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Homological AlgebraLectures by Frauke Bleher

Notes by Roman Aranda

Spring 2018

Preface

This notes are based from a course in Homological Algebra taught by the University of Iowaprofessor Frauke Bleher. As a personal opinion, the course was almost self-contained, withthe only prerequisite of being fluent in basic notions of modern algebra.

The course is roughly divided in two parts. The first part is devoted to understand importantconstructions in category theory such as Hom functors, limits and products; and to closelystudy the category of modules over a given ring and the so called functor categories. Chapter4 is a short discussion of the axioms of set theory with an important result on when functorscan be represented as Hom functors. This part is finished in Chapter 5 with a proof ofa characterization of when two module categories are equivalent, this is the beginning ofMorita Theory.

Second part is tittled Classical Homological Algebra. I am not sure how classic it is but fromthe personal point of view of a topologist, this part can be thought as the algebra behind(or around) the Homology theories seen in a first course of Algebraic Topology. Chapter 6studies the category of chain complexes enough to define the Ext and Tor functors on a fixedmodule category. Chapter 7 is in charge of proving that such Ext and Tor functors can beexpressed as Hom functors between certain categories. This chapter studies sibilings of themodule categories such as abelian, homotopy and derived categories.

Feel free to send me an e-mail about typos, questions or comments regarding this notes [email protected]

Enjoy!

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Contents

I Category Theory 7

1 Cats 9

1.1 Categories and Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.2 Module Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2.1 Products and coproducts . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.2.2 Exactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.3 Opposite categories. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.4 Adjoints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2 Limits 19

2.1 Exactness of limit functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3 Module Categories 31

3.1 Projective Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.2 Injective Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.3 Natural transformations: An interlude. . . . . . . . . . . . . . . . . . . . . . 38

3.4 Flat Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.5 Schannel’s Lemma and Stable module categories . . . . . . . . . . . . . . . . 43

4 Functor Categories 49

5

6 CONTENTS

4.1 The Godel-Bernays System . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.2 Functor categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.2.1 Representable Functors . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5 Morita Theory 57

II Classical Homological Algebra 65

6 Homological Algebra 67

6.1 Chain stuff . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

6.2 Derived Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

6.3 Long Exact Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

6.4 Tor and Ext of abelian groups . . . . . . . . . . . . . . . . . . . . . . . . . . 84

7 Abelian Categories 89

7.1 Abelian Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

7.2 Total complex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

7.3 Homotopy categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

7.3.1 Mapping cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

7.3.2 Triangle Axioms for K(A) . . . . . . . . . . . . . . . . . . . . . . . . 102

7.4 Localization of Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

7.4.1 How to localize triangulated categories? . . . . . . . . . . . . . . . . 110

7.5 Derived categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

7.6 Derived functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

7.6.1 Ext and RHom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

7.6.2 Total tensor product and ⊗L . . . . . . . . . . . . . . . . . . . . . . . 130

Part I

Category Theory

7

Chapter 1

Cats

1.1 Categories and Functors

Definition 1.1.1. Definition of a category. A category C consists of the following data:

(a) a collection of objects denoted by Ob(C),

(b) for every pair X, Y ∈ Ob(C), a set of morphisms denoted by HomC(A,B),

(c) for every triple A,B,C ∈ Ob(C), a law of composition

HomC(A,B)×HomC(B,C)→HomC(A,C)

(f, g) 7→g f

such that

(i) if (A,B) 6= (A′, B′) for A,B,A′, B′ ∈ Ob(C) then HomC(A,B) ∩HomC(A′, B′) =∅;

(ii) for every A ∈ Ob(C), there exists a unique morphism idA ∈ HomC(A,A) (calledthe identity morphism) such that for everyX, Y ∈ Ob(C) and for all f ∈ HomC(X,A),g ∈ HomC(A, Y ), we have that idA f = f and g idA = g;

(iii) the law of composition if associative.

As a notation, we write f : A→ B for a morphism f ∈ Hom(A,B).

Definition 1.1.2. A morphism f : A→ B in C is an isomorphism if and only if there existsg : B → A in C an inverse morphism.

Examples of categories are the category of sets, abelian groups (Ab); for R an associativering with 1, the category of left R−modules (R−mod) and right R−modules (mod−R).

9

10 CHAPTER 1. CATS

Notation. A ∈ Ob(Mod−R) is also written as A = AR; and B ∈ Ob(R−Mod) as B = RB.

Definition 1.1.3. Let C be a category. A subcategory of C is a category C ′ such that

1. Ob(C ′) ⊂ Ob(C)

2. For all A,B ∈ Ob(C ′), HomC′(A,B) ⊂ HomC(A,B) with induced composition lawand,

3. idA ∈ HomC′(A,A) for all A ∈ Ob(C ′).

If moreover HomC′(A,B) = HomC(A,B) for all A,B ∈ Ob(C ′), then C ′ is called a fullsubcategory.

Definition 1.1.4. Let C be a category and f : A→ B be a morphism in C.

(a) f is a monomorphism (or monic) iff ∀X ∈ Ob(C), ∀g, g′ : X → A with f g = f g′,then g = g′.

(b) f is an epimorphism (or epic) iff ∀Y ∈ Ob(C), ∀h, h′ : B → Y with h f = h′ f , thenh = h′.

(c) An object P ∈ Ob(C) is called initial iff for all Y ∈ Ob(C), HomC(P, Y ) has exactlyone element.

(d) An object Q ∈ Ob(C) is terminal iff for all X ∈ Ob(C), HomC(X,Q) has exactly oneelement.

Example 1.1.5. Consider C to be either R−Mod or Mod−R category and take f : A→ Ba morphism. One can show that the notions of monomorphism and epimorphism agree withthe notions of injectivity and surjectivity. Also the 0−module is both initial and terminal.

Definition 1.1.6 (Functor). Let C,D be two categories. A covariant functor (resp. con-travariant functor) F : C → D consists of the following data:

(a) A map F : Ob(C)→ Ob(D)

(b) For all A,B ∈ Ob(C), a map F : HomC(A,B) → HomD(FA, FB) (resp. F :HomC(A,B)→ HomD(FB,FA)).

such that F(idA) = idF(A) and F (f g) = F (g) F (f) (resp. F (f g) = F (g) F (f)).

Example 1.1.7 (Identity functor). The functor IdC : C → C defined by IdC(A) = A andIdC(f) = f for all objects A and morphisms f . This functor is covariant.

Example 1.1.8 (Hom functors). Fix X ∈ Ob(C)

1.2. MODULE CATEGORIES 11

(a) F = HomC(A,−) : C → Sets given by F(C) = HomC(A,C),∀C ∈ Ob(C) and F(f) :HomC(A,C)→ HomC(A,C

′) where F(f) : g 7→ f g is a covariant functor. We denoteF(f) = f∗.

(b) G = HomC(−, A) : C → Sets given by G(C) = HomC(C,A), ∀C ∈ Ob(C) and G(f) :HomC(C,A) → HomC(C

′, A) where G(h) : g 7→ g h is a contravariant functor. Wedenote G(f) = f ∗.

1.2 Module Categories

All rings are associative with 1 and all modules are unital.

Here, the most important functors are the Hom-functors and Tensor products.

(1) Hom-functors. Let C = R − mod or mod − R and let A,B ∈ Ob(C). We writeHomR(A,−) for HomC(A,−). If H is either HomR(A,−) or HomR(−, B), then H : C → Abis an additive functor; i.e. H(f + g) = H(f) +H(g) for all morphisms f, g : X → Y in C.

(2) Tensor product functors. Let A ∈ Ob(Mod − R) and B ∈ Ob(R − Mod). Thefunctors

A⊗R − : R−mod→ Ab

−⊗R B : mod−R→ Ab

are additive covariant functors.

We will now define the tensor product in a categorical way. We are skipping the detailsof the constructions since this is assumed to be known for the class. Recall that a mapϕ : A × B → L for A = AR, B = RB and L a Z−module, is called R−balanced if it isbilinear and ϕ(ar, b) = ϕ(a, rb).

Definition 1.2.1 (Tensor product). Let A = AR, B = RB and L a Z−module. Thetensor product of A and B over R is a Z−module A ⊗R B together with a R−balancedmap h : A × B → A ⊗R B satisfying the following universal mapping property: For allZ−module X with an R−balanced maps ϕ : A × B → X there exists a unique Z−modulehomomorphism φ : A⊗R B → X such that ϕ = φ h.

Theorem. The tensor product exists and is unique up to isomorphism.

More structure.

Definition 1.2.2. LetR, S be rings. An additive abelian group is called an (R, S)−bimodule,written as M = RMS, is an R−module and a module−S such that r(ms) = (rm)s for allr ∈ R, s ∈ S,M ∈M .

12 CHAPTER 1. CATS

Proposition 1.2.3.

1) Let A = SAR, B = RB. The tensor product A⊗RB is a left S−module with s(a⊗ b) =(sa)⊗ b (plus a Z−linear extension).

2) Let A = AR, B = RBS. The tensor product A⊗RB is a right S−module with (a⊗b)s =a⊗ (bs) (plus a Z−linear extension).

Corollary 1.2.4. Let A = SAR, B = RBS. Then the functors

A⊗R − : R−mod→ S −mod−⊗RB : mod−R→Mod− S

are additive covariant functors.

Proposition 1.2.5.

1) Let A = RAS, B = RB. Then HomR(A,B) is a left S−module with (sf)(a) = f(as)

2) Let A = RAS, B = BS. Then HomS(A,B) is a right R−module with (fr)(a) = f(ra).

3) Let A = RA, B = RBS. Then HomR(A,B) is a right S−module with (fs)(a) = f(a).s

4) Let A = AS, B = RBS. Then HomS(A,B) is a left R−module with (rf)(a) = r.f(a).

Corollary 1.2.6. Let A = RAS and B = RBS.

1) HomR(A,−) : R − mod → S − mod and HomS(A,−) : mod − S → mod − S areadditive covariant functors.

2) HomR(−, B) : R − mod → mod − S and HomS(−, B) : mod − S → R − mod areadditive contravariant functors.

1.2.1 Products and coproducts

Let C = R −mod or mod − R, let Λ be an arbitrary index set and (Aλ)λ∈Λ be a family ofmodules in C.

(1) Direct products. The module ΠλAλ with the underlying set (aλ)λ|aλ ∈ Aλ, λ ∈ Λwith component wise addition and R−action, together with projection homomorphisms

pµ :ΠAλ → Aµ

(aλ)λ 7→ aµ

1.2. MODULE CATEGORIES 13

satisfies the following universal mapping property: For all X ∈ Ob(C) with homomorphismsϕµ : X → Aµ for each µ ∈ Λ, there exists a unique map φ : X → ΠλAλ such that pµ φ = ϕµfor each µ.

The pair (ΠλAλ, pµ) is called the product (or direct product) of Aλλ. This is unique upto isomorphism (one can see it with the universal property with a standard argument).

(2) Direct sums or coproducts. The module qλAλ in C with underlying set (aλ)λ|aλ ∈Aλ, λ ∈ Λ with component wise addition and R−action, together with injection homomor-phisms

iµ :Aµ → qAλaµ 7→ (aλ)λ

where aλ = aµ for λ = µ and aλ = 0Aλ for λ 6= mu, satisfies the following universal mappingproperty: For all Y ∈ Ob(C) with homomorphisms ψµ : Amu → Y for each µ ∈ Λ, thereexists a unique map Ψ : qλAλ → Y such that ψµ = Ψ iµ for each µ.

The pair (qλ∈ΛAλ, iλ) is called a coproduct (or direct product) of Aλλ. This is (again)unique up to isomorphism.

One should expect them to behave nicely with respect to the Hom- and tensor functors. Thestudent should be careful to not confuse which goes with which.

Theorem 1.2.7. Let C = R−mod or mod−R. Let Aλλ∈Λ and Bλλ∈Λ be two familiesof modules in C. Fix A,B ∈ Ob(C).

(a)

HomR(A,ΠλBλ)→ΠλHomR(A,Bλ)

f 7→(pλ f)λ

is a Z−module isomorphism.

(b)

HomR(qλAλ, B)→ΠλHomR(Aλ, B)

g 7→(g iλ)λis a Z−module isomorphism.

Theorem 1.2.8. Let Aλλ∈Λ be a familiy of right R−modules and Bλλ∈Λ be a family ofleft R−modules. Fix A = AR and B = RB. The maps,

A⊗R (qλBλ)→qλ (A,Bλ)

a⊗ (bλ)λ 7→(a⊗ bλ)λ

(qλAλ)⊗R B →qλ (Aλ, B)

(aλ)λ ⊗ b 7→(aλ ⊗ b)λ(plus a linear extension) are Z−module isomorphisms.

14 CHAPTER 1. CATS

1.2.2 Exactness

Definition 1.2.9. Suppose F is a covariant (resp. contravariant) functor between modulecategories.

(a) F is left exact is exactness of 0→ Aα−→ B

β−→ C implies 0→ F (A)F (α)−−→ B

F (β)−−→ F (C)is exact.

(resp. exactness of Aα−→ B

β−→ C → 0 implies that 0→ F (C)F (β)−−→ F (B)

F (α)−−→ F (A) isexact)

(b) F is right exact if exactness of Aα−→ B

β−→ F (C) → 0 implies exactness of F (A)α−→

F (B)F (β)−−→ F (C)→ 0

(resp. exactness of 0→ Aα−→ B

β−→ C implies that F (C)F (β)−−→ F (B)

F (α)−−→ F (A)→ 0 isexact)

(c) F is exact if it is both left and right exact.

Homological algebra for algebraist is done around exact functors. It was based in topology(recall universal coeficients theorem) though, so hang it there!

Theorem 1.2.10. Let C = R −mod or mod− R, fix M ∈ Ob(C). Then HomR(M,−) andHomR(−,M) are left exact functors.

Example 1.2.11. The sequence in Z−mod

0→ Z ι→ Q π→ Q/Z→ 0

is exact. But HomZ(−,Z/2) is not an exact functor since

HomZ(Q,Z/2)i∗−→ HomZ(Z,Z/2)→ 0

is not exact because it is equal to 0→ Z/2→ 0.

Theorem 1.2.12. If A = AR and B = RB then A⊗R − and −⊗R B are right exact.

If they are also left exact we call A,B flat modules. As an example that not every module isflat, we can take the ses of the previous example and apply the Z/2⊗Z − functor to obtainthe non-exact sequence

0→ Z/2⊗Z ZidZ/2⊗i−−−−→ Z/2⊗Z Q→ 0

Lemma 1.2.13. Let C = R−mod or mod−R.

1.3. OPPOSITE CATEGORIES. 15

(a) Consider a sequence (*) 0 → Aα−→ B

β−→ C in C. If for all M ∈ Ob(C) we have that

0→ HomR(M,A)α∗−→ HomR(M,B)

β∗−→ HomR(M,C) is exact in Ab, then (*) is exactin C.

(b) Consider a sequence (**) 0 → Aα−→ B

β−→ C in C. If for all N ∈ Ob(C) we have

that 0→ HomR(C,N)α∗−→ HomR(B,N)

β∗−→ HomR(A,N) is exact in Ab, then (**) isexact in C.

1.3 Opposite categories.

This is why some mathematicians only study covariant functors...

Definition 1.3.1 (Opposite category). Let C be a category. The dual category (or oppositecategory) is defined to be the category Cop where Ob(C) = Ob(Cop) and for every A,B ∈Ob(Cop),

HomCop(A,B) =HomC(B,A)

ϕop ↔ϕ

There is a Duality operator

op : C →Cop

A 7→Aϕ 7→ϕop

Since (ψ ϕ)op = ϕop ψop, the duality functor is contravariant. One can check that this isactually an isomorphism of categories.

Remark 1.3.2. A contravariant functor is C → D is a covariant functor Cop → D. Techni-cally, when precomposing with the duality functor, but we tend to omit this.

Definition 1.3.3. Let S be a statement in C. The dual statement of S is the statementSop in C obtained from S by applying the duality functor; i.e., by reversing the arrows andcomposition, and interpreting the result statement in C.

Example 1.3.4.

1. “monic” is dual to “epic”

2. In module categories: “direct product” is dual to “direct sum”, and “kernel” is dualto “cokernel”1.

1coker(α : A→ B) = B/Im(α)

16 CHAPTER 1. CATS

3. “exactnes of sequences” is self dual when writting the definition in terms of kernelsand cokernels.

Remark 1.3.5. It R is a ring the (R−mod)op is not a module category unless R = 0. Thisresult is a little involving, we refer the reader to Fryed’s book on abelian categories.

1.4 Adjoints

Definition 1.4.1. Let C,D be categories and let F : C → D, G : D → C be covariantfunctors. The pair (F,G) is called an adjoint pair, F is called left adjoint and G is calledright adjoint if the following is true:

For all A ∈ Ob(C), C ∈ Ob(D) there exists a bijection

τA,C : HomD(F (A), C)→ HomC(A,G(C))

such that τA,C is natural in each variable. This is that for every f : A → A′ in C andg : C → C ′ in D we have the commutative diagrams

HomD(F (A), C) HomC(A,G(C))

HomD(F (A′), C) HomC(A′, G(C))

τA,C

τA′,C

(Ff)∗ f∗

and

HomD(F (A), C ′) HomC(A,G(C))

HomD(F (A), C ′) HomC(A,G(C ′))

τA,C

g∗ (Gg)∗

τA,C′

These are really usefull, as a canonical example we have the following

Theorem 1.4.2. Let B = RBS.

(a)(B ⊗R −, HomR(B,−)

)is an adjoint pair.

(b)(−⊗RB,HomS(−, B)


Proof. The idea to define the adjunction τA,C is to unwrap the package of definitions andhope for the best: Let A = SA, C = RC, define

τA,C : HomR(B ⊗S A,C)→HomS(A,HomR(B,C))

f 7→f

1.4. ADJOINTS 17

where

f : A→HomR(B,C)

a 7→fa

and

fa : B →Cb 7→f(b⊗ a)

Theorem 1.4.3. Suppose F : R − mod → S − mod and G : S − mod → R − mod arecovariant functors such that (F,G) is an adjoint pair. Then, F is right exact and G is leftexact.

Proof. We prove the statement for F (the one for G is similiar). Let Aα−→ B

β−→ C → 0 be

an exact sequence in R − mod. Consider (*) FAFα−→ FB

Fβ−→ FC → 0 in S − mod. LetN ∈ Ob(S −mod) be arbitrary. We get a sequence in Ab

0 HomS(FC,N) HomS(FB,N) HomS(FA,N)

0 HomR(C,GN) HomR(B,GN) HomR(A,GN)

(Fβ)∗

∼=τC,N

(Fα)∗

∼=τB,N ∼=τA,N

β∗

α∗

The bottom row is exact since the functor HomR(−, GN) is left exact. Since the verticalarrows are isomorphisms and the diagram commutes (by naturality of τ) we get the top rowexact.

Example 1.4.4 (Other than ⊗ and Hom). Let C = sets, D = K − vect where K is a field.Let F : C → D and G : D → C be functors defined as follows:

F (S) = K < S > is the K−vector space with basis S.

F (f) is the linear map sending s 7→ f(s) for all basis elements s ∈ S.

G : D → C is the forgetful functor.

We claim that the pair (F,G) is an adjoint pair. In this case, the adjunction τA,C :HomD(FA,C)→ HomK(A,GC) is given by: S ∈ Ob(C), C ∈ Ob(K − vec). Define

τS,V : HomK(F (S), V )→Maps(S,G(V ))

f 7→f |S

The universal property of free modules gives us that τ is a bijection. Naturality is left as anexercise.

18 CHAPTER 1. CATS

Problem set 1

Problem 1.1. Let C, D be categories, let A,B ∈ Ob(C), and let f ∈ HomC(A,B).

(a) Suppose F : C → D is a functor (covariant or contravariant) and f : A → B is anisomorphism in C. Show that F (f) is an isomorphism in D.

(b) Show that if f is an isomorphism, then f is both a monomorphism and an epimorphismin C. How about the converse? Prove or give a counter-example.

(c) Let C be a concrete category, which means that each object in C is a set with additionalstructure, and each morphism in C is a set map which preserves this structure2. Provethat every injective morphism in C is a monomorphism, and every surjective morphismin C is an epimorphism.

Prove that in R − Mod or Mod − R, every monomorphism is injective and everyepimorphism is surjective. Give an example of a concrete category with non-surjectiveepimorphisms.

Problem 1.2. Let F : R −Mod→ Ab be an additive functor (covariant or contravariant).

Suppose 0 → Aι→ B

p→ C → 0 is a split short exact sequence of (unital) left R−modules.Prove that F (B) ∼= F (A⊕ C) and that F (A⊕ C) ∼= F (A)⊕ F (C).

Problem 1.3.

(a) Give an example of a ring with 1 andR−modulesB, Ajj in whichHomR

(∏j Aj, B

)6∼=∏

j HomR (Aj, B). Explain.

(b) Give an example of a ring with 1 and R−modules A, Bjj in which A⊗R(∏

j Bj

)6∼=∏

j (A⊗R Bj). Explain.

Hint: Try R = Z, using without proof that HomZ(−,Q) and Q⊗Z − are exact functors.

2Strictly speaking, a concrete category is a category C together with a faithful functor C → Sets.

Chapter 2

Limits

Let (I,≤) be a partially ordered set (or POSET). We say (I,≤) is directed if for all i, j ∈ Ithere existe k ∈ I with i ≤ k and j ≤ k. As a remark, many books insist that the index setis directed (see below). We will mention when is important to assume it.

Definition 2.0.1. Let (I,≤) be a POSET and let C be a category.

(a) A direct system in C with index set I consists of a collection of objects in C Fii∈Iand a collection of morphisms ϕij : Fi → Fji≤j in C satisfying ϕii = idFi and whenever

i ≤ j ≤ k we get ϕjk ϕij = ϕik. We denote a direct system by Fi, ϕij.

(b) A inverse system in C with index set I is the dual notion, say it consists of acollection of objects in C ii∈I and a collection of morphisms ψji i≤j:Gj→Gi in Csatisfying ψii = idGi and whenever i ≤ j ≤ k we get ψji ψkj = ψki . We denote a direct

system by Gi, ψji .

Functorially, we can write this as follows:

(c) Define a category J by Ob(J ) = I and HomJ (i, j) = ∅ whenever i 6≤ j and|HomJ (i, j)| = 1 otherwise. Hence, a direct system in C is a covariant functorF : J → C and an inverse system in C is a contravariant functor G : J → C.

Example 2.0.2. C is any category and (I,≤) is a POSET.

1. Let A ∈ Ob(C) and for every i ∈ I define Fi = A and ϕji = idA. The system is calleda constant direct system associated to A, denoted by |A|. Similarly we can define theconstant inverse system.

2. Take I = Z with the usual order ≤. A direct system on I is a sequence F1ϕ21−−→

F2 → F3 · · · in C which commutes; and an inverse system is a reversed sequenceF1

ϕ12←−− F2 ← F3 · · · .

19

20 CHAPTER 2. LIMITS

3. I = 0, 1, 2 with 0 < 1 and 0 < 2 (not directed). A direct system on I is a diagramin C

F0 F1

F2

An inverse system on I is

F1

F2 F0

4. Let I with the trivial order: i ≤ j iff i = j. A direct (resp. inverse) system on I is justa collection Fii of object in C.

Definition 2.0.3. Let C be a category and let (I,≤) be a POSET.

(a) Let Fi, ϕij be a direct system in C on I. A direct limit (or inductive limit) of thissystem is an object lim−→Fi in C together with morphisms αj : Fj → lim−→Fi in C such

that αj ϕij = αi for all i ≤ j ∈ I, satisfying the following universal mapping property:

For any object X ∈ Ob(C) and morphisms fj : Fj → X, ∀j ∈ I with fj ϕij = fiwhenever i ≤ j, there exists a unique morphism φ : lim−→ → X making the followingdiagram commute.

X lim−→Fl

Fi

Fj

∃!φ

fi αi

ϕij

fj αj

(b) Let Fi, ψji be a direct system in C on I. A inverse limit (or projective limit) of thissystem is an object lim←−Fi in C together with morphisms ρj : lim←−Fi → Fj in C such

that ψji ρj = ρi for all i ≤ j ∈ I, satisfying the following universal mapping property:

For any object Y ∈ Ob(C) and morphisms gj : Y → Fj,∀j ∈ I with ψji gj = giwhenever i ≤ j, there exists a unique morphism Ψ : Y → lim←− making the following

21

diagram commute.

Y lim←−Fl

Fj

Fi

∃!Ψ

gj

gi

ρj

ρi

ψji

Theorem 2.0.4. Let C = R−mod or mod−R.

(a) The direct limit of a direct system Fi, ϕiji≤j in C over I exists and it is unique up toisomorphism.

(b) The inverse limit of a inverse system Fi, ψji i≤j in C over I exists and it is unique upto isomorphism.

Proof. (a) Let(qi Fi, ιjj

)be the coproduct (or direct sum) or Fi in C. Let S be the

submodule of qiFi generated by all elements of the form ιj(ϕij(ai)) − ιi(ai) for all i ≤ j,

ai ∈ Fi. Define lim−→Fi :=(qi Fi

)/S and for each j ∈ I define αj : Fj → lim−→Fi by αj = p ιj

where p : qiFi →(qi Fi

)/S is the natural projection. From this, one needs to show that

this object together with the morphisms satisfy the universal mapping property of the directproducts.

(b) Let(ΠiFi, pij

)be the product (or direct product) of Fi. Define lim←−Fi := (ai) ∈

ΠFi|ai = ψji (aj)∀i ≤ j. We can show it is a submodule and then define αj : lim←−FiraFj byαj = pj|lim←−Fi .

Note that if an arbitrary category we have products and coproducts then, the constructionabove can be used and the theorem is true.

Example 2.0.5. As always let C = R−mod or mod−R and (I,≤) be a POSET.

1. Let A ∈ Ob(C) and consider the constant direct (or inverse) system in C over I as-sociated to A. If C is directed, then the directed limit (resp. projective) of |A| isisomorphic to A.

A X

Fi = A

Fj = A

f

id fi

idAid fj


If I is directed, i.e. for all i ≤ j ∈ I there exists k ∈ I with i ≤ k and j ≤ k. We getthen fi = fk = fj and so fi = fj for every i ≤ j. Thus, we can find a unique mapf : A→ X making the diagram commute.

2. J−adic completion of a module.Take R a commutative ring and J ⊂ R an ideal.We get a descending sequence of ideals R = J0 ⊃ J1 ⊃ J2 ⊃ . . . . If M is an R−module,then we get a sequence M = RM ⊃ J1M ⊃ J2M ⊃ . . . of submodules. Let I = Z+

with the usual order ≤. If i ≤ j we have a projection map

ψji : M/J jM →M/J iM

and so an inverse system M/J iM,ψji i≤j. The inverse limit lim−→M/J iM is called the

J−adic completion of M , denoted by M .

Let R = lim−→R/J i. Then R is a commutative ring. Note that R is the completion of R

with respect to the topology on (R,+) with a basis around 0 the set J i|i ≥ 0; i.e., abasis of R is of the form r + J i|i ≥ 0, r ∈ R.Note that M is an R−module. For example, when R = Z, J = (p) for p ∈ Z a primenumber, R is called the ring of p−adic integers denoted by Zp1 or Zp2. In other wordsZp is the inverse limit of Z/pnZn with Qp as a field of fractions.

3. Let I = 0, 1, 2 with 0 < 1 and 0 < 2. Take a direct system

A B

C

f

g

The direct limit is called the push-out. From the construction in theorem 2.0.4, onecan prove that the push-out is isomorphic to

Q :=(B ⊕ C

)/(fa,−ga)|a ∈ A︸︷︷︸

:=K

with morphisms

αB : B →Qb 7→(b, 0) +K

αC : C →Qc 7→(0,−c) +K

We can proceed in a similar fashion with the inverse system

B

C A

f ′

g′

1If you are a number theorist.2If you don’t.

23

The inverse limit is called a pull-back. The pull-back is isomorphic to

P := (b, c) ∈ B ⊕ C|f ′(b) = g′(c)

with morphisms canonical projections.

4. Kernel and cokernel are dual. Let f : X → Y an R−module homomorphism. Bythe above example, the kernel and cokernel are the push-out and pullback of certainsystems, respectively. Say,

X Y

0

f

0

has pushout Y ⊗ 0/(f(x), 0)|x ∈ X ∼= Y/Im(f) = Coker(f).

And the projective system

X

0 Y

f

0

has pullback (x, 0) ∈ X × 0|f(x) = 0 ∼= Ker(f).

5. The direct and inverse limit generalize the notions of product and coproduct in C,respectively. We can see this by endowing I with the trivial order i ≤ j ⇐⇒ i = j.Then

lim−→Fi =qi Filim←−Fi =ΠiFi

What do we do with this notion of limits? We can consider the category of all direct (resp.inverse) systems and study its properties. We can apply the direct limit (resp. inverse)functor to this category and land again in the original category C. Is this functor having aleft (or right) adjoint?

Definition 2.0.6 (Category of direct systems). Let C be a category and let (I,≤) be apartially ordered set.

1. The category of direct systems in C over I has objects all direct systems in C overI. The morphisms are as follows:

A morphism t : Fi, ϕiji≤j → Gi, λiji≤j between two direct systems consists of a

family of morphisms ti : Fi → Gi such that for all i ≤ j ∈ I the following diagramcommutes:

Fi Fj

Gi Gj

ti

ϕij

tj

λij


We denote this category by Dir(I) or DirC(I). In other words, the covariant functorsI → C form a category.

2. The category of inverse systems in C over I, denoted by InvC(I), has objects allinverse systems in C over I. The morphisms are as follows:

A morphism s : Fi, ψji i≤j → Gi, ρjii≤j between two inverse systems consists of a

family of morphisms si : Fi → Gi such that for all i ≤ j ∈ I the following diagramcommutes:

Fj Fi

Gj Gi

sj

ψji

si

λji

In other words, the covariant functors I → C form a category.

If your category C has exact sequences, you can translate exact sequences into your directsystems: a sequence Fi, ϕiji≤j → Gi, λ

iji≤j → Hi, η

iji≤j is exact in DirC(I) iff for all

i ∈ I the sequene Fiti−→ Gi

ηi−→ Hi is exact. We can define a similar notion on InvC(I).

Definition 2.0.7 (Constant and Limit functors). Let C = R − mod or mod − R and let(I,≤) be a POSET. We have the following covariant functors:

(a) | − | : C → DirC(I) (similarly InvC(I) defined by A 7→ |A| for all A ∈ Ob(C) and forevery morphism f ∈MorC(A,A

′) f 7→ |f | where |f |i = f for all i.

(b) lim−→ : DirC(I) → C defined by: given A := Ai, ϕij,B := Bi, ρij ∈ Ob(DirC(I)) and

for all t : A → B morphism in DirC(I)

A 7→ lim−→A :=(qi Ai

)/S

B 7→ lim−→B :=(qi Bi

)/T

t 7→t

where S and T are certain submodules from the construction of the direct limits andt : lim−→A → lim−→B maps (ai)i + S 7→ (ti(ai))i + T .

(c) lim←− : InvC(I) → C defined by: given A := Ai, ψji ,B := Bi, λji ∈ Ob(InvC(I)) and

for all s : A → B morphism in InvC(I)

A 7→ lim←−A ⊂ ΠiAi

B 7→ lim←−B ⊂ ΠiBi

s 7→s

where t : (ai)i 7→ (si(ai))i.

25

Remark 2.0.8. To define the definition of the direct (resp. inverse) limit functor it is enoughto our category C has direct (resp. inverse) limits. To define the map t : lim−→Ai → lim−→Bi

one can use the universal mapping property of the pair(

lim−→Ai, αi)

with the maps βi ti :Ai → lim−→Bl as in the following commutative diagram.

lim−→Al lim−→Bl

Aj Bj

Ai Bi

∃!t

αj

tj

βj

ϕij

αi

ti

βi

ρij

Exercise 1. Draw the diagram above corresponding to the inverse limit functor, assumingC admits inverse limits.

Theorem 2.0.9. Let C be a module category and (I,≤) be a POSET. Then(

lim−→, | − |)

and(| − |, lim←−

)are adjoint pairs.

Proof. We will sketch the construction for the adjunction functions τ for the pair(

lim−→, |−|).

Let A = Ai, ϕij ∈ Ob(DirC(I)) and C ∈ Ob(C). Define

τA,C : HomC(lim−→Ai, C)→HomDirC(I)(A, |C|)f 7→f αii

where (lim−→Ai, αi) is the direct limit of A. The adjuntion map is bijective because of theuniversal mapping property, narutality is left as an exercise.

Theorem 2.0.10. Let F : C → D and G : D → C be covariant functors such that (F,G)is an adjoint pair. If C and D admit limits, then F preserves direct limits and G preservesinverse limits.

Proof. We prove the statement for G. Notice the technique use is similar to the proof oftheorem 1.4.3.

Let Ci, ψij ∈ Ob(InvD(I)) and let(

lim←−Ci, βi)

be its inverse limit in D. Since G is

covariant, G(Ci), G(ψij) ∈ Ob(InvC(I)). To conclude, it is enough to show the universal

mapping property for inverse limits in C for(G(lim←−Ci), G(βj)j

).


Let X ∈ Ob(C) and fj : G(Ci)→ X be morphisms in C. We have the following diagram,

G(

lim←−Ci)

X

G(Ci)

G(Cj)

G(βi)

G(βj)

fi

fj

G(ψji )

We need to show that ∃! β : X → G(lim←−Ci) in C such that G(βi) β = fi ∀i. Using that(F,G) is an adjoint pair, we get the following commutative diagram for all i ≤ j ∈ I.

HomC(X,G(lim←−Ci)

)HomD

(F (X), lim←−Ci

)HomC

(X,G(Cj)

)HomD

(F (X), Cj

)HomC

(X,G(Ci)

)HomD

(F (X), Ci

)

τ(G(βj)

)∗

(βj

)∗

τj(G(ψji )

)∗

(ψji

)∗

τi

The right column above gives us the following commutative diagram in D

lim←−Ci X

Ci

Cj

βi

βj

τi(fi)

τj(fj)

ψji

By the universal mapping property of lim←−Ci, there exists a unique γ ∈ HomD(F (X), lim←−Ci)such that βiγ = τi(fi)∀i ∈ I. Define β := τ−1(γ). By commutativity of the diagrams abovewe obtain,

G(βi) β = G(βi)(τ−1(γ)

= τ−1i

((βi)∗(γ)

)= τ−1

i (β γ)

= τ−1i (τifi) = fi

To end, note that β is unique since γ is unique.

Recall that the pair of functors (tensor, Hom) is an adjoint pair. The previous theoremproves that these tensors preserve certain limits.

2.1. EXACTNESS OF LIMIT FUNCTORS 27

Corollary 2.0.11. If B = SBR then − ⊗S B and B ⊗R − preserve direct limits andHomS(B,−) and HomR(B,−) preserve inverse limits.

NoteHomS(−, B) andHomR(−, B) send direct limits to inverse limits; e.g., HomS(lim−→Ai, B) 'lim←−HomS(Ai, B).

Corollary 2.0.12. Direct limit (possibliy on different index sets) commute and so do inverselimits.

For example, lim−→ilim−→j

and lim−→jlim−→i

are naturally isomorphic functors.

2.1 Exactness of limit functors

Let (I,≤) be a POSET and C be a module category. Recall the construction of lim−→Ai. Wewant an easy criteria to know when the limit functors are exact. But first an technical lemma.Recall that the construction of direct limits in a module category C is given by lim−→Ai =(qiAi)/S where S is the submodule generated by the elements of the form ιj(ϕ

ij(ai))− ιi(ai).

Lemma 2.1.1. Suppose (I,≤) is directed POSET.

(a) lim−→Ai = ιi(bi) + S|i ∈ I, bi ∈ Ai

(b) ιi(bi) + S = 0 + S if and only if there exists h ∈ I h ≥ i with ϕik(bi) = 0.

Proof. (a) Let (ai)i + S ∈ lim−→Ai. By construction (ai)i =∑

i ιi(ai) and if we set I0 ⊂ I tobe the set of indexes i0 with ιi0(ai0) 6= 0, I0 is finite. Since I is directed there exists k ∈ Iwith k ≥ i0 ∀i0 ∈ I0. Thus

(ai)i + S =∑io

[ιi0(ai0) + ιk(ϕ

i0k (ai0))− ιi0(ai0)

]+ S

= ιk(∑

i0

ϕi0k (ai0))

(b) (⇐) By adding a zero (an element of S) we get the desired conclusion since

ιi(ai) + S = ιi(ai) + ιk(ϕik(ak))− ιi(ai) + S = ιk(0) + S = (0) + S

(⇒) Suppose now that ιi(bi) + S = 0 + S, then there exists a finite subset I0 ⊂ I such that∀ i0 ∈ I0 ∃ j0 ≥ i0 and bi0 ∈ Ai0 such that

ιi(ai) =∑i0∈I0

(ιj0(ϕ

i0j0

(bi0))− ιi0(bi0))


Since I is directed, there exists k ≥ j0 ≥ i0 ∀i0 ∈ I0. Then, after adding a zero and combiningterms, we get,

ιk(ϕik(ai)) =ιk(ϕ

ik(ai))− ιi(ai) + ιi(ai)

=(ιk(ϕ

ik(ai))− ιi(ai)

)+∑i0∈I0

(ιj0(ϕ

i0j0

(bi0))− ιi0(bi0))

=∑t∈T

(ιk(ϕ

tk(ct))− ιt(ct)

)In the above, T = i∪ I0∪j0|i0 ∈ I0. Looking at the RHS of the equation we note that ift ∈ T satisfies t < k, then ιt(ct) = 0 and so ct = 0, thus ιk(ϕ

tk(ct))− ιt(ct) = 0. And, if t = k,

then ιk(ϕtk(ck))− ιk(ck) = 0. Hence, RHS is (0) and so LHS. Injectivity of ιk concludes that

ϕik(ai) = 0 ∀i ≤ k as we wanted.

Theorem 2.1.2. If (I,≤) is a directed poset, then lim−→ is exact.

Proof. We already know that lim−→ is always right exact. Suppose t : Ai, ϕij → Bi, λij is a

morphism in DirC(I) with each ti : Ai → Bi injective. We want to show that→t : lim−→Ai →

lim−→Bi is injective. To set notation, let(qi Ai, ιii

)and

(qi Ai, λii

)be the coproducts

of each collection of modules and write

lim−→Ai =(qi∈I

Ai)/S

lim−→Bi =(qi∈I

Bi

)/T

For the corresponding submodules T, S. Take an element in the kernel of→t , using part (a)

of lemma 2.1.1, we have

(0) + T =→t (ιi(ai) + S) = ji(ti(ai)) + T

By part (b) of lemma 2.1.1, there exists k ≥ i such that λik(ti(ai)) = 0 in Bk. But for anyi, 0Bk = λik(ti(ai)) = tk(ϕ

ik(ai)) = 0, which injectivity of tk forces ϕik(ai) = 0 ∀i. Part (b)

concludes ιi(ai) + S = (0) + S, and→t is injective.

Remark 2.1.3. If we wonder about the exactness of lim←−, it is not enough to assume that(I,≤) is exact. For example, consider I = Z+ with the usual order and let p ≥ 3 be a primenumber. Take the inverse system given by the usual projections Z/pnZ, ψnmm≤n and theconstant inverse limit |Z|. Let t : |Z| → Z/pnZ, ψnm be given by tn : Z → Z/pnZ wheretn(a) = a+ pnZ. The morphism t is surjective but

←t : Z→ lim←−Z/pnZ = Zp

a 7→(a+ pnZ

)n≥1

is not surjective since x =(1+p+ · · ·+pn−1 +(pnZ)

)n

does not lie in the image of←t because

(1 − p) is invertible in Zp (it is x) but not in Z. This same map fails to be surjective for

p = 2 also, one can take x = 13∈ Z2.

2.1. EXACTNESS OF LIMIT FUNCTORS 29

So what goes wrong with the map t : |Z| → Z/pnZ, ψnm? The problem is that the preim-ages, say t−1

n (cn) for (cn)n ∈ lim←−Z/pnZ, form an inverse system of sets and the inverse limit(the intersection in this case) could be empty. The Mittag-Leffer condition is one solutionfor this problem:

M-L condition: For I = N, Ai, ψji satisfies the M-L condition if for every i, the sequenceof sets ψji (Aj)|j ≥ i stabilizes as sets in Ai.

One can show that if 0 → Ai, ψji → Bi, ρji → Ci, τ

ji → 0 is a s.e.s. in InvC(I) and

Ai, ψji satisfies the M − L condition, then 0→ lim←−Ai → lim←−Bi → lim←−Ci → 0 is exact.

Problem set 2

Problem 2.1. Let C = R−Mod or Mod−R, let (I,≤) be a directed partially ordered set,and let K be a cofinal subset. A subset K of I is said to be cofinal if for each i ∈ I there isk ∈ K with i ≤ k. Prove the following statements,

(a) If(Aii∈I , ϕiji≤j

)is a directed system in C with index set I, prove that

(Aii∈K , ϕiji≤j

)is a directed system in C with index set K. Moreover, prove that the direct limits inboth these systems are isomorphic. Show that this may be false if I is not directed(look at a pushout).

(b) Dually, if(Aii∈I , ψji i≤j

)is an inverse system in C with index set I, prove that(

Aii∈K , ψji i≤j)

is an inverse system in C with index set K. Moreover, prove thatthe inverse limits in both these systems are isomorphic. Show that this may be false ifI is not directed.

Problem 2.2. Let R, S be rings, let (I,≤) be a partially ordered set.

(a) If(Aii∈I , ϕiji≤j

)is a direct system in R−mod with index set I, prove that there is

an exact sequence in R−mod

qi∈Iqi≤j

Bijf→ q

i∈IAi

p→ lim−→Ai → 0

where Bij = Ai for all i ≤ j.

If(Cii∈I , ψji i≤j

)is an inverse system in R−mod with index set I, prove that there

is an exact sequence in R−mod

0→ lim←−Ciι→ qi∈ICi

g→ qi∈Iqi≤j

Dij

where Dij = Ci for all i ≤ j.


(b) Let F : R−modraS − mod be an additive left exact functor. If F is covariant and preserves directproducts, prove that F preserves inverse limits.

If F is contravariant and converts direct sums into direct products, prove that Fconverts direct limits into inverse limits.

Problem 2.3. Let G be a group and let N be the family of all normal subgroups of finiteindex in G.

(a) If N ′ ⊂ N in N then there is a homomorphism ψN′

N : G/N ′ → G/N . Show that thefamily of all such quotients together with the maps ψN

′N forms an inverse system over

N where N ≤ N ′ iff N ′ ⊂ N .

(b) The inverse limit of the system in (a), lim←−G/N , is denoted by G and is called the

profinite completion3 of G. There is a natural homomorphism f : G→ G sending g 7→(gN)n∈N . Show that f is injective if and only if G is residually finite; i.e.

⋂N∈N N =

1G.

(c) Write down the profinite completion of Z, viewed as a group under addition.

3Since the category of all groups has direct products, this can be built the same way as for modules.

Chapter 3

Module Categories

3.1 Projective Modules

Along this section, C will be a module caterogy over R.

Theorem 3.1.1. Let P ∈ Ob(C). The following are equivalent.

1. Given a diagram with botom exact in C

P

B C 0

α∃γ

β

there exists γ ∈ HomC(P,B) with β γ = α.

2. HomC(P,−) is exact.

3. Every short exact sequence 0→ X → Y → P → 0 splits.

4. P is a direct summand of a free module F , written P |F .

Definition 3.1.2. A module satisfying (1)-(4) is a projective module.

Example 3.1.3. Not every projective module is free. One can take, for example, R = Z×Zand P1 = (a, 0)|a ∈ Z, P2 = (0, b)|b ∈ Z, then P1 and P2 are projective R−modules since

RR = P1 ⊕ P2; but Pi is not free since Z(Pi) ∼= Z and any finitely generated free R−modulehas even rank as a free Z−module.

Proposition 3.1.4.

(a) Every direct summand of a projective module is projective.

31

32 CHAPTER 3. MODULE CATEGORIES

(b) If Pλλ∈Λ is a family of projective modules then qPλ is a projective module.

Definition 3.1.5. Let M be an R − module. A projective resolution of M is an exactsequence in C

. . .→ Pndn→ Pn−1

dn−1→ . . .d2→→ P1

d1→ P0ε→M → 0

where each Pi is a projective module. If all Pi are free, this is called a free resolution.

Theorem 3.1.6. Every module M in C has a projective resolution.

Proof. Let F0 be a free module in C surjecting onto M . By letin K0 be the kernel of thesurjection, we get a s.e.s. 0→ K0

ι0→ F0ε→M → 0. Let F1 be the free module in C surjecting

onto K0, we get a s.e.s. 0 → K1ι1→ F1

p1→ K0 → 0. Inductively, we get for every n ∈ N a

s.e.s. 0→ Knιn→ Fn

pn−1→ Kn−1 → 0. Define dn = ιn−1 pn to obtain a long sequence

. . .→ Pndn→ Pn−1

dn−1→ . . .d2→→ P1

d1→ P0ε→M → 0

It is exact since Im(d1) = Im(ι0 p1) = Im(ι0) = ker(ε) and; for n ≥ 2, Im(dn) =Im(ιn−1 pn) = Im(ιn− 1) and ker(dn−1) = ker(ιn−2 pn−1) = ker(pn−1) = Im(ιn−1).

Definition 3.1.7. Let f : M ′ → M be a morphism in C. We say that f is essentialepimorphism if f(M ′) = M and for any proper submodule M ′′, f(M ′′) 6= M .

Theorem 3.1.8. Let ε : P → M be a surjective morphism in C where P is projective. Thefollowing are equivalent:

(a) ε is an essential epimorphism.

(b) ∀ epimorphism ε′ : Q→M with Q a projective module and ∀ f : Q→ P with ε′ = εf ,we have f is surjective.

0

P M 0

Q

ε

fε′

Proof. (a)⇒ (b): Let Q, ε′, f be as in (b).1 Then, ε(Im(f)) = ε(f(Q)) = ε′(Q) = M and,since ε is essential, Im(f) = P .

(b)⇒ (a): Let L ⊂ P be a proper submodule with ε(L) = M . Let Q be a projective moduleand a surjective homomorphism δ : Q→ L. Let f : Q→ P be the composition of δ followedby the inclusion L → P . f is not surjective since L is proper submodule, but ε f = ε|L δ

1Note that f exists since Q is projective and ε′ is onto.

3.2. INJECTIVE MODULES 33

by construction. This contradicts condition (b) with ε′ := ε|L δ. For fun, the followingcommutative diagram explains contradiction.

0

P M 0

L L

Q

ε

ε|L

δ

f

δ′

ε′=ε|Lδ

Definition 3.1.9. A projective module P together with an essential morphism ε : P → Mis a projective cover of M , denoted (P, ε).

Theorem 3.1.10. If (P, ε) and (P ′, ε′) are two projective covers of M , then there exists anisomorphism f : P → P ′ satisfying ε′ f = ε.

Proof. Since P is projective and ε′ is surjective, there exists f ∈ HomC(P, P ′) with ε = ε′f .Since (P ′, ε′) is a projective cover, f is surjective by theorem 3.1.8. Since P ′ is projective,

the s.e.s. 0 → Ker(f) → Pf→ P ′ → 0 splits and there is a submodule P ′′ ⊂ P such that

f |P ′′ : P ′′ → P ′ is an isomorphism with P = Ker(f)⊕P ′′. By last, since ε(P ′′) = ε′(f(P ′′)) =ε′(P ′) = M and ε is an essential epimorphism, we get P ′′ = P and so Ker(f) = 0, thus f isan isomorphism.

Remark 3.1.11. Projective covers do not always exist. For example, the Z−module Z/2does not admit a projective Z−module cover (see Problem 3.1).

Theorem 3.1.12. If R is a left Artinian ring and M is finitely generated R−module thenM has a projective R−module cover.

Proof. Left to the reader.

3.2 Injective Modules

Theorem 3.2.1. Let E ∈ Obj(C). The following are equivalent.

(a) For every diagram in CE

0 A Bβ

f∃g


there exists g ∈ HomC(B,E) with g α = f .

(b) HomC(−, E) is exact.

(c) Every s.e.s. 0→ E → Y → Z → 0 in C splits.

Definition 3.2.2. An R−module E in C is an injective module if E satisfies (a) in theorem3.2.1.

Proposition 3.2.3.

(a) Every direct summand of a projective module is projective.

(b) If Eλλ∈Λ is a family of injective modules then ΠPλ is a projective module.

Note. To prove (c)⇒ (a) in theorem 3.2.1 assuming (a)⇐⇒ (b) we can use Bear’s Criterion:If E is a left R−module, then E is injective iff for every left ideal I ⊂ R every morphimf : I → E extends to R.

Definition 3.2.4. Let M be a left R−module, m ∈ M and r ∈ R. Then m is divisible byr if there exists m′ ∈ M with m = rm′. M is a divisible left R−module if each m ∈ M isdivisible by each r ∈ R that is not a right zero divisor (i.e. sr 6= 0∀s ∈ R− 0).

Theorem 3.2.5. Every injective left R−module is divisible. If R is a PID, the converseholds.

Example 3.2.6. Take R = K[u, v] the polynomial ring over a field K in two variables and Qits field of fractions. Then Q/R is a divisible R−module that is not an injective R−module.

Theorem 3.2.7. Every Z−module M can be embedded into an injective Z−module.

Main idea of the proof. There exists a free Z−module F =⊕

λ Z together with a surjectiveZ−module homomorphism π : R→M such that

M ∼=(⊕

λ

Z)/Ker(π) →

(⊕λ

Q)/Ker(π)

where the RHS is a divisible Z−module, hence injective.

Theorem 3.2.8. If M is a left (resp. right) R−module, then M can be embedded into aninjective left (resp. right) R−module.

Proof. Suppose M is a left R−module. Let D be an injective Z−module together with aninjective Z−module homomorphism ι : M → D. Define E := HomZ(ZRR, D), so E is a leftR−module. Define f : M → E by f(m)(r) := ι(rm). One can check that f is an injectiveR−module homomorphism.


We claim that, E is an injective left R−module. Using theorems 1.2.10 and 3.2.1, it isenough to check that HomR(−, E) is right exact. Let 0 → A

α→ B be exact in C. We usethe adjoint pair

(R⊗R −, HomZ(R,−)

)to get the following dommutative diagram

HomR(B,E) = HomR

(B,HomZ(R,−)

)HomR

(A,HomZ(R,D)

)= HomR(A,E) 0

HomZ(R⊗R B,D

)HomZ

(R⊗R A,D

)0

HomZ(B,D

)HomZ

(A,D

)0

α∗

∼= ∼=

(idR⊗α)∗

∼= ∼=

α∗

The top square commutes because of the adjoint pair and the top vertical rows are isomor-phisms. The bottom square commutes because R⊗R − ∼= IdR−mod and the bottom verticalrows are isomorphisms. Since D is an injective Z−module, the bottom row is exact. Hencethe top row is exact like we wanted.

Definition 3.2.9. Let M ∈ Ob(C). An injective resolution of M is an exact sequence

0→M → ι→ E0 d0→ E1 d2→ . . .dn−1

→→ En dn→ . . .

where each Ei is an injective module.

Theorem 3.2.10. Every module in C has an injective resolution.

Proof. The proof is analogous to the one of theorem 3.1.6 with the help of theorem 3.2.8.

Definition 3.2.11. Let f : M → N be a morphism in C. We say that f is essentialmonomorphism if f is injective and for any non-zero submodule N ′ ⊂ N , f(M)∩N ′ 6= 0.

Remark 3.2.12. The notions of essential monomorphism and essential epimorphism aredual of each other. The following commutative diagrams makes it easy to see.

0

M ′′ 6= 0

M N 0

M ′ N

0exact

f

exact

ι

f

0

N ′ 6= 0

N N 0

N ′′ N

0exact

π

f

exact

f


In the essential epimorphism diagram (left), if f = f ι, then coker(f) 6= 0. In the essential

monomorphism case (right), f = π f implies kef(f) 6= 0. The following theorems anddefinitions follow naturally from the previous subsection.

Theorem 3.2.13. Let ι : M → E be a monomorphism in C, where E is injective. Thefollowing are equivalent.

(a) ι is an essential monomorphism.

(b) ∀ monomorphism ι′ : M → Q with Q a injective module and ∀ f : E → Q withι′ = ι f , we have f is injective.

0

0 M E

Q

ι

ι′f

Definition 3.2.14. An injective module E together with an essential monomorphism ι :M → E is an injective hull of M .

Theorem 3.2.15. If (E, ι) and (E ′, ι′) are two injective hulls of M , then there exists anisomorphism f : E → E ′ satisfying ι′ = f ι.

Theorem 3.2.16. Every R−module M in C has an injective hull.

Proof. WLOG let C = R−mod. Let D be an injective left R−module together with amonomorphism f : M → D. If f is essential, we are done. Suppose then, f is not essentialand consider S = T ∈ Ob(C)|f(M) ⊂ T ⊂ D, fT = M → T is an essential monom2. S ispartially ordered by inclusion and by assumption f(M) ∈ S. Using Zorn’s lemma S has amaximal element, say E.

We claim that E is an injective left R−module. Since f |E is essential and f is not, we havethat ι : E → D is not an essential monomorphism3. There exists then a submodule of D,say X 6= 0, with E ∩ X = 0. By Zorn’s lemma, we can pick X maximal satisfying suchconditions. Consider the s.e.s. of the quotient

0→ X → Dπ→D/X → 0

d 7→d+X

We will show that there exists a morphism j : D/X → E such that π j = idD/X . Hence,E is isomorphic to a direct summand of D, thus injective; finishing the prove.

2fT is given by fT (m) = f(m).3Since composition of essential monomorphisms is essential.


Let g = π|E : E → D/X, g is injective since E ∩ X = ∅. Moreover, given a submoduleX ( D′ ⊂ D, maximality of X forces D′ ∩ E 6= ∅, say d′ ∈ D′ ∩ E − X. Then, 0 + X 6=d′+X ∈ g(E)∩D′/X; and remark 3.2.12 implies then that g is an essential monomorphism.Now, injectivity of D stated that there exists j : D/X → D an R−module hom with jg = ι.Note that j is a monomorphism since Ker(j) ∩ Im(g) = 0 + X and, since g is essential,Ker(j) jas to be 0 +X. Moreover, the restriction on the image j : D/X → j(D/X) is anessential monomorphism, thus j g : E → j(D/X) is also an essential monomorphism and soj g f : M → j(D/X) is essential. But we know that j g f = fj(D/X), so by maximalityof E in S forces j(D/X) = E. Hence j : D/X → E is an isomorphism, j g = idE andπ j = g j = idD/X , like we wanted.

Definition 3.2.17.

(a) A module P in C is a generator for C if every module in C is isomorphic to a quotientmodule of a direct sum of copies of P .

(b) A module C in C is a cogenerator for C if every module n C is isomorphic to a submoduleof a direct product of copies of C.

For example, R (in fact Rn) is a projective generator for R−mod and Mod−R.

Lemma 3.2.18. Let C be a module in C. C is a cogenerator iff for every 0 6= M ∈ Ob(C),∀m ∈M − 0 there exists a morphism fm : M → C with fm(m) 6= 0.

Proof. (⇒) Let 0 6= M ∈ Ob(C). Then, for some index set Λ, there exists a monomorphismf : M → Π

λ∈ΛC. Let pµ : Π

λ∈ΛC → C be the projection onto the µ−th coordinate. Given

m ∈M−0, we have that f(m) 6= 0 in Πλ∈Λ

C so there exists a µ ∈ Λ with pµ(f(m)) 6= 0 ∈ C.

Let fm := pµ f .

(⇐) Let 0 6= M ∈ Ob(C) and take fm : M → Cm∈M the associated maps. By theuniversal property of the direct product, there exists a unique map f : M → Π

m∈M−0C with

pm f = fm. Note that f is injective.

Corollary 3.2.19. Q/Z is an injective cogenerator for Z−mod.

Proof. Q/Z is a quotient of a divisible Z−module, hence divisible and therefore an injectiveZ−module. Let 0 6= M ∈ Ob(Z − mod) and let m ∈ M − 0. Denote by 〈m〉 ⊂ M thesubmodule generated by m and define g : 〈m〉 → Q/Z by

g(m) =

1√2

+ Z, if o(m) =∞1r

+ Z, if o(m) = r <∞

In any case g(m) 6= 0. We conclude by lemma 3.2.18.


Definition 3.2.20. Let R∗ := HomZ(R,Q/Z), R∗ is an injective R−module in C. A cofreemodule in C is defined to a module that is isomorphic to a direct product of copies of R∗.

Note that cofree modules are injective R−modules.

Theorem 3.2.21. Let E be a module in C. Then E is an injective R−module in C iff E isisomorphic to a direct summand of a cofree R−module in C.

Proof. (⇒) Since R∗ is an injective cogenerator for C.

(⇐) Since direct summands of injective modules are injective.

But why is R∗ an injective cogenerator for C? Let 0 6= M be a module in C and takea Z−module monomorphism ι : M → Π

λ∈ΛQ/Z. By the proof of theorem 3.2.8, there exists

f : M → HomZ(R, Π

λ∈ΛQ/Z

) ∼= Πλ∈Λ

HomZ(R,Q/Z) = Πλ∈Λ

R∗

an injective R−module homomorphism. Hence, R∗ is an injective cogenerator of C.

Remark 3.2.22. If M is a left (resp. right) R−module, then HomZ(M,Q/Z) is a right(resp. left) R−module, called the Pontryagin dual of M .

3.3 Natural transformations: An interlude.

Definition 3.3.1. Let C,D be categories

(a) Let F,G : C → D be functors of the same variance (both covariant/contravariant). Anatural transformation τ : F → G is a collection τA : FA→ GAA∈Ob(C) of morphismsin D such that for every f ∈ HomC(A,A′) we get a commutative diagram.

If covariant FA FA′

GA GA′

Ff

τA τA′

Gf

FA FA′ If contravariant

GA GA′

τA

Ff

τA′

Gf

If additionally, τA is an isomorphism for every object A, we call τ a natural isomor-phism.

(b) The category of covariant functors DC (resp. contravariant functors DC) has objectscovariant (resp. contravariant) functors C → D and morphisms natural transformationsbetween them.

3.4. FLAT MODULES 39

(c) If F,G : C → D are two functors of the same variance, the are naturally isomorphic,written F ∼= G, if there exists a natural isomorphism τ : F → G.

(d) A functor F : C → D is an equivalence of categories if there exists a functor F ′ : D → Cwith F ′ F ∼= idC and F F ′ ∼= idD. Then, F ′ is a quasi-inverse of F .

Remark 3.3.2. Note that HomDC(F,G) may not be a set but a proper class. For this, weneed to allow ”bigger categories” (wait for it).

If F,G : R −Mod → Mod − S are naturally isomorphic, then F is left exact iff G is leftexact; and F is right exact iff G is right exact.

F is an isomorphism of categories if there is a functor G : D → C with G F = idC andF G = idD. Not every equivalence of categories is an isomorphism.

One can check that IdR−mod ∼= R ⊗R − ∼= HomR(R,−) by defining for every C ∈ Ob(R −mod),

σC : C →R⊗R Cx 7→1⊗ x

τC : C →HomR(R,C)

x 7→αx : R→ C

r 7→ rx

3.4 Flat Modules

Definition 3.4.1. A right R−module B = RR (resp. left with B = RB) is flat or R−flat ifB ⊗R − is an exact functor (resp. −⊗R B).

Theorem 3.4.2. Let B = SBR and C = SC. If B is R−flat and C is an injective S−module,then HomS(B,C) is an injective left R−module.

Proof. The functor HomR(−, HomS(B,C)) is, by adjointness, naturally isomorphic to thefunctor HomS(B⊗R−, C) and HomS(B⊗R−, C) = HomS(−, C)(B⊗R−), is a compositionof two exact functors by assumption. Hence HomR(−, HomS(B,C)) is exact and we are doneby theorem 3.2.1.

Theorem 3.4.3.

(a) R = RR is R−flat.

(b) Let Bλλ be a family of right R−modules. Then qλBλ is flat iff every Bλ is flat.


(c) Every projective right R−module is flat.

Proof. Part (c) follows from (a) and (b).(a) Follows since R⊗R − ∼= IdR−mod.(b) Let fλ : Cλ → C ′λ be an S−module homomorphism for each λ ∈ Λ. By universal mappingproperty, there exists a unique S−module homomorphism q

λfλ : q

λCλ → q

λC ′λ. Note that

qλfλ is monic iff each fλ is monic. Now consider an exact sequence of left R−modules

0→ Aα→ A′, we get a commutative diagram of Z−modules,

0(qλBλ

)⊗R A

(qλBλ

)⊗R A′

qλ

(Bλ ⊗R A

)qλ

(Bλ ⊗R A′

)(idqBλ⊗α)∗

∼= ∼=(q(idBλ⊗α)

)∗Hence idqBλ⊗α is monic iff the top row is exact iff the bottom row is exact iff each q(idBλ⊗αis monic, which is (b).

Definition 3.4.4. Let M be a right R−module. A flat resolution of M is an exact sequence

. . .→ Fndn→ Fn−1

dn−1→ . . .d2→→ F1

d1→ F0 →M → 0

where each Fi is a projective module.

Note that since every M has a projective resolution, the previous theorem implies that everyM admits a flat resolution.

Theorem 3.4.5. Let (I,≤) be a directed poset. Let Bi, ϕij be a directed system in Mod−R

over I. If Bi is flat for each i, then lim−→Bi is flat. The converse is not true in general (seeProblem 3.2).

Proof. Let α : A→ A′ be a monic R−module homomorphism between left R−modules. Byassumption, for each i ∈ I idBi ⊗ α : Bi ⊗R A→ Bi ⊗R A′ is a monic Z−module hom. Nowwe use that − ⊗R A and − ⊗R A′ commute with direct limits and that lim−→ is exact (I isdirected) to get a commutative diagram

lim−→(Bi ⊗R A

)lim−→(Bi ⊗R A′

)(

lim−→Bi

)⊗R A

(lim−→Bi

)⊗R A′

−→(idBi⊗α)

∼= ∼=

(id lim−→Bi)⊗α

lim−→ is exact so→

(idBi ⊗ α) is monic and (idlim−→Bi)⊗ α is also monic.

3.4. FLAT MODULES 41

Corollary 3.4.6. If R is an integral domain, then Q = Frac(R) is a flat R−module.

Proof. Let I := (R−0)/ ∼ where r ∼ r′ iff r = r′u for some u ∈ R. Define an order givenby divisibility, say [r] ≤ [r′] ⇐⇒ r|r′. Then (I,≤) is a directed poset since r|rr′ and r′|rr′.For every r ∈ R − 0 define A[r] = 1

rR the locaalization of R along r. Note that A[r]

∼= R

as R−modules, so A[r] is flat. For r|r′, say r′ = ra, define maps ϕ[r][r′] : A[r] → A[r′] given by

ϕ[r][r′](

1rx) = 1

raax = 1

r′ax.

We obtain a direct system(A[r], ϕ

[r][r′]

)r|r′

. It remains to check that Q ∼= lim−→A[r] by showing

that Q satisfies the universal mapping property of the direct limit. This is left as an exercise.

It is nice to note that the construction above holds for any localization S−1R of an integraldomain; i.e. they arise as direct limits of directed systems on S/ ∼⊂ I.

Example 3.4.7. Countexamples.

1. Z is flat and projective Z−module.

2. Q is flat and injective, but not projective Z−module.

3. Q/Z is an injective and not flat Z−module. The inclusion morphism ι : Z → Q showsthe non-flatness since idQ/Z⊗ι : 0 6= Q/Z⊗ZZ→ Q/Z⊗ZQ = 0 is highly not injective.

Definition 3.4.8. Let B = RB. The left R−module B∗ = HomZ(ZBR,Q/Z) is called thecharacter module (or Pontryagin dual) of B.

Lemma 3.4.9. A sequence Aα→ B

β→ C in Mod−R is exact if and only if C∗β∗→ B∗

α∗→ A∗

is exact in R−mod.

Proof. (⇒) Follows since HomZ(−,Q/Z) is exact.

(⇐) Is done by hand: Im(α) ⊂ Ker(β): Otherwise, there would exist an element a ∈ A withβ(α(a)) 6= 0. Using that Q/Z is a co-generator for Z−mod, there is a map f : C → Q/Zin Z−mod with f(β(α(a))) 6= 0; f ∈ C∗. Hence, (α∗ β∗)(f)(a) = f(β(α(a))) 6= 0,contradicting that Im(β∗) = Ker(α∗).

Ker(β) ⊂ Im(α). By contradiction again, suppose there exists b ∈ Ker(β) with b+Im(α) 6=0 + Im(β). Since Q/Z is a co-generatr there exists g : B/Im(α) → Q/Z in Z−mod withg(b + Im(α) 6= 0. In particular, by setting h = g π ∈ B∗, h(b) 6= 0 where π is the

natural projection. Since α∗(h) = h α, h ∈ Ker(α∗) = Im(β∗) and so there exists h ∈ C∗with h = β∗(h) = h β. We have h(b) = g(b + Im(α)) 6= 0, but since b ∈ Ker(β),

h(b) = h(β(b)) = 0, a contradiction.


After this, we cant some kind of Baer criterion for flat modules.

Theorem 3.4.10. Let B = BR. Then B is flat if and only if B∗ is injective.

Proof. (⇒) We already know it since B = ZBR is R−flat, Q/Z is an injective Z−moduleand theorem 3.4.2 implies that R∗ = HomZ(B,Q/Z) is injective, thus flat.

(⇐) Let α : A → A′ be an injective R−module homomorphism. We need to show thatidB ⊗ α is injective. Since B∗ is an injective left R−module, we get an exact sequence,

HomR

(A′, B∗

)HomR

(A,B∗

)0

HomZ(B ⊗R A′,Q/Z

)HomZ

(B ⊗R A,Q/Z

)0

(B ⊗R A′)∗ (B ⊗R A)∗

α∗

∼= ∼=

Here, the bottom line is exact so, by previous lemma, 0→ B⊗RAidB⊗α→ B⊗RA′ is exact.

Theorem 3.4.11 (Baer’s criterion for flatness). Let B = BR. If for each finitely generatedleft ideal I of R with inclusion ι : I → R, the map idB ⊗R ι : B ⊗R I → B ⊗R R is injective.Then B is flat.

Proof. Step 1. Let J be an arbitrary left ideal of R. Define S to be the set of all finitelygenerated left sub-ideals J ′ ⊂ J . S is a partially ordered non empty directed poset. Bytaking the inclusion maps αJ ′ : J ′ → J , one can check that lim−→

J ′∈SJ ′ ∼= J . By assumption, for

every J ′ ∈ S, idB ⊗R ι : B ⊗R J ′ → B ⊗R R is injective. We now use that S is a directedposet so that lim−→J ′

is exact, to see that if we put h = idB ⊗ ιJ in the following diagram, itwill still commute.

lim−→J ′∈S

(B ⊗R J ′

)lim−→J ′∈S

(B ⊗R R

)

B ⊗R lim−→J ′∈S

J ′ B ⊗R R

B ⊗R J

−→(idB⊗ιJ′ )

h=idB⊗ιJ

The top left isomorphism holds since B ⊗R − preserves the direct limits. It follows that forevery ideal J of R the map idB ⊗R ι : B ⊗R J → B ⊗R R is injective.

Step 2. To prove thar BR is flat, we use the previous theorem 3.4.10 to prove instead thatB∗ is injective. By Baer’s criterion, we need to check that for any left ideal with inclusion

3.5. SCHANNEL’S LEMMA AND STABLE MODULE CATEGORIES 43

ιJ : J → R, then HomR(R,B∗)ι∗J→ HomR(J,B∗) → 0 is exact. By adjointness, we get the

commutative diagram,

HomR

(R,B∗

)HomR

(J,B∗

)0

HomZ(B ⊗R R,Q/Z

)HomZ

(B ⊗R J,Q/Z

)0

ι∗J

∼= ∼=

(idB⊗ιJ )∗

But the bottom row is exact by step 1 and exactness of HomZ(−,Q/Z); meaning that B∗ isan injective module.

As an abstract map of ideas, one can think in the following

Flatness Injectivity

Injectivity Projectivity

Pontryagin

Duality

Nice, right?

3.5 Schannel’s Lemma and Stable module categories

Theorem 3.5.1. Let C be a R−module category.

(a) (Schannel’s Lemma) Given two s.e.s. in C 0 → Ki → Pifi→ B → 0 where P1, P2 are

projective, we have that K1 ⊕ P2∼= K2 ⊕ P1.

(b) Dually, given two s.e.s. in C 0 → Bgi→ Ei → Qi → 0 where E1, E2 are injective

modules, then E1 ⊕Q2∼= E2 ⊕Q1. .

Proof. We will sketch part (a), (b) is dual.

Take the pullback of f1 and f2,

X P1

P2 B

π1

π2 f1

f2

where X = (a1, a2) ∈ P1 ⊕ P2|f1a1 = f2a2 and π1(a1, a2) = ai.

We have (this is an exercise) that in any pullback diagram, parallel arrows have isomorphickernels and if fi is an epimorphism then π2−i is an epimorphism (i = 1, 2). We get then, thefollowing s.e.s.

0→ K2−i ∼= Ker(πi) → Xπi→ Pi → 0


This splits since Pi is injective, so P1 ⊕K2∼= X ∼= P2 ⊕K1.

As a comment, one can use the above as a base case for an induction and prove a similarresult with longer exact sequences of projective modules (see Problem 3.3).

0→ Ki → P 1i → P 2

i → . . .→ P ni → B → 0

Definition 3.5.2. Let C = R−mod.

(a) Let ε : P → M be an epimorphism in C, where P is a projective module. Define

Ω(M) := Ker(ε). By Shannel’s lemma this is unique up to “adding on or substractingoff” projective modules.

Let α : M1 → M2 be in C, and let εi : Pi → Mi be epimorphisms in C where Pi isprojective, i = 1, 2.

0 Ω(M1) P1 M1 0

0 Ω(M2) P2 M2 0

Ω(α)

ε1

∃α α

ε2

Since ε2 is surjective and P1 is projective module, there exists (not unique) a morphism

α : P1 → P2 in C such that ε2 α = α ε1. Define Ω(α) := α|Ω(M1)

Note that if x ∈ Ω(M1), then 0 = α(ε1(x)) = ε2(α(x)) and so α(x) ∈ Ker(ε2) =:

Ω(M2). Thus Ω is well defined.

Also note that Ω is unique up to adding morphisms factoring through a projective

module. The latter is true since, given ˜α : P1 → P2 satisfying ε2 ˜α = α ε1, then

ε2(α− ˜α) = 0 and α− ˜α : P1 → Ω(M2). Hence, α|Ω(M1) − ˜α|Ω(M1) factors through P1.

(b) Let ι : M → E be a monomorphism in C where E is injective morule, define Ω−1(M) :=cokerι = E/Im(ι). By the previous result, this is uniqe up to adding on or substractingoff injective modules.

Let α : M1 → M2 be in C and let ιi : Mi → Ei be a monomorphism in C where Ei isinjective (i = 1, 2). We get a diagram,

0 M1 E1 Ω−1(M1) 0

0 M2 E2 Ω−1(M2) 0

ι1

α ∃α Ω−1(α)

ι2

Since ι1 is monic and E2 is an injective module, there exists α : E1 → E2 withα ι1 = ι2 α. Define then,

Ω−1(α) : Ω−1(M1)→Ω−1(M2)

x+ ι1(M − 1) 7→α(x) + ι2(M2)


Ω−1(α) is well defined because of the commutativity of the left side of the diagram.Note that this map is unique up to addition of morphisms factoring through an injec-tive module (E2).

The goal is to make Ω and Ω−1 into functors in some new categories.

Definition 3.5.3 (Stable Module Category).

(a) Stable Module Category, denoted by R-mod, is the category where Ob(R−mod) =Ob(R−mod). For M,N R−modules, define

PHomR(M,N) = α ∈ HomR(M,N)|α factors through a projective module

PHomR(M,N) is a subgroup of HomR(M,N) since if f (resp. g) factors through P(resp. Q), then f − g factors through P ⊕Q. We define the morphisms from M to Nin R-mod to be HomR(M,N) := HomR(M,N)/PHomR(M,N).

Note that in the stable category, projective modules are terminal objects. We get Ωdefines a functor Ω : R−mod→ R−mod.

(b) The category denoted by R−mod, is the category where Ob(R−mod) = Ob(R−mod).For M,N R−modules, define

IHomR(M,N) = α ∈ HomR(M,N)|α factors through an injective module

We define the morphisms from M to N in R−mod to be

HomR(M,N) := HomR(M,N)/PHomR(M,N)

Ω−1 defines a functor Ω−1 : R−mod→ R−mod.

Remark 3.5.4. The functors Ω and Ω−1 are often mixed. This, since in some cases, thefunctors tend to be quasi-inverse of each other.

A ring R is called self-injective is R is an injective lefts R−module. One can show that thefollowing are equivalent notions,

(a) R is self-injective and Noetherian.

(b) R is self-injective and Artinian.

(c) Every injective left R−module is projective.

(d) Every projective left R−module is injective.


A ring satisfying (a)-(d) is called Quasi-Frobenious. Suppose R is Quasi-Frobenious, thenR−mod = R−mod and Ω and Ω−1 are quasi-inverse functors.

To check that they are quasi-inverses, we see that given a s.e.s. 0→ Ω(M)→ Pε→ M → 0

with P projective, since R is Quasi-Frobenious, P has to be injective and so M ∼= Ω−1(Ω(M))

in R−Mod. Similarly, M ∼= Ω(Ω−1(M)) in R−mod.

An example of the above facts, if K is a field, G a finite group, then the group algebra KGis quasi-frobenious.

Problem set 3

Problem 3.1.

(a) Show that the Z−module Z/2Z does not have a provective cover. You can use withoutproof that over a PID, every projective module, whether or not it is finitely generated,is free.

(b) Let R be a left Artinian ring, and let M be a finitely generate left R−module. Provethat M has a projective cover.

Hint: Let L be a finitely generated projective left R−module with epimorphism f :L→M . Let S be the set of all submodules N of Ker(f) such that FN : L/N →M isan essential epimorphism. Use that L is an Artinian R−module to show that S has aminimal element. Call it X. Then show that P = L/X is a projective cover of M .

Problem 3.2. This is an example of a direct system(Bii∈I , ϕiji≤j

)of right R−modules

over a directed partially ordered set (I,≤) such that lim−→Bi is flat but not all Bi are flat.

Let k be a field and let R = k[x, y] be the polynomial ring over k in two commuting variables.

(a) Let m = (x, y) be the maximal ideal of R. Prove that m is not a flat R−module byshowing that the inclusion map ι : m→ R does not stay injective when tensoring withm over R.

Hint: Show that x⊗ y− y⊗ x is not zero in m⊗R m but that x⊗ y− y⊗ x is zero inm×RR. To see that x⊗y−y⊗x is not zero in m⊗Rm, consider the natural surjectionm → m/m2 and use that m ⊗R − is a right exact functor. Show that m ⊗R m2 is a4-dimensional vector space over k and show that the image of x⊗ y− y⊗ x ∈ m⊗R min this vector space is not zero.

(b) Let I = 1, 2 with 1 < 2. Consider the direct system m → R of R−modules, indexedby I. Show that the direct limit of this direct system is isomorphic to R. Since R isflat over R but m is not flat over R, this gives the desired counter-example.


Problem 3.3. [Schannel’s Lemma for longer exact sequences]

(a) Given two exact sequences of R−modules (all left or all right R−modules)

0→ B → E0 → E1 → · · · → En → X → 0

and0→ B → D0 → D1 → · · · → Dn → Y → 0

where all Ei, resp. Di, are injective, prove that

X ⊕Dn ⊕ En−1 ⊕Dn−2 ⊕ · · · ∼= Y ⊕ En ⊕Dn−1 ⊕ En−2 ⊕ · · ·

(b) Given two exact sequences of R−modules (all left or all right R−modules)

0→ K → Pn → Pn−1 → · · · → P0 → B → 0

and0→ L→ Qn → Qn−1 → · · · → Q0 → B → 0

where all P i, resp. Qi, are projective, prove that

K ⊕Qn ⊕ Pn−1 ⊕Qn−2 ⊕ · · · ∼= L⊕ Pn ⊕Qn−1 ⊕ Pn−2 ⊕ · · ·

Chapter 4

Functor Categories

4.1 The Godel-Bernays System

The following axioms are taken from [1]. Upper case letters denote class variables, and lowercase letters denote set variables. 〈u, v〉 denotes ordered pairs.

Axioms for sets:

(1) Y ∈ X → Y is a set

(2) Y = X ↔ ∀u (u ∈ X ↔ u ∈ Y )

(3) ∃x ∀y (¬y ∈ x)

(4) ∀x∀y ∃z ∀w (w ∈ z ↔ w = x ∨ w = y)

(5) ∀x ∃y ∀z (z ∈ y ↔ ∃w(z ∈ w ∧ w ∈ x))

(6) ∃x (∅ ∈ x ∧ ∀y(y ∈ x→ y ∪ y ∈ x))

(7) ∀x ∃y ∀z (z ∈ y ↔ z ⊂ x)

(8) Axiom of replacement:∀X ((∀u∃!v〈u, v〉 ∈ X)→ (∀u∃v(∀t(t ∈ v ↔ ∃w(w ∈ u ∧ 〈w, t〉 ∈ X)))))

Axioms for class formation:

(9) ∃X ∀a(a ∈ X ↔ ∃b∃c(a = 〈b, c〉 ∧ b ∈ c))

(10) ∀X∀Y ∃Z∀u(u ∈ Z ↔ u ∈ X ∧ u ∈ Y )

49

50 CHAPTER 4. FUNCTOR CATEGORIES

(11) ∀X ∃Y ∀u(u ∈ Y ↔ ¬u ∈ X)

(12) ∀X∃Y ∀u(u ∈ Y ↔ ∃u(〈v, u〉 ∈ X))

(13) ∀X∃Y ∀a(a ∈ Y ↔ ∃b, c(〈b, c〉 = a ∧ 〈c, b〉 ∈ X))

(14) ∀X∃Y ∀u(u ∈ Y ↔ ∃a, b, c(〈a, b, c〉 = u ∧ 〈a, c, b〉 ∈ X))

(15) Axiom of choice:∃X∀a(a 6= ∅ → ∃!u(u ∈ a ∧ 〈a, u〉 ∈ X))

(16) Axiom of regularity:∀X(X 6= ∅ → ∃u(u ∈ X ∧ ∀y(y ∈ X → ¬y ∈ u)))

The axioms of GB, with the exception of the Axiom of Choice, are all valid in ZF (Zermelo-Fraenkel). The Axiom of Choice cannot be proved in ZF.

Recall that for C,D categories, the functor category DC is the category of covariant functorsC → D with morphisms being natural transformations between functors. The issue is thatHomDC(F,G) may not be a set. MacLane’s book says that most people base their categorytheory on either GB or ZF with a given universe. Let’s comment on the axioms:

(1) A class has only sets as members.

(2) Is called ”extensionality of classes”.

(3) Is the empty set axiom

(4) The pair set axiom: z = x, y

(5) The union set axiom: y = ∪x

(6) Axiom of infinity: x = ∅, ∅, ∅, . . .

(7) Power set axiom: y = 2x

(8) Is about the rank of a function∀X(∀u∃!v〈u, v〉 ∈ X → v is the range of the functions defined by X on u.

(9) Means X is the collection of all pairs 〈b, c〉 with b ∈ c.

(10) Intersectioin of classes

(11) The complement of a class exists

(12) Y picks out second entry in all pairs of X

(13) Y picks out the reversed ordered pairs of all ordered pairs in X

(15)-(16) Are about ordered triplets.

4.2. FUNCTOR CATEGORIES 51

(17) Axiom of choice: means that there is a class which picks out one member from eachnon-empty set.

(18) Each class X has a member u such that no element of X is a member of u.

Definition 4.1.1. A proper class is a class that is not a set. In other words, the word classmeans either set or a proper class.

Definition 4.1.2. A category C is said to be small if Ob(C) is a set and ∀X, Y ∈ Ob(C)HomC(X, Y ) is a set. A category C is said to be large if Ob(C) is a class and the Hom-setsare also classes.

All the categories previously discussed were large categories with small hom sets.

4.2 Functor categories

If both C,D are small then DC is small. If C is small and D is large, then DC is large.Moreover, if D has small hom sets, then DC has small hom sets.

Example 4.2.1. Suppose D is a category where Ob(D) is a proper class. Define C byOb(C) = ∗ and HomC(∗, ∗) = id∗. Any functor F : C → D is uniquely determined byF (∗). In other words, Ob(DC) = Ob(D) and HomDC(F,G) = HomD(F (∗), G(∗)).

A classical example of a proper class is the Russell class1

R := x|(x is a set ), (¬x ∈ x)

R is a proper class since if R were a set, we would get R ∈ R⇔ ¬R ∈ R, a contradiction.

The problem of Russell’s paradox is still not solved just by defining classes. To see this,define categories C,D by Ob(C) = R and Ob(D) = 0, 1 with only identity morphisms.Again, any functor is uniquely determined by UF = x ∈ R|F (x) = 0. Thus

Ob(DC) = U |(x ∈ U)→ (x ∈ R)

In particular R ∈ Ob(DC), and Ob(DC) is not a class by GB axiom (1). One could say thatDC is a category ”larger than large”.2

Let C,D be large categories.

Definition 4.2.2. Let F : C → D be a covariant functor.

1Russell’s paradox about the set of all sets.2The power class of a class.


(a) F is fully faithful if ∀X, Y ∈ Ob(C), the map

φX,Y : HomC(X, Y )→HomD(FX,FY )

α 7→F (α)

has a 2-sided inverse.

(b) F is dense if there exists a map T : Ob(D) → Ob(C) such that ∀X ′ ∈ Ob(D) we havean isomorphism ρX′ : X ′ → F (TX ′) in D.

Theorem 4.2.3. A covariant functor F : C → D is an equivalence of categories iff F isfully faithful and dense.

Proof. (⇒) By assumption, there is a quasi-inverse functor F ′ : D → C and natural isomor-phisms τ : IdC → F ′ F and σ : IdC → F F ′. Let X, Y ∈ Ob(C), α ∈ HomC(X, Y ). Weget a commutative diagram

F ′(FX) X

F ′(FY ) Y

F ′(Fα) α

τX

τY

(4.1)

Hence α = τ−1Y (F ′(Fα)) τX . This implies that if φY,X(α) = φX,Y (β); i.e. F (α) = F (β),

then α = β. Similarly, using F ′ instead of F , we see that for all X ′, Y ′ objects in D andα′, β′ : X ′ → Y ′ morphisms, we have F ′(α) = F ′(β)⇒ α′ = β′.

Define

ψX,Y : HomD(FX,FY )→HomC(X, Y )

α′ 7→τ−1Y (F ′α′) τX

Then, ψX,Y (φX,Y (α) = τ−1Y (F ′α′) τX

(4.1)= α. Also, φX,Y (ψX,Y (α′) = F (τ−1

Y (F ′α′) τX).

It is enough to show that F (α) = α′, where α = τ−1Y (F ′α′) τX . To check this, we apply

F ′ F to α and get,

F ′(F (α))(4.1)= τY ατ−1

X = F ′(α′)

Hence, F (α) = α′ and F is fully faithful.To see that F is dense, define T : Ob(D) → Ob(C) by T (X ′) = F ′(X ′) and ρX′ : X ′ →F (F ′(X ′) to be σX′ .

(⇐) Suppose now that F is fully faithful and dense. Define a functor F ′ : D → C as follows:For every object X ′ in D, let F ′(X ′) = T (X ′) and for α′ : X ′ → Y ′ a morphism in D considerthe diagram


F (F ′X ′) X ′

F (F ′Y ′) Y ′

α′

ρX′

∼=

ρY ′

∼=

Since F is fully faithful, we can define F ′(α′) : F ′(X ′)→ F ′(Y ′) to be the unique morphismin C such that F (F ′(α′)) = ρY ′ α′ ρ−1

X′ .From F being fully paithful, one can check that F ′ is indeed a functor and ρ = ρX′ givesa natural isomorphism ρ : IdD → F F ′.On a similar fashion, define τ : IdC → F ′ F by τX : X → F ′(FX) being the uniquemorphism in C such that F (τX) = ρFX . Again, τ will the desired natural isomorphism.

4.2.1 Representable Functors

Let C be a large category with small hom sets.

Definition 4.2.4. Let F : C → Sets be a functor.

(a) If F is covariant, F is representable is there exist an objectX in C with F ∼= HomC(X,−).

(b) If F is contravariant, F is representable is there exist an object Y in C with F ∼=HomC(−, Y ).

(c) If the Hom sets in C are always abelian groups, we can use the same definition as in(a), (b) for representable functors F : C → Ab.

Example 4.2.5 (A confusing Representable functor). Take C = R−Mod, A ∈ Ob(Z−Mod)fixed and define F = HomZ(−, A) : R −Mod → Ab. We claim that F is representable; i.e.there exists Y ∈ Ob(R−Mod) such that F ∼= HomR(−, Y ).Define Y = HomZ(ZRR, ZA), Y is a left R−module. Since,

(R ⊗R −, HomZ(R,−)

)is an

adjoint pair, we get:

HomR(−, Y ) = HomR(−, HomZ(R,A)) ∼=HomZ(R⊗R −, A)

=HomZ(−, A) (R⊗R −)∼=F IdR−mod∼=F


Theorem 4.2.6 (Yoneda’s Lemma).

(a) Let F : C → Sets be a covariant functor and X ∈ Ob(C). Then the map

φX : HomSetsC(HomC(X,−), F

)→F (X)

τ 7→τX(IdX)

has a 2-sided inverse. In particular, HomSetsC(HomC(X,−), F

)is a set.

(b) Let F : C → Sets be a contravariant functor and Y ∈ Ob(C). Then the map

ψY : HomSetsC(HomC(−, Y ), F

)→F (Y )

σ 7→σY (IdY )

has a 2-sided inverse. In particular, HomSetsC(HomC(−, Y ), F

)is a set.

Proof. We will prove (a), (b) is similar. Define

φX : F (X)→HomSetsC(HomC(X,−), F

)s 7→τs

where for all Y ∈ Ob(C),

(τs)Y : HomC(X, Y )→F (Y )

f 7→F (f)(s)

One can check that τs is a natural transformation. Moreover,

φX(φX(s)) = φX(τs) = (τs)X(idX) = F (idX)(s) = s

and for every Y ∈ Ob(C) and f ∈ HomC(X, Y ),

φX(φX(τ)Y (f) = φX(τX(idX))Y (f) = F (f)(τX(idX))(4.2)= τY (f∗(idX)) = τY (f)

The equation marked above follows from the following commutative diagram

HomC(X,X) F (X)

HomC(X, Y ) F (Y )

τX

f∗ F (f)

τY

(4.2)

Therefore, φX(φX(τ)) = τ as we wanted.

Corollary 4.2.7. Let C be as in Yoneda’s Lemma.


(a) Define H : C → SetsC by H(X) = HomC(X,−) and, for α : X → Y a morphism in C,

H(α)Z : HomC(Y, Z)→ HomC(X,Z)

f 7→ f α = α∗(f)

Then, H is a contravariant “functor”3 that is fully faithful.

(b) (Dual) Define∨H : C →

∨SetsC by H(Y ) = HomC(−, Y ) and, for β : Y → Z a morphism

in C,

∨H(β)Z : HomC(−, Y )→ HomC(−, Y )

β 7→ β∗

where β∗(f) = β f . Then,∨H is a covariant “functor” that is fully faithful.

Proof of (a). H is contravariant since (β α)∗ = α∗ β∗. To show that H is fully faithful,we will use part (a) of Yoneda’s lemma for F = H(X) and Y as an object in C. We get abijection

HomSetsC((H(Y ), H(X))

)→F (Y ) = HomC(X, Y )

τ 7→τY (idY )

with inverse α 7→ σα, where for Z ∈ Ob(C),

(σα)Z : HomC(Y, Z)→HomC(Y,X)

f 7→F (f)(α)

But,

F (f)(α) =(HomC(X,−)

)(α)

=f∗(α)

=f α=α∗(f)

Hence, (σα)Z = α∗. Meaning σα = H(α) which concludes the faithfulness of H.

Problem set 4

Problem 4.1. A full subcategory C0 of C is said to be a skeleton of C if every object in C isisomorphic to exactly one object in C0. Assuming the Axiom of Choice4, prove that C andC0 are equivalentcategories. Show that C and C0 need not be isomorphic categories (give anexample).

3The Hom sets may get big.4e.g., using Godel-Bernays system.

Chapter 5

Morita Theory

“Tensor products are your friends” Professor Bleher.

The question Morita theory adresses is: When are two module categories equivalent? Themathematician Kiiti Morita was the fist to answer this question [2].

Definition 5.0.1 (Progenerator). A left R−module P is called a progenerator for R−modif P is a finitely generated projective generator for R−mod.

For any n ∈ Z>0, Rn is a progenerator for R−mod.

Theorem 5.0.2. Let F : R−mod→ S −mod be an additive covariant functor such that Fhas an additive quasi-inverse F ′ : S −mod→ R−mod. Then P := F ′(S) is a progeneratorof R−mod and EndR(P )op ∼= S as rings.

Proof. We first see that P is projective via the definition: Given a commutative diagram inR−mod with exact row (left) , we can apply F to get a commutative diagram in S−mod(right), which also has exact botom row.

P

B C 0

αγ

β

F

F (P ) S

F (B) F (C) 0

F (α)

ρ

∼=

∃γ

F (β)

Since S is a projective S−module, there exists γ : S → F (B) satisfying F (β) γ ρ = F (α).Since F is fully faithful, there exists a unique γ such that F (γ) ∼= γ ρ. Hence, β γ = αand P is projective.

57

58 CHAPTER 5. MORITA THEORY

We now prove that P is a generator of the R−module category. Let M be a left R−module,then F (M) ∈ Ob(S − mod) and, since S is a generator of S − mod, there exists an indexset Λ and an epimorphism τ : q

λS → F (M). We get an epimorphism F ′(τ) : F ′(q

λS) →

F ′(F (M)) ∼= M . But (F ′, F ) is an adjoint pair so it commutes with qλ

.

To show that P is finitely generated, we can use the same arguments as above to show thatF (R) ∈ Ob(S −mod) is a projective generator of S−mod. Since S is finitely generated leftS−module, there is n ∈ Z+ and an epimorphism κ : F (R)n → S. Then, F ′(κ) : Rn →F ′(S) ∼= P does the job.

By last, we show the ring isomorphism EndR(P )op ∼= S. Define

ξ : EndR(P )op →SF ′(β) 7→β(1)

This is doable since P = F ′(S) and F ′ being fully faithful allows us to write each α ∈EndR(P ) as α = F ′(β) for a unique β ∈ F (S). One can check that ξ is an additive bijection.To end, ξ is a ring homomorphism since, for any β, γ ∈ EndS(S) ∼= S we get

F ′(β)op F ′(γ) =F ′(γ) F ′(β)

=F ′(γ β)

=γ(β(1))

=γ(β(1) • 1) = β(1) • γ(1)

Notice that the reason of the oposite ring structure in EndR(P ) is that we actually provedthat EndS(S) ∼= EndR(P ) and the last equality on the proof above gives us EndS(S)op ∼= S.

Definition 5.0.3 (Morita Equivalence).

(a) A Morita context consists of two rings R and S, two bimodules P = RPS and Q = SQR

and two bimodule homomorphisms

σ : P ⊗S Q→ R , τ : Q⊗R P → S

(R,R)− and (S, S)−bimodule homomorphism, respectively; satisfying ∀x, y ∈ P ∀u, v ∈Q

σ(x⊗ u).y = xτ(u⊗ y)

v.σ(x⊗ u) = τ(v, x).u

(b) Two rings R and S are said to be Morita equivalent is there exists a Morita context(R, S, P,Q, σ, τ) with σ and τ surjective. We write R ∼M S

59

Theorem 5.0.4 (Morita 1). Let R ∼M S with a Morita context (R, S, P,Q, σ, τ). Then

(1) σ, τ are isomorphisms, and

(2) Q⊗R − : R −mod → S −mod is an additive equivalence with quasi-inverse P ⊗S −.Similarly, −⊗R P : mod−R→ mod− S is an additive equivalence of categories withquasi-inverse −⊗S Q.

(3) P is a progenerator for R −mod and mod − S, and Q is aprogenerator for S −modand mod−R.

(4) S ∼= EndR(P )op and R ∼= EndS(Q)op as rings.

Proof. (3) and (4) follow from (2) and theorem 5.0.2.(1) By assumption, σ, τ are surjective. We will show that σ is injective, τ is similar. Sinceσ is surjective, there exist z =

∑i xi ⊗ ui ∈ P ⊗S Q with σ(z) = 1R. Let t =

∑j yj ⊗ vj ∈

Ker(σ) ⊂ P ⊗S Q. Then,

t = t • σ(z) =∑i,j

yj ⊗(vj • σ(xi ⊗ ui)

)=∑i,j

yj ⊗(τ(vj ⊗ xi) • ui

)=∑i,j

(yjτ(vj ⊗ xi)

)⊗ ui

=∑i,j

σ(yj ⊗ vj) • xi ⊗ ui

=σ(t) • z = 0R • z = 0P⊗SQ

Thus, σ is injective.

(2) Let F = Q⊗R −, F ′ = P ⊗S −. We will show that F ′ F ∼= IdR−mod; F F ′ ∼= IdS−modis similar and so showing that −⊗R P , −⊗S Q are quasi-inverses of each other.Define ρ : F ′ F = P ⊗S Q ⊗R − → IdR−mod as follows; ∀m ∈ Ob(R − mod), ρM :P ⊗S Q⊗R Q→M with ρM = αM (σ ⊗ idM) where

αM : R⊗RM →Mr ⊗m 7→rm

We know that α = αMM is already a natural isomorphism. By (1), σ is an isomorphism.It remains to show that ρ is natural with repect to R−module homomorphisms f : M → N .This is, to check the commutativity of the following diagram,

P ⊗S Q⊗RM M

P ⊗S Q⊗R N N

ρM

(idP⊗SQ)⊗f f

ρN


We have the following equations on simple tensors,

f(ρM(x⊗ u⊗m)) = f(σ(x⊗ u)m) = σ(x⊗ u)f(m)

andρN((idP⊗SQ)⊗ f)(x⊗ u⊗m) = ρN(x⊗ u⊗ f(m)) = σ(x⊗ u)f(m)

The commutativity follows.

Proposition 5.0.5. Let R be a ring, let P be a progenerator for R−mod, and let S =EndR(P )op. Then R ∼M S.

Proof. S acts on the right of P by x.s := s(x), ∀s ∈ S, x ∈ P . S is then a (R, S)−bimodule.Let Q = HomR(RPS, RRR) endowed with a (S,R)−bimodule structure as follows, ∀s ∈S, u ∈ Q, x ∈ P ,

(sur)(x) := u(x.s).r = u(s(x)).r

Define

σ : P ×Q→R(x, u) 7→u(x)

Notice that σ is a S−balanced map, so there exists a unique Z−module homomorphismσ : P ⊗S Q → R satisfying σ(x ⊗ u) = u(x). One can check that σ is a (R,R)−bimodulehomomorphism.We will see that σ is surjective: P is a generator for R−mod and RR is finitely generated leftR−module, so there is an epimorphism π : P n → R for some n ∈ Z+. For each j ∈ 1, . . . , ndenote by ιj : P → P n the jth−injection. Let r ∈ R, and take x ∈ P n with π(x) = r. Then,

σ( n∑i=1

xi ⊗ π ιi)

=∑i

π ιi(xi)

=π(∑

i

ιi(xi))

=π(x) = r

Thus, σ is surjective. In a similar fashion, define τ : Q× P → S by ∀u ∈ Q∀x, y ∈ P ,

τ(u, x)(y) = u(y).x

τ is R−balanced and so there is a unique Z−module homomorphism τ : Q⊗R P → S withτ(u⊗ x)(y) = u(y).x. Again, τ is actually a (S, S)−bimodule homomorphism.

We will show that τ is surjective: Recall P is a finitely generated projective R−module sothere exists m ∈ Z+ so that P is a direct sumand of F := Rm. Let π : F → P , ι : P → F thecorresponding projection and injection satisfying π ι = InP . For every 1 ≤ j ≤ m denoteby ιj : R→ F the j−th inclusion homomorphism and pj : F → R the j−th projection.

61

Let s ∈ S. Then by recalling the definition of τ and the property∑ιj pj = IdF , we have

the following equalities for all y ∈ P

τ(Σmj=1(pj ι)⊗ (s π ιj)(1R)

)(y) =

m∑j=1

(pj ι)(y).s(π(ιj(1R))

)=s(π( m∑i=1

ιj((pj ι)(y)

)))=s(π(ι(y))

)=s(y)

Hence s = τ(Σmj=1(pj ι)⊗ (s π ιj)(1R)

)and τ is onto.

To be able to conclude that (R, S, P,Q, σ, τ) is a Morita context, it remains to show the”shifting property” between σ and τ .Let x, y ∈ P and u, v ∈ Q, by recalling the products in P and Q, we get,

σ(x⊗ u).y =u(x).y

=τ(u⊗ y)(x) = x.τ(u⊗ y)

and (v.σ(x⊗ u)

)(y) = v(y).σ(x⊗ u) =v(y).u(x)

=u(v(y).x

)=u(τ(v ⊗ x)(y)

)=u(y.τ(v ⊗ x)

)=(τ(v ⊗ x).u

)(y)

Corollary 5.0.6. Let R, S be rings, the following are equivalent.

(1) There exists an additive covariand equivalence F : R −mod→ S −mod with additivequasi-inverse.

(2) R ∼M S

(3) There is a progenerator P for R−mod with S ∼= EndR(P )op.

Proof. (1) ⇒ (3) follows from theorem 5.0.2. (2) ⇒ (1) is Morita 1 and (3) ⇒ (2) isproposition 5.0.5.

People ussually recall (1) ⇒ (3) and think of Morita equivalence as a blackbox.

Example 5.0.7. Let R be a ring, n ∈ Z+ and S := Matn(R). We claim that R ∼M S. Tosee this, it is enough to show that there is a progenerator P for R−mod with EndR(P )op ∼=S. Let P = Rn; P is a progenerator because it is free and, after fixing a basis for P ,EndR(P )op ∼= S


Theorem 5.0.8 (Morita II). Let F : R−mod→ S −mod and F ′ : S −mod→ R−mod beadditive covariant functors quasi-inverses of each other. Let P = F ′(S), Q = F (R). Then

F ∼= HomR(P,−) ∼= P ⊗R −F ′ ∼= HomS(Q,−) ∼= Q⊗S −

Proof. The second isomorphism is because(P ⊗R −, HomR(P,−)


Let η : F ′ F → IdR−mod be a natural isomorphism and define ρ : F → HomR(P,−) by

ρM : F (M)→HomR(P,M)

x 7→ηM F ′(ϕx)

where ϕx : S → F (M) is given by ϕx(1S) = x.

Observe that ρM is an isomorphism since is a composition of isomorphisms as follows

F (M) HomS(S, F (M)) Hom(F ′S, F ′FM) HomR(F ′S,M)∼=F ′

∼=(ηM )∗∼=

One can check naturality of ρ to conclude the isomorphism F ∼= HomR(P,−). Similarly onecan show that F ′ ∼= HomS(Q,−). By last, notice that (F, F ′) and (F ′, F ) are adjoint pairsso(F,HomS(R,−)

)and

(F ′, HomR(P,−)

)are adjoint pairs. But

(Q ⊗R −, HomS(Q,−)

)and

(P ⊗S −, HomR(P,−)

)are already adjoint pairs, so since left/right adjoints are unique

up to natural isomorphism, we can conclude that F ∼= Q⊗R − and F ′ ∼= P ⊗S −.

“Sometimes the tensor equivalence of Morita 2 is easier to work with... :)” ProfessorBleher.

Problem set 5

Problem 5.1. Let R be a ring with 1 as usual.

(a) Let IdR−mod be the identity functor on R−mod. Show that the class of all naturaltransformations IdR−mod → IdR−mod is a ring isomorphic to Z(R), the center of R.

(b) If R−mod and S−mod are equivalent categories where the equivalence is given by twoadditive quasi-inverse functors, use part (a) to prove that Z(R) ∼= Z(S) as rings.

Problem 5.2. Let F, F ′ : C → D and G,G′ : D → C be covariant functors.

(a) Prove that if (F,G) and (F ′, G) are adjoint pairs, then F and F ′ are naturally isomor-phic.

63

(b) Prove that if (F,G) and (F,G′) are adjoint pairs, then G and G′ are naturally isomor-phic.

(c) Prove that if F and G are quasi-inverses of each other then (F,G) and (G,F ) areadjoint pairs.

Part II

Classical Homological Algebra

65

Chapter 6

Homological Algebra

Changing gears, we come back to the original/classical question of the course: What is Ho-mological Algebra? For the topologist audience, a good exercise is to relate the constructionsdiscussed along this chapter to the ones known for several homology/cohomology theoriesexisting. Some of them, like the double complex construction for example, could seem morenatural1 to define from the categorical context (see below).

The following section is going to be phrased in the module category, although it can bediscussed in a more general setup (see Chapter 7): Abelian categories.

6.1 Chain stuff

The following will be stated for R−modules although we will see that it can be done forabelian categories (see bellow). Let C be R−mod.

Definition 6.1.1. Let C = R−mod.

(a) A Chain Complex is a family C• = Cn, ∂nn∈Z of objects Cn in C and morphisms∂n : Cn → Cn−1 satisfying ∂n ∂n+1. We cay that x ∈ Cn is in degree n.

(b) A Co-chain Complex is a family C• = Cn, δnn∈Z of objects Cn in C and morphismsδn : Cn → Cn+1 satisfying ∂n ∂n−1. We cay that x ∈ Cn is in degree n.

(c) A complex C• (resp. C•) is bounded below if Cn = 0 (resp. Cn = 0) for all n << 0.C• (resp. C•) is bounded above if Cn = 0 (resp. Cn = 0) for all n >> 0. A complexis bounded if it is bounded both above and bellow.

1At least, some of them seemed more natural for me (J.R.A.C.)

67

68 CHAPTER 6. HOMOLOGICAL ALGEBRA

(d) Given a chain complex C• = Cn, ∂nn∈Z, we define the n-cycles and n-boundaries tobe Zn(C•) = Ker(∂n) and Bn(C•) = Im(∂n+1), respectively. The n-th Homology ofC• is the R-module

Hn(C•) := Zn(C•)/Bn(C•)

(e) Given a chain complex C• = Cn, δnn∈Z, we define the n-cocycles and n-coboundariesto be Zn(C•) = Ker(δn) and Bn(C•) = Im(δn−1), respectively. The n-th Co-homology of C• is the R−module

Hn(C•) := Zn(C•)/Bn(C•)

(f) A Chain map f : C• → D• consists of a family of morphisms fn : Cn → Dn in Cwith the condition ∂n,D fn = fn−1 ∂n,C .

A Co-chain map f : C• → D• consists of a family of morphisms fn : Cn → Dn inC with the condition δnD fn = fn+1 δnC .

Remark 6.1.2.

1. Note that a chain map f : C• → D• induces a Z−homomorphism on homology

Hn(f) : Hc(C•)→Hn(D•)

[x] 7→[fn(x)]

Similarly with co-chains maps and cohomology.

2. Given C• = cn, ∂n a chain complex over C, we can define a cochain complex C• =Cn, δn by Cn := Cn and δn := ∂−n.

3. The homology groups (resp. cohomology groups) of chain complexes (resp. cochain)are actually objects in C.

Definition 6.1.3. Let f, f ′ : C• → D• be a chain maps. We say that f and f ′ are chainhomotopic , written f ∼ f ′, if for all n ∈ Z there in a morphism hn : Cn → Dn+1 such thatfn − f ′n = ∂n+1,D hn − hn−1 ∂n,C ∀n ∈ Z.

Cn+1 Cn Cn−1

Dn+1 Dn Dn−1

∂n+1,C ∂n,C

hnfnf ′n

hn−1

∂n+1,D ∂n,D

The collection of maps h = hn is called a homotopy operator We say that C• and D•are homotopy equivalent if there are cochain maps f : C• → D• and g : D• → C• withg f ∼ idC and f g ∼ idD. We denote them as C• ∼ D•.

Similarly we can define cochain homotopic maps and cochain homotopy equivalentcochain complexes.

6.1. CHAIN STUFF 69

Theorem 6.1.4. Let f, f ′ : C• → D• be chain homotopic maps. Then for each n ∈ Z,Hn(f) = Hn(f ′). In particular, if C• ∼ D• then Hn(Cn) ∼= Hn(D) ∀n. Similarly forcochains.

Proof. :)

Theorem 6.1.5 (Snake Lemma). Consider the following commutative diagram in C withexact rows and commuting squares

M ′ M M ′′ 0

0 N ′ N N ′′

f1

α′

g1

α α′′

f2 g2

Then we get a well-defined morphism in C, called the connecting morphism,

∂ : Ker(α′′)→Coker(α′)m′′ 7→f−1(α(m)) + Im(α′) ∀m ∈ g−1

1 (m′′)

Moreover, we get an exact sequence in C:

0 0 0

Ker(α′) Ker(α) Ker(α′′)

M ′ M M ′′

N ′ N N ′′

Coker(α′) Coker(α) Coker(α′′)

0 0 0

f1|Ker(α′) g1|Ker(α)

∂

f1

α′

g1

α α′′

f2 g2

f1 g1

Proof. The proof is standard.

Theorem 6.1.6 (Long exact homology sequence). Let 0 → C ′•f→ C•

g→ C ′′•0 be a s.e.s. ofchain complexes2. Then we get a long exact sequence in C

· · · → Hn(C ′•)Hn(f)→ Hn(C•)

Hn(g)→ Hn(C ′′• )∂n→ Hn−1(C ′•)→ · · ·

2This means exact on each degree.


Proof. Let n ∈ Z. We have a well defined morphism in C:

ϕn : Cn/Im(∂n+1)→Ker(∂n−1

x+ Im(∂n+1) 7→∂n(x)

Notice that ker(ϕn) = Hn(C•) and Coker(ϕn) = Ker(∂n−1)/Im(∂n) = Hn−1(C•). The prooffollows from the snake lemma by taking α = ϕ.

Theorem 6.1.7 (Long exact Co-homology sequence). Let 0→ C ′•f→ C•

g→ C ′′•0 be a s.e.s.of co-chain complexes. Then we get a long exact sequence in C

· · · → Hn(C ′•)Hn(f)→ Hn(C•)

Hn(g)→ Hn(C ′′•)δn→ Hn+1(C ′•)→ · · ·

6.2 Derived Functors

In the following pages, we will define the left and right Turaev functors which extend thenotions of our old friends3 the Ext and Tor functors. We will have two constructions of theExt, depending if one decides to take projective or injective resolutions (see bellow). Thespoiler is that they agree.

Theorem 6.2.1 (Comparison theorem). Given a diagram in C where the row are chaincomplexes as above

X• : · · ·X2 X1 X0 M 0

X ′• : · · ·X ′2 X ′1 X ′0 M ′ 0

∂2 ∂1 ε

α

∂′2 ∂′1 ε′

If all Xn are projective modules (n ≥ 0) and the bottom row is exact, then there exists achain map f• : X• → X ′• satisfying f−1 = α.

Moreover, f is unique up to homotopy; i.e., if f ′• : X• → X ′• is another chain map satisfyingf ′−1 = α, then f ∼ f ′.

Proof. Let X−1 = M , ∂0 = ε, Xi = 0 ∂i = 0 ∀i ≤ −2, and X ′−1 = M ′, ∂′0 = ε′, X ′i = 0 ∂′i = 0∀i ≤ −2. Define fi : Xi → X ′i to be the zero map for i ≤ −2 and f−1 = α. Suppose byinduction that we have constructed fk : Xk → X ′k for all k ≤ n satisfying fk−1 ∂k = ∂′k fkfor k ≤ n. We have that ∂′n fn ∂n+1 = fn−1 ∂n ∂n+1 = 0, so we get the following diagram

Xn+1

X ′n+1 Ker(∂′n) = Im(∂′n+1) 0

fn∂n+1

∂′n+1

3For the topology persons reading this.

6.2. DERIVED FUNCTORS 71

Since Xn+1 is projective, there exists a map fn+1 : Xn+1 → X ′n+1 in C satisfying fn ∂n+1 =∂′n+1 fn+1. The statement follows by induction.

Now suppose we have another chain map f ′ : X• → X ′• with f ′−1 = α. Define hi : Xi → X ′i+1

as follows: hi = 0 for i ≤ −1. This works as a homotopy operator because

f−1 − f−1 = α− α = 0 = ∂′0 h−1 + h−2 ∂−1

Assume by induction we have found hk : Xk → X ′k+1 for all k ≤ n satisfying fk − f ′k =∂′k+1 hk + hk−1 ∂k.

· · · Xn+2 Xn+1 Xn Xn−1 · · ·

· · · X ′n+2 X ′n+1 X ′n X ′n−1 · · ·

∂n+2

fn+2f ′n+2

hn hn−1

∂′n+2

Let g := fn+1 − f ′n+1 − hn ∂n+1. We have the following equations

∂′n+1 g =∂′n+1 fn+1 − ∂′n+1 f ′n+1 − ∂′n+1 hn ∂n+1

=fn ∂n+1 − f ′n ∂n+1 − ∂′n+1 hn ∂n+1

=(fn − f ′n) ∂n+1 − ∂′n+1 hn ∂n+1

=(∂′n+1 hn + hn−1 ∂n) ∂n+1 − ∂′n+1 hn ∂n+1

=∂′n+1 hn ∂n+1 − ∂′n+1 hn ∂n+1

=0

We get then the diagram

Xn

X ′n+2 Ker(∂′n+1) = Im(∂′n+2) 0

fn+1−f ′n+1−hn∂n+1

∂′n+2

But Xn is projective so there is a map hn+1 : Xn+1 → X ′n+2 in C such that ∂′n+2 hn+1 =fn+1 − f ′n+1 − hn ∂n+1, as we wanted.

Theorem 6.2.2 (Dual Comparison theorem). Given a diagram in C where the row arecochain complexes as above

X• : 0M X0 X1 X2 · · ·

X ′• : 0M ′ X ′0 X ′1 X ′2 · · ·

ι δ0 δ1

β

δ′0 δ′1

If all Xn are injective modules (n ≥ 0) and the bottom row is exact, then there exists acochain map f • : X ′• → X• satisfying f−1 = β.

Moreover, if f ′• : X ′• → X• is another chain map satisfying f ′−1 = β, then f ∼ f ′.


Definition 6.2.3 (Left Turaev Functor). Let T : C → Ab be an additive covariant functor.For each M ∈ Ob(C), choose a projective resolution

· · · → Pn∂n→ Pn−1 → · · · → P1

∂1→ P0ε→M → 0

Let(PM)• be the truncated complex

(PM)• : · · · → Pn

∂n→ Pn−1 → · · · → P1∂1→ P0

ε→M → 0

For any n ≥ 0 define LnT : C → Ab as follows: For M ∈ Ob(C), LnT (M) := Hn

(T (PM)•

)and; for a morphism in C α : M → M ′, le f :

(PM)• →

(PM ′

)• be a chain map associated

to α by the comparison theorem. Define(LnT

)α := Hn(Tf).

Theorem 6.2.4. Let T : C → Ab be as in the definition. For every n ≥ 0, LnT : C → Ab isan additive covariant functor, called the nth left derived functor of T or nth Left TuraevFunctor.

Proof. By the uniqueness part of the comparison theorem, it follows that (LnT )α is well-defined. The rest is an exercise.

Theorem 6.2.5. Let T and LnT be as above for n ≥ 0. Suppose LnT : C → Ab is anothernth Left Turaev functor of T obtained by choosing possibly different projective resolutions foreach object M ∈ Ob(C). Then LnT and LnT are naturally isomorphic ∀n ≥ 0. In particular

(LnT )M ∼= (LnT )M .

Proof sketch. For all M ∈ Ob(C), let(PM)• be the truncated complex (resp.

(PM)• )

associated to the projective resolution of M used to define (LnT )M (resp. (LnT )M). By

the comparison theorem, there are chain maps on both directions g :(PM)• →

(PM)• and

h :(PM)• →

(PM)• associated to idM .

Define σM : (LnT )M → (LnT )M by σM := Hn(Tg) and τM : (LnT )M → (LnT )M byτM := Hn(Th). Note that via the comparison theorem, h g ∼ id(PM )• and g h ∼ id(PM )•

.Hence,

τM σM =Hn(T (h g))

=Hn

(idT (PM )•big)

=id(LnT )M

and σM τM = id(LnT )M .

To check that σ is a natural transformation let α : M →M ′ in C and take f : (PM)• → (PM ′)•(resp. f : (PM)• → (PM ′)•) be a chain map associated to α by the comparison theorem. We


get a diagram

(LnT )M (LnT )M

(LnT )M ′ (LnT )M ′

σM=Hn(Tg)

Hn(Tf) Hn(T f)

σM′=Hn(Tg′)

To show this commutes we need to check

Hn(T (f g)) = Hn(T (g′ f))

Note that, f g, g′ f : (PM)• → (PM ′)•, where f g is associated to α idM = α and g′ fis associated to idM ′ α = α. Thus, by the comparison theorem we conclude f g ∼ g′ fand so the desired equality.

Definition 6.2.6 (Tor Functor). Let M ′ = M ′R be a right module. Define TorRn (M ′,−) :=

Ln(M ′ ⊗R −) : R−Mod→ Ab.

Remark 6.2.7. We can describe Tor “explicitly”. Let M = RM be a left module, and

choose a projective resolution · · · → P1∂1→ P0

ε→M → 0. After tensoring with M ′ we obtain(M ′ ⊗R (PM)•

): · · · →M ′ ⊗R P2

idM′⊗∂2→ P1idM′⊗∂1→ M ′ ⊗R P0 → 0

Thus, TorRn (M ′,M) = Ker(idM′ ⊗ ∂n

)/Im

(idM′ ⊗ ∂n+1

).

(a) Since M ′ ⊗R − is a right exact functor we get that

TorR0 (M ′,M) =M ′ ⊗RM/Im(idM′ ⊗ ∂1)

=M ′ ⊗R P/ker(idM′ ⊗ ε)∼=Im(idM ′ ⊗ ε)=M ′ ⊗RM

(b) If M is a projective left R-module, then 0 → MidM→ M → 0 is a projective resolution

of M , so the higher Tors are zero; i.e., TorRn (M ′,M) = 0 ∀n ≥ 1 and TorR0 (M ′,M) =M ′ ⊗RM .

Definition 6.2.8 (Right Turaev Functor of a covariant T ). Let T : C → Ab be an additive

covariant functor. For each M ∈ Ob(C) choose an injective resolution 0 → Mι→ E0 δ0→

E1 → · · · . Let E•M be the associated truncated cochain complex E•M : 0→ E0 δ0→ E1 → · · · .

Define (RnT )M := Hn(TE•M). For α : M → M ′ in C, let f : E•M → E•M ′ be a cochain mapassociated to α by the dual comparison theorem. Define (RnT )α := Hn(Tf).

Theorem 6.2.9. Let T : C → Ab be as above. Then for every n ≥ 0 RnT : C → Abis an additive covariant functor called the Right Turaev functor of T. Up to naturalisomorphism, RnT is independent of the choices of injective resolutions.


Definition 6.2.10 (Right Turaev Functor of a contravariant T ). Let T : C → Ab be acontravariant additive functor. For every object M in C choose a projective resolutionwith the associated truncated chain complex (PM)•. For every n ≥ 0 define (RnT )M :=Hn(T (PM)•). For all α : M → M ′ morphism in C let f : (PM)• → (PM ′)• be an associatedchain map by the comparison theorem. Then Tf : T (PM ′)• → T (PM)• is in Ab and define(RnT )α := Hn(Tf).

Theorem 6.2.11. Let T : C → Ab be an additive contravariant functor. For every n ≥ 0RnT : C → Ab is an additive contravariant functor called the Right Turaev functor ofT. Up to natural isomorphism, RnT is independent of the choices of projective resolutions.

Remark 6.2.12. Notice the change of indexing in the contravariant case of the right Turaev

functor. For a projective resolution of M · · · → P1∂1→ P0

ε→M → 0, we have

T (PM)• : 0→ TP0T∂1→ TP1

T∂2→ · · ·

and so (RnT )M = ker(T∂n+1)/Im(T∂n).

Definition 6.2.13 (Ext functor). Let C be R-mod or Mod-R and fix M ′ ∈ Ob(C). Definefor n ≥ 0 ExtnR(−,M ′) := RnHomR(−,M ′).

Remark 6.2.14. If M ∈ Ob(C), we choose a projective resolution · · · → P1∂1→ P0

ε→M → 0.By appying HomR(−,M ′) to the truncated resolution we obtain

0→ HomR(P0,M′)

∂∗1→ HomR(P1,M′)

∂∗2→ · · ·

concluding ExtnR(M,M ′) = ker(∂∗n+1)/Im(∂∗n).

(a) HomR(−,M ′) is left exact so

Ext0R(M,M ′) =Ker(∂∗1)

=Im(ε∗)∼=HomR(M,M ′)

(b) If M is projective, then a truncated projective resolution of M is · · · → 0 → M → 0with M in degree zero. Thus, ExtnR(M,M ′) = 0 ∀n ≥ 1.

(c) We write right Turaev functors (like Ext) with a super index and left Turaev functors(like Tor) with an under one to recall that the corresponding constrictions arise fromthe cohomology or homology of a truncated sequence applied to T .

Example 6.2.15 (Ext and Tor of Z/mZ). Take R = Z, M0 ∈ Ob(Z − mod) and letM = Z/mZ for some fixed m ≥ 2.

6.3. LONG EXACT SEQUENCES 75

1. A projective resolutioin of M is given by 0 → Z ×m→ Z π→ Z/mZ → 0. We get a

truncated chain complex 0→ Z ×m→ Z→ 0, where Z is in degree 0 and 1. To computeTorZn(M0,M

′) we complete the diagram

1 0

0 M0 ⊗Z Z M0 ⊗Z Z 0

0 M0 M0 0

idM0⊗(×m)

∼= ∼=

×m

Hence, TorZ0 (M0,M) = M0/mM0, TorZ1 (M0,M) ∼= x ∈M0 : mx = 0 and TorZn(M0,M) =0 ∀n ≥ 2.

2. To compute ExtnZ(M,M0) we note the isomorphism HomZ(Z,M0) ∼= M0 via f 7→ f(1)and that (×m)∗ : g 7→ g(×m). The following diagram commutes

0 1

0 HomZ(Z,M0) HomZ(Z,M0) 0

0 M0 M0 0

∼=

(×m)∗

∼=

×m

Hence, Ext0Z(M,M0) = x ∈M0 : mx = 0, Ext1Z(M,M0) = M0/mM0 and ExtnZ(M,M0) =0 ∀n ≥ 2.

6.3 Long Exact Sequences

Along this section, C will denote a module category.

For the following lemma, it is important to recall what a horseshoe is.

Figure 6.1: The shoe of a horse.


Lemma 6.3.1 (Horseshoe Lemma). Take a diagram in C as bellow, where the two columnsare projective resolutions of M ′, M ′′ with corresponding truncated complexes (PM ′)• and(PM ′′)•, and the bottom rows are exact.

......

P ′1 P ′′1

P ′0 P ′′0

0 M ′ M M ′′ 0

0 0

∂′1 ∂′′1

ε′ ε′′

α β

Then there exists a projective resolution of M · · · → P1∂1→ P0

ε→ M → 0 with truncatedcomplex (PM)• and there exist chain maps f, g associated to α, β such that the sequence

0→ (PM ′)•f→ (PM)•)

g→ (PM ′′)• → 0

is exact.

Proof. For the proof, we will show that if we start with the horseshoe part of the followingdiagram in C where the left and right columns and the row are exact and P ′,P ′′ are projective,then we can fill in the new parts so that the middle columns, and the top and middle rowsare exact, P is projective and all squares commute. This will be enough to conclude thelemma.

0 0 0

0 K ′ K K ′′ 0

0 P ′ P P ′′ 0

0 L′ L L′′ 0

0 0

λ0|K′ µ0|K

∂′

λ0

∂

µ0

∂′′

λ µ


We build P and the parts by hand. Define P = P ′ ⊕ P ′′ and λ0, µ0 the cannonical inclusionand projection maps λ0 : x 7→ (x, 0) µ0 : (x, y) 7→ y. By construction, the sequence0 → P ′ → P → P ′′ → 0 (splits) is exact. Since µ is onto and P ′′ is projective, there existsτ : P ′′ → L a morphism in C satisfying µ τ = ∂′′. Define ∂ : P = P ′ ⊕ P ′′ → L by∂(x, y) = λ(∂′(x)) + τ(y). ∂ is a R−module homomorphism since it is sum and compositionof morphisms. Define K := ker(∂), the top squares commute and the top row is exact byconstruction. Moreover, we have

∂(λ0(x)) = ∂(x, 0) = λ(∂′(x))

and

µ(∂(x, y)) =µ(λ(∂′(x)) + τ(y)

)=µ(λ∂′(x)) + µ(τ(y))

=0 + µ(τ(y))

=∂′′(y)

=∂′′(µ0(x, y))

Hence, the complete diagram commutes, and we have completed the diagram as desired.

To conclude the proof, we can proceed by induction on n, starting with L′ = M ′, L = M ,L′′ = M ′′, P ′ = P ′0, P0 = P , P ′′0 = P ′′ so K ′ = ker(ε) and K ′′ = ker(ε′′). We then fit thediagram (see bellow) we just completed into the big horseshoe, using that the left and rightcolumns are exact; i.e. ker(∂′n) = Im(∂′n+1) and ker(∂′′n) = Im(∂′′n+1), we can apply ourinductive argument.

P ′1 P ′′1

K ′ K K ′′

P ′0 P P ′′0

M ′ M M ′′

∂′′1

Lemma 6.3.2 (Dual Horseshoe Lemma). Take a diagram in C as bellow, where the twocolumns are injective resolutions of M ′, M ′′ with corresponding truncated complexes (EM ′)

•

and (EM ′′)•, and the bottom rows are exact.


......

(E ′)1 (E ′′)1

(E ′)0 (E ′′)0

0 M ′ M M ′′ 0

0 0

(δ′)0 (δ′′)0

ι′

α β

ι′′

Then there exists a injective resolution of M →E1→E2 → · · · with truncated complex (EM)•

and there exist chain maps f, g associated to α, β such that the sequence

0→ (EM ′)• f→ (EM)•)

g→ (EM ′′)• → 0

is exact.

Theorem 6.3.3. Let 0 → M ′ α→ Mβ→ M ′′ → 0 be a short exact sequence in C. Let

T : C → Ab be an additive covariant functor. Then we have a long exact sequence in Ab:

· · · (LnT )M ′ (LnT )M (LnT )M ′′ (Ln−1T )M ′ · · ·

· · · (L1T )M ′′ (L0T )M ′ (L0T )M (L0T )M ′′ 0

(LnT )α (LnT )β ∂n (Ln−1T )α

∂1 (L0T )α (LnT )β

In particular, L0T : C → Ab is right exact. If C = R −mod and M0 is a right module, thenwe get a long exact Tor-sequence in Ab:

· · · TorRn (M0,M′) TorRn (M0,M) TorRn (M0,M

′′) TorRn−1(M0,M′) · · ·

· · · TorR1 (M0,M′′) M0 ⊗RM ′ M0 ⊗RM M0 ⊗RM ′′ 0

Proof. This follows from the Horseshoe Lemma (lem 6.3.1) and the long exact sequence inhomology (thm 6.1.6). Let (PM ′)• and (PM ′′)• be truncated chain complexes associated toprojective resolutions of M ′ and M ′′, respectively. By the Horseshoe Lemma, there exist aprojective resolution with trunctated chain complex (PM)• and chain maps f, g associated to

α, β, respectively, such that 0→ (PM ′)•f→ (PM)•)

g→ (PM ′′)• → 0 is a short exact sequencesof chain complexes.

For all n ≥ 0 we have a short exact sequence 0 → PM ′,nfn→ PM,n

gn→ PM ′′,n → 0 whichsplits because PM ′′,n is projective. Problem 1.2 shows that the functor T applied to an


split sequence is also exact; thus 0 → T (PM ′)•Tf→ T (PM)•)

Tg→ T (PM ′′)• → 0 is also exact.Applying then the long exact homology sequence we obtain the desired long exact sequenceof LnT .

Corollary 6.3.4. Let M0 be a right R−module. The following are equivalent:

(a) M0 is R−flat.

(b) TorRn (M0,M) = 0 ∀n ≥ 1, ∀M = RM .

(c) TorR1 (M0,M) = 0 ∀M = RM .

Proof. (b) =⇒ (c): is obvious.

(c) =⇒ (a): Follows from the long exact Tor-sequence (Theorem 6.3.3). Suppose 0 →M ′ α→M

β→M ′′ → 0 is a short exact sequence in R−mod, then

0 = TorR1 (M0,M′′)

∂1→M0 ⊗RM ′ id⊗α→ M0 ⊗RMid⊗β→ M0 ⊗RM ′′ → 0

is also exact. In particular, M0 ⊗R − is an exact functor thus M0 is flat.

(a) =⇒ (b): Let M = RM with a truncated projective projective resolution (PM)• :

· · · → P1∂1→ P0

∂0=0→ 0. Since M0 is R−flat, we have that the chain complex

M0 ⊗R (PM)• : · · · →M0 ⊗R P1id⊗∂1→ M0 ⊗R P0→0

has zero homology except (maybe) when n = 0; i.e. TorRn (M0,M) = 0 ∀n ≥ 1.

Theorem 6.3.5. Let 0→M ′ α→Mβ→M ′′ → 0 be a short exact sequence in C.

(a) If T : C → Ab is an additive covariant functor, we get a long exact sequence in Ab.

0 (R0T )M ′ (R0T )M (R0T )M ′′ (R1T )M ′ · · ·

· · · (RnT )M ′ (RnT )M (RnT )M ′′ (Rn+1T )M ′ · · ·

(R0T )α (R0T )β δ0 (R1T )α

(RnT )α (RnT )β δn

In particular (R0T ) : C → Ab is left exact.

(b) If T : C → Ab is an additive contravariant functor, we get a long exact sequence inAb.

0 (R0T )M ′′ (R0T )M (R0T )M ′ (R1T )M ′′ · · ·

· · · (RnT )M ′′ (RnT )M (RnT )M ′ (Rn+1T )M ′′ · · ·

(R0T )β (R0T )α δ0 (R1T )β

(RnT )β (RnT )α δn

In particular (R0T ) : C → Ab is left exact.


Similarly, if we fix M0 ∈ Ob(C), then we get a long exact Ext−sequence

0 HomR(M ′′,M0) HomR(M,M0) HomR(M ′,M0) Ext1R(M ′′,M0) · · ·

· · · ExtnR(M ′′,M0) ExtnR(M,M0) ExtnR(M ′,M0) Extn+1R (M ′′,M0) · · ·

β∗ α∗ δ0

β∗ α∗ δn

Corollary 6.3.6. Let M0 ∈ Ob(C). The following are equivalent:

(a) M0 is injective.

(b) ExtnR(M0,M) = 0 ∀n ≥ 1 and ∀M ∈ Ob(C).

(c) Ext1R(M0,M) = 0 ∀M ∈ Ob(C).

Proof. Proceed as in Corollary 6.3.4 recalling that M0 is injective iff HomR(−,M0) is exactin Theorem 3.2.1.

We also get a long exact Tor and Ext sequences in the ”other variable”.

Proposition 6.3.7. Let 0→M0α→M1

β→M2 → 0 be a short exact sequence in C.

(a) For every M = RM we have a long exact sequence in Ab:

· · · TorRn (M0,M) TorRn (M1,M) TorRn (M2,M) TorRn−1(M0,M) · · ·

· · · TorR1 (M2,M) M0 ⊗RM M1 ⊗RM M2 ⊗RM 0

(b) For every M = RM we have a long exact sequence in Ab:

0 HomR(M,M0) HomR(M,M1) HomR(M,M2) Ext1R(M,M0) · · ·

· · · ExtnR(M,M0) ExtnR(M,M1) ExtnR(M,M2) Extn+1R (M,M0) · · ·

α∗ β∗ δ1

α∗ β∗ δn

Proof. Let M = RM . Take (PM)• be the truncated complex associated to a projectiveresolution of M . Since PM,n is projective, hence flat, we get a short exact sequence of chaincomplexes

0→M0 ⊗R (PM)•α⊗id(PM )•→ M1 ⊗R (PM)•

β⊗id(PM )•→ M2 ⊗R (PM)• → 0

We use the long exact homology sequence to get (a).

For (b), since Pn is projective, we get a short exact sequence of co-chain complexes

0→ HomR((PM)•,M0)α∗→ HomR((PM)•,M1)

β∗→ HomR((PM)•,M2)→ 0

Then we use long exact co-homology sequence to finish it.


Theorem 6.3.8.

(a) Let M ′ = M ′R and M = RM . Then, for every n ≥ 0,

Ln(M ′ ⊗R −)M ∼= Ln(−⊗RM)M ′

(b) For C = R−mod or mod−R, and M,M ′ ∈ Ob(C). Then for every n ≥ 0,

Rn(HomR(−,M ′)

)M ∼= Rn

(HomR(M,−)

)M ′

Remark 6.3.9. Theorem 6.3.8 stated that the order is not important when computing Extand Tor of the pair (M,M’). Part (a) translates like

Hn

(M ′ ⊗R (PM)•

) ∼= Hn

((P ′M)• ⊗RM

)Part (b) gives us

Hn(HomR((PM)•,M

′) ∼= Hn(HomR(M, (EM ′)•

)Proof of the Theorem. We introduce the method of dimension shifting so read carefully.

Define ExtnR(M,M ′) = Rn(HomR(M,−)

)M ′ and extnR(M,M ′) = Rn

(HomR(−,M ′)

)M .

Both Ext and ext have long exact cohomology sequences in both variables4. Consider ashort exact sequence in C 0 → M1

ι→ Pπ→ M → 0 where P is projective. We get a exact

sequences of abelian groups

· · · → Extn−1R (P,M ′)→ Extn−1

R (M1,M′)→ ExtnR(M,M ′)→ ExtnR(P,M ′)→ · · ·

and

· · · → extn−1R (P,M ′)→ extn−1

R (M1,M′)→ extnR(M,M ′)→ extnR(P,M ′)→ · · ·

If n ≥ 2, Extn−1R (P,M ′) = 0 because P is projective, so

0→ Extn−1R (M1,M

′)→ ExtnR(M,M ′)→ 0

and Extn−1R (M1,M

′) ∼= ExtnR(M,M ′) ∀n ≥ 2. Similarly, we get extn−1R (M1,M

′)→ extnR(M,M ′)∀n ≥ 2 using that HomR(P,−) is exact.

Therefore, by induction, we only have to consider n = 1. The n = 0 case follows since R0 isthe original Hom functor in both cases.

For n = 1, we get exact sequences

0 HomR(M,M ′) HomR(P,M ′) HomR(M1,M′) Ext1R(M,M ′) Ext1R(P,M ′) = 0

0 HomR(M,M ′) HomR(P,M ′) HomR(M1,M′) ext1R(M,M ′) Ext1R(P,M ′) = 0

π∗ ι∗ α

π∗ ι∗ β

4We have seen it for Ext and a similar argument works for ext.


We get,

Ext1R(M,M ′) ∼=HomR(M1,M′)/ker(α)

∼=HomR(M1,M′)/Im(ι∗)

∼=HomR(M1,M′)/ker(β)

∼=ext1R(M,M ′)

Corollary 6.3.10.

(a) If M ′ of M is flat then TorRn (M ′,M) = 0 ∀n ≥ 1.

(b) If M ′ is injective of M is projective then ExtnR(M,M ′) = 0 ∀n ≥ 1.

Theorem 6.3.11. For every n ≥ 0

(a) TorRn (∐

k∈K Ak, B) ∼=∐

k∈K TorRn (Ak, B) and TorRn (A,

∐k∈K Bk) ∼=

∐k∈K Tor

Rn (A,Bk)

(b) If Ak, ϕkl is a directed system of right R−modules over a directed index set K, thenTorRn (lim−→Ak, B) ∼= lim−→TorRn (Ak, B).

Similarly, if Bk, ψkl is a direct system of left R−modules over a directed index set K,

then TorRn (A, lim−→Bk) ∼= lim−→TorRn (A,Bk).

Proof. (a) The n = 0 case is Theorem 1.2.8. Let’s show TorR1 (∐

k∈K Ak, B) ∼=∐

k∈K TorR1 (Ak, B),

the other is similar. For each k ∈ K choose a short exact sequence 0→Mk → Pk → Ak → 0where Pk is a projective module. We get a short exact sequence 0 →

∐Mk →

∐Pk →∐

Ak → 0 where∐Pk is projective. We get then a commutative diagram with exact rows,

0 TorR1 (∐

k Pk, B) TorR1 (∐

k Ak, B)(∐

kMk

)⊗R B

(∐k Pk

)⊗R B

0∐

k TorR1 (Pk, B)

∐k Tor

R1 (Ak, B)

∐k

(Mk ⊗R B

) ∐k

(Pk ⊗R B

)

0 δ

γ=α|...

f

∃α∼= ∃β∼=

0∐k δk

∐k fk

To cook γ: we note that(∐

fk)α β = β f δ = 0. Since the left two terms in each row

are zero, there is a well defined map γ that makes the diagram commute, this will be therestriction of α. Thus, by the five lemma we conclude the isomorphism TorR1 (

∐k∈K Ak, B) ∼=∐

k∈K TorR1 (Ak, B).

For n > 1, we use dimension shifting since TorRn (M ′, B) ∼= TorRn−1(M ′, B) where 0→ B1 →P → B → 0 is a short exact sequence with P is projective.

(b) The same idea is used here. For n = 0 is Theorem 1.2.8 and for n > 1 we use a dimensionshifting argument.


Now we will show TorR1 (Ak, B) ∼= lim−→TorR1 (Ak, B). Again, for each k choose a short exactsequence 0→ Mk → Pk → Ak → 0 where Pk is projective, hence flat. For k ≤ l in K, sincePk is projective, there is a map ψkl making the right square commute.

0 Mk Pk Ak 0

0 Ml Pl Al 0

αkl ∃ψkl

Let αkl = ψkl |Mk. We get a short exact sequence of direct systems, and hence of direct limits5

0 → lim−→Mk → lim−→Pk → lim−→Ak → 0 where lim−→Pk is flat, since Pk is flat for each k. Thus,Tor(lim−→Pk) = 0, getting a commutatice diagram with exact rows:

0 TorR1 (lim−→Pk, B) TorR1 (lim−→Ak, B)(

lim−→Mk

)⊗R B

(lim−→Pk

)⊗R B

0 lim−→TorR1 (Pk, B) lim−→TorR1 (Ak, B) lim−→(Mk ⊗R B

)lim−→(Pk ⊗R B

)

0

γ=α|... ∃α∼= ∃β∼=

We cook γ as in (a).

Theorem 6.3.12. For n ≥ 0 we have

ExtnR(∐

k

Ak, B) ∼= ∏

k

ExtnR(Ak, B)

andExtnR

(A,∏k

Bk

) ∼= ∏k

ExtnR(A,Bk)

Proof by intimidation. After proving the last two theorems, this proof should be ok. Again,n = 0 is Theorem 1.2.7, dimension shifting reduces the problem to check only when n = 1.For n = 1, we use a short exact sequence for each k ∈ K to get a short exact sequence0 →

∐Mk →

∐Pk →

∐Ak → 0 where

∐k Pk is projective. We get then a commutative

diagram with exact rows:

HomR

(∐Pk, B

)HomR

(∐Mk, B

)Ext1R

(∐Ak, B

)Ext1R

(∐Pk, B

)0

∏HomR(Pk, B)

∏HomR(Mk, B)

∏Ext1R(Ak, B)

∏Ext1R(Pk, B

)0

∃α∼= ∃β∼= γ

Dually to the previous theorem, we cook γ and finish with the 5-lemma.

5recall K is directed.


6.4 Tor and Ext of abelian groups

Recall Example 6.2.15 where we computed for a Z−module A the Tor groups TorZn(A,Z/mZ)m > 0. We showed,

TorZ0 (A,Z/mZ) ∼= A/mA TorZ1 (A,Z/mZ) ∼= x ∈ A : mx = 0 TorZn = (A,Z/mZ) = 0∀n ≥ 2

Also recall that if M is a Z−module then t(M) := x ∈M : ∃r ∈ Z −0 with rx = 0 is asubmodule of M , called the torsion submodule of M . We say is a torsion abelian groupif M = t(M). We say M is torsion free if t(M) = 0.

Proposition 6.4.1. For every A, B Z−modules, TorZ1 (A,B) is a torsion abelian group andTorZn(A,B) = 0 ∀n ≥ 2.

Proof. Let Bα be the collection of all finitely generated submodules of B. Note that it isa directed index set under inclusion and B ∼= lim−→Bα.

By Theorem 6.3.11, we have for each n ≥ 0 that TorZn(A,B) ∼= lim−→TorZn(A,Bα) = 0. Sincelim−→ of torsion abelian groups is torsion, and since lim−→ 0 = 0, we can restrict our argument tothe case when B is a finitely generated Z−module. Thus,

B ∼= Zr ⊕ Z/m1Z⊕ · · · ⊕ Z/mkZ

and

TorZn(A,B) ∼= TorZn(A,Zr)⊕k⊕i=1

TorZn(A,Z/miZ)

The first summand is zero for n ≥ 1 since Zr is flat Z−module, the second summand is zerofor n ≥ 2 by Exampl 6.2.15. For n = 1, Example 6.2.15 implies that TorZ1 (A,B) is a torsionabelian group.

Proposition 6.4.2.

(a) If B is a torsion abelian group, then TorZ1 (Q/Z, B) ∼= B.

(b) Let A be a Z−module. Then A is torsion free iff TorZ1 (A,−) = 0 iff TorZ1 (−, A) = 0.

Proof. (a) Consider the short exact sequence 0 → Z ι→ Q π→ Q/Z → 0. We get an exactsequence

0 ∼= TorZ1 (Q, B)→ TorZ1 (Q/Z, B)→ Z⊗Z B → Q⊗Z B ∼= 0

Where 0 ∼= TorZ1 (Q, B) because Q is flat, and Q⊗ZB ∼= 0 because B is torsion. This finishespart (a).

(b) We will show that A is torsion free iff TorZ1 (A,−) = 0.

6.4. TOR AND EXT OF ABELIAN GROUPS 85

(⇒) Again, we have A ∼= lim−→Aα for the collection Aα of finitely generated submodules ofA. Notice that each Aα must be a finitely generated torsion free Z−module, hence a freeZ−module. Then,

TorZ1 (A,−) = lim−→TorZ1 (Aα,−) = lim−→ 0 = 0

Similarly TorZ1 (−, A) = 0.

(⇐) In particular, for each m ∈ Z+ 0 = TorZ1 (A,Z/mZ) = x ∈ A : mx = 0. Hence A istorsion free.

Similarly, one shows that TorZ1 (Z/mZ, A) = x ∈ A : mx = 0 to see that A is torsion freewhen TorZ1 (−, A) = 0.

Proposition 6.4.3. Let A, B be Z−modules. Then ExtnZ(A,B) = 0 ∀n ≥ 2.

Proof. Let E be an injective Z−module such that there is a short exact sequence

0→ Bι→ E

π→ Q→ 0

Note that E is injective Z−module hence divisible6. Then, since Q is a quotient of a divisibleZ−module, Q is also divisible and so is an injective Z−module.

We have shown that 0→ E → Q→ 0 is an injective resolution of B with truncated complex

E• : 0→ Eπ→ Q→ 0

E in degree 0 and Q in degree 1. Then,

HomZ(A,E•) : 0→ HomZ(A,E)π∗→ HomZ(A,Q)→ 0

and so ExtnZ(A,B) = 0 ∀n ≥ 2.

Example 6.4.4. If A is a torsion abelian group then Ext1Z(A,Z) ∼= A∗ = HomZ(A,Q/Z) isthe Pontryagin dual. The reason of this isomorphism is that, since 0→ Z→ Q→ Q/Z→ 0is an injective resolution of Z so we get an exact sequence

HomZ(A,Q)→ HomZ(A,Q/Z)→ Ext1Z(A,Z)→ Ext1Z(A,Q)

where HomZ(A,Q) = 0 since A is torsion, Ext1Z(A,Q) = 0 since Q is an injective Z−module,and HomZ(A,Q/Z) = A∗ by definition (see Remark 3.2.22).

To end, let’s discuss the behavior of Ext with respect to extensions.

Let R be a ring. Let A, C be left R−modules. An extension of C by A is a short exactsequence in R−mod:

ξ : 0→ Aι→ E

π→ C → 0

6A left R−module M is divisible iff for every non-zero divisor r ∈ R every element m ∈M can be writtenas m = r.m′ for some m′ ∈M . It can be shown that for abelian groups, being divisible is equivalent to beinginjective.


Two extensions ξ, ξ′ of C by A are equivalent if there is a commutative diagram in R−mod:

ξ : 0 A E C 0

ξ′ : 0 A E ′ C 0

ι π

f

ι′ π′

Note that, by the 5-lemma, f is an isomorphism.

Define e(C,A) to be the set of equivalence classes of extensions of C by A. If [ξ] ∈ e(C,A),then there exists [φξ] ∈ Ext1R(C,A) as follows: Take a projective resolution of C

· · · P2 P1 P0 C 0

· · · 0 A E C 0

∂2 ∂1

φξ

∂0

ψξ

ι π

By the comparison Theorem, there exists ψξ and φξ making this diagram commute. We get,φξ ∂2 = 0 and so φξ ∈ ker(∂∗2), thus [φξ] ∈ Ker(∂∗2)/Im(∂∗1) = Ext1R(C,A).

We get a map h : e(C,A) → Ext1R(C,A) given by [ξ] 7→ [φξ]. One can check that h iswell-defined. Moreover, one can define an addition on e(C,A) called the Baer sum andshow this makes h into a group homomorphism between additive abelian groups.

Problem set 6

Problem 6.1. Let n ∈ Z+, and let R = Z/m.

(a) Let A = Z/d where d|m, and let B be an arbitrary R−module. Determine TorRn (A,B)for all n ≥ 0.

Hint: Show that · · · d→ Rm/d→ R

d→ Rε→ A→ 0 is a perioric free resolution.

(b) Let C be an arbitrary R−module, and let D = Z/p where p|m. Determine ExtnR(C,D)for all n ≥ 0 in terms of HomR(C,R). Moreover, show that if p2|m, then ExtnR(D,D) ∼=D for all n.

Problem 6.2. A minimal projective resolution of an R−module M is a projective resolution

· · ·→Pndn→ Pn−1 → · · · → P1

d1→ P0d0→M

such that (P0, d0) is a projective cover of M and (Pn, dn) is a projective cover of ker(dn−1)for n ≥ 1. Define ΩnM = ker(dn−1) for n ≥ 1.

6.4. TOR AND EXT OF ABELIAN GROUPS 87

Dually, a minimal injective resolution of an R−module M is an injective resolution

0→Md−1

→ E0d0→ E1 → · · · → En dn→ En+1 → · · ·

such that (E0, d0) is an injective hull of M and (En, dn−1) is an injective hull of coker(dn−2)for n ≥ 1. Define Ω−nM = coker(dn−2) for n ≥ 1.

Suppose R is a self-injective finite dimensional algebra over a field k. Let M,N be finitelygenerated left R−modules. Prover for n ≥ 1

ExtnR(M,N) ∼= HomR(ΩnM,N) and ExtnR(M,N) ∼= HomR(M,Ω−nN)

Chapter 7

Abelian Categories

We’ve developed some intuition on what Ext and Tor functors do. In this chapter, we willintroduce abelian and additive categories and extend the work done in Chapter 6 for modulecategories. Freyd-Mitchell’s embedding theorem will permit us to use “local” arguments todo the corresponding extensions. At the end, we will dedicate the rest of the time to set thecorrect environment to talk about Ext and Tor as Hom functors between specific categories.

Be ready for more diagrams.

7.1 Abelian Categories

Studying abelian categories is the first step to see the Ext and Tor functors as Hom functors...

Definition 7.1.1. Let C, D be categories1

(a) C is called an Ab-category if for all X, Y ∈ Ob(C) HomC(X, Y ) is an additive abeliangroup such that composition distributes over addition.

(b) Suppose C, D are Ab-categories and let F : C → D be a covatiant functor. ThenF is called an additive functor if for all X, Y ∈ Ob(C), the map HomC(X, Y ) →HomD(FX,FY ) given by α 7→ Fα is a group homomorphism under +.

(c) C is called an additive category if C is an Ab-category and C has a zero object (i.e.an object that is both initial and terminal) and for all A,B ∈ Ob(C) there exists aproduct

(A×B, πA, πB) in C, not necessarily the cartisian product.

R−mod and Mod−R are additive categories.

1Larger with small Hom sets.

89

90 CHAPTER 7. ABELIAN CATEGORIES

Example 7.1.2. Let D = R−mod or Mod−R, let C(D) be the category of cochain com-plexes and cochain maps. Then C(D) is an Ab-category, the zero object is the zero complex,and we have arbitrary products and coproducts defined on on each degree. In particularC(D) is an additive category.

Definition 7.1.3. Let C be an additive category, let f : A→ B be a morphism in C.

1. A kernel of f is a pair (K, ι) where K ∈ Ob(C), ι : K → A is a morphism in C suchthat (K, ι) is universal with respect to the property f ι = 0; i.e., if (X, g) is anotherpair with X ∈ Ob(C) and g : X → A in C satisfying f g = 0, then there is a uniquemorphism α : X → K in C with g = ι α.

K A B

X

ι f

g∃!α

Note that ι is monic.

2. A cokernel of f is a pair (Q, π) where Q ∈ Ob(C), π : B → Q is a morphism in C suchthat (Q, π) is universal with respect to the property π f = 0; i.e., if (Y, h) is anotherpair with Y ∈ Ob(C) and h : B → Y in C satisfying h f = 0, then there is a uniquemorphism β : Q→ Y in C with h = β π.

A B Q

X

f π

h∃!β

Note that π is epic.

Lemma 7.1.4. Let C be an additive category, let f : A→ B be a morphism in C.

(a) If β : B → Y is monic, then ker(f) = ker(β f).

(b) If α : X → A is epic, then coker(f) = coker(f α).

Proof. (a) Let (K, ι) be the kernel of f . Then (β f) ι = β (f ι) = 0 and if (X, g) issuch that g : X → A is in C with (β f) g = 0 = β 0, since β is monic, this means thatf g = 0. Hence, there exists a unique morphism α : X → K in C with g = ι α.

(b) is similar.

Example 7.1.5. Let D be a module category, let C = C(D) be the category of chaincomplexes. Let C• = Cn, δnC be a chain complex. A subcomplex of C• is a cochaincomplex A• = An, δnA where An ⊂ Cn is a submodule and δnA = δnC |An for all n ∈ Z.

7.1. ABELIAN CATEGORIES 91

Given a subcomplex A• of C•, we can build the quotient complex Q• == Qn, δnQ whereQn = Cn/An and δnQ : Cn/An → Cn+1/An+1 is given by δnQ(x+ An) = δnC(x) + An+1.

Let f : C• → D• be a cochain map, then kef(fn)n∈Z forms a subcomplex of C• denoted byker(f). Note that the homomorphisms ιn : ker(fn) → Cn form a cochain map ι : ker(f)→C•. Then it is easy to check that (ker(f), ι) is a kernel of f .

Similarly, coker(fn)n∈Z forms a quotient complex of D•, denoted by coker(f). Notice thatthe projection hom’s πn : Dn → coker(fn) form a cochain map π : D• → coker(f). Thenone can check that (coker(f), π) is a cokernel of f .

Definition 7.1.6. A category A is called abelian category if (a)-(d) is satisfied, where

(a) A is an additive category.

(b) Every morphism in A has a kernel and a cokernel.

(c) Every monomorphism is the kernel of its cokernel.

(d) Every epimorphism is the cokernel of its kernel.

Remark 7.1.7. (c) means that if f : A→ B is a monomorphism inA, and (Q, π) = coker(f)and (K, ι) = ker(π). Then there is a unique isomorphism w : A→ K with f = ι w.

A B Q

K

f

∃!w

π

ι

Important: (a)-(d) imply:

(e) Every morphism f : A→ B in A can be factored as

A I Be

f

m

where e is epic and m is monic. This is a Problem 7.1, which should follow only fromconditions (a)-(d) and definitions of kernel and cokernel.

Lemma 7.1.8. Let A be an abelian category.

(a) Let f : A → B be a morphism in A, let (K,m) = ker(coker(f)) and (C, e) =coker(ker(f)). Then there exist unique maps α : A → K and β : C → B satisfy-ing f = m α and f = β e.


(b) If m : Y → Z is monic, λ : X → Y is such that coker(m) = coker(mλ), then λ isepic.

If e : X → Y is epic, µ : Y → Z is such that ker(e) = ker(µ e) then µ is monic.2

Proof. (a) Let (X, ι) = ker(f) and (Y, π) = coker(f), we get the following diagram

K

X A B Y

C

m

ι f

e

π

Since πf = 0, the universal property for (K,m) gives us the existance of a unique morphismα : A → K in A with f = m α. Again, f ι = 0 and so the universal property of (C, e)gives us a unique morphism β : C → B in A with f = β e.

(b) By property (e) of Ab categories, we can factor λ as

X I Yε

λ

µ

where ε is epic and µ is monic. Then

coker(m µ) =coker(m µ ε)=coker(m λ)

=coker(m)

The first equality holds since ε is epic. Now, since m, m µ are monic, they are the kernelsof their cokernels. We get that (Y,m) = (I,m µ), where equality means, by the universalproperties of kernel and cokernel, that there is a unique isomorphism w : I → Y satisfyingm w = m µ

Y Z

I

m

mµ∃!w

But m is monic so w = µ and so µ is an isomorphism. In particular λ = µ ε is epic since εis epic.

Theorem 7.1.9. Let A be an abelian category, let f : A → B be a morphism in A. Let(X, ι) = ker(f), (Y, π) = coker(f), (K,m) = ker(π), (C, e) = coker(ι). Let α, β be as inpart (a) of Lemma 7.1.8. Then

X A K B Yι α

f

m π

2Compare this with the universal property of the quotient.

7.1. ABELIAN CATEGORIES 93

where α is epic (K,α) = coker(ι) = (C, e) and

X A C B Yι e

f

β π

where β is monic and (C, β) = ker(π) = (K,m). Moreover, there is an isomorphism λ :C → K with m λ = β and λ e = α. In other words ker(coker(f)) = coker(ker(f)).

Proof. We have that

coker(m α) = coker(f) = (Y, π) = coker(ker(π)) = coker(m)

The third equality holds because π is epic. Then, part (b) of Lemma 7.1.8 and m beingmonic imply that α is epic.

In a similar way, since α is epic and m is monic, we get

(K,α) = coker(ker(α)) = coker(ker(m α)) = coker(ker(f)) = coker(ι) = (C, e)

Then second part of (b) in Lemma 7.1.8 forces β to be monic and (C, β) = ker(π) =(K,m). In particular, by the universal properties of kernel and cokernel, there are uniqueisomorphisms λ, ν : C → K with m λ = β and ν e = α. Then

m λ e = β e = f = m α = m ν e

But m is monic and e is epic so λ = ν

We can use the previous theorem to define the image of a morphism, and so exactness ofsequences as follows.

Definition 7.1.10. Let A be an abelian category.

(a) Let f : A→ B be a morphism inA, define the image of f to be Im(f) := ker(coker(f)) =(I,m)

A B

I

f

∃!e epicm monic

(b) A sequence Aα→ B

β→ C is morphisms in A is said to be exact at B is Im(α) = ker(β).

(c) A subcategory B of A is called an abelian subcategory if B is abelian and exactsequences in B stay exact when viewed as sequences in A.


Example 7.1.11 (Warning). Let R 6= 0 be a ring, A = R−mod, and X ∈ Ob(A), X 6= 0.

Define B by Ob(B) = X and HomB(X,X) = idX. Note that XidX→ X

idX→ X is exact inB but not in A. Thus, B is abelian but not an abelian subcategory of A.

Theorem 7.1.12 (Fryed-Mitchell Embedding Theorem). Let A be a small abelian category,then there exists a ring R (possibly zero) and an exact fully faithful additive functor F :A → R − mod which embeds A into a full subcategory of R−mod; i.e., HomA(A,B) ∼=HomR(FA, FB) as abelian subgroups.

The exactness part of F says that the image is an abelian subcategory. We will concentrate insomething else, homotopy and derived categories, so we will skip the proof of this theorem,using them in various situations in the rest of the notes. The trick when using Fryed-Mitchell’s Embedding Theorem is to restrict our arguments to small categories containingenough objects. See Remark 7.2.8 for an example of the latter.

Problem set 7

Problem 7.1. Use the definition of abelian category from the note to prove that everymorphism f : A → B in A factors as f = m α for an epimorphism α : A → K and amonomorphism M : K → B, where K is a suitable object in A.

Hint: Look at (K,m) = ker(coker(f)). Prove that there is a unique α : A → K withf = m α. Show that α is epic.

7.2 Total complex

Theorem 7.2.1. Let A be an abelian category. Then C(A), the category of cochain com-plexes and cochain maps, is an abelian category.

Proof sketch. To see C(A) is additive, look at when we did it for R−mod and note that weonly used that A = R−mod is additive.

We now show that every morphism in C(A) has a kernel. Let f : C• → D• be a cochainmap in C(A). Take ker(fn) = (Kn, ιn), we have that

fn+1 δnC ιn = δnC (fn ιn) = δnC 0 = 0

By the universal property of (Kn+1, ιn+1), there is a unique morphism in A δnK : Kn → Kn+1

7.2. TOTAL COMPLEX 95

making the diagram commute

Kn Kn+1

Cn Cn+1

Dn Dn+1

δnK

ιn ιn+1

δnC

fn fn+1

δnD

We have,

ιn+1 δnK δn−1K =δnC ιn δn−1

K

=δnC δn−1C ιn−1

=0 ιn−1 = 0

Since ιn−1 is monic, we get δnK δn−1K = 0. We have built then a cochain complex K =

Kn, δnK and a chain map ι = ιn. One can check that the pair (K, ι) is indeed the kernelof f . In a similar way, every morphism f has a cokernel, using the pairs coker(fn) = (Qn, πn).

By last,

f is monic ⇔∀α, β : X• → C•, with f α = f β =⇒ α = β

⇔∀n,∀αn, βn : Xn → Cn, with fn αn = fn βn =⇒ αn = βn

⇔fn is monic

Thus, f is monic iff fn is monic which implies that fn is the kernel if its cokernel and, byconstruction, implies that f is the kernel of its cokernel. We proceed similarly with the dualstatement for when f is epic.

Remark 7.2.2. For every n ∈ Z we get an additive functor Hn : C(A) → A as follows:For C• = Cn, δnC in C(A), let (Zn(C•), ιn) = ker(δnC) and (Bn(C•), jn) = Im(δn−1

C ). Let(Hn(C•), π) = coker

(gn : Bn(C•) → Zn(C•)

)where gn is the unique morphism from the

universal property of ker(δnC).

Definition 7.2.3. A bicomplex (or double complex) in A is a collection of objectsCp,qp,q∈Z in A together with morphisms dp,qh : Cp,q → Cp+1,q and dp,qv : Cp,q → Cp,q+1

such that dp,qh dp−1,qh = 0, dp,qv dp,q−1

v = 0 and, each dp+1,qv dp,qh = 0 + dp,q+1

h dp,qv = 0; i.e.,each square anticommutes.

Cp,q Cp+1,q

Cp,q+1 Cp+1,q+1

dp,qh

dp,qv dp+1,qv

dp,q+1h

The bicomplex C•,• = Cp,q, dp,qh , dp,qv is said to be bounded is along each diagonal p+q = n,there are only finitely many non-zero objects Cp,q.

The double complex arising from a good cover of a smooth manifold satisfies Cp,q = 0whenever p < 0 of q < 0, thus is bounded.


Remark 7.2.4. Due to the anticommutativity, d•,qv : C•,q → C•,q+1 is not a cochain map. Noworries, we can allways to a sign trick and define f •,q : C•,q → C•,q+1 to be fp,q = (−1)pdp,qvto get a cochain map. In particular, we can identify the category of bicomplexes in A withthe category C(C(A))3. The real usefulness of a bicomplex is the ”natural” total complexstructure we can define from it.

Definition 7.2.5 (Total complexes). Let C•,• be a bicomplex in A. We can define TotΠ(C)•

and Tot⊕(C)• by

TotΠ(C)n = Πp+q=nCp,q Tot⊕(C)n = qp+q=nCp,q

and dn : TotΠ(C)n → TotΠ(C)n+1 by

(xp,q)p+q=n 7→(dr−1,sh xr−1,s + dr,s−1

v xr,s−1)r+s=n+1

Remark 7.2.6. If C•,• is not bounded, then Totπ(C)• and Tot⊕(C)• may or may not exist.For example, if A is the abelian category of finite abelian groups, this does not exist.

If A is an abelian category such that all direct products exists, then A is called complete.Similarly, cocomplete categories have all coproducts. For example, R−mod is complete andcocomplete.

Definition 7.2.7 (Shift functor). Let p ∈ Z, for all C• ∈ Ob(C(A)) define C[p]• by C[p]n =Cp+n and dnC[p] = (−1)pdp+nC . We shift C• p degrees to the left. If f : C• → D• is in C(A),

define f [p] : C[p]• → D[p]• by f [p]n = fn+p.

We obtain an additive functor [p] : C(A) → C(A) called the shift (or translation) functorof degree p.

Note that Hn(C[p]•) = Hn+p(C•).

Remark 7.2.8. Let A be an abelian category. Then the snake lemma is valid in A.

Why: Let C be the smallest full abelian subcategory of A containing the objects in a snakediagram. Then, C is a small category and, by Freyd-Mitchell, C can be embedded intoR−mod for some ring R as a full abelian subcategory. Snake lemma is true in R−mod sothen is also true in C, and so in A.

This allows us to talk about long exact cohomology sequences in C(A) as follows: If 0 →C ′•

f→ C•g→ C ′′• → 0 is a short exact sequence in C(A), then we have a lonh exact

cohomology sequence in A

· · · → Hn(C ′•)Hn(f)→ Hn(C•)

Hn(g)→ Hn(C ′′•)δn→ Hn+1(C ′•)→ · · ·

3It is not a typo.

7.3. HOMOTOPY CATEGORIES 97

Remark 7.2.9. We can identify A with a full abelian subcategory of C(A) by identifying

C ∈ Ob(A) with C• : · · · 0→ C → 0→ · · · , C in degree zero, and Cf→ D in A with

C• : · · · 0 C 0 · · ·

D• : · · · 0 D 0 · · ·0 f 0

7.3 Homotopy categories

Let A be an abelian category. A morphism f : C• → D• in C(A) is said to be homotopicto zero, written f ∼ 0, if for all n ∈ Z there is a morphism kn : Cn → Dn−1 in A such thatfn = dn−1

D kn + kn+1 dnC .

Define Ht(C•, D•) = f ∈ HomC(A)(C•, D•) : f ∼ 0. Then Ht(C•, D•) is a subgroup of

HomC(A)(C•, D•) under addition. Moreover, if f : C• → D•, g : D• → E•, e : B• → C• are

cochain maps with f ∼ 0, then g f ∼ 0 and f e ∼ 0. One can see that g f ∼ 0 via anexplicit computation: suppose fn = dn−1

D kn + kn+1 dnC . Then

gn fn =gn dn−1D kn + gn kn+1 dnC

=dn−1E (gn−1 kn) + (gn kn+1) dnC

Hence g f is null-homotopic.

Definei a new category K(A) called the Homotopy category of C(A) by Ob(K(A)) =Ob(C(A)) and for every C•, D• ∈ Ob(K(A)),

HomK(A)(C•, D•) := HomK(A)(C

•, D•)/Ht(C•, D•)

Note that having well defined compositions follows fromthe fact that if f ∼ 0, then gf ∼ 0and f e ∼ 0.

Remark 7.3.1. K(A) is an additive category. We also have a functor

F : C(A)→K(A)

C• 7→C•

f 7→[f ]

K(A) is not abelian in general, so we need something to replace exactnes...

Decorations. For an abelian category A, denote by Cb(A), C+(A) and C−(A) the full sub-categories of C(A) consisting of all bounded, bounded below and bounded above complexes,respectively. Denote by Kb(A), K+(A) and K−(A) the corresponding homotopy categories,all identifiable with full subcategories of K(A).


Lemma 7.3.2. The cohomologies of cochain complexes in A defines for all n ∈ Z are welldefines additive functors

Hn : K(A)→ A

Proof. We can check it only on maps: When A = R−mod, we already proved that f ∼ 0⇒ Hn(f) = 0, thus Hn is well defined. When A is arbitrary, we can use Fryed-MitchellEmbedding Theorem to use the previous case.

Definition 7.3.3. A cochain map f : X• → Y • in C(A) is called quasi-isomorphism (orqis) if for every n ∈ Z, Hn(f) : Hn(X•)→ Hn(Y •) is an isomorphism in A.

A morphism f in K(A) is called a quasi-isomorphism if Hn(f) is an isomorphism in A forevery n ∈ Z.

Note that if f = f +Ht(X•, Y •), then f is a qis in K(A) iff f is a qis in C(A).

Big Picture

Let A be a module category and take X, Y ∈ Ob(A). We know that HomA(X,−) andHomA(−, Y ) are left exact functors, but usually not right exact. If we want to get right exactfunctors, the remedy is to construct right derived functors ExtnA(X,−) and ExtnA(−, Y ) forevery n ≥ 0.

If 0 → Aα→ B

β→ C → 0 is a short exact sequence in A, then we would get a long Extsequence

0 HomA(X,A) HomA(X,B) HomA(X,C) Ext1A(X,A) · · ·α∗ β∗ δ0

Similarly for −⊗R Y and TorRn (−, Y ).

Recall that to define Extn and Torn we used a projective and injective resolutions; i.e. weused chain complexes. The idea is to define a new category D(A), called the derivedcategory,such that

ExtnA(X,A) := Hn[HomA(X, I•A)

]= HomD(A)(X

•, IA[n]•)

where I•A is a truncated injective resolution of A.

We would like HnHomA(X, I•A) to be a homomorphism group HomD(A)(X•, IA[n]•), where


X• : 0→ X → 0 X in degree zero. The idea of the construction goes as follows:

C(A) abelian

K(A) additive triangulated hom’s modulo homotopy

D(A) additive triangulated qis become isomorphisms

7.3.1 Mapping cones

Definition 7.3.4 (Mapping Cones). Let f : X• → Y • be in C(A). Define the mappingcone of f to be the cochain complex M(f)•, M(f)n = Xn+1 ⊕ Y n with(

−dn+1X 0

−fn+1 dnY

)

We get a short exact sequence in C(A)

0→ Y •α(f)→ M(f)•

β(f)→ X[1]• → 0

where

α(f)n : Y n ( 01 )→ Xn+1 ⊕ Y n

β(f)n : Xn+1 ⊕ Y n (−1 0 )→ Xn+1

We get the following lemma

Lemma 7.3.5. The long exact cohomology sequence associated to the mapping cone of f is

· · · → Hn−1(Y •)Hn−1(α(f))→ Hn−1(M(f)•)

Hn−1(β(f))→ Hn−1(X[1]•) = Hn(X•)Hn(f)=δn−1

→ Hn(Y •)→ · · ·

Proof. We prove this in A = R−mod and then use Fryed-Mitchell’s Embedding Theorem.For that, we only need to show that the connecting homomorphism δn−1 agrees with Hn(f).Recall the Snake Lemma construction

?

• • •

• • •

?

δ


Let z ∈ Zn(X•); i.e. z ∈ X• and dnX(z) = 0, then z = β(f)n−1 ( −z0 ) and

dn−1M(f) ( −z0 ) =

(−dnX(−z)−fn(−z)

)=(

0−fn(−z)

)= α(f)n (fn(z))

Thus, δn−1([z]) = [fn(z)] = Hn(f)([z]).

Corollary 7.3.6 (A reason why not to stay in C(A)). Using the same notation as above,there exists a map φ : X[1]• →M(α(f))• in C(A) such that the following diagram commutesin K(A); i.e. up to homotopy!

Y • M(f)• X[1]• Y [1]•

Y • M(f)• M(α(f))• Y [1]•

α(f) β(f)

φ

−f [1]

α(f) α(α(f)) β(α(f))

Proof. We have M(α(f))n = Y n+1 ⊕Xn+1 ⊕ Y n and

DnM(α(f)) =

(−dn+1

Y 0 0

0 −dn+1X 0

−1 −fn+1 dnY

)

Define φn : Xn+1 → Y n+1 ⊕ Xn+1 ⊕ Y n = M(α(f))n by φn =(fn+1

−10

). Define also ψn :

M(α(f))n := Y n+1 ⊕ Xn+1 ⊕ Y n → Xn+1 by ψn = ( 0 −1 0 ). We have that ψn φn =

( 0 −1 0 ) ·(fn+1

−10

)= idXn+1 .

Moreover, we get that φn ψn ∼ idM(α(f)). For this, define sn : M(α(f))n = Y n+1 ⊕Xn+1 ⊕Y n → Y n ⊕Xn ⊕ Y n−1 = M(α(f))n−1 by sn =

(0 0 −10 0 00 0 0

), and check that

idM(α(f))n − φn ψn =(

1 fn+1 00 0 00 0 1

)=(

0 0 dn+1Y

0 0 00 0 1

)+(

1 fn+1 −dn+1Y

0 0 00 0 0

)=dn−1

M(α(f)) sn + sn+1 dnM(α(f))

Hence φn ψn ∼ idM(α(f)).

By last, we check commutativity of the diagram in K(A) by computing

ψn α(α(f))n = ( 0 −1 0 ) ·(

0 01 00 1

)= ( −1 0 ) = β(f)n

and

β(α(f))n φn = ( −1 0 0 ) ·(fn+1

−10

)= −fn+1 = −f [1]n


Corollary 7.3.7. f : X• → Y • is a qis in K(A) iff M(f)• is acyclic (i.e. Hn(M(f)•) = 0∀n).

Proof. Recall that f is a qis iff Hn(f) is an isomorphism in A for every n ∈ Z. The resultnow follows from the Long exact cohomology sequence of M(f)• in Lemma 7.3.5.

Definition 7.3.8 (Triangle). Let A be an abelian category.

1. A triangle in K(A) is a sequence of morphisms X•u→ Y •

v→ Z•w→ X[1]• in K(A). A

morphism of triangles is a commutative diagram in K(A):

X• Y • Z• X[1]•

X ′• Y ′• Z ′• X ′[1]•

φ

u v

µ ν

w

φ[1]

u′ v′ w′

This is called an isomorphism of triangles if φ, µ, ν are isomorphisms in K(A).

2. A distinguished triangle in K(A) is a triangle X•u→ Y •

v→ Z•w→ X[1]• that is

isomorphic to a triangle of the form X ′•f→ Y ′•

α(f)→ M(f)•β(f)→ X ′[1]•.

The short hand notation for a triangle X•u→ Y •

v→ Z•w→ X[1]• in K(A) is

Z•

X• Y •

w

+1u

v

Lemma 7.3.9. If X•u→ Y •

v→ Z•w→ X[1]• is a distinguished triangle in K(A) then we

hace a long exact cohomology sequence in A,

· · · → Hn(X•)Hn(u)→ Hn(Y •)

Hn(v)→ Hn(Z•)Hn(w)→ Hn+1(X•)

Hn+1(u)→ · · ·

In other words, for distinguished triangles we have

Z•

X• Y •

w

+1u

v =⇒H∗(Z•)

H∗(X•) H∗(Y •)

H∗(w)

+1H∗(u)

H∗(v)

Proof. We have shown this for triangles of the form X ′•f→ Y ′•

α(f)→ M(f)•β(f)→ X ′[1]•. The

general case follows since Hn is an additive functor, so it sends isomorphisms in K(A) toisomorphisms in K(A).


7.3.2 Triangle Axioms for K(A)

Let A be an abelian category. Then the collection of distinguished triangles in K(A) satisfiesthe triangle axioms (TR0)-(TR5):

(TR0) A triangle isomorphix to a distinguished triangle is isomorphic.

(TR1) For any X• ∈ Ob(K(A)), X•idX•→ X•→0→X[1]• is a distinguished triangle.

(TR2) Any f : X• → Y • can be embedded into a distinguished triangleX•f→ Y •→Z•→X[1]•.

(TR3) X•u→ Y •

v→ Z•w→ X[1]• is a distinguished triangle ⇐⇒ Y •

v→ Z•w→ X[1]•

−u[1]→ Y [1]•

is a distinguished triangle.

(TR4) Given two distinguished triangles X•u→ Y •

v→ Z•w→ X[1]• and X ′•

u′→ Y ′•v′→ Z ′•

w′→

X ′[1]• and a commutative diagramX• Y •

X ′• Y ′•

u

φ µ

u′

, there exists a morphism of triangles

X• Y • Z• X[1]•

X ′• Y ′• Z ′• X ′[1]•

φ

u v

µ ν

w

φ[1]

u′ v′ w′

(TR5) (Octahedral axiom) Given distinguished triangles:

X•u→ Y •→Z ′•→X[1]•

Y •v→ Z•→X ′•→Y [1]•

X•vu→ Z•→Y ′•→X[1]•

there exists a distinguished triangle Z ′•f→ Y ′•

g→ X ′•h→ Z ′[1]• such that the following

diagram commutes in K(A):

X• Y • Z ′• X[1]•

X• Z• Y ′• X[1]•

Y • Z• X ′• Y [1]•

Z ′• Y ′• X ′• Z ′[1]•

u

v f

u

vu

g u[1]

v

f g h


Visualization of (TR5): Given distingued the distinguished triangles red, orange andgreen, there is a blue distinguised triangle making all the faces commute of the Octahedroncommute.

Y ′•

Z ′• X ′•

X• Z•

Y •

+1

g

+1

f

+1

h

+1vu

u v

Theorem 7.3.10. The distinguished triangles in K(A) satisfy the triangle axioms (TR0)-(TR5).

Proof. (TR0) and (TR2) follow by definition and (TR3) is Lemma 7.3.5. (TR1) follows bynoticing that 0 : 0• → X• has mapping cone M(0)• = X• with a distinguished triangle

M(0)•

0• X•+1

id . By (TR3), we get a distinguished triangle

0•

X• X•

+1

idX•

.

We will now sketch (TR4). Assume Z• = M(u)•, Z ′• = M(u′)•. We have that µ u = u′ φin K(A), thus there is a homotopy kn : Xn → Y n−1 with

µn un − u′n φn = dn−1Y ′ k

n + kn+1 dnX

Define ν : Zn = Xn+1 ⊕ Y n → Z ′n = X ′n+1 ⊕ Y ′n by ν =(

φn+1 0−kn+1 µn

). One can check that ν

defines a cochain map and that makes the desired diagram commute in K(A):

X• Y • Z• X[1]•

X ′• Y ′• Z ′• X ′[1]•

φ

u v

µ ν

w

φ[1]

u′ v′ w′

We now explain (TR5), recall the visualization of (TR5) described above. We build the newdistinguished triangle by hand: let Z ′• := M(u)•, X ′• := M(v)• and Y ′• := M(v u)•, anddefine fn : Z ′n = Xn+1 ⊕ Y n → Y ′n = Xn+1 ⊕ Zn and g : Y ′n = Xn+1 ⊕ Zn → X ′n =Y n+1 ⊕ Zn by fn = ( 1 0

0 vn ) and gn =(un+1 0

0 1

).

Define h = α(u)[1] β(v), we get hn : X ′n = Y n+1 ⊕ Zn → Z ′[1]n+1 = Xn+2 ⊕ Y n+1 is givenby the matrix hn = ( 0 0

−1 0 ).


To check the commutativity of the octahedron, one only needs to multiply the corresponding

matrices. To check thatY ′•

Z ′• X ′•g

f

+1

h

is also a distinguished triangle, it is enough to

check the following isomorphism in K(A) (up to homotopy):

Z ′• Y ′• X ′• Z ′[1]•

Z ′• Y ′• M(f)• Z ′[1]•

α(f) g

ψ

h

f α(f) β(f)

φ

Define φn : M(f)n = (Xn+2 ⊕ Y n+1) ⊕ (Xn+1 ⊕ Zn) → X ′n = Y n+1 ⊕ Zn and ψn : X ′n →M(f)n are given by

φ =(

0 1 u[1] 00 0 0 1

)ψ =

(0 01 00 00 1

)To conclude the isomorphism one can check that φ ψ ∼ idX′• and ψ φ ∼ idM(F )• .

Definition 7.3.11 (Triangulated category). A triangulated category is an additive cat-egory C together with

(1) an additive automorphism T : C → C of categories called shift or translation functor(sometimes T=[1]), and

(2) a collection of triangles in C called distinguished triangles such that the triangleaxioms (TR0)-(TR5) are satisfied.

Here, a triangle is a sequence of morphisms Xu→ Y

v→ Zw→ T (X) in C and a morphism of

triangles is a commutative diagram in C

X Y Z T (X)

X ′ Y ′ Z ′ T (X ′)

φ

u v

µ ν

w

T (φ)

u′ v′ w′

which is an isomorphism iff φ, µ, ν are isomorphisms in C.

Notation. We sometimes write a triangle Xu→ Y

v→ Zw→ T (X) in C as

Z

X Y

w

+1

u

v


Example 7.3.12. We have shown in Theorem 7.3.10 that K(A) is a triangulated categoryfor any abelian category A.

In this context, the triangles will replace the short exact sequences.

Definition 7.3.13.

(a) Let (C, T ) and (C ′, T ′) be two triangulated categories. An additive covariant functorF : C → C ′ is called ∂−functor (or a functor of triangulated categories) if F T = T ′Fand F sends distinguished triangles in C to distinguished triangles in C ′.

(b) Let A be an abelian category. An additive covariant (resp. contravariant) functorH : C → A is called a covatiant (resp. contravariant) cohomological functor if foreach distinguished triangle X

u→ Yv→ Z

w→ T (X) there is a long exact sequence in AIf covariant,

· · · → H(T i(X))H(T i(u))→ H(T i(Y ))

H(T i(v))→ H(T i(Z))H(T i(w))→ H(T i+1(X))

H(T i+1(u))→ · · ·

If contravariant,

· · · → H(T i(Z))H(T i(v))→ H(T i(Y ))

H(T i(u))→ H(T i(X))H(T i−1(w))→ H(T i−1(Z))

H(T i−1(w))→ · · ·

Example 7.3.14. Take A be an abelian category, then H0 : K(A) → A is a covariantcohomological functor since H0(X[i]•) = H i(X•) ∀i ∈ Z, ∀X• ∈ Ob(A).

Remark 7.3.15. Let (C, T ) be a triangulated category, let A be an abelian category and letH : C → A be an additive covariant (resp. contravariant) functor. Then by property (TR3),H is a cohomological functor iff for every distinguised triangle X

u→ Yv→ Z

w→ T (X), the

sequence HXHu→ HY

Hv→ HZ (resp. HZHv→ HY

Hu→ X ) is exact in A.

Proposition 7.3.16. Let (C, T ) be a triangulated category.

(a) It Xu→ Y

v→ Zw→ T (X) is a distinguished triangle in C, then v u = 0 and w v = 0.

(b) For all M ∈ Ob(C), HomC(M,−) (resp. HomC(−,M)) is a covariant (resp. con-travariant) cohomological functor C → Ab.

Proof. (a) By (TR3)4, it suffices to show v u = 0. By (TR1), the sequence XidX→

X→0→T (X) is a distinguished triangle. Then, by (TR4) there exists ν : 0 → Z in Cmaking the diagram commute

X X 0 T (X)

X Y Z T (X)

idX

u ν

u v w

4(TR3) is your friend!


Then, v u = ν 0 = 0, as we wanted.

(b) We will show the result for HomC(M,−). Let Xu→ Y

v→ Zw→ T (X) be a distinguished

triangle in C. We need to show that HomC(M,X)u∗→ HomC(M,Y )

v∗→ HomC(M,Z) isexact in Ab; i.e. kerv∗ = Imu∗. By part (a), we have v∗ u∗ = (v u)∗ = 0 so Imu∗ ⊂ kerv∗.

Let f : M → Y be a morphism in C with v∗(f) = 0; i.e. v f = 0. Take the distinguished

triangle MidM→ M→0→T (M) and using (TR3) and (TR4) we get a morphism g : M → X

in C making the following diagram in C commute,

M M 0 T (X)

X Y Z T (X)

g

idX

f 0 T (g)

u v w

Thus, f = u g = u∗(g); i.e. f ∈ Im(u∗).

Problem 7.2. Let A be an abelian category and let f : X• → Y • be a morphism in C(A).Then we get a distinguished triangle

X• Y • M(f) = X[1]• ⊕ Y • X[1]•f α(f) β(f)

Recall that α(f) = ( 01 ), β(f) = ( −1 0 ) and dnM(f) =

(dn+1X 0

−fn+1 dnY

)and note that α(f) f =(

0f

)6= ( 0

0 ).

Show that(

0f

)∼ ( 0

0 ) in K(A).

Lemma 7.3.17 (5-Lemma for distinguished triangles). Let (C, T ) be a triangulated category.Given a morphism of distinguished triangles in C

X Y Z T (X)

X ′ Y ′ Z ′ T (X ′)

φ

u v

µ ν

w

T (φ)

u′ v′ w′

If φ and µ are isomorphisms in C, then so is ν.

Proof. Apply HomC(Z′,−) to the morphism of triangles to obtain a commutative diagram

in Ab with exact rows.

HomC(Z′, X) HomC(Z

′, Y ) HomC(Z′, Z) HomC(Z

′, T (X)) HomC(Z′, TY )

HomC(Z′, X ′) HomC(Z

′, Y ′) HomC(Z′, Z ′) HomC(Z

′, X ′) HomC(Z′, T (Y ′))

φ∗

u∗ v∗

µ∗ ν∗

w∗

T (φ)∗

T (u)∗

T (µ)∗

u′∗ v′∗ w′∗ T (u′)∗

7.4. LOCALIZATION OF CATEGORIES 107

φ and µ are isomorphisms in C so φ∗, µ∗, T (φ)∗ and T (µ)∗ are isomorphisms in Ab. Then,by the 5-Lemma in Ab, we get that ν∗ is an isomorphism. Hence there exists ξ : Z ′ → Z inC with ν ξ = ν∗(ξ) = idZ′ . Now we apply HomC(−, Z) to the morphism of distinguishedtriangles to obtain that nu∗ is an isomorphism in Ab. Hence there exists ε : Z ′ → Z in Cwith ε ν = ν∗(ε) = idZ . We get,

ε = ε (ν ξ) = (ε ν) ξ = ξ

Thus, ν is an isomorphism in C.

7.4 Localization of Categories

In order to introduce Derived categories , we need to learn how to localize elements. Thismimics the well known localization for rings (see bellow).

Definition 7.4.1 (Multiplicative systems). Let C be a category, let S be a collection ofmorphisms in C. S is called a multiplicative system if (S1)-(S4) are satisfied:

(S1) idX ∈ S ∀X ∈ Ob(C).

(S2) For any f, g ∈ S such that g f is defined, then g f ∈ S.

(S3) Given a diagram in C,Z

X Y

fs

with s ∈ S, we can complete this to a commutative

diagram in CW Z

X Y

g

t

fs

for some t ∈ S.

Ditto5 for the diagrams with all arrows reversed.

(S4) Given f, g : X → Y in C, TFAE:

(i) There exists a morphism s : X ′ → X in S with f s = g s.(ii) There exists a morphism t : Y → Y ′ in S with t f = t g.

Definition 7.4.2 (Localization of categories). Let C be a category, let S be a multiplicativesystem of morphisms in C. The localization of C with respect to S is a category CS togetherwith a functor Q : C → CS called the localizationi functor, satisfying:

(i) Q(s) is an isomorphism in CS for every s ∈ S and,

(ii) ∀D and ∀ functor F : C → D such that F (s) is an isomorphism in CS for each s ∈ S,there is a unique functor G : CS → D with G Q = F .

5the same


Moreover, if C is an additive category then CS is an additive category.

Proposition 7.4.3. Let C be a category, let S be a multiplicative system of morphisms inC. Then the localization CS can be obtained as follows:

Ob(CS) = Ob(C) and HomCS(X, Y ) =

X ′

X Ys

f : X ′ ∈ Ob(C), s ∈ S, f ∈ C

/∼where

X ′

X Y

s f ∼ X ′′

X Y

t g if there is X ′′′ ∈ Ob(C) and morphisms u : X ′′′ → X ′

in S, h : X ′′′ → X ′′ in C such that the following diagram commutes

X ′′′

X ′ X ′′

X Y

u h

s

ft

g

i.e. s u = t h and f u = g h.

The composition of two morphismsX ′

X Y

S3s f andY ′

Y Z

S3t g in CS is defined as fol-

lows: By (S3), there exists a commutative diagram in C

Z ′

X ′ Y ′

X Y Z

S3u h

s f t g

Define

X ′

X Y

s f Y ′

Y Z

t g =Z ′

X Z

su gh

Proof sketch. One can check (exercise) that CS is indeed a category, in particular that thecomposition of morphisms is well defined. Note here that the identity of X ∈ Ob(CS) = Ob(C)

is represented by.X

X X

idX idX .

We will describe Q and show the universal property of (CS, Q). Define Q : C → CS by

Q(X) = X and for f ∈ HomC(X, Y ) define Q(f) =X

X Y

idX f . As an exercise, show that

Q is a functor.

7.4. LOCALIZATION OF CATEGORIES 109

Let s : X → Y be in S, then Q(s) =X

X Y

idX s and Q(s)−1 =X

Y X

s idX since

X

X Y

idX s X

Y X

s idX =

X

X X

X Y X

s s

=X

X X

and

X

Y X

s idX X

X Y

idX s =

X

X X

Y X Y

s s

=X

Y Y

s s ∼ Y

Y Y

The last equivalence follows by taking u = idX and h = s in the definition of the equivalencerelation to get a commutative diagram in C

X

X Y

Y Y

s

s

s

=⇒ X

Y Y

s s ∼ Y

Y Y

Hence Q sends morphisms in S to isomorphisms in CS.

We will now show the universal property of (CS, Q). Let F : C → D be a functor suchthat F (s) is an isomorphism in D for every morphism s ∈ S. Then, define G : CS → D byG(X) = F (X) ∀X ∈ Ob(CS) = Ob(C) and

G

X ′

X Y

s f

= G

X ′

X Y

f

X ′

X Y

s

−1 = F (f) F (s)−1

One can check that G is the unique functor satisfying G Q = F .

By last, we need to convince ourselves that if C is additive then CS is additive. It is clear thatzero object is Q(0). As an exercise, show that if X, Y are objects in C and (XΠY, pX , pY ) isthe product of X and Y in C, then (Q(XΠY ), Q(pX), Q(pY )) is the product of X and Y inCS. We will explain why CS is an Ab−category; i.e. HomCS(X, Y ) is an abelian group:

Given two morphisms in CS,X ′

X Y

s f andX ′′

X Y

t g , by property (S3) there exists

u ∈ S and a commutative diagram

X ′′′

X ′ Y

X

u v

s

t


Define then

X ′

X Y

s f +X ′′

X Y

t g :=X ′′′

X Y

su fu +X ′′′

X Y

tv gv =X ′′′

X Y

su fu+gv

Proposition 7.4.4. Let C be a category, S be a multiplicative system of morphisms in C.Let C ′ be a full subcategory of C, and S ′ = S ∩ C ′. Suppose S ′ is a multiplicative system forC ′ such that one of the conditions (i) or (ii) is satisfied:

(i) For all s : X → X ′ in S with X ∈ Ob(C ′), there is f : X ′′ → X in C such thatX ′′ ∈ Ob(C ′) and s f ∈ S.

(ii) For all t ∈ X ′ → X in S with X ′ ∈ Ob(C ′), there is a morphism g : X → X ′′ in C suchthat X ′′ ∈ Ob(C ′) and g t ∈ S.

Then the localization C ′S′ is a full subcategory of CS in the localization sense; i.e, the naturalfunctor C ′S′ → CS is fully faithful.

Proof. The proof is a exercise.

7.4.1 How to localize triangulated categories?

Definition 7.4.5. Let (C, T ) be a triangulated category, let S be a multiplicative system ofmorphisms in C. We say that S is compatible with the triangulation if S satisfies (S5) and(S6):

(S5) T (s) ∈ S, ∀s ∈ S.

(S6) Given distinguished triangles Xu→ Y

v→ Zw→ T (X) and X ′

u′→ Y ′v′→ Z ′

w′→ T (X ′) in Ctogether with morphisms in S, f : X → X ′ and g : Y → Y ′, such that g u = u′ f ,there exists h : Z → Z ′ in S such that we have the morphism of triangles

X Y Z T (X)

X ′ Y ′ Z ′ T (X ′)

f

u

v

g h

w

T (f)

u′ v′ w′

Proposition 7.4.6. Let (C, T ) be a triangulated category, let S be a multiplicative systemof morphisms in C that is compatible with the triangulation. Then CS has a unique structureof triangulated categories satisfying,

7.5. DERIVED CATEGORIES 111

Q : C → CS is a ∂−functor, and

For every ∂−functor F : C → D of triangulated categories such that F (s) is an iso-morphism in D for each s ∈ S, there is a unique ∂−functor G : CS → D such thatG Q = F .

Proof idea. Define the distinguished triangles in CS to be those who are isomorphic to imagesof distinguished triangles in C under Q. Properties (TR0)-(TR5) hold in CS since theyalready hold in C. By (S5) and (S6), Q is a ∂−functor and the universality is a standardargument.

7.5 Derived categories

Theorem 7.5.1. Let A be an abelian category, let (C, T ) be a triangulated category, and letH : C → A be a cohomological covariant functor. Define

S =s morphism in C : H(T i(s)) is an isomorphism ∀i ∈ Z

Then S is a multiplicative system that is compatible with the triangulation.

Proof. We need to check (S1)-(S6).

(S1) Let X ∈ Ob(C), we have that XidX→ X

0→ 00→ TX is a distinguished triangle. Since H

is a cohomological functor, we get an exact sequence

· · · 0→H(T iX)H(T idX)→ H(T iX)→ 0 · · ·

which implies that H(T iidX) is an isomorphism for every i ∈ Z. Thus idX ∈ S.

(S2) Let f, ginS such that gf is defined. For each i ∈ Z, H (T i(g f)) = H (T ig)H (T if),thus an isomorphism. Hence g f ∈ S.

(S5) We have the following chain of equivalences: s ∈ S iff H(T is) is an isomorphism forevery i ∈ Z iff H(T i(Ts)) is an isomorphism for every i ∈ Z iff T (s) ∈ S.

(S3) Given a diagram in CZ

X Y

fs

(TR2) implies that there is a distinguished triangle Zs→ Y

f→ Ug→ TZ and (TR2)+(TR3)

give us a distinguished triangle Wt→ X

fu→ Uv→ TW .


Note that s ∈ S imples that H(T is) is an isomorphism and, since

· · ·H(T i−1U)→ H(T iZ)H(T is)→ H(T iY )→ H(T iU)→ · · ·

is exact for every i, we conclude that H(T iU) = 0. Then

· · ·H(T i−1U) = 0→ H(T iW )H(T it)→ H(T iX)→ H(T iU) = 0→ · · ·

forcing H(T it) to be an isomorphism for every i ∈ Z, thus t ∈ S.

By construction, we have a commutative diagram

W X U TW

Z Y U TZ

t

u

fu v

s f g

By (TR3) and (TR4) there is a morphism w : W → Z in C with sw = ut, as we wanted.

(S6) By (TR4) it is enough to show that given a morphism of distinguished triangles

X Y Z T (X)

X ′ Y ′ Z ′ T (X ′)

f

u v

g h

w

T (f)

u′ v′ w′

then f, g ∈ S imply that h ∈ S.

H is a cohomological functor so, for every i ∈ Z, we get a commutative diagram in A withexact rows:

H(T iX) H(T iY ) H(T iZ) H(T i+1(X)) H(T i+1(Y ))

H(T iX ′) H(T iY ′) H(T iZ ′) H(T i+1(X ′)) H(T i+1(Y ′))

H(T if)∼= H(T ig)∼= H(T ih) H(T i+1(f))∼= H(T i+1(g))∼=

The isomorphisms come from the assumption that f, g ∈ S. By the 5-lemma, H(T ih) is anisomorphism. Hence h ∈ S.

(S4) Since C is additive, it suffices to show the following statement: Given f : X → Y in CTFAE

1. There exists s : Y → Y ′ in S with s f = 0.

2. There exists t : X ′ → X in S with f t = 0.


(1 =⇒ 2) Take ff→ Y

s→ X ′. By (TR2) and (TR3) there is a distinguished triangleZ

u→ Ys→ Y ′

v→ TZ. Apply the cohomological functor HomC(X,−) to get an exactsequence in Ab:

HomC(X,Z)u∗→ HomC(X, Y )

s∗→ HomC(X, Y′)

We have that s f = 0 so s∗(f) = 0 and there is g : X → Z in C with u∗(g) = u g = f .

By (TR2), (TR3) there is a distinguished triangle X ′t→ X

g→ Z→TX ′. Since s ∈ S,H(T iZ) = 0 for all i ∈ Z. In particular H(T it) is an isomorphism for every i and t ∈ S.Appy now the HomC(−, Y ) functor to the previous triangle to get an exact sequence in Ab:

HomC(Z, Y )g∗→ HomC(X, Y )

t∗→ HomC(X′, Y )

We get f t = t∗(f) = t∗(g∗(u)) = 0. This proves (2).

(2 =⇒ 1) is similar.

Corollary 7.5.2. Let A be an abelian category, let K(A) be the homotopy category of C(A).Let Qis be the collection of quasi-isomorphisms in K(A). Then Qis is a multiplucativesystem of morphisms in K(A) that is compatible with the triangulation.

Proof. Qis agrees with S in the previous theorem.

Definition 7.5.3. Let A be an abelian category. The derived category of A, denoted byD(A), is defined by D(A) := K(A)Qis.

Remark 7.5.4. The additive functor Hn(s) : K(A) → A (n ∈ Z) satisfies Hn(s) is aisomorphism for all s ∈ Qis. By the universal property of Q, Hn factors through D(A) in aunique way. We denote the resulting additive functor again by

Hn : D(A)→ A

Let K−(A) (resp. K+(A), resp Kb(A)) be the full subcategory of K(A) consisting ofbounded above (resp. bounded below, resp. bounded) complexes. Define

D−(A) =K−(A)Qis

D+(A) =K+(A)Qis

Db(A) =Kb(A)Qis

Proposition 7.5.5. Let A be an abelian category,

(a) D−(A) (resp. D+(A), resp. Db(A)) is equivalent to the full subcategory of D(A)consisting of complexes X• with Hn(X•) = 0 ∀n >> 0 (resp. Hn(X•) = 0 ∀n << 0,resp. Hn(X•) = 0 ∀|n| >> 0).


(b) The composition of functors

A → C(A)→ K(A)Q→ D(A)

identifies A with the full subcategory of D(A) consisting of complexes X• with Hn(X•) =0 ∀n 6= 0.

Proof. This one is a little tedious to write down, although it follows from the earlier resultabout localizing full subcategories.

Remark 7.5.6. Let Q : K(A)→ D(A) be the localization functor. Then

(a) For all X• ∈ Ob(K(A)), Q(X•) = 0 in D(A) ⇔ X• is quasi-isomorphic to 0 in K(A).

(b) Let f : X• → Y • in C(A) with corresponding f = f + Ht(X•, Y •) in K(A). ThenQ(f) = 0 in D(A) ⇔ there is s : X ′• → X• quasi-isomorphism with f s ∼ 0 ⇔ thereis t : Y • → Y ′• quasi-isomorphism with t f ∼ 0.

Proof of Remark 7.5.6.

(a) (⇒) is immediate.

(⇐) Suppose thatQ(X•) ∼= 0• inD(A). This means that there exists a morphismX ′•

X• 0•s f

with inverseX ′′•

0• X•t g .

By the definition of idX• in D(A), there exist Z•, u and a qis h such that f u = t hin K(A) and s u = g h in K(A). For every n ∈ Z we have that Hn(X ′′•) = 0 sincet : X ′′• → 0 is a qis, then

Hn(Z•) Hn(X•)

Hn(X ′′•) = 0

Hn(su)

∼=Hn(h) Hn(g)

This implies that Hn(h) = 0 = Hn(g) and so Hn(su) = 0 and Hn(X•) = 0 for every n ∈ Z.This concludes that 0• : X• → 0• is a qis, as we wanted.

(b) We will show the first equivalence, the second one follows from (S4). (⇒) is immediate.

(⇐) Q(f) = 0 in D(A) means that for some qis in K(A) s′

X•

X• X•

f ∼ X ′′•

X• Y •s′ 0


Then, there exist X ′•, s, g in K(A) such that the following diagram commutes,

X ′•

X• X ′′•

X• Y •

s g

0

s′

In particular, f s = 0 g = 0 and so f s ∼ 0, as we wanted.

Remark 7.5.7. We have seen that for morphisms f, g : X• → Y • in C(A), f ∼ g ⇒Q(f) = Q(g) in D(A). However, the converse is false in general. Consider for example

X• : · · · → 0→ 0→ Z ×2→ Z π→ Z/2→ 0→ 0→ · · ·

X• is acyclic; i.e. Hn(X•) = 0 for all n ∈ Z. Let f = idX• , let t• : X• → 0• be t = 0. t is aqis and t f = 0. In particular, Q(f) = 0 in D(A = Ab) but f 6∼ 0.


(a) P ∈ Ob(A) is called projective is HomA(P,−) is exact; i.e. for all diagram in A:

P

X Y 0

α

f

there exists β ∈ HomA(P,X) such that f β = α.

(b) E ∈ Ob(A) is called injective if HomA(−, E) is exact.

From now on, we will get some slopy in the sense of not stating Frey-Mitchell embeddingTheorem. One needs to be carefull that projective objects in A do not allways map toprojective modules in R−mod.


(a) Let f : P • → X• in C(A) such that X• is acyclic (i.e. Hn(X•) = 0 ∀n) and P • is abounded above complex of projective objects. Then f ∼ 0.

(b) Let f : X• → E• in C(A) such that X• is acyclic and E• is a bounded below complexof injective objects. Then f ∼ 0.

Proof. One should compare this with the Comparison Theorem.


We will prove (a). Without loss of generality we assume that P n = 0 ∀n > 0. Definekn : P n → Xn−1 to be zero for n > 0. By induction, suppose we have constructed kn forn ≥ n0.

P n0−2 P n0−1 P n P n0+1

Xn0−2 Xn0−1 Xn0 Xn0+1

fn0−2 fn0−1

kn0 kn0+1

We have

dn0−1X

(fn0−1 − kn0 dn0−1

P

)=dn0−1

X fn0−1 − dn0−1X kn0 dn0−1

P

=dn0−1X fn0−1 − 0

=fn0 dn0−1P

=(dn0−1X kn0 + kn0+1 dn0

P

) dn0−1

P

We get, a diagram

P n0−1

Xn0−2 Ker(dn0−1X = Im(dn0−2

X ) 0

fn0−1−kn0dn0−1P

dn0−2X

exact

Since P n0−1 is projective, there is a map kn0−1 : P n0−1 → Xn0−2 with dn0−2X kn0−1 =

fn0−1 − kn0 dn0−1P . By induction, we obtain the homotopy that gives us f ∼ 0.

(b) is similar.


(a) Let s : X• → P • be a quasi-isomorphism an P • be a bounded above complex of projectiveobjects. Then s has a right homotopy inverse; i.e., there is t : P • → X• with s t ∼idP •. Moreover, if X• is also a bounded above complex of projective objects we also gett s ∼ idX .

(b) Let s : E• → X• be a quasi-isomorphism an E• be a bounded below complex of injectiveobjects. Then s has a left homotopy inverse; i.e., there is t : X• → E• with t s ∼idE•. Moreover, if X• is also a bounded below complex of injective objects we also gets t ∼ idX .

Proof. We prove (a). We use the notation for mapping cone as in module categories.

Since s is a qis, the map M(s)• is acyclic (use Fryed-Mitchell’s Theorem for this) and weget then a cochain map P • →M(s)• given in degree n by ( 0

1 ) : P n →M(s)n = Xn+1 ⊕ P n.

Recall that dM(s) =(−dn+1

X 0

−sn+1 dnP

).


By the previous lemma, ( 01 ) : P • →M(s)• is homotopic to zero. Let the homotopy be such

that ( −tn

un ) : P n →M(s)n−1 = Xn ⊕ P n−1 satisfies

( 01 ) = dn−1

M(s) ( −tn

un ) + ( −tn

un ) dnP

Then, 0 = dnX tn − tn+1 dnP , and so t : P • → X• is a cochain map. Also,

idPn = sn tn + dn−1P un + un+1 dnP

This implies that s t ∼ idP where un : P n → P n−1 is the corresponding homotopy.

Now suppose that X• is also a bounded complex of projective objects. Then s, s t are qisimplies that t• : P • → X• is a qis. Using the previous paragraph, t has a right homotopyinverse, say η; i.e. t η ∼ idP . Then

η ∼ (s t) η = s (t η) ∼ s

Hence t s ∼ idX as we wanted.

Theorem 7.5.11. Let A be an abelian category. Suppose there exists a full additive subcat-egory I such that ∀ X ∈ Ob(A) ∃ E ∈ Ob(I) and a monomorphism f : X → E. Then

(a) ∀ X• ∈ Ob(K+(A)) ∃ E• ∈ Ob(K+(I)) and a qis f : X• → E•.

(b) The natural functor D+(I)→ D+(A) is an equivalence of categories.

Proof. We will prove (b) assuming (a). Note that K+(I) is a triangulated category andQis ∩ K+(I) form a multiplicative system that is compatible with the triangulation. Toshow that the natural functor is fully faithful, we will show the following (coming from anearlier result on how to localize full categories):

(?) If s : X ′• → X• is a qis in K+(A) and X ′• ∈ Ob(K+(I)), then there is f : X• → X ′′• inK+(I) such that X ′′• ∈ Ob(K+(I)) and f s is a qis.

Note that (?) follows directly from (a) and (?) implies that the natural functor α : D+(I)→D+(A) is fully faithful. To show that α is dense, let X• ∈ Ob(D+(A)) = Ob(K+(A)). By(a), there exists E• ∈ Ob(K+(I)) = Ob(D+(I)) and a qis f : X• → E•. But f becomes anisomorphism when viewed as a morphism in D+(A). Hence f is dense.

We now prove (a). Without loss of generality, assume that Xn = 0 for n < 0. Define En = 0,fn = 0 for n < 0. By our assumption, there exists E0 ∈ Ob(I) and a monomorphismf 0 : X0 → E0.

X• : · · · 0 0 X0 X1 X2 · · ·

E• : · · · 0 0 E1

f 0 0 f0

d0X d1X


Suppose we have constructed En, fn, dn−1E . Consider the induced morphisms dnX : coker(dn−1

X )→Xn+1 and fn : coker(dn−1

X )→ coker(dn−1E ) and build their pushout in A.

coker(dn−1X ) Xn+1

coker(dn−1E ) Zn+1

fn

dnX

kn+1

ln

By hypothesis, there exists an object En+1 in I and a monomorphism gn+1 : Zn+1 → En+1.Define fn+1 := gn+1 kn+1 and dnE := gn+1 ln πn.

coker(dn−1X ) Xn+1

coker(dn−1E ) Zn+1

En En+1

fn

dnX

kn+1

fn+1

ln

gn−1πn

We get dnE dn−1E = gn+1 ln (πn dn−1

E ) = gn+1 ln 0 = 0.

We will prove by induction that fn is a monomorphism, dn dn−1X = dn−1

E fn−1 and thatHn−1(f) : Hn−1(X•) → Hn−1(E•) is an isomorphism. For this, we prove it in R−mod andthen apply Freyd-Mitchell. We get a commutative diagram,

0 Hn(X) Xn/Im(dn−1X ) Xn+1

0 ker(ln) En/Im(dn−1E ) Zn+1

0 Hn(E•) En/Im(dn−1E ) En+1

fn

dnX

kn+1

fn+1gn+1

ln

gn−1

dnE

where dnE = gn+1 ln and dnE = dnE πn. Moreover, by construction of the push-out, we have

Zn+1 = E/Im(dn−1E ⊕Xn+1/

(fn(x) + Im(dn−1

E ),−dnX(x))

: x ∈ Xn

We show that fn+1 is a monomorphism. Let x ∈ Xn+1 with fn+1(x) ∈ Im(dnE); i.e. there isx0 ∈ En with

gn+1(kn+1(x)) = fn+1(x) = gn+1(ln(x0 + Im(dn−1E )))

But gn+1 is monic so kn+1(x) = (ln(x0 + Im(dn−1E )) and so there exists y ∈ Xn with(

x0 + Im(dn−1E ),−x

)=(fn(y) + Im(dn−1

E ),−dnX(y)). Then x = dnX(y) and x + Im(dnX) =

0 + Im(dnX), thus fn+1 is monic.


Using the commutativity of the diagram, we get fn+1 dnX = dnE fn; both having domainXn/Im(dn−1

X ). We can conclude that fn+1 dnX = dnE fn.

By last, we show that Hn(f) is an isomorphism. Since fn is monic, Hn(f) is monic. toshow that it is epic, let z ∈ En with dnE(z) = 0. Hence, gn+1(ln(z + Im(dn−1

E ))) = 0. Butgn+1 is monic so ln(z + Im(dn−1

E )) = 0 and so there is a ∈ Xn with(z + Im(dn−1

E ), 0)

=(fn(a) + Im(dn−1

E ),−dnX(a)). This forces dnX(a) = 0 and so [a] ∈ Hn(X•). Hence, z +

Im(dn−1E ) = fn(a) + Im(dn−1

E ) and [z] = Hn(f)([a]), as we wanted.

We also have a dual theorem.

Theorem 7.5.12. Let A be an abelian category. Suppose P is a full additive subcategorysuch that for all X ∈ Ob(A) there is P ∈ Ob(P) and an epimorphism f : P → X. Then

(a) ∀ X• ∈ Ob(K−(A)) ∃ P • ∈ Ob(K−(P)) and a qis f : P • → X•, and

(b) the natural functor D−(P)→ D−(A) is an equivalence of categories.


(a) We say that A has enough injectives ∀ X ∈ Ob(A) ∃ an injective object E in A anda monomorphism f : X → E in A.

(b) We say that A has enough projectives ∀ X ∈ Ob(A) ∃ an projective object P in Aand a epimorphism f : P → X.

Theorem 7.5.14. Let A be an abelian category.

(a) If A has enough injectives, let I be the full subcategory of injective objects. Then thenatural functor K+(I)→ D+(A) is an equivalence of categories.

(b) If A has enough projectives, let P be the full subcategory of projective objects. Thenthe natural functor K−(I)→ D−(A) is an equivalence of categories.

Proof. We prove (a). It suffices to show that each qis in K+(I) is an isomorphism inK+(I) = D+(I). But part (b) of Lemma 7.5.10 states that every qis s : E• → E ′• in C(I)has a 2-sided homotopy inverse; i.e., s defines an isomorphism in K(I).

7.6 Derived functors

Motivation. Let A and B be two abelian categories, let F : A → B be an additive functor.Then F induces an additive functor

F : C(A)→ C(B)


Since F sens homotopies to homotopies, F induces an additive functor

F : K(A)→ K(B)

Since F commutes with the shift functor [1] and sends mapping cones to mapping cones, italso send distinguished triangles in K(A) to distinguished triangles K(B). Thus F : K(A)→K(B) is a covariant ∂−functor.

The goal of this section is to extend F in some way to a ∂−functor

D(A)→ D(B)

There is an easy case, when F sends qis in K(A) to qis in K(B); e.g., if F is an exactfunctor. In sich case, the universal property of localization functors gives us the existance ofa unique functor F : D(A)→ D(B) that makes the following diagram commute.

K(A) K(B)

D(A) D(B)

F

QA QB

∃!F

Let’s do it!

Definition 7.6.1. Let A be an abelian category. Let K∗(A) be a full triangulated subcat-egory of K(A). We call K∗(A) a localizing subcategory of K(A) if the natural functor

K∗(A)K∗(A)∩Qis → D(A)

is fully faithful. In this case, we define D∗(A) := K∗(A)K∗(A)∩Qis.

Example 7.6.2. K−(A), K+(A) and Kb(A) are localizing subcategories of K(A). Inter-sections of localizing subcategories are localizing.

Definition 7.6.3 (Derived functors). Let A, B be abelian categoeis. Let K∗(A) be alocalizing subcategory of K(A), let Q denote the localization functors K∗(A) → D∗(A),resp. K(B)→ D(B). Let F : K∗(A)→ K(B) be a covariant ∂−functor.

K∗(A) D∗(A)

K(B) D(B)

F

Q

Q

(a) A right derived functor of F is a covariant ∂−functor R∗F : D∗(A) → D(B)together with a natural transformation ξ : Q F → R∗F Q satisfying the followinguniversal property.

(?) For every covariant ∂−functor G : D∗(A) → D(B) with a natural transformationζ : Q F → G Q, there is a unique natural transformation η : R∗F → G such that(η Q) ξ = ζ.


(b) A left derived functor of F is a covariant ∂−functor L∗F : D∗(A)→ D(B) togetherwith a natural transformation ξ : L∗F Q → Q F satisfying the dual universalproperty of (?); i.e. (G, ζ) factors through a unique η : G→ L∗F .

If R∗F (resp. L∗F ) exists, it is unique up to natural isomorphism.

Example 7.6.4. Let A be an abelian category. Let I be the sull subcategory of injectiveobjects. Let F : K+(I) → K(B) be a covariant ∂−functor. We know that Q : K+(I) →D+(I) is an isomorphism of categories, so we can form QFQ−1:

D+(I) K+(I) K(B) D(B)Q−1

QFQ−1

∼=F Q

Then QF = (QFQ−1)Q and we get that QFQ−1 is both the left and right derived functorof F .

In a dual fashion, if P is the full subcategory of A of projective objects and F : K−(P) →K(B) is a covariant ∂−functor, then QFQ−1 is both the left and right derived functor of Fwhere

D+(P) K+(P) K(B) D(B)Q−1

QFQ−1

∼=F Q

Notation: If there is no confusion, we write RF for R∗F and LF for L∗F . We define thefollowing hyper derived functors: RiF := H i(RF ) and LiF := H−i(LF ).

Theorem 7.6.5 (Existance of derived functors). Let A, B be abelian categories.

(a) Let F : K+(A) → K(B) be a covariant ∂−functor. If A has enough injectives, thenthe right derived functor R+F : D+(A) → D(B) exists. Moreover, if E• is a boundedbelow complex of injectives, then (R+F )(E•) ∼= Q(FE•).

(b) Let F : K−(A) → K(B) be a covariant ∂−functor. If A has enough projectives, thenthe right derived functor L−F : D−(A) → D(B) exists. Moreover, if P • is a boundedabove complex of projectives, then (L−F )(P •) ∼= Q(FP •).

Proof. We prove (a). We use that we know that the natural functor α : K+(I) → D+(A)given by α = Q |K+(I) where Q is the localization functor K+(A)→ D+(A), is an equivalenceof categories. Let β : D+(A) → K+(I) be a quasi-inverse of α and let τ : idD+(A) → α βbe a natural isomorphism. Define R+F as the composition,

D+(A) K+(I) K(B) D(B)β

R+F

F Q


We claim the isomorphism HomK+(A) (X•, βQX•) ∼= HomD+(A) (QX•, αβQX•) via the mapf 7→ Q(f). Assume this claim for now.

We now define ξ : Q F → (R+F ) Q = Q F β Q as follows: Let X• ∈ Ob(K+(A)),there exists a unique map fX• ∈ HomK+(A) (X•, βQX•) such that Q(FX•) = τQX• . DefineξX• := (Q F )(fX•). One can check that naturality of τ implies that ξ is natural.

We now check the universal property. Let G : D+(A) → D(B) be a covariant ∂−functorwith natural transformation ζ : QF → GQ. We need to define η : R+F = QF β → G.Let X• ∈ Ob(D+(A). Recalling that Q(fX•)

−1 = τ−1QX• , we can take

Q(F (βX•)) G(Q(βX•)) = G(Q(β(QX•))) G(QX•) = GX•ζβX•

ηX•

G(Q(fX• )−1)

=G(τ−1QX• )

Since ζ, τ are natural transformations, also η is natural. Moreover, ηQX• ζX• = ξX• and ηis unique with respect to this property. Hence, (R+F, ξ) is the right derived functor of F .

By last, let E• be a bounded below complex of injectives. Since β α ∼= idK+(I), we get

(R+F )(E•) =Q(F (βR•))

=Q(F ((β α)(E•)))∼=Q(F (E•))

It remains to show the isomorphism HomK+(A) (X•, βQX•) ∼= HomD+(A) (QX•, αβQX•)via the map ϕ : f 7→ Q(f).

For this, observe first that we can define hom set in D+(A) also as

HomD+(A) (U•, V •) =

U• V •

W •f t: f, t,W • in K+(A), t qis

/ ∼

whereU• V •

W •f t∼ U• V •

W ′•g uif there is a commutative diagram

U• V •

W • W ′•

Z•

gf

tu

h v qis

In the context of the isomorphism, Q(f) = Q(g) means

X• βQX•

βQX• βQX•

Z•

gf u

t t qis


In particular, t f = t g in K+(A). Also, since t : βQX• → Z• is a qis in K+(I), t has aleft homotopy inverse s. Then s t f = s t g in K+(A) and so f = g in K+(A). Henceϕ is a monomorphism.

LetX• βQX•

W •f tbe an arbitrary element of HomD+(A) (QX•, αβQX•). Let s : W • →

βQX• be as in the previous paragraph; i.e. s is the left homotopy inverse of t in K+(I). Weconclude

X• βQX•

W •f t∼

X• βQX•

βQX•sf

= Q(s f)

Concluding that ϕ is epimorphism. ThereforeHomK+(A) (X•, βQX•) ∼= HomD+(A) (QX•, αβQX•).

Theorem 7.6.6 (Generalized Existance Theorem). Let A, B be abelian categories. LetK∗(A) be a localizing subcategory of K(A). Let F : K∗(A) → K(B) be a covariant∂−functor. Let K ′ be a full triangulated subcategory of K∗(A) satisfiying

(i) If X ∈ Ob(K∗(A)), then we can choose X ′• ∈ Ob(K ′) and a qis X• → X ′•.

(ii) If X ′• ∈ Ob(K ′) is acyclic, then F (X ′•) is acyclic.

Then R∗F : D∗(A)→ D(B) exists. Moreover, the natural transformation ξ : QF → R∗FQsatisfies that ξX′• is an isomorphism in D(B) for all X ′• ∈ Ob(K ′).

Proof. Let s : X ′• → Y ′• be a qis in K ′. There exists a distinguished triangle X ′•s→ Y ′• →

Z ′• → X ′[1]• in K ′, which makes Z ′• acyclic. Condition (ii) implies that F (Z ′•) is acyclicand so F (s) is a qis. We get a diagram

K ′ KQis∩K′

K(B) D(B)

Q

F ∃!FQ

which we can complete with a unique ∂−functor F by the universal property of Q upstairs.We get, F Q = Q F .

We now use Theorem 7.5.11 as follows. If s : X ′• → X• is a qis in K∗(A) with X ′• in K ′

then, by (i), there exists a qis f : X• → X ′′• with X ′′• ∈ Ob(K ′). Then, f s : X ′• → X ′′•

is in Qis ∩ K ′. By Theorem 7.5.11, the natural functor α : K ′Qis∩K′ → D∗(A) is fullyfaithfull. Using (i) again, we see that α is dense, hence an equivalence of categories. Letβ : D∗(A)→ K ′Qis∩K′ be a quasi-inverse for α.

Define R∗F : D∗(A) → D(B) by R∗F = F β. Using a similar argument as in Theorem7.6.5, one can show that R∗F is the right derived functor of F .


7.6.1 Ext and RHom

Definition 7.6.7. Let A be an abelian category. Let X•, Y • ∈ Ob(D(A)). For every n ∈ Z,define the nth hyper ext of X• and Y • to be

Extn(X•, Y •) = ExtnD(A) (X•, Y •) := HomD(A) (X•, Y [n]•)

Remark 7.6.8. Recall thatD+(A) is a full subcaterogy ofD(A). So ifX•, Y • ∈ Ob(D+(A)),then Extn(X•, Y •) = ExtD+(A) (X•, Y [n]•).

Since A can be embedded as a full subcategory into D(A), we get a definition of Extn(X, Y )for every X, Y ∈ Ob(A). One of our goals is to show that this coincides with the classicaldefinition of ExtnA(X, Y ) using an injective resolution of Y (provided that A has enoughinjectives).

Lemma 7.6.9. Let A be an abelian category. Let 0→ X•f→ Y •

g→ Z• → 0 be a short exactsequence in C(A). Then there is h ∈ HomD(A)(Z

•, X[1]•) such that

X•f→ Y •

g→ Z•h→ X[1]•

is a distinguished triangle in D(A).

Proof. Consider the distinguished triangle in K(A)

X•f→ Y •

α(f)→ M(f)•β(f)→ X[1]•

We write our maps as in R−mod using Frey-Mitchell’s Theorem. We have, M(f)n = Xn+1⊕Y n,

dnM(f) =(−dn+1

X 0

−fn+1 dnY

), α(f) = ( 0

1 ) , β(f) = ( −1 0 )

Define s : M(f)• = Xn+1⊕Y n → Z• by s = ( 0 gn ). Since gn+1 fn+1 = 0 and g is a cochainmap, we get that s is a cochain map as follows,

sn+1 dnM(f) = ( gn+1fn+1 gn+1dnY )

= ( 0 gn+1dnY )

= ( 0 dnZgn )

=dnZ sn

We will now show that s is a qis. Consider the long exact cohomology sequences associated

to 0→ X•f→ Y •

g→ Z• → 0 and the distinguished triangle X•f→ Y •

α(f)→ M(f)•β(f)→ X[1]•:

Hn(X•) Hn(Y •) Hn(M(f)•) Hn+1(X•) Hn+1(Y •)

Hn(X•) Hn(Y •) Hn(Z•) Hn+1(X•) Hn+1(Y •)

Hn(f)

Hn(α(f))

Hn(s)

Hn(β(f))

−id

Hn+1(f)

−id

Hn(f) Hn(g) ∂n Hn+1(f)


The first two and the last square commute by construction. To see that −Hn(βf) = ∂n Hn(s) = ∂n Hn(( 0 g )), we need to do some work.

We want to compute ∂n(Hn(( 0 g ))[z]) where z = ( z1z2 ) ∈M(f)n = Xn+1⊕Y n and dnM(f)(z) =0. This means (

−dn+1X (z1)

−fn+1(z1)+dnY (z2)

)= ( 0

0 )

or dn+1X (z1) = 0 and dnY (z2) = fn+1(z1). Recall the Snake Lemma construction for ∂n([gn(z2)]).

Y n3 z2 gn(z2) ∈Zn

Xn+13 z1 Y n+13 fn+1(z1)

gn

dnY

fn+1

Thus

∂(Hn(( 0 g ))([z])) =∂([gn(z2)]

)=[z1]

=−Hn(β(f))([z])

since β(f) = ( −1 0 ). Hence, −Hn(βf) = ∂n Hn(s) and the diagram commutes. By the5-lemma, Hn(s) = Hn(( 0 g )) is an isomorphism for all n ∈ Z, thus s is a qis.

To complete the Lemma, we recall that X•f→ Y •

α(f)→ M(f)•β(f)→ X[1]• is a distinguished

triangle in K(A) and D(A) so the following diagram

X• Y • M(f)• X[1]•

X• Y • Z• X[1]•

f α(f) β(f)

s

f g β(f)s−1

commutes in D(A) since s is an isomorphism in D(A). Hence, the bottom row is a distin-guished triangle in D(A). Define h := β(f) s−1.

One could say that Lemma 7.6.9 is the real reason why distinguished triangles replace shortexact sequences in derived categories.

Proposition 7.6.10. Let A be an abelian category. Let 0 → X•f→ Y •

g→ Z• → 0 be ashort exact sequence in C(A). Then for every V • ∈ Ob(D(A)), we get long exact hyper extsequences:

· · · → Exti(V •, X•)→Exti(V •, Y •)→Exti(V •, Z•)→ Exti+1(V •, X•) · · ·

and· · · → Exti(Z•, V •)→Exti(Y •, V •)→Exti(X•, V •)→ Exti+1(Z•, V •) · · ·


Proof. We use the long exact cohomology sequence with respect to the distinguishe triangle

X•f→ Y •

g→ Z•h→ X[1]• in D(A) from Lemma 7.6.9 and that HomD(A)(V

•,−) andHomD(A)(−, V •) are cohomological functors.

Total Hom functor

Let A be an abelian category, let X•, Y • ∈ Ob(C(A)). Define a bicomplex C•,• as follows:Cp,q := HomA(X−p, Y q),

dp,qh := (d−p−1X )∗ : HomA(X−p, Y q)→HomA(X−p−1, Y q)

f 7→f d−p−1X

dp,qv := (−1)p+q+1(dqY )∗ : HomA(X−p, Y q)→HomA(X−p, Y q+1)

f 7→(−1)p+q+1dqY f

With this definition, one can check that we get a bicomplex as in Definition 7.2.3; i.e.dh dh = 0, dv dv = 0 and dh dv + dv dh = 0.

Define now the Total Hom complex , Hom•(X•, Y •) := TotΠ(C•,•); in other words,

Homn(X•, Y •) =∏

p+q=n

HomA(X−p, Y q) =∏i∈Z

HomA(X i, Y i+n)

Following Definition 7.2.5, the diferential in Hom•(X•, Y •) is given by,

dn :(f−p,q

)p+q=n

7→(f−r+1,s d−rX + (−1)r+sds−1

Y f−r,s−1)r+s=n+1

for fp,q ∈ HomA(X−p, Y q). Or

dn :(f i)i∈Z 7→

(f i+1 diX + (−1)n+1di+nY f i

)i∈Z

for f i ∈ HomA(X i, Y i+n).

We can compute the cocycles and coboundaries of the Total Hom complex.

n-cocycles

ZnH•(X•, Y •) =ker(dn)

=(f i)i∈Z : f i+1 diX = (−1)ndi+nY f i

=(f i)i∈Z : f i+1 diX = diY [n] f i

=HomC(A)(X

•, Y [n]•)


n-coboundaries:

BnHom•(X•, Y •) =Im(dn−1)

=(f i+1 diX + (−1)i−1+ndi−1+n

Y f i)i∈Z : fi ∈ HomA(X i, Y i+1+n)

=(f i+1 diX + di−1

Y [n] fi)i∈Z

: fi ∈ HomA(X i, Y i+1+n)

=Ht(X•, Y [n]•)

Therefore,HnHom•(X•, Y •) = HomK(A)(X

•, Y [n]•) (?)

Note thatHom• : K(A)op×K(A)→ K(Ab) is a bi-∂−functor as in the following definition.

Definition 7.6.11. Let C1, C2, D be categories. A bifunctor C1×C2 → D consists of a mapF : Ob(C1) × Ob(C2) → Ob(D) and, for every Xi, Yi ∈ Ob(Ci), a map F : HomC1(X1, Y1) ×HomC2(X2, Y2)→ HomD (F (X1, X2) , F (Y1, Y2)) such that

F (X1,−) : C2 → D and F (−, X2) : C1 → D

are covariant functors, and for every fi ∈ HomCi(Xi, Yi) we have

F (Y1, f2) F (f1, X2) = F (f1, Y2) F (X1, f2)

A bifunctor F is called additive, or exact, or ∂−functor, if this is the case with respect toeach variable.

Construction of RHom

We want to compute the Right Turaev Functor of Hom•, say R+Hom• : D(A)op×D+(A)→D(Ab). The construction goes as follows.

Suppose A is an abelian category having enough injective objects. Let I be the full subcat-egory of A of injective objects. Fix X• ∈ Ob(K(A)), then Hom•(X•,−) : K+(A)→ K(A)is a ∂−functor. We can build the Right Turaev Functor

R+IIHom

•(X•,−) : D+(A)→ D(Ab)

as follows,

R+IIHom

•(X•,−) : D+(A) K+(Ab) D(Ab)β

∼=Q

localize

Fix Y • ∈ Ob(D+(A)). Then

R+IIHom

•(−, Y •) := Q Hom•(−, βY •) : K(A)op → D(Ab)


is a ∂−functor. We get then a bi-∂−functor

R+IIHom

• : K(A)op ×D+(A)→ D(Ab)

Fix again Y • ∈ Ob(D+(A)), we will show that R+IIHom

•(−, Y •) send qis in K+(A) to qis inK(Ab); i.e. to isomorphisms in D(Ab).

By Corollary 7.3.7, s : A• → B• is a qis if and only if M(s)• is acyclic, so it suffices to showthat R+

IIHom•(−, Y •) sends acyclic complexes to acyclic complexes. Let X• ∈ Ob(K(A))

be acyclic. We have for each n ∈ Z,

HnR+IIHom

•(X•, Y •) =Hn (QHom•(X•, βY •))

=Hn (Hom•(X•, βY •))

(1)=HomK(A)(X

•, βY [n]•)

(2)=0

Where (1) was discussed in equation (?) of Subsection 7.6.1, and (2) holds by Lemma 7.5.9since X• is acyclic and βY [n]• ∈ Ob(K+(I)). This proves that R+

IIHom•(−, Y •) send qis

in K+(A) to qis in K(Ab). Thus, R+IIHom

•(−, Y •) passes to a functor, namelly the RightTuraev Functor

RIR+IIHom

•(−, Y •) : D(A)op → D(Ab)

We get then a derived bifunctor called

R+Hom• = RIR+IIHom

• : D(A)op ×D+(A)→ D(Ab)

Note that R+Hom•(X•, Y •) = RIQHom•(X•, βY •) ∼= Hom•(X•, βY •). Here, βY • works

like in R−mod when when replacing Y with an injective resolution. Also note that ifY •inOb(D+(I)), then βY • ∼= Y • and R+Hom•(X•, Y •) ∼= Hom•(X•, Y •).

If A has enough projective objects, let P be the full subcategory of projective objects.In a similar fashion, we get Right Derived bifunctors:

R−I Hom• = RIIRIHom

• : D−(A)op ×D(A)→ D(Ab)

using the equivalence of categories D−(A)β→ K−(P).

Theorem 7.6.12 (Yoneda). Let A be an abelian category.

(a) If A has enough injectives, let X• ∈ Ob(D(A)) and let Y • ∈ Ob(D+(A)). Then, forevery i ∈ Z,

H i(R+Hom•(X•, Y •)

) ∼= ExtiD(A)(X•, Y •) := HomD(A)(X

•, Y [i]•)

(b) If A has enough projectives, let X• ∈ Ob(D−(A)) and let Y • ∈ Ob(D(A)). Then, forevery i ∈ Z,

H i(R−Hom•(X•, Y •)

) ∼= ExtiD(A)(X•, Y •)


Proof. We prove (a). We have the following chain of isomorphisms,

H i(R+Hom•(X•, Y •)

) ∼=H i (Hom•(X•, βY •))

(?)∼=HomK(A)(X•, βY [i]•)

(3)∼=HomD(A)(QX•, αβY [i]•)

∼=HomD(A)(X•, Y [i]•) = ExtiD(A)(X

•, Y •)

The isomorphism of (3) really comes from the localization functor Q and has been provenwhen Y • = X• during the proof of Theorem 7.6.5. We will sketch the general case now. Inthis setup I is the full subcategory of A consisting of injective objects, α : K+(I)→ D+(A)is given by α = Q |K+(I) where Q is the localization functor K+(A) → D+(A), α is anequivalence of categories; and β : D+(A)→ K+(I) is a quasi-inverse of α. Define

ϕ : HomK(A)(X•, βY [i]•)→HomD(A)(QX

•, αβY [i]•)

f 7→Q(f)

We use that

HomD(A) (QX•, αβY [i]•) =

X• βY [i]•

W •f t: f, t,W • in K(A), t qis

/ ∼

whereX• βY [i]•

W •f t∼

X• βY [i]•

W ′•f ′ t′iff there is a commutative diagram in

K(A)

X• βY [i]•

W • W ′•

Z•

f ′f

tt′

g u qis

We claim that ϕ : f 7→ Qf is an isomorphism. Suppose Qf = Qg, we get then the followingcommutative diagram in K(A),

X• βY [i]•

ββY [i]• βY [i]•

Z•

gf

t ∃t qis

But βY [i]• is in Ob(K+(I)) so t has a left homotopy inverse by Lemma 7.5.10. This forcesf = g in K(A), and ϕ is injective.


We now check surjectivity. LetX• βY [i]•

W •f t qis

be an arbitrary element ofHomD(A) (QX•, αβY [i]•).

Because βY [i]• ∈ Ob(K+(I)), t has a left homotopy inverse, say s : W • → βQX•. We con-clude

X• βY [i]•

W •f t∼

X• βY [i]•

βY [i]•sf

= Q(s f)

Finishing the proof.

Corollary 7.6.13. Let A be an abelian category, let X, Y ∈ Ob(A). Let X• (resp. Y •) bethe corresponding complex in D(A); i.e. X• : · · · 0 → X → 0 → · · · , X in degreee zero(resp. Y •). If A has enough injectives (resp. projectives) then for every i ≥ 0,

ExtiD(A)(X•, Y •) ∼= ExtiA(X, Y )

where ExtiA(X, Y ) is the classic exti defined by taking an injective resolution of Y (resp aprojective resolution of Y ).

Proof. We prove this when A has enough injectives. Let 0 → Yι→ E0

d0E→ E1d1E→ E2

d2E→ · · ·be an injective resolution of Y with corresponding truncated comples E•Y . Then Y •

ι→ E•Y isa qis by definition; also since •Y ∈ Ob(K+(I)) we get that E•Y

∼= βE•Y . The previous theoremshows that,

ExtiD(A)(X•, Y •) ∼=ExtiD(A)(X

•, E•Y )

7.6.12∼= H i(R+(Hom•(X•, E•Y )

)∼=H i (Hom•(X•, E•Y ))∗∼=H i (HomA(X,E•Y ))

ExtiA(X, Y )

The isomorphism * follows from the contruction of X•.

7.6.2 Total tensor product and ⊗L

Let R be a ring with 1. Let X• ∈ Ob(C(Mod − R)), Y ∈ Ob(C(R − Mod)). Define abicomplex C•,• by Cp,q := Xp ⊗R Y q,

dp,qh := dpX ⊗ idY q : Xp ⊗R Y q → Xp+1 ⊗R Y q

dp,qv := (−1)pidXp ⊗ dqY : Xp ⊗R Y q → Xp ⊗R Y q+1


Define X• ⊗R Y • := Tot⊕(C•,•); i.e. (X• ⊗R Y •)n =⊕

p+q=nXp ⊗R Y q, and

dn : (X• ⊗R Y •)n →(X• ⊗R Y •)n+1

(zp,q)p+q=n 7→((dr−1X ⊗ idsY

)(zr−1,s) +

((−1)ridrX ⊗ ds−1

Y

)(zr,s−1)

)r+s=n+1

for zp,q ∈ Xp ⊗R Y q.

We get a bi-∂−functor

−⊗R − : K(Mod−R)×K(R−Mod)→ K(Ab)

We want to construct a Left Turaev bi-functor

−⊗LR − : D−(Mod−R)×D−(R−Mod)→ D(Ab)

Construction of −⊗LR −

We know R−mod has enough projective objects; i.e., every module is the quotient of aprojective module. Let RP be the full subcategory of R−mod of projective modules. Wehave seen in Theorem 7.5.14 that for Q : K−(R−mod)→ D−(R−mod), the restriction

α = Q|K−(RP) : K−(RP)→ D−(R−mod)

is an equivalence of categories. Let γ : D−(R−mod)→ K−(RP) be a quasi-inverse.

Fix X• ∈ Ob(K(Mod − R)). Then X• ⊗R − : K−(R − mod) → K(Ab) has a left derivedfunctor

L−II(X• ⊗R −) : D−(R−mod)→ D(Ab)

defined as follows:

D−(R−mod) K−(RP) K(Ab) D(Ab)γ

L−II(X•⊗R−)

X•⊗R− Q

One can check that

L−II(−⊗R −) : K(R−mod)×D−(R−mod)→ D(Ab)

is a bi-∂−functor.

Now fix Y • ∈ Ob(D−(R−mod)). Then

L−II(−⊗R Y•) : K−(mod−R)→ D(Ab)


is a ∂−functor. One can show, using spectral sequences (more involved than for RHom),that L−II(−⊗R Y •) sends acyclic complexes to acyclic complexes. Hence L−II(−⊗R Y •) sendsqis to isomorphisms in D(Ab), and so it passes to a functor, the Left Derived Functor

L−I(L−II(−⊗R Y

•))

: D−(mod−R)→ D(Ab)

We get then a left Turaev bi-functor

−⊗LR − := L−I(L−II(−⊗R −)

)LD−(mod−R)×D−(R−mod)→ D(Ab)

Note that L−I(L−II(−⊗R −)

)is naturally isomorphic to L−II

(L−I (−⊗R −)

).

Hom to compute −⊗LR −

For X• ∈ Ob(D−(mod−R)) and Y • ∈ Ob(D−(R−mod)), let P • ∈ Ob(K−(mod−R)) andQ ∈ Ob(K−(R−mod)) be complexes quasi-isomorphic to X• and Y •, respectivelly. Then

X• ⊗LR Y • ∼= P • ⊗R Y • ∼= X• ⊗R Q•

Recall the hyper derived functors defined in Example 7.6.4

Li(−⊗R −) := H−i(−⊗LR −)

Proposition 7.6.14. Let A ∈ Ob(mod − R) with associated one term complex A• : · · · →0 → A → 0 → · · · , and let B ∈ Ob(R − mod) with associated one term complex B•. Foreach i ≥ 0,

TorRi (A,B) ∼= H−i (A•, B•)

Proof. Consider a projective resolution of B in R−mod · · ·dP,3→ P2

dP,2→ P1

dP,1→ P0ε→ B → 0

with corresponding truncated chain complex

PB,• : · · ·dP,3→ P2

dP,2→ P1

dP,1→ P0 → 0

Viewed as a cochain complex, PB,0 = P0, PB,−1 = P1, etcetera. By definition, ε induces a qis

PB,• · · · P1 P0 0 · · ·

B• · · · 0 B 0 · · ·

ε 0 ε 0

Then, viewing PB,• as a cochain complex, we get

H−i(A• ⊗LR B•) ∼=H−i(A• ⊗R PB,•)=H−i(A⊗R PB,•)=H i (A⊗R PB,•)=:TorRi (A,B)

Bibliography

[1] P.J. Cohen, Set theory and the continuum hypotesis. W.A. Benjamin, Inc., New York,1966.

[2] Morita, Kiiti (1958). “Duality for modules and its applications to the theory of rings withminimum condition”. Science reports of the Tokyo Kyoiku Daigaku. Section A. 6 (150):83–142. ISSN 0371-3539

133

Index

adjointleft, 16pair, 16right, 16

Baer sum, 86Baer’s criterion for flatness, 42bicomplex, 95, 126bifunctor, 127bimodule, 11

category, 9Ab-, 89abelian, 91additive, 89complete, 96Derived, 113full sub-, 10Homotopy, 97large, 51of direct systems, 23of inverse systems, 24opposite, 15R-mod, 10small, 51sub-, 10triangulated, 104

Chainmap, 68

chaincomplex, 67homotopy, 68

class, 39, 49proper, 51

Co-chainmap, 68

co-chaincomplex, 67

Co-homology, 68long exact sequence, 70

cokernel, 90Comparison Theorem, 70, 115

dual version, 71coproduct, 13

Derivedcategory, 98, 107Functor, 72Functors, 98left functor, 121, 132right functor, 120

dimension shifting, 81, 82direct

limit, 20system, 19

direct product, 12direct sum, 13directed

POSET, 19double complex, 95, 126dual, 23

operator, 15statement, 15

enoughinjectives, 119projectives, 119

equivalenceof categories, 39

essential, 35epimorphism, 32monomorphism, 35

exact, 28, 93functor, 14left, 14right, 14

134

INDEX 135

Existance of derived functors, 121Ext Functor, 74extension of modules, 85

flatmodule, 39resolution, 40

Freyd-Mitchell Embedding Theorem, 94functor, 9, 10

∂−, 105additive, 11, 89bi, 127bi-∂, 127category, 38cohomological, 105contravariant, 10covariant, 10dense, 52exact, 14, 73, 74fully faithful, 52, 54Hom, 10, 11representable, 53shift, 96, 104tensor product, 11

Generalized Existance Theorem, 123generator, 37

co, 37pro-, 57

Homology, 68long exact sequence, 69

Homotopyequivalent, 68, 97Null-, 97operator, 68, 71

Horseshoe lemma, 76hyper

derived functor, 121ext, 124

image, 93inductive limit, 20injective

hull, 36module, 34

object, 115resolution, 35

inverselimit, 20system, 19

kernel, 90

localization of categories, 107localizing subcategory, 120

mapping cone, 99module

category, 10cofree, 38divisible, 34flat, 39projective, 31

Moritacontext, 58equivalent, 58

morphism, 9epi-, 10epic, 10iso-, 9monic, 10mono, 10

multiplicative system, 107

naturalisomorphism, 38map, 16transformation, 38

objectinitial, 10terminal, 10

Pontryagin dual, 38, 41, 85progenerator, 57projective

cover, 33limit, 20module, 70object, 115resolution, 32

pull-back, 23

136 INDEX

push-out, 22

qis, 98Quasi-Frobenious, 46quasi-inverse, 39quasi-isomorphism, 98

Schannel’s Lemma, 43set, 49Snake lemma, 69Stable Module Category, 45

Tensor product, 11Tor Functor, 73torsion submodule, 84Total complex, 96Total Hom complex, 126triangle, 101

distinguished, 101, 104Turaev Functor

Left, 72, 131Right, 73, 74, 127, 128

Yoneda’s lemma, 54

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