F (X ′)

// F (X)

LF (Y ) // F (Y ′) // F (Y )

commutes.

Proof. Part (1) holds if we only assume that the colimits

RF (X) = colims:X→X′ F (X ′) and RF (Y ) = colims′:Y→Y ′ F (Y ′)

exist. Namely, to give a morphism RF (X) → RF (Y ) between the colimits is thesame thing as giving for each s : X → X ′ in Ob(X/S) a morphism F (X ′)→ RF (Y )compatible with morphisms in the category X/S. To get the morphism we choosea commutative diagram

X

f

s// X ′

f ′

Y

s′ // Y ′

with s, s′ in S as is possible by MS2 and we set F (X ′) → RF (Y ) equal to thecomposition F (X ′) → F (Y ′) → RF (Y ). To see that this is independent of thechoice of the diagram above use MS3. Details omitted. The proof of (2) is dual.

Lemma 14.4.05SB Assumptions and notation as in Situation 14.1. Let s : X → Y bean element of S.

(1) RF is defined at X if and only if it is defined at Y . In this case the mapRF (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 mapLF (s) : LF (X)→ LF (Y ) between values is an isomorphism.

Proof. Omitted.

Lemma 14.5.05SU Assumptions and notation as in Situation 14.1. Let X be an objectof 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 isa 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 isa canonical isomorphism LF (X)[n]→ LF (X[n]) between values.

Proof. Omitted.

Lemma 14.6.05SC 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 D′. Similarly for LF .

Proof. Say RF is defined at X,Y with values A,B. Let RF (f) : A → B bethe induced morphism, see Lemma 14.3. We may choose a distinguished triangle(A,B,C,RF (f), b, c) in D′. We claim that C is a value of RF at Z.To see this pick s : X → X ′ in S such that there exists a morphism α : A→ F (X ′)as in Categories, Definition 22.1. We may choose a commutative diagram

X

f

s// X ′

f ′

Y

s′ // Y ′

with s′ ∈ S by MS2. Using that Y/S is filtered we can (after replacing s′ by somes′′ : Y → Y ′′ in S) assume that there exists a morphism β : B → F (Y ′) as inCategories, Definition 22.1. Picture

A

RF (f)

α// F (X ′) //

F (f ′)

A

RF (f)

Bβ // F (Y ′) // B

It may not be true that the left square commutes, but the outer and right squarescommute. The assumption that the ind-object F (Y ′)s′:Y ′→Y is essentially con-stant means that there exists a s′′ : Y → Y ′′ in S and a morphism h : Y ′ → Y ′′

such that s′′ = h s′ and such that F (h) equal to F (Y ′)→ B → F (Y ′)→ F (Y ′′).Hence after replacing Y ′ by Y ′′ and β by F (h) β the diagram will commute (bydirect computation with arrows).Using MS6 choose a morphism of triangles

(s, s′, s′′) : (X,Y, Z, f, g, h) −→ (X ′, Y ′, Z ′, f ′, g′, h′)with s′′ ∈ S. By TR3 choose a morphism of triangles

(α, β, γ) : (A,B,C,RF (f), b, c) −→ (F (X ′), F (Y ′), F (Z ′), F (f ′), F (g′), F (h′))

By Lemma 14.4 it suffices to prove thatRF (Z ′) is defined and has value C. Considerthe category I of Lemma 5.8 of trianglesI = (t, t′, t′′) : (X ′, Y ′, Z ′, f ′, g′, h′)→ (X ′′, Y ′′, Z ′′, f ′′, g′′, h′′) | (t, t′, t′′) ∈ S

To show that the system F (Z ′′) is essentially constant over the category Z ′/Sis equivalent to showing that the system of F (Z ′′) is essentially constant over I

DERIVED CATEGORIES 48

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

colimIMorD′(W,F (X ′′)) MorD′(W,A)oo

colimIMorD′(W,F (Y ′′))

OO

MorD′(W,B)

OO

oo

colimIMorD′(W,F (Z ′′))

OO

MorD′(W,C)

OO

oo

colimIMorD′(W,F (X ′′[1]))

OO

MorD′(W,A[1])

OO

oo

colimIMorD′(W,F (Y ′′[1]))

OO

MorD′(W,B[1])

OO

oo

where the horizontal arrows are given by composing with (α, β, γ). Since filteredcolimits are exact (Algebra, Lemma 8.8) the left column is an exact sequence. Thusthe 5 lemma (Homology, Lemma 5.20) tells us the

colimIMorD′(W,F (Z ′′)) −→MorD′(W,C)

is bijective. Choose an object (t, t′, t′′) : (X ′, Y ′, Z ′)→ (X ′′, Y ′′, Z ′′) of I. Applyingwhat we just showed to W = F (Z ′′) and the element idF (X′′) of the colimit we finda unique morphism c(X′′,Y ′′,Z′′) : F (Z ′′) → C such that for some (X ′′, Y ′′, Z ′′) →(X ′′′, Y ′′′, Z ′′) in I

F (Z ′′)c(X′′,Y ′′,Z′′)−−−−−−−−→ C

γ−→ F (Z ′)→ F (Z ′′)→ F (Z ′′′) equals F (Z ′′)→ F (Z ′′′)

The family of morphisms c(X′′,Y ′′,Z′′) form an element c of limIMorD′(F (Z ′′), C)by uniqueness (computation omitted). Finally, we show that colimI F (Z ′′) = C viathe morphisms c(X′′,Y ′′,Z′′) which will finish the proof by Categories, Lemma 22.9.Namely, let W be an object of D′ and let d(X′′,Y ′′,Z′′) : F (Z ′′)→W be a family ofmaps corresponding to an element of limIMorD′(F (Z ′′),W ). If d(X′,Y ′,Z′) γ = 0,then for every object (X ′′, Y ′′, Z ′′) of I the morphism d(X′′,Y ′′,Z′′) is zero by theexistence of c(X′′,Y ′′,Z′′) and the morphism (X ′′, Y ′′, Z ′′) → (X ′′′, Y ′′′, Z ′′) in Isatisfying the displayed equality above. Hence the map

limIMorD′(F (Z ′′),W ) −→MorD′(C,W )

(coming from precomposing by γ) is injective. However, it is also surjective becausethe element c gives a left inverse. We conclude that C is the colimit by Categories,Remark 14.4.

Lemma 14.7.05SD Assumptions and notation as in Situation 14.1. Let X,Y be objectsof D.

(1) If RF is defined at X and Y , then RF is defined at X ⊕ Y .(2) If D′ 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 ⊕ Yand 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 D′ is Karoubian. SinceS is a saturated system S is the set of arrows which become invertible under theadditive localization functor Q : D → S−1D, see Categories, Lemma 27.21. Thusfor any s : X → X ′ and s′ : Y → Y ′ in S the morphism s⊕ s′ : X ⊕ Y → X ′ ⊕ Y ′is an element of S. In this way we obtain a functor

X/S × Y/S −→ (X ⊕ Y )/SRecall that the categories X/S, Y/S, (X ⊕ Y )/S are filtered (Categories, Remark27.7). By Categories, Lemma 22.12 X/S × Y/S is filtered and F |X/S : X/S → D′(resp. G|Y/S : Y/S → D′) is essentially constant if and only if F |X/S pr1 : X/S ×Y/S → D′ (resp. G|Y/S pr2 : X/S × Y/S → D′) is essentially constant. Below wewill 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 → D′ is essentially constant. By Homology, Lemma 30.3 (and this iswhere we use that D′ is Karoubian) we see that F |X/S pr1 ⊕ F |Y/S pr2 beingessentially constant implies F |X/S pr1 and F |Y/S pr2 are essentially constantproving 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 → X ′, Z → Y ′ and s : X → X ′, s′ : Y → Y ′

in S such thatX

s

X ⊕ Y

oo // Y

s′

X ′ Zoo // Y ′

commutes. This proves there is a map Z → X ′ ⊕ Y ′ in (X ⊕ Y )/S, i.e., we getpart (1) of Categories, Definition 17.1. To prove part (2) it suffices to prove thatgiven t : X ⊕ Y → Z and morphisms si ⊕ s′i : Z → X ′i ⊕ Y ′i , i = 1, 2 in (X ⊕ Y )/Swe can find morphisms a : X ′1 → X ′, b : X ′2 → X ′, c : Y ′1 → Y ′, d : Y ′2 → Y ′

in S such that a s1 = b s2 and c s′1 = d s′2. To do this we first choose anyX ′ and Y ′ and maps a, b, c, d in S; this is possible as X/S and Y/S are filtered.Then the two maps a s1, b s2 : Z → X ′ become equal in S−1D. Hence we canfind a morphism X ′ → X ′′ in S equalizing them. Similarly we find Y ′ → Y ′′ in Sequalizing cs′1 and ds′2. Replacing X ′ by X ′′ and Y ′ by Y ′′ we get as1 = bs2and c s′1 = d s′2.The proof of the corresponding statements for LF are dual.

Proposition 14.8.05SE 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 −→ D′ of triangulated categories.(3) Elements of S with either source or target in E are morphisms of E.(4) The functor S−1

E E → S−1D is a fully faithful exact functor of triangulatedcategories.

DERIVED CATEGORIES 50

(5) Any element of SE = Arrows(E) ∩ S is mapped to an isomorphism by RF .(6) We obtain an exact functor

RF : S−1E E −→ D

′.

(7) If D′ 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 followingCategories, 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 thatS−1E E → S−1D is exact follows from the fact that a triangle in S−1

E E is distinguishedif and only if it is isomorphic to the image of a distinguished triangle in E , seeproof of Proposition 5.5. Part (5) follows from Lemma 14.4. The factorization ofRF : E → D′ through an exact functor S−1

E 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 sub-category of S−1D and is an exact functor on this triangulated category. Picture:

D

Q

F// D′

S−1D S−1E E

fully faithfulexact

ooRF

<<

Definition 14.9.05SV In Situation 14.1. We say F is right derivable, or that RFeverywhere 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 −→ D′, (resp. LF : S−1D −→ D′),

see Proposition 14.8. In most interesting situations it is not the case that RF Qis 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 weonly know how to prove F is right derivable by showing that RF can be computedby evaluating F at judiciously chosen objects of the triangulated category D. Thiswarrants a definition.

Definition 14.10.05SX 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.

Lemma 14.11.05SY Assumptions and notation as in Situation 14.1. Let X be anobject 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

Lemma 14.12.05SZ 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 forLF .

Proof. By Lemma 14.6 we know that RF is defined at Z and that RF appliedto the triangle produces a distinguished triangle. Consider the morphism of distin-guished 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.

Lemma 14.13.05T0 Assumptions and notation as in Situation 14.1. Let X,Y beobjects of D. If X ⊕ Y computes RF , then X and Y compute RF . Similarly forLF .

Proof. If X ⊕Y computes RF , then RF (X ⊕Y ) = F (X)⊕F (Y ). In the proof ofLemma 14.7 we have seen that the functor X/S×Y/S → (X⊕Y )/S, (s, s′) 7→ s⊕s′is cofinal. We will use this without further mention. Let s : X → X ′ be an elementof S. Then F (X)→ F (X ′) has a section, namely,F (X ′)→ F (X ′ ⊕ Y )→ RF (X ′ ⊕ Y ) = RF (X ⊕ Y ) = F (X)⊕ F (Y )→ F (X).

where we have used Lemma 14.4. Hence F (X ′) = F (X)⊕ E for some object E ofD′ such that E → F (X ′⊕Y )→ RF (X ′⊕Y ) = RF (X ⊕Y ) is zero (Lemma 4.12).Because RF is defined at X ′⊕Y with value F (X)⊕F (Y ) we can find a morphismt : X ′ ⊕ Y → Z of S such that F (t) annihilates E. We may assume Z = X ′′ ⊕ Y ′′and t = t′ ⊕ t′′ with t′, t′′ ∈ S. Then F (t′) annihilates E. It follows that F isessentially constant on X/S with value F (X) as desired.

Lemma 14.14.05T1 Assumptions and notation as in Situation 14.1.(1) If for every object X ∈ Ob(D) there exists an arrow s : X → X ′ in S such

that X ′ computes RF , then RF is everywhere defined.(2) If for every object X ∈ Ob(D) there exists an arrow s : X ′ → X in S such

that X ′ computes LF , then LF is everywhere defined.

Proof. This is clear from the definitions.

Lemma 14.15.06XN Assumptions and notation as in Situation 14.1. If there exists asubset I ⊂ Ob(D) such that

(1) for all X ∈ Ob(D) there exists s : X → X ′ in S with X ′ ∈ I, and(2) for every arrow s : X → X ′ in S with X,X ′ ∈ I the map F (s) : F (X) →

F (X ′) is an isomorphism,then RF is everywhere defined and every X ∈ I computes RF . Dually, if thereexists a subset P ⊂ Ob(D) such that

(1) for all X ∈ Ob(D) there exists s : X ′ → X in S with X ′ ∈ P, and(2) for every arrow s : X → X ′ in S with X,X ′ ∈ P the map F (s) : F (X) →

F (X ′) 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 →X ′ in S with X ′ ∈ I are cofinal in the category X/S. Assumption (2) implies thatF is constant on this cofinal subcategory. Clearly this implies that F : (X/S)→ D′is essentially constant with value F (X ′) for any s : X → X ′ in S with X ′ ∈ I.

Lemma 14.16.05T2 Let A,B, C be triangulated categories. Let S, resp. S′ be a satu-rated multiplicative system in A, resp. B compatible with the triangulated structure.Let F : A → B and G : B → C be exact functors. Denote F ′ : A → (S′)−1B thecomposition of F with the localization functor.

(1) If RF ′, RG, R(G F ) are everywhere defined, then there is a canonicaltransformation of functors t : R(G F ) −→ RG RF ′.

(2) If LF ′, LG, L(G F ) are everywhere defined, then there is a canonicaltransformation of functors t : LG LF ′ → L(G F ).

Proof. In this proof we try to be careful. Hence let us think of the derived functorsas the functors

RF ′ : S−1A → (S′)−1B, R(G F ) : S−1A → C, RG : (S′)−1B → C.

Let us denote QA : A → S−1A and QB : B → (S′)−1B the localization functors.Then F ′ = QB F . Note that for every object Y of B there is a canonical map

G(Y ) −→ RG(QB(Y ))in other words, there is a transformation of functors t′ : G→ RG QB . Let X bean object of A. We have

R(G F )(QA(X)) = colims:X→X′∈S G(F (X ′))t′−→ colims:X→X′∈S RG(QB(F (X ′)))= colims:X→X′∈S RG(F ′(X ′))= RG(colims:X→X′∈S F

′(X ′))= RG(RF ′(X)).

The system F ′(X ′) is essentially constant in the category (S′)−1B. Hence we maypull 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 trans-formation 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 functorbetween abelian categories.

Situation 15.1.05T4 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 multiplicativesystem Qis(A).

DERIVED CATEGORIES 53

(2) The right derived functor of F : K+(A) → D+(B) relative to the multi-plicative system Qis+(A).

(3) The left derived functor of F : K(A)→ D(B) relative to the multiplicativesystem Qis(A).

(4) The left derived functor of F : K−(A)→ D−(B) relative to the multiplica-tive 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.

Lemma 15.2.05T5 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 functorof K(A) → D(B) if and only if X computes the right derived functor ofK+(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 functorof K(A) → D(B) if and only if X computes the left derived functor ofK−(A)→ D−(B).

Proof. LetX be an object ofK+(A). Consider a quasi-isomorphism s : X → X ′ inK(A). By Lemma 11.5 there exists quasi-isomorphism X ′ → X ′′ with X ′′ boundedbelow. 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 toparts (1) and (2).

Given an object A of an abelian category A we get a complexA[0] = (. . .→ 0→ A→ 0→ . . .)

where A is placed in degree zero. Hence a functor A → K(A), A 7→ A[0]. Let ustemporarily say that a partial functor is one that is defined on a subcategory.

Definition 15.3.0157 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.

Lemma 15.4.05T7 Let A be an abelian category. Let P ⊂ Ob(A) be a subset containing0 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 Pn ∈ P and Pn → Kn surjective for all n and Pn = 0 for n > a.

DERIVED CATEGORIES 54

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

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

Pn

α

// Pn+1

α

// Pn+2

α

// . . .

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

is commutative, such that dj+1dj = 0 for j ≥ n, such that α induces isomorphismsHj(K•) → Ker(dj)/ Im(dj−1) for j > n, and such that α : Ker(dn) → Ker(dnK) issurjective. Then we choose a surjection

Pn−1 −→ Kn−1 ×Kn Ker(dn) = Kn−1 ×Ker(dnK

) Ker(dn)

with Pn−1 in P. This allows us to extend the diagram above to

Pn−1

α

// Pn

α

// Pn+1

α

// Pn+2

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+ 1since we can just take P j = 0 for j > a. Hence we see that proceeding by descendinginduction we produce a complex P • with Pn = 0 for n > a consisting of objectsfrom 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.

Lemma 15.5.05T6 Let A be an abelian category. Let I ⊂ Ob(A) be a subset containing0 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.

Lemma 15.6.05T8 In Situation 15.1. Let I ⊂ Ob(A) be a subset with the followingproperties:

(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 aquasi-isomorphism with K• bounded below. Then we can find a quasi-isomorphismK• → I• with I• bounded below and each In ∈ I, see Lemma 15.5. Hence wesee that these resolutions are cofinal in the category A[0]/Qis+(A). To finish theproof it therefore suffices to show that for any quasi-isomorphism A[0] → I• withI• bounded above and In ∈ I we have F (A)[0] → F (I•) is a quasi-isomorphism.To see this suppose that In = 0 for n < n0. Of course we may assume that n0 < 0.Starting with n = n0 we prove inductively that Im(dn−1) = Ker(dn) and Im(d−1)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 thatI0/ 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 thenon one continues as before to show that Im(dn−1) = Ker(dn) is an element of I forall n > 0. Applying F to each of the short exact sequences mentioned above andusing (2) we observe that F (A)[0]→ F (I•) is an isomorphism as desired.

Lemma 15.7.05T9 In Situation 15.1. Let P ⊂ Ob(A) be a subset with the followingproperties:

(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”.

Lemma 16.1.05TC 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•)) = 0for all i < a.

(2) If RF is defined at K• and τ≤aK•, then Hi(RF (τ≤aK•)) = Hi(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 ourassumption on K•. Hence in the category K•/Qis+(A) the quasi-isomorphismss : K• → L• with Ln = 0 for n < a are cofinal. Thus RF is the value of theessentially constant ind-object F (L•) for these s it follows that Hi(RF (K•)) = 0for i < 0.

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 trianglethen shows what we want.

Definition 16.2.015A Let F : A → B be an additive functor between abelian cate-gories. Assume RF : D+(A) → D+(B) is everywhere defined. Let i ∈ Z. The ithright 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 theright derived functor unless the functor is left exact.

Lemma 16.3.05TD Let F : A → B be an additive functor between abelian categoriesand 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. Thenit is also true that A[0] → τ≥0K

• is a quasi-isomorphism. Hence in the categoryA[0]/Qis+(A) the quasi-isomorphisms s : A[0] → K• with Kn = 0 for n < 0 arecofinal. Thus it is clear that Hi(RF (A[0])) = 0 for i < 0. Moreover, for such an sthe sequence

0→ A→ K0 → K1

is exact. Hence if F is left exact, then 0 → F (A) → F (K0) → F (K1) is exact aswell, 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 weobtain a distinguished triangle (A[0], B[0], C[0], a, b, c) in K+(A). From the longexact cohomology sequence (and the vanishing for i < 0 proved above) we deducethat 0 → R0F (A) → R0F (B) → R0F (C) is exact. Hence R0F is left exact. Ofcourse this also proves that if F → R0F is an isomorphism, then F is left exact.

Lemma 16.4.015C Let F : A → B be an additive functor between abelian categoriesand 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 isomorphismand 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) = 0for all i > 0.

Proof. If A is right acyclic for F , then RF (A[0]) = F (A)[0] and in particularF (A) → R0F (A) is an isomorphism and RiF (A) = 0 for i 6= 0. Conversely, ifF (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.If F is left exact then F = R0F , see Lemma 16.3.

DERIVED CATEGORIES 57

Lemma 16.5.015D Let F : A → B be a left exact functor between abelian categories andassume RF : D+(A) → D+(B) is everywhere defined. Let 0 → A → B → C → 0be 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 ofacyclic objects in Lemma 16.4.

Lemma 16.6.05TE Let F : A → B be an additive functor between abelian categoriesand assume RF : D+(A)→ D+(B) is everywhere defined.

(1) The functors RiF , i ≥ 0 come equipped with a canonical structure of aδ-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 , thenRiF, δi≥0 is a universal δ-functor, see Homology, Definition 12.3.

Proof. The functorA → 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 ex-act. Finally, the functor H0 : D+(B) → B is a homological functor, see Definition11.3. 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 inLemma 16.4.

Lemma 16.7 (Leray’s acyclicity lemma).015E Let F : A → B be an additive func-tor between abelian categories and assume RF : D+(A) → D+(B) is everywheredefined. Let A• be a bounded below complex of F -acyclic objects. The canonicalmap

F (A•) −→ RF (A•)is an isomorphism in D+(B), i.e., A• computes RF .

Proof. First we claim the lemma holds for a bounded complex of acyclic objects.Namely, it holds for complexes with at most one nonzero object by definition.Suppose that A• is a complex with 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

see Homology, Section 15. Thus a distinguished triangle (σ≥a+1A•, A•, σ≤aA

•). Byinduction hypothesis the two outer complexes compute RF . Then the middle onedoes too by Lemma 14.12.

DERIVED CATEGORIES 58

Suppose that A• is a bounded below complex of acyclic objects. To show thatF (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 shortexact 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

•)(σ≥i+2F (A•), F (A•), σ≤i+1F (A•))

(RF (σ≥i+2A•), RF (A•), RF (σ≤i+1A

•))Using the last two we obtain a map of exact sequences

Hi(σ≥i+2F (A•)) //

Hi(F (A•)) //

α

Hi(σ≤i+1F (A•)) //

β

Hi+1(σ≥i+2F (A•))

Hi(RF (σ≥i+2A

•)) // Hi(RF (A•)) // Hi(RF (σ≤i+1A•)) // Hi+1(RF (σ≥i+2A

•))

By the results of the first paragraph the map β is an isomorphism. By inspectionthe objects on the upper left and the upper right are zero. Hence to finish the proofit suffices to show that Hi(RF (σ≥i+2A

•)) = 0 and Hi+1(RF (σ≥i+2A•)) = 0. This

follows immediately from Lemma 16.1.

Proposition 16.8.05TA 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 functorRF : D+(A) −→ D+(B)

see (14.9.1). Moreover, any bounded below complex A• whose terms areacyclic for RF computes RF .

(2) If every object of A is quotient of an object acyclic for LF , then LF isdefined 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 areacyclic for LF computes LF .

Proof. Assume every object of A injects into an object acyclic for RF . Let I bethe set of objects acyclic for RF . Let K• be a bounded below complex in A. ByLemma 15.5 there exists a quasi-isomorphism α : K• → I• with I• bounded belowand In ∈ I. Hence in order to prove (1) it suffices to show that F (I•)→ F ((I ′)•) isa quasi-isomorphism when s : I• → (I ′)• is a quasi-isomorphism of bounded belowcomplexes of objects from I, see Lemma 14.15. Note that the cone C(s)• is anacyclic bounded below complex all of whose terms are in I. Hence it suffices toshow: given an acyclic bounded below complex I• all of whose terms are in I thecomplex F (I•) is acyclic.Say In = 0 for n < n0. Setting Jn = Im(dn) we break I• into short exact se-quences 0 → Jn → In+1 → Jn+1 → 0 for n ≥ n0. These sequences induce

DERIVED CATEGORIES 59

distinguished triangles (Jn, In+1, Jn+1) in D+(A) by Lemma 12.1. For each k ∈ Zdenote Hk the assertion: For all n ≤ k the right derived functor RF is definedat Jn and RiF (Jn) = 0 for i 6= 0. Then Hk holds trivially for k ≤ n0. IfHn holds, then, using Proposition 14.8, we see that RF is defined at Jn+1 and(RF (Jn), RF (In+1), RF (Jn+1)) is a distinguished triangle of D+(B). Thus thelong exact cohomology sequence (11.1.1) associated to this triangle gives an exactsequence

0→ R−1F (Jn+1)→ R0F (Jn)→ F (In+1)→ R0F (Jn+1)→ 0and gives that RiF (Jn+1) = 0 for i 6∈ −1, 0. By Lemma 16.1 we see thatR−1F (Jn+1) = 0. This proves that Hn+1 is true hence Hk holds for all k. We alsoconclude that

0→ R0F (Jn)→ F (In+1)→ R0F (Jn+1)→ 0is short exact for all n. This in turn proves that F (I•) is exact.The proof in the case of LF is dual.

Lemma 16.9.015F 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 complexesand quasi-isomorphisms into quasi-isomorphisms. Details omitted.

17. Triangulated subcategories of the derived category

06UP LetA be an abelian category. In this section we look at certain strictly full saturatedtriangulated subcategories D′ ⊂ D(A).Let B ⊂ A be a weak Serre subcategory, see Homology, Definition 10.1 and Lemma10.3. We let DB(A) the full subcategory of D(A) whose objects are

Ob(DB(A)) = X ∈ Ob(D(A)) | Hn(X) is an object of B for all n

We also define D+B (A) = D+(A) ∩ DB(A) and similarly for the other bounded

versions.

Lemma 17.1.06UQ Let A be an abelian category. Let B ⊂ A be a weak Serre subcat-egory. The category DB(A) is a strictly full saturated triangulated subcategory ofD(A). Similarly for the bounded versions.

Proof. It is clear that DB(A) is an additive subcategory preserved under the trans-lation functors. If X ⊕ Y is in DB(A), then both Hn(X) and Hn(Y ) are kernelsof maps between maps of objects of B as Hn(X ⊕ Y ) = Hn(X) ⊕Hn(Y ). Henceboth X and Y are in DB(A). By Lemma 4.16 it therefore suffices to show thatgiven 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) andthe definition of a weak Serre subcategory (see Homology, Definition 10.1) showthat Hn(Z) is an object of B for all n. Thus Z is an object of DB(A).

DERIVED CATEGORIES 60

We continue to assume that B is a weak Serre subcategory of the abelian categoryA. Then B is an abelian category and the inclusion functor B → A is exact. Hencewe obtain a derived functor D(B) → D(A), see Lemma 16.9. Clearly the functorD(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 ofDB(A)and as distinguished triangles in DB(A) are exactly the distinguished triangles ofD(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) → D+

B (A), D−(B) → D−B (A),and Db(B) → Db

B(A) for the bounded versions. A key question in many cases iswhether the displayed functor is an equivalence.

Now, suppose that B is a Serre subcategory of A. In this case we have the quotientfunctor A → A/B, see Homology, Lemma 10.6. In this case DB(A) is the kernel ofthe 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.

Lemma 17.2.06XL 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 objectof 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 Awhose kernel and cokernel are in B (equivalently si becomes an isomorphism in thequotient category) and f i : Y i → Xi+1 is a morphism of A. By induction we willconstruct a commutative diagram

(X ′)1 // (X ′)2 // . . .

X0

<<

X1

OO

X2

OO

. . .

Y 0

s0

OO

f0

<<

Y 1

s1

OO

f1

::

Y 2

s2

OO

f2

<<

. . .

where the vertical arrows Xi → (X ′)i become isomorphisms in the quotient cate-gory. Namely, we first let (X ′)1 = Coker(Y 0 → X0 ⊕X1) (or rather the pushoutof the diagram with arrows s0 and f0) which gives the first commutative diagram.Next, we take (X ′)2 = Coker(Y 1 → (X ′)1 ⊕X2). And so on. Setting additionally(X ′)n = Xn for n ≤ 0 we see that the map (X•, d•)→ ((X ′)•, (d′)•) is an isomor-phism of complexes in A/B. Hence we may assume dn : Xn → Xn+1 is given by amap 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 anisomorphism in the quotient category and gi : Xi → Zi+1 is a morphism. By

DERIVED CATEGORIES 61

induction we will construct a commutative diagram

. . . Z−2 Z−1 Z0

. . . X−2

t−2

OO

g−2

99

X−1

t−1

OO

g−1

;;

X0

t0

OO

. . . (X ′)−2

OO

// (X ′)−1

OO ;;

where the vertical arrows (X ′)i → Xi become isomorphisms in the quotient cate-gory. Namely, we take (X ′)−1 = X−1 ×Z0 X0. Then we take (X ′)−2 = X−2 ×Z−1

(X ′)−1. And so on. Setting additionally (X ′)n = Xn for n ≥ 0 we see that themap ((X ′)•, (d′)•) → (X•, d•) is an isomorphism of complexes in A/B. Hence wemay assume dn : Xn → Xn+1 is given by a map dn : Xn → Xn+1 in A for alln ∈ Z.

In this case we know the compositions dn dn−1 are zero in A/B. If for n > 0 wereplace Xn by

(X ′)n = Xn/∑

0<k≤nIm(Im(Xk−2 → Xk)→ Xn)

then the compositions dn dn−1 are zero for n ≥ 0. (Similarly to the secondparagraph above we obtain an isomorphism of complexes (X•, d•)→ ((X ′)•, (d′)•).)Finally, for n < 0 we replace Xn by

(X ′)n =⋂

n≤k<0(Xn → Xk)−1 Ker(Xk → Xk+2)

and we argue in the same manner to get a complex in A whose image in A/B isisomorphic to the given one.

Lemma 17.3.06XM Let A be an abelian category. Let B ⊂ A be a Serre subcategory.Suppose that the functor v : A → A/B has a left adjoint u : A/B → A such thatvu ∼= id. Then

D(A)/DB(A) = D(A/B)and similarly for the bounded versions.

Proof. The functor D(v) : D(A) → D(A/B) is essentially surjective by Lemma17.2. For an object X of D(A) the adjunction mapping cX : uvX → X maps to anisomorphism in D(A/B) because vuv ∼= v by the assumption that vu ∼= id. Thus ina distinguished triangle (uvX,X,Z, cX , g, h) the object Z is an object of DB(A) aswe see by looking at the long exact cohomology sequence. Hence cX is an element ofthe multiplicative system used to define the quotient category D(A)/DB(A). ThusuvX ∼= X in D(A)/DB(A). For X,Y ∈ Ob(A)) the map

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

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.

DERIVED CATEGORIES 62

Lemma 17.4.0FCL 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) thereexists X ′ ⊂ X, X ′ ∈ Ob(B) which surjects onto Y . Then the functor D−(B) →D−B (A) of (17.1.1) is an equivalence.

Proof. Let X• be a bounded above complex of A such that Hi(X•) ∈ Ob(B) forall 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) anddi(Ei) ⊂ Di+1 +Ei+1 we see that setting Y i = Di +Ei gives a subcomplex whoseterms 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 isalso injective. For n 0 we can take En equal to 0. By descending induction wecan choose Ei for all i with the desired property. Namely, given Ei+1, Ei+2, . . . wechoose Ei ⊂ Xi such that di(Ei) = (Di+1 +Ei+1)∩ Im(di). This is possible by ourassumption in the lemma combined with the fact that (Di+1 +Ei+1)∩ Im(di) is inB as B is a Serre subcategory of A.

The claim above implies the lemma. Essential surjectivity is immediate from theclaim. 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 existsa quasi-isomorphism s : V • → X• into a bounded above complex of A such thatsf is homotopic to zero. Choose a homotopy hi : U i → Xi−1 between 0 and sf .Apply the claim with Bi = hi+1(U i+1) + si(V i). The resulting map s′ : V • → Y •

is a quasi-isomorphism as well and s′ f is homotopic to zero as is clear from thefact that hi factors through Y i−1. This proves faithfulness. Fully faithfulness isproved in the exact same manner.

18. Injective resolutions

013G In this section we prove some lemmas regarding the existence of injective resolutionsin abelian categories having enough injectives.

Definition 18.1.013I Let A be an abelian category. Let A ∈ Ob(A). An injectiveresolution 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.

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 complexI• together with a map α : K• → I• of complexes such that

DERIVED CATEGORIES 63

(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 injectiveresolution of an object A of A is almost the same thing as an injective resolutionof the complex A[0].

Lemma 18.2.013J 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.

Lemma 18.3.013K 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 objectof A. Next, choose an injection I0/A → I1 into an injective object of A. Denoted0 the induced map I0 → I1. Next, choose an injection I1/ Im(d0) → I2 into aninjective object of A. Denote d1 the induced map I1 → I2. And so on. By Lemma18.2 part (2) follows from part (3). Part (3) is a special case of Lemma 15.5.

Lemma 18.4.013R Let A be an abelian category. Let K• be an acyclic complex. LetI• be bounded below and consisting of injective objects. Any morphism K• → I• ishomotopic to zero.

Proof. Let α : K• → I• be a morphism of complexes. Assume that αj = 0 forj < n. We will show that there exists a morphism h : Kn+1 → In such thatαn = h d. Thus α will be homotopic to the morphism of complexes β defined by

βj =

0 if j ≤ nαn+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 notethat αn|dn−1(Kn−1) = 0 since αn−1 = 0. Since K• is acyclic we have dn−1(Kn−1) =Ker(Kn → Kn+1). Hence we can think of αn as a map into In defined on thesubobject Im(Kn → Kn+1) of Kn+1. By injectivity of the object In we can extendthis to a map h : Kn+1 → In as desired.

DERIVED CATEGORIES 64

Remark 18.5.05TF Let A be an abelian category. Using the fact that K(A) is atriangulated category we may use Lemma 18.4 to obtain proofs of some of thelemmas 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•)

rrHomK(A)(C(α)•[−1], I•)

is an exact sequence of abelian groups, see Lemma 4.2. At this point Lemma18.4 guarantees that the outer two groups are zero and hence HomK(A)(L•, I•) =HomK(A)(K•, I•).

Lemma 18.6.013P Let A be an abelian category. Consider a solid diagram

K•α//

γ

L•

βI•

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 tohomotopy.

(2) If α is injective in every degree then we can find a β which makes thediagram commute.

Proof. The “correct” proof of part (1) is explained in Remark 18.5. We also givea direct proof here.We first show that (2) implies (1). Namely, let α : K → L•, π, s be as in Lemma9.6. Since α is injective by (2) there exists a morphism β : L• → I• such thatγ = 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 alldegrees ≤ n − 1 compatible with differentials and such that γj = βj αj for allj ≤ n− 1. Consider the commutative solid diagram

Kn−1 //

γ

α

Kn

γ

α

Ln−1 //

β

Ln

In−1 // In

DERIVED CATEGORIES 65

Thus we see that the dotted arrow is prescribed on the subobjects α(Kn) anddn−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, usingthe 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 thatwe can set βn = 0 for all n 0 since I• is bounded below – in this way startingthe induction).It remains to prove the equality (18.6.1). The reader is encouraged to argue thisfor 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. Takean object T of A and a morphism x : T → Ln whose image is contained in thesubobject α(Kn)∩ dn−1(Ln−1). Since α is injective we see that x = α x′ for somex′ : T → Kn. Moreover, since x lies in dn−1(Ln−1) we see that dn x = 0. Henceusing injectivity of α again we see that dn x′ = 0. Thus x′ gives a morphism[x′] : 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 concludethat [x′] = 0. This of course means exactly that the image of x′ is contained indn−1(Kn−1) and we win.

Lemma 18.7.013S Let A be an abelian category. Consider a solid diagram

K•α//

γ

L•

βiI•

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 ho-motopy 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 homotopicto β2 π, then we deduce that β1 ∼ β2 because π s is the identity. Hence we mayassume αn : Kn → Ln is the inclusion of a direct summand for all n. Thus we geta short exact sequence of complexes

0→ K• → L• →M• → 0which is termwise split and such that M• is acyclic. We choose splittings Ln =Kn ⊕Mn, so we have βni : Kn ⊕Mn → In and γn : Kn → In. In this case thecondition on βi is that there are morphisms hni : Kn → In−1 such that

γn − βni |Kn = d hni + hn+1i d

Thus we see thatβn1 |Kn − βn2 |Kn = d (hn1 − hn2 ) + (hn+1

1 − hn+12 ) d

Consider the map hn : Kn⊕Mn → In−1 which equals hn1−hn2 on the first summandand zero on the second. Then we see that

βn1 − βn2 − (d hn + hn+1) d

DERIVED CATEGORIES 66

is a morphism of complexes L• → I• which is identically zero on the subcomplexK•. Hence it factors as L• →M• → I•. Thus the result of the lemma follows fromLemma 18.4.

Lemma 18.8.05TG Let A be an abelian category. Let I• be bounded below complexconsisting 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 homotopicto γ. This proves that the map is surjective. Let b be an element of the left handside which maps to zero in the right hand side. Then b is the homotopy class ofa 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 zeroalso, i.e., b = 0.

Lemma 18.9.013T 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 existsa commutative diagram in Comp+(A)

0 // A• //

B• //

C• //

0

0 // I•1 // I•2 // I•3 // 0

where the vertical arrows are injective resolutions and the rows are short exactsequences of complexes. In fact, given any injective resolution A• → I• we mayassume I•1 = I•.

Proof. Step 1. Choose an injective resolution A• → I• (see Lemma 18.3) or usethe given one. Recall that Comp+(A) is an abelian category, see Homology, Lemma13.9. Hence we may form the pushout along the injective map A• → I• to get

0 // A• //

B• //

C• //

0

0 // I• // E• // C• // 0Note that the lower short exact sequence is termwise split, see Homology, Lemma27.2. Hence it suffices to prove the lemma when 0 → A• → B• → C• → 0 istermwise 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 resolutionsf1 : A• → I•1 and f3 : C• → I•3 . (If A• is a complex of injectives, then useI•1 = A•.) We may assume f3 is injective in every degree. By Lemma 18.6 we mayfind a morphism δ′ : I•3 → I•1 [1] such that δ′ f3 = f1[1] δ (equality of morphismsof complexes). Set In2 = In1 ⊕ In3 . Define

dnI2=(dnI1

(δ′)n0 dnI3

)

DERIVED CATEGORIES 67

and define the maps Bn → In2 to be given as the sum of the maps An → In1 andCn → In3 . Everything is clear.

19. Projective resolutions

0643 This section is dual to Section 18. We give definitions and state results, but we donot reprove the lemmas.

Definition 19.1.0644 Let A be an abelian category. Let A ∈ Ob(A). An projectiveresolution of A is a complex P • together with a map P 0 → A such that:

(1) We have Pn = 0 for n > 0.(2) Each Pn 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 complexP • together with a map α : P • → K• of complexes such that

(1) We have Pn = 0 for n 0, i.e., P • is bounded above.(2) Each Pn is an projective object of A.(3) The map α : P • → K• is a quasi-isomorphism.

Lemma 19.2.0645 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.

Lemma 19.3.0646 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 Pn = 0 for n > a such that each αn : Pn →Kn is surjective.

Proof. Dual to Lemma 18.3.

Lemma 19.4.0647 Let A be an abelian category. Let K• be an acyclic complex. LetP • be bounded above and consisting of projective objects. Any morphism P • → K•

is homotopic to zero.

Proof. Dual to Lemma 18.4.

Remark 19.5.0648 LetA 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.

DERIVED CATEGORIES 68

Lemma 19.6.0649 Let A be an abelian category. Consider a solid diagram

K• L•αoo

P •

OO

β

==

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 tohomotopy.

(2) If α is surjective in every degree then we can find a β which makes thediagram commute.

Proof. Dual to Lemma 18.6.

Lemma 19.7.064A Let A be an abelian category. Consider a solid diagram

K• L•αoo

P •

OO

βi

==

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 ho-motopy are homotopic.

Proof. Dual to Lemma 18.7.

Lemma 19.8.064B Let A be an abelian category. Let P • be bounded above complexconsisting of projective objects. Let L• ∈ K(A). Then

MorK(A)(P •, L•) = MorD(A)(P •, L•).

Proof. Dual to Lemma 18.8.

Lemma 19.9.064C 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 existsa commutative diagram in Comp+(A)

0 // P •1 //

P •2 //

P •3 //

0

0 // A• // B• // C• // 0where the vertical arrows are projective resolutions and the rows are short exactsequences of complexes. In fact, given any projective resolution P • → C• we mayassume P •3 = P •.

Proof. Dual to Lemma 18.9.

Lemma 19.10.064D Let A be an abelian category. Let P •, K• be complexes. Letn ∈ 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.

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Then HomK(A)(P •,K•) = HomD(A)(P •,K•) = 0.

Proof. The first equality follows from Lemma 19.8. Note that there is a distin-guished 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 isLemma 19.4.

Lemma 19.11.064E 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 arehomotopic.

Proof. Consider the cone C• = C(α)• with map i : L• → C•. Note that i β iszero 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 functorsof an additive functor between an abelian category having enough injectives and ageneral abelian category.

Lemma 20.1.05TH 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)→ Dinto any triangulated category D.

(2) I is right acyclic for any additive functor F : A → B into any abeliancategory 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 wesee that α has a left inverse. Hence the category I•/Qis+(A) is essentially constantwith 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 and14.10.

Lemma 20.2.05TI Let A be an abelian category with enough injectives.(1) For any exact functor F : K+(A) → D into a triangulated category D the

right derived functorRF : D+(A) −→ D

is everywhere defined.(2) For any additive functor F : A → B into an abelian category B the right

derived functorRF : D+(A) −→ D+(B)

is everywhere defined.

DERIVED CATEGORIES 70

Proof. Combine Lemma 20.1 and Proposition 16.8 for the second assertion. To seethe first assertion combine Lemma 18.3, Lemma 20.1, Lemma 14.14, and Equation(14.9.1).

Lemma 20.3.0159 Let A be an abelian category with enough injectives. Let F : A → Bbe 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 thatby 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.

Lemma 20.4.015B Let A be an abelian category with enough injectives. Let F : A → Bbe a left exact functor.

(1) For any short exact sequence 0 → A• → B• → C• → 0 of complexes inComp+(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 thatby 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 co-homology 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.

21. Cartan-Eilenberg resolutions

015G This section can be expanded. The material can be generalized and applied inmore cases. Resolutions need not use injectives and the method also works in theunbounded case in some situations.

Definition 21.1.015H Let A be an abelian category. Let K• be a bounded belowcomplex. A Cartan-Eilenberg resolution of K• is given by a double complex I•,•and a morphism of complexes ε : 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.

DERIVED CATEGORIES 71

(2) We have Ip,q = 0 if q < 0.(3) The complex Ip,• is an injective resolution of Kp.(4) The complex Ker(dp,•1 ) is an injective resolution of Ker(dpK).(5) The complex Im(dp,•1 ) is an injective resolution of Im(dpK).(6) The complex Hp

I (I•,•) is an injective resolution of Hp(K•).

Lemma 21.2.015I Let A be an abelian category with enough injectives. Let K• be abounded below complex. There exists a Cartan-Eilenberg resolution of K•.

Proof. Suppose that Kp = 0 for p < n. Decompose K• into short exact sequencesas follows: Set Zp = Ker(dp), Bp = Im(dp−1), Hp = Zp/Bp, and consider

0→ Zn → Kn → Bn+1 → 00→ Bn+1 → Zn+1 → Hn+1 → 00→ Zn+1 → Kn+1 → Bn+2 → 00→ Bn+2 → Zn+2 → Hn+2 → 0

. . .

Set Ip,q = 0 for p < n. Inductively we choose injective resolutions as follows:(1) Choose an injective resolution Zn → Jn,•Z .(2) Using Lemma 18.9 choose injective resolutions Kn → In,•, Bn+1 → Jn+1,•

B ,and an exact sequence of complexes 0 → Jn,•Z → In,• → Jn+1,•

B → 0compatible with the short exact sequence 0→ Zn → Kn → Bn+1 → 0.

(3) Using Lemma 18.9 choose injective resolutions Zn+1 → Jn+1,•Z , Hn+1 →

Jn+1,•H , and an exact sequence of complexes 0 → Jn+1,•

B → Jn+1,•Z →

Jn+1,•H → 0 compatible with the short exact sequence 0→ Bn+1 → Zn+1 →Hn+1 → 0.

(4) Etc.Taking as maps d•1 : Ip,• → Ip+1,• the compositions Ip,• → Jp+1,•

B → Jp+1,•Z →

Ip+1,• everything is clear.

Lemma 21.3.015J Let F : A → B be a left exact functor of abelian categories. LetK• be a bounded below complex of A. Let I•,• be a Cartan-Eilenberg resolution forK•. The spectral sequences (′Er, ′dr)r≥0 and (′′Er, ′′dr)r≥0 associated to the doublecomplex F (I•,•) satisfy the relations

′Ep,q1 = RqF (Kp) and ′′Ep,q2 = RpF (Hq(K•))

Moreover, these spectral sequences are bounded, converge to H∗(RF (K•)), and theassociated induced filtrations on Hn(RF (K•)) are finite.

Proof. We will use the following remarks without further mention:(1) As Ip,• is an injective resolution of Kp we see that RF is defined at Kp[0]

with value F (Ip,•).(2) As Hp

I (I•,•) is an injective resolution of Hp(K•) the derived functor RF isdefined at Hp(K•)[0] with value F (Hp

I (I•,•)).(3) By Homology, Lemma 25.4 the total complex Tot(I•,•) is an injective res-

olution 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 toH∗(Tot(L•,•)),and induce finite filtrations on Hn(Tot(L•,•)), see Homology, Lemma 25.3. Since

DERIVED CATEGORIES 72

Tot(L•,•) = Tot(F (I•,•)) = F (Tot(I•,•)) computes Hn(RF (K•)) we find the finalassertion of the lemma holds true.Computation of the first spectral sequence. We have ′Ep,q1 = Hq(Lp,•) in otherwords

′Ep,q1 = Hq(F (Ip,•)) = RqF (Kp)as desired. Observe for later use that the maps ′dp,q1 : ′Ep,q1 → ′Ep+1,q

1 are themaps RqF (Kp)→ RqF (Kp+1) induced by Kp → Kp+1 and the fact that RqF is afunctor.Computation of the second spectral sequence. We have ′′Ep,q1 = Hq(L•,p) =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 aninjective object of A. Hence we can split the differentials, i.e., each differential isa split surjection onto a direct summand. It follows that the same is true afterapplying F . Hence ′′Ep,q1 = F (Hq(I•,p)) = F (Hq

I (I•,p)). The differentials on thisare (−1)q times F applied to the differential of the complex Hp

I (I•,•) which is aninjective resolution of Hp(K•). Hence the description of the E2 terms.

Remark 21.4.015K The spectral sequences of Lemma 21.3 are functorial in the complexK•. This follows from functoriality properties of Cartan-Eilenberg resolutions. Onthe other hand, they are both examples of a more general spectral sequence whichmay be associated to a filtered complex of A. The functoriality will follow from itsconstruction. 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. Supposethat A,B, C be abelian categories. Let F : A → B and G : B → C be left exactfunctors. 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. Thenthere exists a canonical transformation

t : R(G F ) −→ RG RFof functors from D+(A) to D+(C), see Lemma 14.16. This transformation need notalways be an isomorphism.

Lemma 22.1.015M Let A,B, C be abelian categories. Let F : A → B and G : B → Cbe left exact functors. Assume A, B have enough injectives. The following areequivalent

(1) F (I) is right acyclic for G for each injective object I of A, and(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 Lemma14.16) by the maps

R(G F )(A•) = (G F )(I•) = G(F (I•)))→ RG(F (I•)) = RG(RF (A•))

DERIVED CATEGORIES 73

where the arrow is an isomorphism by Lemma 16.7.

Lemma 22.2 (Grothendieck spectral sequence).015N With assumptions as in Lemma22.1 and assuming the equivalent conditions (1) and (2) hold. Let X be an object ofD+(A). There exists a spectral sequence (Er, dr)r≥0 consisting of bigraded objectsEr of C with dr of bidegree (r,−r + 1) and with

Ep,q2 = RpG(Hq(RF (X)))Moreover, this spectral sequence is bounded, converges to H∗(R(G F )(X)), andinduces a finite filtration on each Hn(R(G F )(X)).For an object A of A we get Ep,q2 = RpG(RqF (A)) converging to Rp+q(G F )(A).

Proof. We may represent X by a bounded below complex A•. Choose an injectiveresolution A• → I•. Choose a Cartan-Eilenberg resolution F (I•) → I•,• usingLemma 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 additivesubcategory of A whose objects are the injective objects of A. It turns out thatK+(I) and D+(A) are equivalent in this case (see Proposition 23.1). For manypurposes it therefore makes sense to think of D+(A) as the (easier to grok) categoryK+(I) in this case.Proposition 23.1.013V Let A be an abelian category. Assume A has enough injectives.Denote I ⊂ A the strictly full additive subcategory whose objects are the injectiveobjects of A. The functor

K+(I) −→ D+(A)is exact, fully faithful and essentially surjective, i.e., an equivalence of triangulatedcategories.Proof. It is clear that the functor is exact. It is essentially surjective by Lemma18.3. Fully faithfulness is a consequence of Lemma 18.8.

Proposition 23.1 implies that we can find resolution functors. It turns out that wecan prove resolution functors exist even in some cases where the abelian categoryA is a “big” category, i.e., has a class of objects.Definition 23.2.013W Let A be an abelian category with enough injectives. A resolu-tion functor5 for A is given by the following data:

(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•).

Lemma 23.3.05TJ Let A be an abelian category with enough injectives. Given aresolution functor (j, i) there is a unique way to turn j into a functor and i into a2-isomorphism producing a 2-commutative diagram

K+(A)

$$

j// K+(I)

zzD+(A)

5This is likely nonstandard terminology.

DERIVED CATEGORIES 74

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 morphismj(α) : j(K•)→ j(L•) in K+(I) such that

K•α

//

iK•

L•

iL•

j(K•)

j(α) // j(L•)

is commutative in K+(A). To see this either use Lemmas 18.6 and 18.7 or theequivalent Lemma 18.8. The uniqueness implies that j is a functor, and the com-mutativity of the diagram implies that i gives a 2-morphism which witnesses the2-commutativity of the diagram of categories in the statement of the lemma.

Lemma 23.4.013X 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 func-tors.

Proof. Consider the set of all objects K• of K+(A). (Recall that by our conven-tions any category has a set of objects unless mentioned otherwise.) By Lemma18.3 every object has an injective resolution. By the axiom of choice we can choosefor each K• an injective resolution iK• : K• → j(K•).

Lemma 23.5.014W Let A be an abelian category with enough injectives. Any resolutionfunctor j : K+(A)→ K+(I) is exact.

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

K•[1]

iK•[1]

K•[1]

iK• [1]

j(K•[1])ξK• // j(K•)[1]

By Lemmas 18.6 and 18.7 there exists a unique dotted arrow ξK• in K+(I) mak-ing the diagram commute in K+(A). We omit the verification that this gives afunctorial isomorphism. (Hint: use Lemma 18.7 again.)Let (K•, L•,M•, f, g, h) be a distinguished triangle of K+(A). We have to showthat (j(K•), j(L•), j(M•), j(f), j(g), ξK•j(h)) is a distinguished triangle ofK+(I).Note that we have a commutative diagram

K•f

//

L•g//

M•h

//

K•[1]

j(K•)

j(f) // j(L•)j(g) // j(M•)

ξK•j(h) // j(K•)[1]

in K+(A) whose vertical arrows are the quasi-isomorphisms iK , iL, iM . Hence wesee that the image of (j(K•), j(L•), j(M•), j(f), j(g), ξK• j(h)) in D+(A) is iso-morphic 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)) isa distinguished triangle in K+(I).

DERIVED CATEGORIES 75

Lemma 23.6.05TK Let A be an abelian category which has enough injectives. Let jbe a resolution functor. Write Q : K+(A)→ D+(A) for the natural functor. Thenj = j′ Q for a unique functor j′ : D+(A)→ K+(I) which is quasi-inverse to thecanonical functor K+(I)→ D+(A).

Proof. By Lemma 11.6 Q is a localization functor. To prove the existence of j′ itsuffices to show that any element of Qis+(A) is mapped to an isomorphism underthe functor j, see Lemma 5.6. This is true by the remarks following Definition23.2.

Remark 23.7.013Y Suppose that A is a “big” abelian category with enough injectivessuch as the category of abelian groups. In this case we have to be slightly morecareful in constructing our resolution functor since we cannot use the axiom ofchoice with a quantifier ranging over a class. But note that the proof of the lemmadoes show that any two localization functors are canonically isomorphic. Namely,given quasi-isomorphisms i : K• → I• and i′ : K• → J• of a bounded belowcomplex K• into bounded below complexes of injectives there exists a unique(!)morphism a : I• → J• in K+(I) such that i′ = i a as morphisms in K+(I).Hence the only issue is existence, and we will see how to deal with this in the nextsection.

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 reasonsfor 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 subcategoryof A consisting of injective objects. Consider the category InjRes(A) of arrowsα : K• → I• where K• is a bounded below complex of A, I• is a bounded belowcomplex of injectives of A and α is a quasi-isomorphism. In other words, α is aninjective 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•.

Lemma 24.1.0141 Let A be an abelian category. Assume A has functorial injectiveembeddings, 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 thenfor any object A we have 0 → A → 0 which gives a direct sum decompositionJ(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 ′(A) =Ker(J(A)→ J(0)) if needed.

DERIVED CATEGORIES 76

Let K• be a bounded below complex of A. Say Kp = 0 if p < B. We are going toconstruct 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 complexof I•,•. First we set Ip,q = 0 whenever q < 0. Next, we set Ip,0 = J(Kp) andαp : Kp → Ip,0 the functorial embedding. Since J is a functor we see that I•,0 is acomplex and that α is a morphism of complexes. Each αp is injective. And Ip,0 = 0for p < B because J(0) = 0. Next, we set Ip,1 = J(Coker(Kp → Ip,0)). Againby functoriality we see that I•,1 is a complex. And again we get that Ip,1 = 0 forp < B. It is also clear that Kp maps isomorphically onto Ker(Ip,0 → Ip,1). As ourthird 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 theconstruction above is functorial. This verification is omitted.Suppose we have a functor inj such that s inj = id. For every object K• ofComp+(A) we can write

inj(K•) = (iK• : K• → j(K•))This provides us with a resolution functor as in Definition 23.2.

Remark 24.2.05TL Suppose inj is a functor such that s inj = id as in part (2) ofLemma 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 mapinj(α) in the category InjRes(A). It induces a commutative diagram

K•α //

iK

L•

iL

j(K)•inj(α) // j(L)•

of morphisms of complexes. Hence, looking at the proof of Lemma 23.3 we see thatthe 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•))Hence we see that j matches t inj in this case, i.e., the diagram

Comp+(A)tinj

//

&&

K+(I)

K+(A)j

::

is commutative.

Remark 24.3.0142 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). Butthe proof of Lemma 24.1 does apply to Mod(OX). Thus we obtain

j : K+(Mod(OX)) −→ K+(I)

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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.

Lemma 25.1.05TN Let A be an abelian category with enough injectives Let F : A → Bbe 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

D+(A)

RF $$

j′ // K+(I)

FzzD+(B)

where j′ is the functor from Lemma 23.6.

Proof. By Lemma 20.1 we have RF (K•) = F (j(K•)).

Remark 25.2.0158 In the situation of Lemma 25.1 we see that we have actually liftedthe right derived functor to an exact functor F j′ : D+(A) → K+(B). It isoccasionally 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 enoughinjectives, then so does the category Filf (A) in some sense. One can use thisobservation to compute in the filtered derived category of A.The category Filf (A) is an example of an exact category, see Injectives, Remark9.6. 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 complexA → B → C in Filf (A) is exact if the sequence gr(A) → gr(B) → gr(C) is exactin A. This implies that A → B and B → C are strict morphisms, see Homology,Lemma 19.15.

Definition 26.1.015P Let A be an abelian category. We say an object I of Filf (A) isfiltered injective if each grp(I) is an injective object of A.

Lemma 26.2.05TP Let A be an abelian category. An object I of Filf (A) is filteredinjective if and only if there exist a ≤ b, injective objects In, a ≤ n ≤ b of A andan isomorphism I ∼=

⊕a≤n≤b In such that F pI =

⊕n≥p In.

Proof. Follows from the fact that any injection J → M of A is split if J is aninjective object. Details omitted.

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Lemma 26.3.05TQ 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 I ′ be the object of Filf (A) whose underlying object of A is F pI and withfiltration given by FnI ′ = 0 for n > p and FnI ′ = I ′ = F pI for n ≤ p. Note thatI ′ → I is a strict monomorphism too. The fact that u is a strict monomorphismimplies that F pI → A/F p+1(A) is injective, see Homology, Lemma 19.13. Choosea splitting s : A/F p+1A→ F pI in A. The induced morphism s′ : A→ I ′ is a strictmorphism of filtered objects splitting the composition I ′ → I → A. Hence we canwrite A = I ′ ⊕ Ker(s′) and I = I ′ ⊕ Ker(s′|I). Note that Ker(s′|I) → Ker(s′) is astrict monomorphism and that Ker(s′|I) is a filtered injective object. By inductionon the length of the filtration on I the map Ker(s′|I)→ Ker(s′) is a split injection.Thus we win.

Lemma 26.4.05TR Let A be an abelian category. Let u : A→ B be a strict monomor-phism of Filf (A) and f : A→ I a morphism from A into a filtered injective objectin Filf (A). Then there exists a morphism g : B → I such that f = g u.

Proof. The pushout f ′ : I → I qA B of f by u is a strict monomorphism, seeHomology, Lemma 19.10. Hence the result follows formally from Lemma 26.3.

Lemma 26.5.05TS Let A be an abelian category with enough injectives. For anyobject A of Filf (A) there exists a strict monomorphism A→ I where I is a filteredinjective object.

Proof. Pick a ≤ b such that grp(A) = 0 unless p ∈ a, a + 1, . . . , b. For eachn ∈ a, a + 1, . . . , b choose an injection un : A/Fn+1A → In with In an injectiveobject. Set I =

⊕a≤n≤b In with filtration F pI =

⊕n≥p In and set u : A→ I equal

to the direct sum of the maps un.

Lemma 26.6.05TT Let A be an abelian category with enough injectives. For anyobject A of Filf (A) there exists a filtered quasi-isomorphism A[0]→ I• where I• isa complex of filtered injective objects with In = 0 for n < 0.

Proof. First choose a strict monomorphism u0 : A→ I0 of A into a filtered injec-tive 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,choose a strict monomorphism u2 : Coker(u1) → I2 into a filtered injective objectof A. Denote d1 the induced map I1 → I2. And so on. This works because each ofthe sequences

0→ Coker(un)→ In+1 → Coker(un+1)→ 0

is short exact, i.e., induces a short exact sequence on applying gr. To see this useHomology, Lemma 19.13.

Lemma 26.7.05TU 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• andB[0]→ J• where I•, J• are complexes of filtered injective objects with In = Jn = 0

DERIVED CATEGORIES 79

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 concludethat a : A → I0, b : B → J0 and all the morphisms dnI , dnJ are strict, see Ho-mology, Lemma 19.15. We will inductively construct the maps fn in the followingcommutative diagram

Aa//

f

I0 //

f0

I1 //

f1

I2 //

f2

. . .

Bb // J0 // J1 // J2 // . . .

Because A→ I0 is a strict monomorphism and because J0 is filtered injective, wecan find a morphism f0 : I0 → J0 such that f0 a = b f , see Lemma 26.4. Thecomposition d0

J b f is zero, hence d0J f0 a = 0, hence d0

J f0 factors througha unique morphism

Coker(a) = Coim(d0I) = Im(d0

I) −→ J1.

As Im(d0I)→ I1 is a strict monomorphism we can extend the displayed arrow to a

morphism f1 : I1 → J1 by Lemma 26.4 again. And so on.

Lemma 26.8.05TV 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-isomorphismsA[0] → I• and C[0] → J• where I•, J• are complexes of filtered injective objectswith In = Jn = 0 for n < 0, then there exists a commutative diagram

0 // A[0] //

B[0] //

C[0] //

0

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

Proof. As A[0] → I• and C[0] → J• are filtered quasi-isomorphisms we concludethat a : A→ I0, c : C → J0 and all the morphisms dnI , dnJ are strict, see Homology,Lemma 13.4. We are going to step by step construct the south-east and the southarrows in the following commutative diagram

Bβ//

b

Cc//

b

J0

δ0

// J1

δ1

// . . .

A

α

OO

a // I0 // I1 // I2 // . . .

As A → B is a strict monomorphism, we can find a morphism b : B → I0 suchthat 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 thediagram. As c is a strict monomorphism and I1 is filtered injective we can findδ0 : J0 → I1, see Lemma 26.4. Because B → C is a strict epimorphism and

DERIVED CATEGORIES 80

because B → I0 → I1 → I2 is zero, we see that C → I1 → I2 is zero. Hence d1I δ0

is zero on C ∼= Im(c). Hence d1I δ0 factors through a unique morphism

Coker(c) = Coim(d0J) = Im(d0

J) −→ I2.

As I2 is filtered injective and Im(d0J)→ J1 is a strict monomorphism we can extend

the displayed morphism to a morphism δ1 : J1 → I2, see Lemma 26.4. And so on.We set M• = I• ⊕ J• with differential

dnM =(dnI (−1)n+1δn

0 dnJ

)Finally, the map B[0]→M• is given by b⊕ c β : M → I0 ⊕ J0.

Lemma 26.9.05TW 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• boundedbelow, 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 mayassume that Kn = 0 for n < 0. Consider the short exact sequences

0→ Ker(d0K)→ K0 → Coim(d0

K)→ 00→ Ker(d1

K)→ K1 → Coim(d1K)→ 0

0→ Ker(d2K)→ K2 → Coim(d2

K)→ 0. . .

of the exact category Filf (A) and the maps ui : Coim(diK)→ Ker(di+1K ). For each

i ≥ 0 we may choose filtered quasi-isomorphismsKer(diK)[0]→ I•ker,i

Coim(diK)[0]→ I•coim,i

with Inker,i, Incoim,i filtered injective and zero for n < 0, see Lemma 26.6. By Lemma26.7 we may lift ui to a morphism of complexes u•i : I•coim,i → I•ker,i+1. Finally, foreach i ≥ 0 we may complete the diagrams

0 // Ker(diK)[0] //

Ki[0] //

Coim(diK)[0] //

0

0 // I•ker,iαi // I•i

βi // I•coim,i // 0

with the lower sequence a termwise split exact sequence, see Lemma 26.8. For i ≥ 0set di : I•i → I•i+1 equal to di = αi+1 u•i βi. Note that di di−1 = 0 becauseβi αi = 0. Hence we have constructed a commutative diagram

I•0 // I•1 // I•2 // . . .

K0[0] //

OO

K1[0] //

OO

K2[0] //

OO

. . .

Here the vertical arrows are filtered quasi-isomorphisms. The upper row is a com-plex of complexes and each complex consists of filtered injective objects with nononzero objects in degree< 0. Thus we obtain a double complex by setting Ia,b = Ibaand using

da,b1 : Ia,b = Iba → Iba+1 = Ia+1,b

DERIVED CATEGORIES 81

the map dba and using for

da,b2 : Ia,b = Iba → Ib+1a = Ia,b+1

the map dbIa. Denote Tot(I•,•) the total complex associated to this double complex,

see Homology, Definition 18.3. Observe that the maps Kn[0]→ I•n come from mapsKn → 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, wesee that gr(Tot(I•,•)) = Tot(gr(I•,•)). Thus we can use Homology, Lemma 25.4 toconclude that gr(K•)→ gr(Tot(I•,•)) is a quasi-isomorphism as desired.

Lemma 26.10.05TX Let A be an abelian category. Let K•, I• ∈ K(Filf (A)). AssumeK• 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 forj < n. We will show that there exists a morphism h : Kn+1 → In such thatαn = h d. Thus α will be homotopic to the morphism of complexes β defined by

βj =

0 if j ≤ nαn+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 notethat αn dn−1

K = 0 since αn−1 = 0. Since K• is filtered acyclic we see that dn−1K

and dnK are strict and that

0→ Im(dn−1K )→ Kn → Im(dnK)→ 0

is an exact sequence of the exact category Filf (A), see Homology, Lemma 19.15.Hence we can think of αn as a map into In defined on Im(dnK). Using that Im(dnK)→Kn+1 is a strict monomorphism and that In is filtered injective we may lift thismap to a map h : Kn+1 → In as desired, see Lemma 26.4.

Lemma 26.11.05TY Let A be an abelian category. Let I• ∈ K(Filf (A)) be a boundedbelow complex consisting of filtered injective objects.

(1) Let α : K• → L• in K(Filf (A)) be a filtered quasi-isomorphism. Then themap

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)

DERIVED CATEGORIES 82

is a distinguished triangle in K(Filf (A)) (Lemma 9.14) and C(α)• is a filteredacyclic complex (Lemma 13.4). Then

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

qqHomK(Filf (A))(C(α)•[−1], I•)

is an exact sequence of abelian groups, see Lemma 4.2. At this point Lemma26.10 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 representa = 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• suchthat β α is homotopic to γ. This proves that the map is surjective. Let b be anelement of the left hand side which maps to zero in the right hand side. Then bis the homotopy class of a morphism β : L• → I• such that there exists a filteredquasi-isomorphism α : K• → L• with β α homotopic to zero. Then part (1) showsthat β is homotopic to zero also, i.e., b = 0.

Lemma 26.12.015Q Let A be an abelian category with enough injectives. Let If ⊂Filf (A) denote the strictly full additive subcategory whose objects are the filteredinjective objects. The canonical functor

K+(If ) −→ DF+(A)

is exact, fully faithful and essentially surjective, i.e., an equivalence of triangulatedcategories. Furthermore the diagrams

K+(If )

grp

// DF+(A)

grp

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

K+(If )

forget F

// DF+(A)

forget F

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

are commutative, where I ⊂ A is the strictly full additive subcategory whose objectsare the injective objects.

Proof. The functorK+(If )→ DF+(A) is essentially surjective by Lemma 26.9. Itis fully faithful by Lemma 26.11. It is an exact functor by our definitions regardingdistinguished triangles. The commutativity of the squares is immediate.

Remark 26.13.015R We can invert the arrow of the lemma only if A is a category inour sense, namely if it has a set of objects. However, suppose given a big abeliancategory A with enough injectives, such as Mod(OX) for example. Then for anygiven set of objects Aii∈I there is an abelian subcategory A′ ⊂ A containing allof them and having enough injectives, see Sets, Lemma 12.1. Thus we may use thelemma above for A′. 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 cannotuse the letter F for the functor since this would conflict too much with our use of

DERIVED CATEGORIES 83

the letter F to indicate filtrations.) Note that T induces an additive functorT : Filf (A)→ Filf (B)

by the rule T (A,F ) = (T (A), F ) where F pT (A) = T (F pA) which makes sense asT is left exact. (Warning: It may not be the case that gr(T (A)) = T (gr(A)).) Thisinduces 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 14.2for 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 thispoint we can redo the discussion of Section 20 to define the filtered right derivedfunctors(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 wecan define RT even if T is just additive. Namely, we first choose a quasi-inversej′ : DF+(A) → K+(If ) of the equivalence of Lemma 26.12. By Lemma 4.18 wesee that j′ is an exact functor of triangulated categories. Next, we note that for afiltered injective object I we have a (noncanonical) decomposition

(26.13.3)015T I ∼=⊕

p∈ZIp, with F pI =

⊕q≥p

Iq

by Lemma 26.2. Hence if T is any additive functor T : A → B then we get anadditive functor(26.13.4)05U0 Text : If → Filf (B)by setting Text(I) =

⊕T (Ip) with F pText(I) =

⊕q≥p T (Iq). Note that we have

the property gr(Text(I)) = T (gr(I)) by construction. Hence we obtain a functor(26.13.5)05U1 Text : K+(If )→ K+(Filf (B))which commutes with gr. Then we define (26.13.2) by the composition(26.13.6)05U2 RT = Text j′.Since RT : D+(A) → D+(B) is computed by injective resolutions as well, seeLemmas 20.1, the commutation of T with gr, and the commutative diagrams ofLemma 26.12 imply that(26.13.7)015U grp RT ∼= RT grp

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 notmake sense to say that RT restricts to a δ-functor on them. (This can be repaired

DERIVED CATEGORIES 84

by thinking of these categories as exact categories and formulating the notion of aδ-functor from an exact category into a triangulated category.) But it does makesense, and it is true by construction, that RT is an exact functor on the triangulatedcategory KF+(A).

Lemma 26.14.015W Let A,B be abelian categories. Let T : A → B be a left exact func-tor. 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 Band dr of bidegree (r,−r + 1) and with

Ep,q1 = Rp+qT (grp(K•))

Moreover, this spectral sequence is bounded, converges to R∗T (K•), and inducesa finite filtration on each RnT (K•). The construction of this spectral sequence isfunctorial in the object K• of Comp+(Filf (A)) and the terms (Er, dr) for r ≥ 1 donot depend on any choices.

Proof. Choose a filtered quasi-isomorphism K• → I• with I• a bounded be-low complex of filtered injective objects, see Lemma 26.9. Consider the complexRT (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, Section24. By Homology, Lemma 24.2 we have Ep,q1 = Hp+q(grp(T (I•))). By Equa-tion (26.13.3) we have Ep,q1 = Hp+q(T (grp(I•))), and by definition of a filteredinjective resolution the map grp(K•) → grp(I•) is an injective resolution. HenceEp,q1 = Rp+qT (grp(K•)).

On the other hand, each In has a finite filtration and hence each T (In) has afinite filtration. Thus we may apply Homology, Lemma 24.11 to conclude thatthe spectral sequence is bounded, converges to Hn(T (I•)) = RnT (K•) moreoverinducing finite filtrations on each of the terms.

Suppose that K• → L• is a morphism of Comp+(Filf (A)). Choose a filteredquasi-isomorphism L• → J• with J• a bounded below complex of filtered injectiveobjects, see Lemma 26.9. By our results above, for example Lemma 26.11, thereexists a diagram

K• //

L•

I• // J•

which commutes up to homotopy. Hence we get a morphism of filtered complexesT (I•)→ T (J•) which gives rise to the morphism of spectral sequences, see Homol-ogy, Lemma 24.4. The last statement follows from this.

Remark 26.15.015X As promised in Remark 21.4 we discuss the connection of thelemma above with the constructions using Cartan-Eilenberg resolutions. Namely,let T : A → B be a left exact functor of abelian categories, assume A has enoughinjectives, and let K• be a bounded below complex of A. We give an alternativeconstruction of the spectral sequences ′E and ′′E of Lemma 21.3.

First spectral sequence. Consider the “stupid” filtration on K• obtained by set-ting F p(K•) = σ≥p(K•), see Homology, Section 15. Note that this stupid in thesense that d(F p(K•)) ⊂ F p+1(K•), compare Homology, Lemma 24.3. Note that

DERIVED CATEGORIES 85

grp(K•) = Kp[−p] with this filtration. According to Lemma 26.14 there is a spec-tral sequence with E1 term

Ep,q1 = Rp+qT (Kp[−p]) = RqT (Kp)as in the spectral sequence ′Er. Observe moreover that the differentials Ep,q1 →Ep+1,q

1 agree with the differentials in ′E1, see Homology, Lemma 24.3 part (2) andthe description of ′d1 in the proof of Lemma 21.3.Second spectral sequence. Consider the filtration on the complex K• obtained bysetting F p(K•) = τ≤−p(K•), see Homology, Section 15. The minus sign is necessaryto 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 withE1 term

Ep,q1 = Rp+qT (H−p(K•)[p]) = R2p+qT (H−p(K•)) = ′′Ei,j2

with i = 2p+q and j = −p. (This looks unnatural, but note that we could just havewell 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.) Weleave it to the reader to see that the differentials match up.Actually, given a Cartan-Eilenberg resolution K• → I•,• the induced morphismK• → Tot(I•,•) into the associated total complex will be a filtered injective reso-lution for either filtration using suitable filtrations on Tot(I•,•). This can be usedto 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.

Definition 27.1.06XQ Let A be an abelian category. Let i ∈ Z. Let X,Y be objectsof D(A). The ith extension group of X by Y is the group

ExtiA(X,Y ) = HomD(A)(X,Y [i]) = HomD(A)(X[−i], Y ).

If A,B ∈ Ob(A) we set ExtiA(A,B) = ExtiA(A[0], B[0]).

Since HomD(A)(X,−), resp. HomD(A)(−, Y ) is a homological, resp. cohomologi-cal functor, see Lemma 4.2, we see that a distinguished triangle (Y, Y ′, Y ′′), resp.(X,X ′, X ′′) leads to a long exact sequence

. . .→ ExtiA(X,Y )→ ExtiA(X,Y ′)→ ExtiA(X,Y ′′)→ Exti+1A (X,Y )→ . . .

respectively. . .→ ExtiA(X ′′, Y )→ ExtiA(X ′, Y )→ ExtiA(X,Y )→ Exti+1

A (X ′′, Y )→ . . .

Note that since D+(A), D−(A), Db(A) are full subcategories we may computethe Ext groups by Hom groups in these categories provided X, Y are contained inthem.In case the category A has enough injectives or enough projectives we can computethe Ext groups using injective or projective resolutions. To avoid confusion, recallthat having an injective (resp. projective) resolution implies vanishing of homologyin all low (resp. high) degrees, see Lemmas 18.2 and 19.2.

Lemma 27.2.06XR Let A be an abelian category. Let X•, Y • ∈ Ob(K(A)).

DERIVED CATEGORIES 86

(1) Let Y • → I• be an injective resolution (Definition 18.1). Then

ExtiA(X•, Y •) = HomK(A)(X•, I•[i]).

(2) Let P • → X• be a projective resolution (Definition 19.1). Then

ExtiA(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 categoryitself. First we observe the following.

Lemma 27.3.06XS 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

and Hj(Y ) = 0 for j < b, we have ExtnA(X,Y ) = 0 for n < b− a and

Extb−aA (X,Y ) = HomA(Ha(X), Hb(Y ))

(2) Let A,B ∈ Ob(A). For i < 0 we have ExtiA(B,A) = 0. We haveExt0

A(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 aquasi-isomorphism, f : L• → Y •[n] a morphism, and Li = 0 for i < a. Note thatf maps Li to Y i+n. Thus f = 0 if n < b − a because always either Li or Y i+n iszero. 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 → A′ → A′′ → 0 is a shortexact sequence of objects of A. Then 0 → A[0] → A′[0] → A′′[0] → 0 leads to adistinguished triangle in D(A) (see Lemma 12.1) hence a long exact sequence ofExt groups

0→ Ext0A(B,A)→ Ext0

A(B,A′)→ Ext0A(B,A′′)→ Ext1

A(B,A)→ . . .

Similarly, given a short exact sequence 0 → B → B′ → B′′ → 0 we obtain a longexact sequence of Ext groups

0→ Ext0A(B′′, A)→ Ext0

A(B′, A)→ Ext0A(B,A)→ Ext1

A(B′′, A)→ . . .

We may view these Ext groups as an application of the construction of the derivedcategory. It shows one can define Ext groups and construct the long exact sequenceof Ext groups without needing the existence of enough injectives or projectives.There is an alternative construction of the Ext groups due to Yoneda which avoidsthe use of the derived category, see [Yon60].

Definition 27.4.06XT Let A be an abelian category. Let A,B ∈ Ob(A). A degree iYoneda extension of B by A is an exact sequence

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

DERIVED CATEGORIES 87

in A. We say two Yoneda extensions E and E′ of the same degree are equivalent ifthere exists a commutative diagram

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

0 // A //

id

OO

id

Z ′′i−1//

OO

. . . // Z ′′0 //

OO

B //

id

OO

id

0

0 // A // Z ′i−1// . . . // Z ′0 // 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 equivalencerelation. Although it is instructive to prove this directly this will also follow fromLemma 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 elementδ(E) ∈ Exti(B,A) as the morphism δ(E) = fs−1 : B[0] → A[i] where s is thequasi-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 classcharacterizes the equivalence class of the Yoneda extension.

Lemma 27.5.06XU Let A be an abelian category with objects A, B. Any element inExtiA(B,A) is δ(E) for some degree i Yoneda extension of B by A. Given twoYoneda extensions E, E′ of the same degree then E is equivalent to E′ if and onlyif δ(E) = δ(E′).

Proof. Let ξ : B[0]→ A[i] be an element of ExtiA(B,A). We may write ξ = fs−1

for some quasi-isomorphism s : L• → B[0] and map f : L• → A[i]. After replacingL• by τ≤0L

• we may assume that Lj = 0 for j > 0. Picture

L−i−1 // L−i //

. . . // L0 // B // 0

A

Then setting Zi−1 = (L−i+1 ⊕ A)/L−i and Zj = L−j for j = i − 2, . . . , 0 we seethat we obtain a degree i extension E of B by A whose class δ(E) equals ξ.It is immediate from the definitions that equivalent Yoneda extensions have thesame class. Suppose that E : 0 → A → Zi−1 → Zi−2 → . . . → Z0 → B → 0 andE′ : 0 → A → Z ′i−1 → Z ′i−2 → . . . → Z ′0 → B → 0 are Yoneda extensions withthe same class. By construction of D(A) as the localization of K(A) at the set ofquasi-isomorphisms, this means there exists a complex L• and quasi-isomorphisms

t : L• → (. . .→ 0→ A→ Zi−1 → . . .→ Z0 → 0→ . . .)

DERIVED CATEGORIES 88

andt′ : L• → (. . .→ 0→ A→ Z ′i−1 → . . .→ Z ′0 → 0→ . . .)

such that s t = s′ t′ and f t = f ′ t′, see Categories, Section 27. Let E′′be the degree i extension of B by A constructed from the pair L• → B[0] andL• → A[i] in the first paragraph of the proof. Then the reader sees readily thatthere exists “morphisms” of degree i Yoneda extensions E′′ → E and E′′ → E′ asin the definition of equivalent Yoneda extensions (details omitted). This finishesthe proof.

Lemma 27.6.06XV Let A be an abelian category. Let A, B be objects of A. ThenExt1

A(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.

Lemma 27.7.0EWW Let A be an abelian category and let p ≥ 0. If ExtpA(B,A) = 0for any pair of objects A, B of A, then ExtiA(B,A) = 0 for i ≥ p and any pair ofobjects 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(E′) and δ(E′′) where

E′ : 0→ C → Zp−1 → . . .→ Z0 → B → 0

andE′′ : 0→ A→ Zi−1 → Zi−2 → . . .→ Zp → C → 0

Since δ(E′) ∈ ExtpA(B,C) = 0 we conclude.

Lemma 27.8.0EWX Let A be an abelian category. Assume Ext2A(B,A) = 0 for any

pair of objects A, B of A. Then any object K of Db(A) is isomorphic to the directsum of its cohomologies: K ∼=

⊕Hi(K)[−i].

Proof. Choose a, b such that Hi(K) = 0 for i 6∈ [a, b]. We will prove the lemmaby induction on b − a. If b − a ≤ 0, then the result is clear. If b − a > 0, then welook at the distinguished triangle of truncations

τ≤b−1K → K → Hb(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

HomD(A)(Hb(K)[−b], (τ≤b−1K)[1]) =⊕

i<bExtb−i+1

A (Hb(K), Hi(K))

Since ExtiA(B,A) = 0 for i ≥ 2 and any pair of objects A,B of A by our assumptionand Lemma 27.7 we are done.

DERIVED CATEGORIES 89

28. K-groups

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

Definition 28.1.0FCN Let D be a triangulated category. We denote K0(D) the zerothK-group of D. It is the abelian group constructed as follows. Take the free abeliangroup on the objects on D and for every distinguished triangle X → Y → Z imposethe relation [Y ]− [X]− [Z] = 0.

Observe that this implies that [X[n]] = (−1)n[X] because we have the distinguishedtriangle (X, 0, X[1], 0, 0,−id[1]).

Lemma 28.2.0FCP Let A be an abelian category. Then there is a canonical identifica-tion K0(Db(A)) = K0(A) of zeroth K-groups.

Proof. Given an object A of A denote A[0] the object A viewed as a complexsitting in degree 0. If 0 → A → A′ → A′′ → 0 is a short exact sequence, then weget a distinguished triangle A[0] → A′[0] → A′′[0] → A[1], see Section 12. Thisshows that we obtain a map K0(A) → K0(Db(A)) by sending [A] to [A[0]] withapologies for the horrendous notation.On the other hand, given an object X of Db(A) we can consider the element

c(X) =∑

(−1)i[Hi(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 factorsthrough a map c : K0(Db(A))→ K0(A).We want to show that the two maps above are mutually inverse. It is clear thatthe composition K0(A) → K0(Db(A)) → K0(A) is the identity. Suppose that X•is a bounded complex of A. The existence of the distinguished triangles of “stupidtruncations” (see Homology, Section 15)

σ≥nX• → σ≥n−1X

• → Xn−1[−n+ 1]→ (σ≥nX•)[1]and induction show that

[X•] =∑

(−1)i[Xi[0]]

in K0(Db(A)) (with again apologies for the notation). It follows that the composi-tion K0(A)→ K0(Db(A)) is surjective which finishes the proof.

Lemma 28.3.0FCQ Let F : D → D′ be an exact functor of triangulated categories.Then F induces a group homomorphism K0(D)→ K0(D′).

Proof. Omitted.

Lemma 28.4.0FCR Let H : D → A be a homological functor from a triangulatedcategory to an abelian category. Assume that for any X in D only a finite numberof the objects H(X[i]) are nonzero in A. Then H induces a group homomorphismK0(D)→ K0(A) sending [X] to

∑(−1)i[H(X[i])].

Proof. Omitted.

Lemma 28.5.0FCS Let B be a weak Serre subcategory of the abelian category A. Thenthere are canonical maps

K0(B) −→ K0(DbB(A)) −→ K0(B)

DERIVED CATEGORIES 90

whose composition is zero. The second arrow sends the class [X] of the object X tothe element

∑(−1)i[Hi(X)] of K0(B).

Proof. Omitted.

Lemma 28.6.0FCT Let D, D′, D′′ be triangulated categories. Let

⊗ : D ×D′ −→ D′′

be a functor such that for fixed X in D the functor X ⊗ − : D′ → D′′ is an exactfunctor and for fixed X ′ in D′ the functor − ⊗X ′ : D → D′′ is an exact functor.Then ⊗ induces a bilinear map K0(D)×K0(D′)→ K0(D′′) which sends ([X], [X ′])to [X ⊗X ′].

Proof. Omitted.

29. Unbounded complexes

06XW A reference for the material in this section is [Spa88]. The following lemma is usefulto find “good” left resolutions of unbounded complexes.

Lemma 29.1.06XX Let A be an abelian category. Let P ⊂ Ob(A) be a subset. AssumeP contains 0, is closed under (finite) direct sums, and every object of A is a quotientof an element of P. Let K• be a complex. There exists a commutative diagram

P •1

// P •2

// . . .

τ≤1K• // τ≤2K

• // . . .

in the category of complexes such that(1) the vertical arrows are quasi-isomorphisms and termwise surjective,(2) P •n is a bounded above complex with terms in P,(3) the arrows P •n → P •n+1 are termwise split injections and each cokernel

P in+1/Pin is an element of P.

Proof. We are going to use that the homotopy category K(A) is a triangulatedcategory, see Proposition 10.3. By Lemma 15.4 we can find a termwise surjectivemap of complexes P •1 → τ≤1K

• which is a quasi-isomorphism such that the termsof P •1 are in P. By induction it suffices, given P •1 , . . . , P

•n to construct P •n+1 and

the maps P •n → P •n+1 and P •n+1 → τ≤n+1K•.

Choose a distinguished triangle P •n → τ≤n+1K• → C• → P •n [1] in K(A). Applying

Lemma 15.4 we choose a map of complexes Q• → C• which is a quasi-isomorphismsuch that the terms of Q• are in P. By the axioms of triangulated categories wemay fit the composition Q• → C• → P •n [1] into a distinguished triangle P •n →P •n+1 → Q• → P •n [1] in K(A). By Lemma 10.7 we may and do assume 0→ P •n →P •n+1 → Q• → 0 is a termwise split short exact sequence. This implies that theterms of P •n+1 are in P and that P •n → P •n+1 is a termwise split injection whosecokernels are in P. By the axioms of triangulated categories we obtain a map of

DERIVED CATEGORIES 91

distinguished triangles

P •n //

P •n+1//

Q• //

P •n [1]

P •n // τ≤n+1K

• // C• // P •n [1]

in the triangulated category K(A). Choose an actual morphism of complexes f :P •n+1 → τ≤n+1K

•. The left square of the diagram above commutes up to homotopy,but as P •n → P •n+1 is a termwise split injection we can lift the homotopy and modifyour choice of f to make it commute. Finally, f is a quasi-isomorphism, becauseboth P •n → P •n and Q• → C• are.At this point we have all the properties we want, except we don’t know that themap f : P •n+1 → τ≤n+1K

• is termwise surjective. Since we have the commutativediagram

P •n

// P •n+1

τ≤nK

• // τ≤n+1K•

of complexes, by induction hypothesis we see that f is surjective on terms in alldegrees except possibly n and n + 1. Choose an object P ∈ P and a surjectionq : P → Kn. Consider the map

g : P • = (. . .→ 0→ P1−→ P → 0→ . . .) −→ τ≤n+1K

•

with first copy of P in degree n and maps given by q in degree n and dK q indegree n + 1. This is a surjection in degree n and the cokernel in degree n + 1 isHn+1(τ≤n+1K

•); to see this recall that τ≤n+1K• has Ker(dn+1

K ) in degree n + 1.However, since f is a quasi-isomorphism we know that Hn+1(f) is surjective. Henceafter replacing f : P •n+1 → τ≤n+1K

• by f ⊕ g : P •n+1⊕P • → τ≤n+1K• we win.

In some cases we can use the lemma above to show that a left derived functor iseverywhere defined.

Proposition 29.2.0794 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 isa quotient of an element of P,

(2) for any bounded above acyclic complex P • of A with Pn ∈ P for all n thecomplex 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).

Proof. By (1) and Lemma 15.4 for any bounded above complex K• there existsa quasi-isomorphism P • → K• with P • bounded above and Pn ∈ P for all n.Suppose that s : P • → (P ′)• is a quasi-isomorphism of bounded above complexesconsisting of objects of P. Then F (P •)→ F ((P ′)•) is a quasi-isomorphism becauseF (C(s)•) is acyclic by assumption (2). This already shows that LF is defined on

DERIVED CATEGORIES 92

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

P •1

// P •2

// . . .

τ≤1K• // τ≤2K

• // . . .

as in Lemma 29.1. Note that the map colimP •n → K• is a quasi-isomorphismbecause colimits over N in A are exact and Hi(P •n) = Hi(K•) for n > i. We claimthat

F (colimP •n) = colimF (P •n)(termwise colimits) is LF (K•), i.e., that colimP •n computes LF . To see this, byLemma 14.15, it suffices to prove the following claim. Suppose that

colimQ•n = Q•α−−→ P • = colimP •n

is a quasi-isomorphism of complexes, such that each P •n , Q•n is a bounded abovecomplex whose terms are in P and the maps P •n → τ≤nP

• and Q•n → 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 betweenthe complexes P •n and Q•n. However, for each n we know that the solid arrows inthe diagram

R•

P •n

L•oo // Q•n

τ≤nP

• τ≤nα // τ≤nQ•

are quasi-isomorphisms. Because quasi-isomorphisms form a multiplicative systemin K(A) (see Lemma 11.2) we can find a quasi-isomorphism L• → P •n and mapof complexes L• → Q•n 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 firstparagraph of the proof we see that F (R•)→ F (P •n) and F (R•)→ F (Q•n) are quasi-isomorphisms. Thus we obtain a isomorphisms Hi(F (P •n)) → Hi(F (Q•n)) fittinginto the commutative diagram

Hi(F (P •n)) //

Hi(F (Q•n))

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

Hi(F (P •)) = Hi(F (colimP •n)) = Hi(colimF (P •n)) = colimHi(F (P •n))

DERIVED CATEGORIES 93

and similarly for Q• we conclude that Hi(α) : Hi(F (P •) → Hi(F (Q•) is an iso-morphism and the claim follows.

Lemma 29.3.070F Let A be an abelian category. Let I ⊂ Ob(A) be a subset. AssumeI contains 0, is closed under (finite) products, and every object of A is a subobjectof an element of I. Let K• be a complex. There exists a commutative diagram

. . . // τ≥−2K• //

τ≥−1K•

. . . // I•2 // I•1

in the category of complexes such that(1) the vertical arrows are quasi-isomorphisms and termwise injective,(2) I•n is a bounded below complex with terms in I,(3) the arrows I•n+1 → I•n are termwise split surjections and Ker(Iin+1 → Iin)

is an element of I.

Proof. This lemma is dual to Lemma 29.1.

30. Deriving adjoints

0FNC Let F : D → D′ and G : D′ → D be exact functors of triangulated categories.Let S, resp. S′ be a multiplicative system for D, resp. D′ compatible with thetriangulated structure. Denote Q : D → S−1D and Q′ : D′ → (S′)−1D′ thelocalization functors. In this situation, by abuse of notation, one often denotes RFthe partially defined right derived functor corresponding to Q′ F : D → (S′)−1D′and the multiplicative system S. Similarly one denotes LG the partially definedleft derived functor corresponding to Q G : D′ → S−1D and the multiplicativesystem S′. Picture

DF

//

Q

D′

Q′

S−1D RF // (S′)−1D′

and

D′G

//

Q′

D

Q

(S′)−1D′ LG // S−1D

Lemma 30.1.0FND In the situation above assume F is right adjoint to G. Let K ∈Ob(D) and M ∈ Ob(D′). If RF is defined at K and LG is defined at M , thenthere is a canonical isomorphism

Hom(S′)−1D′(M,RF (K)) = HomS−1D(LG(M),K)

This isomorphism is functorial in both variables on the triangulated subcategoriesof S−1D and (S′)−1D′ where RF and LG are defined.

Proof. Since RF is defined at K, we see that the rule which assigns to a s : K → Iin S the object F (I) is essentially constant as an ind-object of (S′)−1D′ with valueRF (K). Similarly, the rule which assigns to a t : P →M in S′ the object G(P ) is

DERIVED CATEGORIES 94

essentially constant as a pro-object of S−1D with value LG(M). Thus we have

Hom(S′)−1D′(M,RF (K)) = colims:K→I Hom(S′)−1D′(M,F (I))= colims:K→I colimt:P→M HomD′(P, F (I))= colimt:P→M colims:K→I HomD′(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 holdsby the definition of morphisms in D(B), see Categories, Remark 27.15. The thirdequality holds by Categories, Lemma 14.10. The fourth equality holds because Fand G are adjoint. The fifth equality holds by definition of morphism in D(A), seeCategories, Remark 27.7. The sixth equality holds by Categories, Lemma 22.10.We omit the proof of functoriality.

Lemma 30.2.0DVC Let F : A → B and G : B → A be functors of abelian categoriessuch that F is a right adjoint to G. Let K• be a complex of A and let M• be acomplex of B. If RF is defined at K• and LG is defined at M•, then there is acanonical isomorphism

HomD(B)(M•, RF (K•)) = HomD(A)(LG(M•),K•)

This isomorphism is functorial in both variables on the triangulated subcategoriesof 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 unboundedderived categories. Namely, without having the unbounded derived functors, thelemma could not even be stated.

Lemma 30.3.09T5 Let F : A → B and G : B → A be functors of abelian categoriessuch 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 onthe unbounded derived category.

Definition 31.1.070H Let A be an abelian category. A complex I• is K-injective if forevery 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 ias the translate of an acyclic complex is acyclic.

Lemma 31.2.070I Let A be an abelian category. Let I• be a complex. The followingare equivalent

(1) I• is K-injective,

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(2) for every quasi-isomorphism M• → N• the mapHomK(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 coho-mological 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• wheres : M• → N• is a quasi-isomorphism and f : M• → I• is a map. By (2) thiscorresponds 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.

Lemma 31.3.090X Let A be an abelian category. Let (K,L,M, f, g, h) be a distin-guished triangle of K(A). If two out of K, L, M are K-injective complexes, thenthe third is too.

Proof. Follows from the definition, Lemma 4.2, and the fact that K(A) is a trian-gulated category (Proposition 10.3).

Lemma 31.4.070J Let A be an abelian category. A bounded below complex of injectivesis K-injective.

Proof. Follows from Lemmas 31.2 and 18.8.

Lemma 31.5.0BK6 Let A be an abelian category. Let T be a set and for each t ∈ T letI•t be a K-injective complex. If In =

∏t Int exists for all n, then I• is a K-injective

complex. Moreover, I• represents the product of the objects I•t in D(A).

Proof. Let K• be an complex. Observe that the complex

C :∏

bHom(K−b, Ib−1)→

∏b

Hom(K−b, Ib)→∏

bHom(K−b, Ib+1)

has cohomology HomK(A)(K•, I•) in the middle. Similarly, the complex

Ct :∏

bHom(K−b, Ib−1

t )→∏

bHom(K−b, Ibt )→

∏b

Hom(K−b, Ib+1t )

computes HomK(A)(K•, I•t ). Next, observe that we have

C =∏

t∈TCt

as complexes of abelian groups by our choice of I. Taking products is an exact func-tor on the category of abelian groups. Hence ifK• is acyclic, then HomK(A)(K•, I•t ) =0, hence Ct is acyclic, hence C is acyclic, hence we get HomK(A)(K•, I•) = 0. Thuswe find that I• is K-injective. Having said this, we can use Lemma 31.2 to concludethat

HomD(A)(K•, I•) =∏

t∈THomD(A)(K•, I•t )

and indeed I• represents the product in the derived category.

DERIVED CATEGORIES 96

Lemma 31.6.070Y Let A be an abelian category. Let F : K(A)→ D′ be an exact func-tor of triangulated categories. Then RF is defined at every complex in K(A) whichis quasi-isomorphic to a K-injective complex. In fact, every K-injective complexcomputes RF .Proof. By Lemma 14.4 it suffices to show that RF is defined at a K-injectivecomplex, i.e., it suffices to show a K-injective complex I• computes RF . Anyquasi-isomorphism I• → N• is a homotopy equivalence as it has an inverse byLemma 31.2. Thus I• → I• is a final object of I•/Qis(A) and we win.

Lemma 31.7.070K 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) →D′ of triangulated categories has a right derived functor

RF : D(A) −→ D′

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-injectivecomplexes. 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 iso-morphism. This is clear because I• → J• is a homotopy equivalence, i.e., anisomorphism in K(A), by Lemma 31.2.

The following lemma can be generalized to limits over bigger ordinals.Lemma 31.8.070L Let A be an abelian category. Let

. . .→ I•3 → I•2 → I•1

be an inverse system of complexes. Assume(1) each I•n is K-injective,(2) each map Imn+1 → Imn is a split surjection,(3) the limits Im = lim Imn exist.

Then the complex I• is K-injective.Proof. We urge the reader to skip the proof of this lemma. Let M• be anacyclic complex. Let us abbreviate Hn(a, b) = HomA(Ma, Ibn). With this nota-tion HomK(A)(M•, I•) is the cohomology of the complex∏m

limnHn(m,m−2)→

∏m

limnHn(m,m−1)→

∏m

limnHn(m,m)→

∏m

limnHn(m,m+1)

in the third spot from the left. We may exchange the order of∏

and lim and eachof the complexes∏

m

Hn(m,m− 2)→∏m

Hn(m,m− 1)→∏m

Hn(m,m)→∏m

Hn(m,m+ 1)

is exact by assumption (1). By assumption (2) the maps in the systems

. . .→∏m

H3(m,m− 2)→∏m

H2(m,m− 2)→∏m

H1(m,m− 2)

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.

DERIVED CATEGORIES 97

Lemma 31.9.08BJ Let A and B be abelian categories. Let u : A → B and v : B → Abe 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, whenthe functor has bounded cohomological dimension.

Lemma 32.1.07K6 Let A be an abelian category. Let d : Ob(A)→ 0, 1, 2, . . . ,∞ bea function. Assume that

(1) every object of A is a subobject of an object A with d(A) = 0,(2) d(A⊕B) ≤ maxd(A), d(B) for A,B ∈ A, and(3) if 0→ A→ B → C → 0 is short exact, then d(C) ≤ maxd(A)− 1, d(B).

Let K• be a complex such that n + d(Kn) tends to −∞ as n → −∞. Then thereexists a quasi-isomorphism K• → L• with d(Ln) = 0 for all n ∈ Z.

Proof. By Lemma 15.5 we can find a quasi-isomorphism σ≥0K• →M• withMn =

0 for n < 0 and d(Mn) = 0 for n ≥ 0. Then K• is quasi-isomorphic to the complex. . .→ K−2 → K−1 →M0 →M1 → . . .

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 ′ = Coker(Kn →M ⊕Kn+1). Consider the map of complexes

K• :

Kn−1

// Kn

// Kn+1

// Kn+2

(K ′)• : Kn−1 // M // M ′ // Kn+2

It is clear that K• → (K ′)• is a quasi-isomorphism. Moreover, it is clear thatd((K ′)n) = 0 and

d((K ′)n+1) ≤ maxd(Kn)− 1, d(M ⊕Kn+1) ≤ maxd(Kn)− 1, d(Kn+1)and the other values are unchanged.To finish the proof we carefuly choose the order in which to do the elementaryreplacements so that for every integer m the complex σ≥mK

• is changed only afinite number of times. To do this set

ξ(K•) = maxn+ d(Kn) | d(Kn) > 0

DERIVED CATEGORIES 98

andI = n ∈ Z | ξ(K•) = n+ d(Kn) and d(Kn) > 0

Our assumption that n + d(Kn) tends to −∞ as n → −∞ and the fact thatd(Kn) = 0 for n >> 0 implies ξ(K•) < +∞ and that I is a finite set. It is clearthat ξ((K ′)•) ≤ ξ(K•) for an elementary transformation as above. An elementarytransformation changes the complex in degrees ≤ ξ(K•) + 1. Hence if we can findfinite sequence of elementary transformations which decrease ξ(K•), then we win.However, note that if we do an elementary transformation starting with the smallestelement n ∈ I, then we either decrease the size of I, or we increase min I. Sinceevery element of I is ≤ ξ(K•) we see that we win after a finite number of steps.

Lemma 32.2.07K7 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)

(a) Hi(RF (τ≤aE)→ Hi(RF (E)) is an isomorphism for i ≤ a,(b) Hi(RF (E))→ Hi(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 → A′ withA′ acyclic. Then we see that Rn+1F (A) = RnF (A′/A) = 0 by the long exactcohomology sequence. Hence we conclude that Rn+1F = 0. Continuing like thisusing 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, . . . givenby d(A) = max0 ∪ i | RiF (A) 6= 0. The first assumption of Lemma 32.1 is ourassumption (1). The second assumption of Lemma 32.1 follows from the fact thatRF (A⊕B) = RF (A)⊕RF (B). The third assumption of Lemma 32.1 follows fromthe long exact cohomology sequence. Hence for every complex K• there exists aquasi-isomorphism K• → L• into a complex of objects right acyclic for F . Thisproves statement (3).We claim that if L• → M• is a quasi-isomorphism of complexes of right acyclicobjects for F , then F (L•)→ F (M•) is a quasi-isomorphism. If we prove this claimthen we get statements (1) and (2) of the lemma by Lemma 14.15. To prove theclaim pick an integer i ∈ Z. Consider the distinguished triangle

σ≥i−n−1L• → σ≥i−n−1M

• → Q•,

i.e., let Q• be the cone of the first map. Note that Q• is bounded below and thatHj(Q•) is zero except possibly for j = i−n−1 or j = i−n−2. We may apply RF toQ•. Using the second spectral sequence of Lemma 21.3 and the assumed vanishingof cohomology (2) we conclude that Hj(RF (Q•)) is zero except possibly for j ∈i−n−2, . . . , i−1. Hence we see thatRF (σ≥i−n−1L

•)→ RF (σ≥i−n−1M•) induces

DERIVED CATEGORIES 99

an isomorphism of cohomology objects in degrees ≥ i. By Proposition 16.8 we knowthat 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 forF . By part (2) RF (E) is represented by the complex F (L•) and RF (σ≥cL•) isrepresented by σ≥cF (L•). Consider the distinguished triangle

Hb−n(L•)[n− b]→ τ≥b−nL• → τ≥b−n+1L

•

of Remark 12.4. The vanishing established above gives that Hi(RF (τ≥b−nL•))agrees with Hi(RF (τ≥b−n+1L

•)) for i ≥ b. Consider the short exact sequence ofcomplexes

0→ Im(Lb−n−1 → Lb−n)[n− b]→ σ≥b−nL• → τ≥b−nL

• → 0Using the distinguished triangle associated to this (see Section 12) and the vanishingas before we conclude that Hi(RF (τ≥b−nL•)) agrees with Hi(RF (σ≥b−nL•)) fori ≥ 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 getthe vanishing of Hi(RF (E)) for i ≤ a − 1 from part (4)(a). Similarly, we haveτ≥b+1E = 0 and hence we get the vanishing of Hi(RF (E)) for i ≥ b+ n from part(4)(b).

Lemma 32.3.07K8 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)

(a) Hi(LF (τ≤a+n−1E)→ Hi(LF (E)) is an isomorphism for i ≤ a,(b) Hi(LF (E))→ Hi(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.

Definition 33.1.090Z Let D be a triangulated category. Let (Kn, fn) be a system ofobjects of D. We say an object K is a derived colimit, or a homotopy colimit of thesystem (Kn) if the direct sum

⊕Kn exists and there is a distinguished triangle⊕

Kn →⊕

Kn → K →⊕

Kn[1]

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.

DERIVED CATEGORIES 100

By TR3 a derived colimit, if it exists, is unique up to (non-unique) isomorphism.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 trian-gulated category where all homotopy colimits exist.

The nonuniqueness makes it hard to pin down the derived colimit. In More onAlgebra, Lemma 81.4 the reader finds an exact sequence

0→ R1 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.

Remark 33.2.0CRH Let D be a triangulated category. Let (Kn, fn) be a system ofobjects of D. We may think of a derived colimit as an object K of D endowed withmorphisms in : Kn → K such that in+1 fn = in and such that there exists amorphism c : K →

⊕Kn with the property that⊕Kn

1−fn−−−→⊕

Knin−→ K

c−→⊕

Kn[1]

is a distinguished triangle. If (K ′, i′n, c′) is a second derived colimit, then thereexists an isomorphism ϕ : K → K ′ such that ϕ in = i′n and c′ ϕ = c. Theexistence of ϕ is TR3 and the fact that ϕ is an isomorphism is Lemma 4.3.

Remark 33.3.0CRI 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 ofthe first system and let (L, jn, d) be a derived colimit of the second system withnotation as in Remark 33.2. Then there exists a morphism a : K → L such thata in = jn and d a = (an[1]) c. This follows from TR3 applied to the definingdistinguished triangles.

Lemma 33.4.0CRJ Let D be a triangulated category. Let (Kn, fn) be a system of objectsof 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//

id

hocolimKni

Kni

// hocolimKn

commutes for all i.

Proof. Let gi : Kni→ Kni+1 be the composition fni+1−1 . . . fni

. We constructcommutative diagrams⊕

iKni 1−gi

//

b

⊕iKni

a

⊕nKn

1−fn //⊕nKn

and

⊕nKn 1−fn

//

d

⊕nKn

c

⊕iKni

1−gi //⊕iKni

as follows. Let ai = a|Knibe the inclusion of Kni

into the direct sum. In otherwords, a is the natural inclusion. Let bi = b|Kni

be the map

Kni

1, fni, fni+1fni

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

DERIVED CATEGORIES 101

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

Kj

fni−1...fj−−−−−−−−→ Kni

We let dj = d|Kjbe zero if j 6= ni for any i and we let dni

be the natural inclusionof Kni

into the direct sum. In other words, d is the natural projection. By TR3these 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 byLemma 4.3. The other way around we get the morphisms a c and b d. Considerthe morphism h = (hj) :

⊕Kn →

⊕Kn given by the rule: for ni−1 < j < ni we

sethj : Kj

1, fj , fj+1fj , ..., fni−1...fj−−−−−−−−−−−−−−−−−−−−→ Kj ⊕ . . .⊕Kni

Then the reader verifies that (1−f)h = id−a c and h (1−f) = id− bd. Thismeans that id−ψ ϕ has square zero by Lemma 4.5 (small argument omitted). Inother words, ψ ϕ differs from the identity by a nilpotent endomorphism, hence isan isomorphism. Thus ϕ and ψ are isomorphisms as desired.

Lemma 33.5.0A5L 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 K•iindexed by a countable index set I the termwise direct sum

⊕K•i is the direct sum

of K•i in D(A).

Proof. Let L• be a complex. Suppose given maps αi : K•i → L• in D(A). Thismeans there exist quasi-isomorphisms si : M•i → K•i of complexes and maps ofcomplexes fi : M•i → L• such that αi = fis

−1i . By assumption the map of com-

plexess :⊕

M•i −→⊕

K•i

is a quasi-isomorphism. Hence setting f =⊕fi we see that α = fs−1 is a map in

D(A) whose composition with the coprojection K•i →⊕K•i is αi. We omit the

verification that α is unique.

Lemma 33.6.093W Let A be an abelian category. Assume colimits over N exist andare 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

0→⊕

An →⊕

An → colimAn → 0

where the first map in degree n is given by 1− fn.

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 ofthe transition maps. Take the colimit to get the sequence of the lemma.

Lemma 33.7.0949 Let A be an abelian category. Let L•n be a system of complexes ofA. Assume colimits over N exist and are exact in A. Then the termwise colimitL• = colimL•n is a homotopy colimit of the system in D(A).

DERIVED CATEGORIES 102

Proof. We have an exact sequence of complexes

0→⊕

L•n →⊕

L•n → L• → 0

by Lemma 33.6. The direct sums are direct sums in D(A) by Lemma 33.5. Thusthe result follows from the definition of derived colimits in Definition 33.1 and thefact that a short exact sequence of complexes gives a distinguished triangle (Lemma12.1).

Lemma 33.8.0CRK 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 ahomological functor commuting with countable direct sums. Then H(hocolimKn) =colimH(Kn) for any system of objects of D.

Proof. Write K = hocolimKn. Apply H to the defining distinguished triangle toget ⊕

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 bedefined later) commutes with derived colimits.

Lemma 33.9.094A Let D be a triangulated category with countable direct sums. LetK ∈ D be an object such that for every countable set of objects En ∈ D the canonicalmap ⊕

HomD(K,En) −→ HomD(K,⊕

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

colim HomD(K,Ln) −→ HomD(K,L)

is a bijection.

Proof. Consider the defining distinguished triangle⊕Ln →

⊕Ln → L→

⊕Ln[1]

Apply the cohomological functor HomD(K,−) (see Lemma 4.2). By elementaryconsiderations concerning colimits of abelian groups we get the result.

34. Derived limits

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

Definition 34.1.08TC Let D be a triangulated category. Let (Kn, fn) be an inversesystem of objects of D. We say an object K is a derived limit, or a homotopy limitof the system (Kn) if the product

∏Kn exists and there is a distinguished triangle

K →∏

Kn →∏

Kn → K[1]

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 limKn.

DERIVED CATEGORIES 103

By TR3 a derived limit, if it exists, is unique up to (non-unique) isomorphism.Moreover, by TR1 a derived limit R limKn exists as soon as

∏Kn exists. The

derived category D(Ab) of the category of abelian groups is an example of a trian-gulated category where all derived limits exist.The nonuniqueness makes it hard to pin down the derived limit. In More onAlgebra, Lemma 81.3 the reader finds an exact sequence

0→ R1 lim Hom(L,Kn[−1])→ Hom(L,R limKn)→ lim Hom(L,Kn)→ 0describing the Homs into a derived limit in terms of the usual Homs.

Lemma 34.2.07KC Let A be an abelian category with exact countable products. Then(1) D(A) has countable products,(2) countable products

∏Ki in D(A) are obtained by taking termwise products

of any complexes representing the Ki, and(3) Hp(

∏Ki) =

∏Hp(Ki).

Proof. Let K•i be a complex representing Ki in D(A). Let L• be a complex. Sup-pose given maps αi : L• → K•i in D(A). This means there exist quasi-isomorphismssi : K•i → M•i of complexes and maps of complexes fi : L• → M•i such thatαi = s−1

i fi. By assumption the map of complexes

s :∏

K•i −→∏

M•i

is a quasi-isomorphism. Hence setting f =∏fi we see that α = s−1f is a map

in D(A) whose composition with the projection∏K•i → K•i 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.

Lemma 34.3.0BK7 Let A be an abelian category with countable products and enoughinjectives. Let (Kn) be an inverse system of D+(A). Then R limKn exists.

Proof. It suffices to show that∏Kn exists in D(A). For every n we can represent

Kn by a bounded below complex I•n of injectives (Lemma 18.3). Then∏Kn is

represented by∏I•n, see Lemma 31.5.

Lemma 34.4.070M Let A be an abelian category with countable products and enoughinjectives. Let K• be a complex. Let I•n be the inverse system of bounded belowcomplexes of injectives produced by Lemma 29.3. Then I• = lim I•n exists, is K-injective, and the following are equivalent

(1) the map K• → I• is a quasi-isomorphism,(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 I•n is K-injective by Lemma 31.4. Choose direct sum decom-positions Ipn+1 = Cpn+1 ⊕ Ipn for all n ≥ 1. Set Cp1 = Ip1 . The complex I• = lim I•nexists because we can take Ip =

∏n≥1 C

pn. Fix p ∈ Z. We claim there is a split

short exact sequence0→ Ip →

∏Ipn →

∏Ipn → 0

of objects of A. Here the first map is given by the projection maps Ip → Ipn and thesecond map by (xn) 7→ (xn − fpn+1(xn+1)) where fpn : Ipn → Ipn−1 are the transition

DERIVED CATEGORIES 104

maps. The splitting comes from the map∏Ipn →

∏Cpn = Ip. We obtain a termwise

split short exact sequence of complexes

0→ I• →∏

I•n →∏

I•n → 0

Hence a corresponding distinguished triangle in K(A) and D(A). By Lemma 31.5the products are K-injective and represent the corresponding products in D(A). Itfollows that I• represents R lim I•n (Definition 34.1). Moreover, it follows that I•is K-injective by Lemma 31.3. By the commutative diagram of Lemma 29.3 weobtain a corresponding commutative diagram

K• //

R lim τ≥−nK•

I• // R lim I•n

in D(A). Since the right vertical arrow is an isomorphism (as derived limits aredefined on the level of the derived category and since τ≥−nK• → I•n is a quasi-isomorphism), the lemma follows.

Lemma 34.5.090Y Let A be an abelian category having enough injectives and exactcountable 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 isomor-phism for all K in D(A). Consider the defining distinguished triangle

R lim τ≥−nK →∏

τ≥−nK →∏

τ≥−nK → (R lim τ≥−nK)[1]

By Lemma 34.2 we have

Hp(∏

τ≥−nK) =∏

p≥−nHp(K)

It follows in a straightforward manner from the long exact cohomology sequence ofthe displayed distinguished triangle that Hp(R lim τ≥−nK) = Hp(K).

35. Operations on full subcategories

0FX0 Let T be a triangulated category. We will identify full subcategories of T withsubsets 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 allobjects 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 areisomorphic to direct summands of objects of A,

(3) add(A) be the full subcategory of T consisting of all objects which areisomorphic 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 fitinto 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 ofT whose single object is E. Then we can consider things like add(E[−1, 2]) and soon and so forth. We warn the reader that this notation is not universally accepted.

DERIVED CATEGORIES 105

Lemma 35.1.0FX1 Let T be a triangulated category. Given full subcategories A, B, Cwe have (A ? B) ? C = A ? (B ? C).

Proof. If we have distinguished triangles A → X → B and X → Y → C then byAxiom TR4 we have distinguished triangles A→ Y → Z and B → Z → C.

Lemma 35.2.0FX2 Let T be a triangulated category. Given full subcategories A, B wehave 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 theinclusion. The equality follows trivially from this.

Lemma 35.3.0FX3 Let T be a triangulated category. Given full subcategories A, B thefull subcategories add(A) ? add(B) and smd(add(A)) are closed under direct sums.

Proof. Namely, if A → X → B and A′ → X ′ → B′ are distinguished trianglesand A,A′ ∈ add(A) and B,B′ ∈ add(B) then A ⊕ A′ → X ⊕ X ′ → B ⊕ B′ is adistinguished triangle with A⊕ A′ ∈ add(A) and B ⊕ B′ ∈ add(B). The result forsmd(add(A)) is trivial.

Lemma 35.4.0FX4 Let T be a triangulated category. Given a full subcategory A forn ≥ 1 the subcategory

Cn = smd(add(A)?n) = smd(add(A) ? . . . ? add(A))defined above is a strictly full subcategory of T closed under direct sums and directsummands and Cn+m = smd(Cn ? Cm) for all n,m ≥ 1.

Proof. Immediate from Lemmas 35.1, 35.2, and 35.3.

Remark 35.5.0FX5 Let F : T → T ′ be an exact functor of triangulated categories.Given a full subcategory A of T we denote F (A) the full subcategory of T ′ whoseobjects 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.

Remark 35.6.0FX6 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,

DERIVED CATEGORIES 106(⋃Ai)?n

=⋃A?ni .

We omit the trivial verifications.

Lemma 35.7.0FX7 Let A be an abelian category. Let D = D(A). Let E ⊂ Ob(A) bea 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 ifi ∈ [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 fori 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 insmd(add(E [a, b])?(b−a+1)).

Proof. We will use Lemma 35.4 without further mention. We will prove (2) whichtrivially implies (1). We use induction on b− a. If b− a = 0, then K is isomorphicto Hi(K)[−a] in D and the result is immediate. If b− a > 0, then we consider thedistinguished triangle

τ≤b−1K• → K• → Kb[−b]

and we conclude by induction on b− a. We omit the proof of (3) and (4).

Lemma 35.8.0FX8 Let T be a triangulated category. Let H : T → A be a homologicalfunctor to an abelian category A. Let a ≤ b and E ⊂ Ob(T ) be a subset such thatHi(E) = 0 for E ∈ E and i 6∈ [a, b]. Then for X ∈ smd(add(E [−m,m])?n) we haveHi(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 fora 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 ofadd(A) is different).Let D be a triangulated category. Let E be an object of D. Denote 〈E〉1 the strictlyfull subcategory of D consisting of objects in D isomorphic to direct summands offinite direct sums ⊕

i=1,...,rE[ni]

of shifts of E. It is clear that in the notation of Section 35 we have〈E〉1 = smd(add(E[−∞,∞]))

For n > 1 let 〈E〉n denote the full subcategory of D consisting of objects of Disomorphic to direct summands of objects X which fit into a distinguished triangle

A→ X → B → A[1]where A is an object of 〈E〉1 and B an object of 〈E〉n−1. In the notation of Section35 we have

〈E〉n = smd(〈E〉1 ? 〈E〉n−1)

DERIVED CATEGORIES 107

Each of the categories 〈E〉n is a strictly full additive (by Lemma 35.3) subcategory ofD preserved under shifts and under taking summands. But, 〈E〉n is not necessarilyclosed under “taking cones” or “extensions”, hence not necessarily a triangulatedsubcategory. This will be true for the subcategory

〈E〉 =⋃

n〈E〉n

as will be shown in the lemmas below.

Lemma 36.1.0FX9 Let T be a triangulated category. Let E be an object of T . Forn ≥ 1 we have〈E〉n = smd(〈E〉1 ? . . . ? 〈E〉1) = smd(〈E〉1?n) =

⋃m≥1

smd(add(E[−m,m])?n)

For n, n′ ≥ 1 we have 〈E〉n+n′ = smd(〈E〉n ? 〈E〉n′).

Proof. The left equality in the displayed formula follows from Lemmas 35.1 and35.2 and induction. The middle equality is a matter of notation. Since 〈E〉1 =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 finalstatement follows from the remark and the corresponding statement of Lemma35.4.

Lemma 36.2.0ATG Let D be a triangulated category. Let E be an object of D. Thesubcategory

〈E〉 =⋃

n〈E〉n =

⋃n,m≥1

smd(add(E[−m,m])?n)

is a strictly full, saturated, triangulated subcategory of D and it is the smallest suchsubcategory of D containing the object E.

Proof. The equality on the right follows from Lemma 36.1. It is clear that 〈E〉 =⋃〈E〉n contains E, is preserved under shifts, direct sums, direct summands. If

A ∈ 〈E〉a and B ∈ 〈E〉b and if A → X → B → A[1] is a distinguished triangle,then X ∈ 〈E〉a+b by Lemma 36.1. Hence

⋃〈E〉n is also preserved under extensions

and it follows that it is a triangulated subcategory.Finally, let D′ ⊂ D be a strictly full, saturated, triangulated subcategory of Dcontaining E. Then D′[−∞,∞] ⊂ D′, add(D) ⊂ D′, smd(D′) ⊂ D′, and D′ ?D′ ⊂D′. In other words, all the operations we used to construct 〈E〉 out of E preserveD′. Hence 〈E〉 ⊂ D′ and this finishes the proof.

Definition 36.3.09SJ 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 〈E〉 = D.

(2) We say E is a strong generator of D if 〈E〉n = 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.

Lemma 36.4.09SK Let D be a triangulated category. Let E,K be objects of D. Thefollowing are equivalent

(1) Hom(E,K[i]) = 0 for all i ∈ Z,(2) Hom(E′,K) = 0 for all E′ ∈ 〈E〉.

DERIVED CATEGORIES 108

Proof. The implication (2) ⇒ (1) is immediate. Conversely, assume (1). ThenHom(X,K) = 0 for all X in 〈E〉1. Arguing by induction on n and using Lemma4.2 we see that Hom(X,K) = 0 for all X in 〈E〉n.

Lemma 36.5.09SL Let D be a triangulated category. Let E be an object of D. If E isa classical generator of D, then E is a generator.

Proof. Assume E is a classical generator. Let K be an object of D such thatHom(E,K[i]) = 0 for all i ∈ Z. By Lemma 36.4 Hom(E′,K) = 0 for all E′ in 〈E〉.However, since D = 〈E〉 we conclude that idK = 0, i.e., K = 0.

Lemma 36.6.0FXA Let D be a triangulated category which has a strong generator. LetE be an object of D. If E is a classical generator of D, then E is a strong generator.

Proof. Let E′ be an object of D such that D = 〈E′〉n. Since D = 〈E〉 we see thatE′ ∈ 〈E〉m for some m ≥ 1 by Lemma 36.2. Then 〈E′〉1 ⊂ 〈E〉m hence

D = 〈E′〉n = smd(〈E′〉1 ? . . . ? 〈E′〉1) ⊂ smd(〈E〉m ? . . . ? 〈E〉m) = 〈E〉nmas desired. Here we used Lemma 36.1.

Remark 36.7.0ATH Let D be a triangulated category. Let E be an object of D. LetT be a property of objects of D. Suppose that

(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 〈E〉.

37. Compact objects

09SM Here is the definition.

Definition 37.1.07LS Let D be an additive category with arbitrary direct sums. Acompact object of D is an object K such that the map⊕

i∈IHomD(K,Ei) −→ HomD(K,

⊕i∈I

Ei)

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

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

Lemma 37.2.09QH Let D be a (pre-)triangulated category with direct sums. Thenthe 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 ofD withX and Y compact.Then it follows from Lemma 4.2 and the five lemma (Homology, Lemma 5.20) thatZ is a compact object too. It is clear that if X ⊕ Y is compact, then X, Y arecompact objects too. Hence Dc is a saturated triangulated subcategory. Since D isKaroubian by Lemma 4.14 we conclude that the same is true for Dc.

DERIVED CATEGORIES 109

Lemma 37.3.09SN Let D be a triangulated category with direct sums. Let Ei, i ∈ I bea family of compact objects of D such that

⊕Ei generates D. Then every object

X of D can be written asX = hocolimXn

where X1 is a direct sum of shifts of the Ei and each transition morphism fits intoa distinguished triangle Yn → Xn → Xn+1 → Yn[1] where Yn is a direct sum ofshifts of the Ei.

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 witha 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, andϕ : Ei[m] → Xn is a morphism such that Ei[m] → Xn → X is zero. Choosea distinguished triangle Yn → Xn → Xn+1 → Yn[1] and let Xn+1 → X be anymorphism such that Xn → Xn+1 → X is the given one; such a morphism exists byour choice of Yn. We obtain a morphism hocolimXn → X by the construction ofour 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 theconstruction of Yn shows that the composition Ei[m] → Xn → Xn+1 is zero. Inother words, the composition Ei[m] → C → hocolimXn is zero. This means thatour morphism Ei[m]→ C comes from a morphism Ei[m]→ X[−1]. The construc-tion of X1 then shows that such morphism lifts to hocolimXn and we conclude thatour morphism Ei[m]→ C is zero. The assumption that

⊕Ei generates D implies

that C is zero and the proof is done.

Lemma 37.4.09SP With assumptions and notation as in Lemma 37.3. If C is acompact object and C → Xn is a morphism, then there is a factorization C → E →Xn where E is an object of 〈Ei1 ⊕ . . .⊕ Eit〉 for some i1, . . . , it ∈ I.

Proof. We prove this by induction on n. The base case n = 1 is clear. If n > 1consider the composition C → Xn → Yn−1[1]. This can be factored through someE′[1] → Yn−1[1] where E′ is a finite direct sum of shifts of the Ei. Let I ′ ⊂ I bethe finite set of indices that occur in this direct sum. Thus we obtain

E′ //

C ′ //

C //

E′[1]

Yn−1 // Xn−1 // Xn

// Yn−1[1]

By induction the morphism C ′ → Xn−1 factors through E′′ → Xn−1 with E′′ anobject of 〈

⊕i∈I′′ Ei〉 for some finite subset I ′′ ⊂ I. Choose a distinguished triangle

E′ → E′′ → E → E′[1]then E is an object of 〈

⊕i∈I′∪I′′ Ei〉. By construction and the axioms of a tri-

angulated category we can choose morphisms C → E and a morphism E → Xn

fitting into morphisms of triangles (E′, C ′, C) → (E′, E′′, E) and (E′, E′′, E) →(Yn−1, Xn−1, Xn). The composition C → E → Xn may not equal the given mor-phism C → Xn, but the compositions into Yn−1 are equal. Let C → Xn−1 be

DERIVED CATEGORIES 110

a morphism that lifts the difference. By induction assumption we can factor thisthrough a morphism E′′′ → Xn−1 with E′′ an object of 〈

⊕i∈I′′′ Ei〉 for some finite

subset I ′ ⊂ I. Thus we see that we get a solution on considering E ⊕ E′′′ → Xn

because E ⊕ E′′′ is an object of 〈⊕

i∈I′∪I′′∪I′′′ Ei〉.

Definition 37.5.09SQ 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 objectssuch that

⊕Ei generates D.

The following proposition clarifies the relationship between classical generators andweak generators.

Proposition 37.6.09SR Let D be a triangulated category with direct sums. Let E be acompact 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 = 〈E〉. It follows formallyfrom the assumption that D is compactly generated and Lemma 36.4 that E is agenerator for D.

The converse is more interesting. Assume that E is a generator for D. Let X be acompact 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 throughXn for some n (Lemma 33.9). Thus X is a direct summand of Xn. By Lemma 37.4we see that X is an object of 〈E〉 and the lemma is proven.

38. Brown representability

0A8E A reference for the material in this section is [Nee96].

Lemma 38.1.0A8F [Nee96, Theorem3.1].

Let D be a triangulated category with direct sums which is com-pactly generated. Let H : D → Ab be a contravariant cohomological functor whichtransforms direct sums into products. Then H is representable.

Proof. Let Ei, i ∈ I be a set of compact objects such that⊕

i∈I Ei generatesD. 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

X1 =⊕

(i,a)Ei

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 elementa1 defines a natural transformation H1 → H.

We are going to inductively construct Xn and transformations an : Hn → Hwhere Hn = HomD(−, Xn). Namely, we apply the procedure above to the functorKer(Hn → H) to get an object

Kn+1 =⊕

(i,k), k∈Ker(Hn(Ei)→H(Ei))Ei

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and a transformation HomD(−,Kn+1) → Ker(Hn → H). By Yoneda’s lemma thecomposition HomD(−,Kn+1) → Hn gives a morphism Kn+1 → Xn. We choose adistinguished 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. SinceH 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 distin-guished triangle ⊕

Xn →⊕

Xn → X →⊕

Xn[1]

we obtain an exact sequence∏H(Xn)←

∏H(Xn)← H(X)

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 construc-tion of X1. On the other hand, by construction of Xn → Xn+1 the kernel ofHomD(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

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, thesubcategory D′ is a strictly full, saturated, triangulated subcategory of D (detailsomitted; see proof of Lemma 6.3). Moreover, as bothH and HomD(−, X) transformdirect sums into products, we see that direct sums of objects of D′ are in D′. Thusderived colimits of objects of D′ are in D′. Since Ei is preserved under shifts, wesee that Ei is an object of D′ for all i. It follows from Lemma 37.3 that D′ = Dand the proof is complete.

Proposition 38.2.0A8G [Nee96, Theorem4.1].

Let D be a triangulated category with direct sums which iscompactly generated. Let F : D → D′ be an exact functor of triangulated categorieswhich transforms direct sums into direct sums. Then F has an exact right adjoint.

Proof. For an object Y of D′ consider the contravariant functor

D → Ab, W 7→ HomD′(F (W ), Y )

This is a cohomological functor as F is exact and transforms direct sums intoproducts as F transforms direct sums into direct sums. Thus by Lemma 38.1 wefind an object X of D such that HomD(W,X) = HomD′(F (W ), Y ). The existenceof the adjoint follows from Categories, Lemma 24.2. Exactness follows from Lemma7.1.

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39. Admissible subcategories

0CQP A reference for this section is [BK89, Section 1].

Definition 39.1.0FXB 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 ofD such that Hom(A,X) = 0 for all A ∈ Ob(A). The left orthogonal ⊥A of A is thefull subcategory consisting of the objects X of D such that Hom(X,A) = 0 for allA ∈ Ob(A).

Lemma 39.2.0CQQ Let D be a triangulated category. Let A ⊂ D be a full subcategoryinvariant 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 along exact sequence as in (3.5.1). Assume (1) and let A ∈ Ob(A). Then we considerthe 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.

Lemma 39.3.0FXC Let D be a triangulated category. Let A ⊂ D be a full subcategoryinvariant under all shifts. Then both the right orthogonal A⊥ and the left orthogonal⊥A of A are strictly full, saturated6, triangulated subcagories of D.

Proof. It is immediate from the definitions that the orthogonals are preservedunder taking shifts, direct sums, and direct summands. Consider a distinguishedtriangle

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 inA⊥. This is immediate from Lemma 39.2.

Lemma 39.4.0CQR Let D be a triangulated category. Let A be a full triangulatedsubcategory of D. For an object X of D consider the property P (X): there exists adistinguished 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 twoout 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 Pholds for X1 and X2. Choose distinguished triangles

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

6Definition 6.1.

DERIVED CATEGORIES 113

as in condition P . Since Hom(A1, A2) = Hom(A1, X2) by Lemma 39.2 there is aunique 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 henceQ1, Q2 ∈ Ob(A⊥). As Q1 → Q2 → Q3 → Q1[1] is a distinguished triangle we seethat Q3 ∈ Ob(A⊥) by Lemma 39.3. Since A is a full triangulated subcategory, wesee 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 Lemma4.11.

Lemma 39.5.0CQS Let D be a triangulated category. Let A ⊂ D be a full triangulatedsubcategory. 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 → X ′[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, thenA = ⊥(A⊥).

Proof. Assume (1) and denote v : D → A the right adjoint. Let X ∈ Ob(D). SetA = v(X). We may extend the adjunction mapping A → X to a distinguishedtriangle A→ X → B → A[1]. Since

HomA(A′, A) = HomA(A′, v(X)) = HomD(A′, X)

for A′ ∈ 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). ChooseA → X → B → A[1] as in (2). Set v(X) = A. Let f : X → Y be a morphism inD. Choose A′ → Y → B′ → A′[1] as in (2). Since Hom(A,A′) = Hom(A, Y ) by

DERIVED CATEGORIES 114

Lemma 39.2 there is a unique morphism f ′ : A→ A′ such that the diagramA

f ′

// X

f

A′ // Y

commutes. Hence we can set v(f) = f ′ to get a functor. To see that v is adjoint tothe inclusion morphism use Lemma 39.2 again.Proof of the final statement. In other to prove that A is saturated we may replaceA by the strictly full subcategory having the same isomorphism classes as A; detailsomitted. Assume A is strictly full. If we show that A = ⊥(A⊥), then A will besaturated by Lemma 39.3. Since the incusion A ⊂ ⊥(A⊥) is clear it suffices toprove the other inclusion. Let X be an object of ⊥(A⊥). Choose a distinguishedtriangle A → X → B → A[1] as in (2). As Hom(X,B) = 0 by assumption we seethat A ∼= X ⊕ B[−1] by Lemma 4.11. Since Hom(A,B[−1]) = 0 as B ∈ A⊥ thisimplies B[−1] = 0 and A ∼= X as desired.

Lemma 39.6.0CQT Let D be a triangulated category. Let A ⊂ D be a full triangulatedsubcategory. 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→ K[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, thenA = (⊥A)⊥.

Proof. Omitted. Dual to Lemma 39.5.

Definition 39.7.0FXD LetD be a triangulated category. A right admissible subcategoryof D is a strictly full triangulated subcategory satisfying the equivalent conditionsof Lemma 39.5. A left admissible subcategory of D is a strictly full triangulatedsubcategory satisfying the equivalent conditions of Lemma 39.6. A two-sided ad-missible 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. LetXn → Xn−1 → . . .→ X0

be a complex in D. In this section we consider the problem of constructing a“totalization” of this complex.

Definition 40.1.0D7Z Let D be a triangulated category. LetXn → 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.

DERIVED CATEGORIES 115

(3) If n > 1, then it is a choice of a Postnikov system for Xn−1 → . . . → X0and 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 givenmorphism Xn → Xn−1.

Given a morphism

(40.1.1)0D80

Xn//

Xn−1 //

. . . // X0

X ′n // X ′n−1

// . . . // X ′0

between complexes of the same length in D there is an obvious notion of a morphismof Postnikov systems.

Here is a key example.

Example 40.2.0D8Z Let A be an abelian category. Let . . . → A2 → A1 → A0 be achain 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 systemas in Definition 40.1 for . . . → X2 → X1 → X0. Here we are using the obviousextension of Postnikov systems for an infinite complex of D(A). Finally, if colimitsover 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 Definition40.1 we can consider the maps

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

Y0 // Y1[1]

// Y2[2]

// Y3[3]

X1[1]

+1aa

X2[2]+1oo

+1cc

X3[3]+1oo

+1cc

We encourage the reader to think of Yn[n] as obtained fromX0, X1[1], . . . , Xn[n]; forexample if the maps Xi → Xi−1 are zero, then we can take Yn[n] =

⊕i=0,...,nXi[i].

Postnikov systems do not always exist. Here is a simple lemma for low n.

Lemma 40.3.0D81 Let D be a triangulated category. Consider Postnikov systems forcomplexes of length n.

(1) For n = 0 Postnikov systems always exist and any morphism (40.1.1) ofcomplexes extends to a unique morphism of Postnikov systems.

(2) For n = 1 Postnikov systems always exist and any morphism (40.1.1) ofcomplexes extends to a (nonunique) morphism of Postnikov systems.

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(3) For n = 2 Postnikov systems always exist but morphisms (40.1.1) of com-plexes 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 casen = 1 follows immediately from TR1 (existence of triangles) and TR3 (extendingmorphisms to triangles). For the case n = 2 we argue as follows. Set Y0 = X0. Bythe case n = 1 we can choose a Postnikov system

Y1 → X1 → Y0 → Y1[1]Since the compositionX2 → X1 → X0 is zero, we can factorX2 → X1 (nonuniquely)as X2 → Y1 → X1 by Lemma 4.2. Then we simply fit the morphism X2 → Y1 intoa distinguished triangle

Y2 → X2 → Y1 → Y2[1]to get the Postnikov system for n = 2. For n > 2 we cannot argue similarly, as wedo not know whether the composition Xn → Xn−1 → Yn−1 is zero in D.

Lemma 40.4.0D82 Let D be a triangulated category. Given a map (40.1.1) considerthe condition(40.4.1)0DW1 Hom(Xi[i− j − 1], X ′j) = 0 for i > j + 1Then

(1) If we have a Postnikov system for X ′n → X ′n−1 → . . . → X ′0 then property(40.4.1) implies that

Hom(Xi[i− j − 1], Y ′j ) = 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 isnothing to prove as Y ′0 → X ′0 is an isomorphism. Say the result holds for j− 1. Weconsider the distinguished triangle

Y ′j → X ′j → Y ′j−1 → Y ′j [1]The long exact sequence of Lemma 4.2 gives an exact sequenceHom(Xi[i− j − 1], Y ′j−1[−1])→ Hom(Xi[i− j − 1], Y ′j )→ Hom(Xi[i− j − 1], X ′j)From the induction hypothesis and (40.4.1) we conclude the outer groups are zeroand we win.Proof of (2). For n = 1 the existence of morphisms has been established in Lemma40.3. For n > 1 by induction, we may assume given the map of Postnikov systemsof length n− 1. The problem is that we do not know whether the diagram

Xn//

Yn−1

X ′n // Y ′n−1

is commutative. Denote α : Xn → Y ′n−1 the difference. Then we do know thatthe composition of α with Y ′n−1 → X ′n−1 is zero (because of what it means to be amap of Postnikov systems of length n− 1). By the distinguished triangle Y ′n−1 →

DERIVED CATEGORIES 117

X ′n−1 → Y ′n−2 → Y ′n−1[1], this means that α is the composition of Y ′n−2[−1]→ Y ′n−1with a map α′ : Xn → Y ′n−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 commutativityguarantees we can choose the dotted arrow fitting into the diagram

Yn−1[−1]

// Yn //

Xn//

Yn−1

Y ′n−1[−1] // Y ′n // X ′n // Y ′n−1

by TR3.

Lemma 40.5.0FXE Let D be a triangulated category. Given a map (40.1.1) assume weare given Postnikov systems for both complexes. If

(1) Hom(Xi[i], Y ′n[n]) = 0 for i = 1, . . . , n, or(2) Hom(Yn[n], X ′n−i[n− i]) = 0 for i = 1, . . . , n, or(3) Hom(Xj−i[−i+ 1], X ′j) = 0 and Hom(Xj , X

′j−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] //

uu

. . . // Yn[n]

rrY ′n[n]

The arrows are the composition of the morphism Yn[n]→ Y ′n[n] and the morphismYi[i] → Yn[n]. The arrow Y0 → Y ′n[n] is determined as it is the compositionY0 = X0 → X ′0 = Y ′0 → Y ′n[n]. Since we have the distinguished triangle Y0 →Y1[1]→ X1[1] we see that Hom(X1[1], Y ′n[n]) = 0 guarantees that the second verticalarrow is unique. Since we have the distinguished triangle Y1[1]→ Y2[2]→ X2[2] wesee that Hom(X2[2], Y ′n[n]) = 0 guarantees that the third vertical arrow is unique.And so on.Proof of (2). The composition Yn[n] → Y ′n[n] → Xn[n] is the same as the com-position Yn[n] → Xn[n] → X ′n[n] and hence is unique. Then using the distin-guished triangle Y ′n−1[n − 1] → Y ′n[n] → X ′n[n] we see that it suffices to showHom(Yn[n], Y ′n−1[n− 1]) = 0. Using the distinguished triangles

Y ′n−i−1[n− i− 1]→ Y ′n−i[n− i]→ X ′n−i[n− i]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 nit suffices to show that the dotted arrow in the morphism of triangles

Yn−1[−1]

// Yn //

Xn//

Yn−1

Y ′n−1[−1] // Y ′n // X ′n // Y ′n−1

is unique. By Lemma 4.8 part (5) it suffices to show that Hom(Yn−1, X′n) = 0

and Hom(Xn, Y′n−1[−1]) = 0. To prove the first vanishing we use the distinguished

DERIVED CATEGORIES 118

triangles Yn−i−1[−i]→ Yn−i[−(i− 1)]→ Xn−i[−(i− 1)] for i > 0 and induction oni to see that the assumed vanishing of Hom(Xn−i[−i + 1], X ′n) is enough. For thesecond we similarly use the distinguished triangles Y ′n−i−1[−i − 1] → Y ′n−i[−i] →X ′n−i[−i] to see that the assumed vanishing of Hom(Xn, X

′n−i[−i]) is enough as

well.

Lemma 40.6.0D83 Let D be a triangulated category. Let Xn → Xn−1 → . . .→ X0 bea complex in D. If

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

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

Proof. We argue by induction on n. The cases n = 0, 1, 2 follow from Lemma40.3. 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 oflength n is that we have to find a morphism Xn → Yn−1 such that the compositionXn → Yn−1 → Xn−1 is equal to the given map Xn → Xn−1. Considering thedistinguished 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 compositionXn → Xn−1 →Yn−2 is zero. Since we know that Xn → Xn−1 → Xn−2 is zero we can apply thedistinguished 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 ofLemma 40.4 part (1) the reader easily sees this follows from the condition statedin the lemma.The statement on isomorphisms follows from the existence of a map between thePostnikov systems extending the identity on the complex proven in Lemma 40.4part (2) and Lemma 4.3 to show all the maps are isomorphisms.

41. Essentially constant systems

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

Lemma 41.1.0G39 Let D be a triangulated category. Let (Ai) be an inverse system inD. Then (Ai) is essentially constant (see Categories, Definition 22.1) if and onlyif there exists an i and for all j ≥ i a direct sum decomposition Aj = A⊕ Zj suchthat (a) the maps Aj′ → Aj are compatible with the direct sum decompositions andidentity on A, (b) for all j ≥ i there exists some j′ ≥ j such that Zj′ → Zj is zero.

Proof. Assume (Ai) is essentially constant with value A. Then A = limAi andthere exists an i and a morphism Ai → A such that (1) the composition A→ Ai →A is the identity on A and (2) for all j ≥ i there exists a j′ ≥ j such that Aj′ → Ajfactors as Aj′ → Ai → A → Aj . From (1) we conclude that for j ≥ i the mapsA→ Aj and Aj → Ai → A compose to the identity on A. It follows that Aj → Ahas a kernel Zj and that the map A ⊕ Zj → Aj is an isomorphism, see Lemmas

DERIVED CATEGORIES 119

4.12 and 4.11. These direct sum decompositions clearly satisfy (a). From (2) weconclude that for all j there is a j′ ≥ j such that Zj′ → Zj is zero, so (b) holds.Proof of the converse is omitted.

Lemma 41.2.0G3A Let D be a triangulated category. LetAn → Bn → Cn → An[1]

be an inverse system of distinguished triangles in D. If (An) and (Cn) are essen-tially constant, then (Bn) is essentially constant and their values fit into a distin-guished 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 limn′≥nAn′ → A and sim-ilarly for B and C.

Proof. After renumbering we may assume that An = A ⊕ A′n and Cn = C ⊕ C ′nfor inverse systems (A′n) and (C ′n) which are essentially zero, see Lemma 41.1. Inparticular, the morphism

C ⊕ C ′n → (A⊕A′n)[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 → Cδ−→ 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 commutativediagram

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

HomD(B,D) //

HomD(A,D) //

. . .

. . . // colim HomD(Cn, D) // colim HomD(Bn, D) // colim HomD(An, D) // . . .

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

Lemma 41.3.0G3B Let A be an abelian category. Let An be an inverse system ofobjects of D(A). Assume

(1) there exist integers a ≤ b such that Hi(An) = 0 for i 6∈ [a, b], and(2) the inverse systems Hi(An) of A are essentially constant for all i ∈ Z.

Then An is an essentially constant system of D(A) whose value A satisfies thatHi(A) is the value of the constant system Hi(An) for each i ∈ Z.

DERIVED CATEGORIES 120

Proof. By Remark 12.4 we obtain an inverse system of distinguished trianglesτ≤aAn → An → τ≥a+1An → (τ≤aAn)[1]

Of course we have τ≤aAn = Ha(An)[−a] in D(A). Thus by assumption these forman essentially constant system. By induction on b − a we find that the inversesystem τ≥a+1An is essentially constant, say with value A′. By Lemma 41.2 we findthat An is an essentially constant system. We omit the proof of the statement oncohomologies (hint: use the final part of Lemma 41.2).

Lemma 41.4.0G3C Let D be a triangulated category. LetAn → Bn → Cn → An[1]

be an inverse system of distinguished triangles. If the system Cn is pro-zero (essen-tially constant with value 0), then the maps An → Bn determine a pro-isomorphismbetween the pro-object (An) and the pro-object (Bn).

Proof. For any object X of D consider the exact sequencecolim 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 as-sumption 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 andCategories, Remark 22.7.

Lemma 41.5.0G3D Let A be an abelian category.An → Bn

be an inverse system of maps of D(A). Assume(1) there exist integers a ≤ b such that Hi(An) = 0 and Hi(Bn) = 0 for

i 6∈ [a, b], and(2) the inverse system of maps Hi(An)→ Hi(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).

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

Hp(An) αn−−→ Hp(Bn) βn−−→ Hp(Cn) δn−→ Hp+1(An) εn−→ Hp+1(Bn)Namely, for every n we can find an m > n such that Im(βm) maps to zero inHp(Cn) because we may choose m such that Hp(Bm) → Hp(Bn) factors throughαn : Hp(An) → Hp(Bn). For a similar reason we may then choose k > m suchthat Im(δk) maps to zero in Hp+1(Am). Then Hp(Ck) → Hp(Cn) is zero be-cause Hp(Ck) → Hp(Cm) maps into Ker(δm) and Hp(Cm) → Hp(Cn) annihilatesKer(δm) = Im(βm).

DERIVED CATEGORIES 121

42. Other chapters

Preliminaries(1) Introduction(2) Conventions(3) Set Theory(4) Categories(5) Topology(6) Sheaves on Spaces(7) Sites and Sheaves(8) Stacks(9) Fields(10) Commutative Algebra(11) Brauer Groups(12) Homological Algebra(13) Derived Categories(14) Simplicial Methods(15) More on Algebra(16) Smoothing Ring Maps(17) Sheaves of Modules(18) Modules on Sites(19) Injectives(20) Cohomology of Sheaves(21) Cohomology on Sites(22) Differential Graded Algebra(23) Divided Power Algebra(24) Differential Graded Sheaves(25) Hypercoverings

Schemes(26) Schemes(27) Constructions of Schemes(28) Properties of Schemes(29) Morphisms of Schemes(30) Cohomology of Schemes(31) Divisors(32) Limits of Schemes(33) Varieties(34) Topologies on Schemes(35) Descent(36) Derived Categories of Schemes(37) More on Morphisms(38) More on Flatness(39) Groupoid Schemes(40) More on Groupoid Schemes(41) Étale Morphisms of Schemes

Topics in Scheme Theory(42) Chow Homology

(43) Intersection Theory(44) Picard Schemes of Curves(45) Weil Cohomology Theories(46) Adequate Modules(47) Dualizing Complexes(48) Duality for Schemes(49) Discriminants and Differents(50) de Rham Cohomology(51) Local Cohomology(52) Algebraic and Formal Geometry(53) Algebraic Curves(54) Resolution of Surfaces(55) Semistable Reduction(56) Derived Categories of Varieties(57) Fundamental Groups of Schemes(58) Étale Cohomology(59) Crystalline Cohomology(60) Pro-étale Cohomology(61) More Étale Cohomology(62) The Trace Formula

Algebraic Spaces(63) Algebraic Spaces(64) Properties of Algebraic Spaces(65) Morphisms of Algebraic Spaces(66) Decent Algebraic Spaces(67) Cohomology of Algebraic Spaces(68) Limits of Algebraic Spaces(69) Divisors on Algebraic Spaces(70) Algebraic Spaces over Fields(71) Topologies on Algebraic Spaces(72) Descent and Algebraic Spaces(73) Derived Categories of Spaces(74) More on Morphisms of Spaces(75) Flatness on Algebraic Spaces(76) Groupoids in Algebraic Spaces(77) More on Groupoids in Spaces(78) Bootstrap(79) Pushouts of Algebraic Spaces

Topics in Geometry(80) Chow Groups of Spaces(81) Quotients of Groupoids(82) More on Cohomology of Spaces(83) Simplicial Spaces(84) Duality for Spaces(85) Formal Algebraic Spaces(86) Restricted Power Series

DERIVED CATEGORIES 122

(87) Resolution of Surfaces RevisitedDeformation Theory

(88) Formal Deformation Theory(89) Deformation Theory(90) The Cotangent Complex(91) Deformation Problems

Algebraic Stacks(92) Algebraic Stacks(93) Examples of Stacks(94) Sheaves on Algebraic Stacks(95) Criteria for Representability(96) Artin’s Axioms(97) Quot and Hilbert Spaces(98) Properties of Algebraic Stacks(99) Morphisms of Algebraic Stacks

(100) Limits of Algebraic Stacks(101) Cohomology of Algebraic Stacks

(102) Derived Categories of Stacks(103) Introducing Algebraic Stacks(104) More on Morphisms of Stacks(105) The Geometry of Stacks

Topics in Moduli Theory(106) Moduli Stacks(107) Moduli of Curves

Miscellany(108) Examples(109) Exercises(110) Guide to Literature(111) Desirables(112) Coding Style(113) Obsolete(114) GNU Free Documentation Li-

cense(115) Auto Generated Index

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