homological algebra - nicholas camacho · 2019-09-23 · chapter 4. classical homological algebra {...

Homological Algebra

Lectures by Frauke Bleher

Typed by: Nicholas Camacho

Contents

Preface 5

Chapter 1. Preliminaries 71. Categories and Functors 72. Natural Transformations 93. Hom Functors and Tensor Product Functors 104. Products and Coproducts 125. Adjoints 146. Direct and Inverse Limits 16

Chapter 2. Module Categories 277. Projective Modules 278. Injective Modules 309. Generators and Nongenerators 3310. Flat Modules 3511. Character Modules 3812. Schanuel’s Lemma 4013. Stable Module Categories 4214. Morita Theory 43

Chapter 3. More on Categories 4915. Godel-Bernays System 4916. Functor Categories 5017. Equivalences of Categories 5118. Yoneda’s Lemma 52

Chapter 4. Classical Homological Algebra – Tor and Ext 5519. Basic Definitions 5520. Snake Lemma & Long Exact Homology/Cohomology Sequence 5621. Left Derived Functors 5922. Tor 6123. Right Derived Functors 6224. Ext 6325. Examples of Tor and Ext in Z-Mod 6426. Horseshoe Lemma 6527. Long Exact Sequences for Left and Right Derived Functors 6728. Properties of Tor and Ext 7129. Tor for abelian groups 7430. Ext for abelian groups 7531. Ext and extensions 76

3

4 CONTENTS

Chapter 5. Abelian Categories and Homotopy Categories 7732. Abelian Categories 7733. Category of Cochain Complexes, C(A) 8134. Bicomplexes and Total Complexes 8335. Shift/Translation Functor 8436. Homotopy Categories 8537. Mapping Cones 8638. Projective and Injective Objects 89

Chapter 6. Triangulated Categories 9139. Triangles and Distinguished Triangles in K(A) 9140. The Triangle Axioms for K(A) 9241. Triangle Axioms satisfied in K(A) 9342. Triangulated Categories 9443. Localization of Categories 9644. Localization of Triangulated Categories 99

Chapter 7. Derived Categories 10345. Derived Categories 10346. Enough Injectives or Projectives 10747. Derived Functors 10748. Existence Theorem for Derived Functors 10949. Ext and RHom: Hyperext 11150. Long exact hyperext sequences 11351. Total Hom complex 11352. Right Derived Bifunctor of Total Hom Complex 11553. Yoneda’s Result on HiRHom 11654. Total Tensor Product Complex 11955. Left Derived Bifunctor of Total Tensor Product 11956. Connection with Classical Tor 120

Index 121

Preface

This document contains lecture notes from the graduate course HomologicalAlgebra, taught by Dr. Frauke Bleher at the University of Iowa in Spring 2018. Itook the liberty of breaking up the notes into my own chapters, and I also changedthe order in which a few of the sections were actually presented. However, all thecontent comes directly from lectures. If you find any typos, no matter how small,please let me know.

Nicholas CamachoJuly 2018

5

CHAPTER 1

Preliminaries

1. Categories and Functors

Definition 1.1. A category C consists of the following data:

(a) a collection of objects denoted by Ob(C),(b) ∀ pair A,B ∈ Ob(C), a set of morphisms, denoted by HomC(A,B),(c) ∀ triple A,B,C ∈ Ob(C), a law of composition

HomC(A,B)×HomC(B,C) −→ HomC(A,C)

(f, g) 7−→ g fsuch that

(i) if (A,B) 6= (A′, B′) for A,B,A′, B′ ∈ Ob(C) then

HomC(A,B) ∩HomC(A′, B′) = ∅;

(ii) ∀A ∈ Ob(C),∃! morphism idA ∈ HomC(A,A) (called the identity mor-phism) such that: ∀X,Y ∈ Ob(C), ∀f ∈ HomC(X,A), ∀g ∈ HomC(A, Y )

idA f = f and g idA = g;

(iii) the law of composition is associative.

Notation. We write f : A→ B for a morphism f ∈ HomC(A,B).

Definition 1.2. A morphism f : A→ B in C is called an isomorphism if andonly if there exists g : B → A in C with g f = idA and f g = idB .

Examples 1.

(1) C = Sets: Objects = sets; Morphisms = set maps; law of composition =composition of set maps.

(2) R= associative ring with 1(a) C = R-Mod: Objects = left (unital) R-modules; Morphisms = R-

module homomorphisms.(b) C = Mod-R: Objects = right (unital) R-modules; Morphisms =

same.(3) Ab := Z-Mod.

Definition 1.3. Let C be a category. A subcategory of C is a category C′ suchthat

• Ob(C′) ⊆ Ob(C)• ∀A,B ∈ Ob(C′) : HomC′(A,B) ⊆ HomC(A,B) with the induced composi-

tion law, and idA ∈ HomC′(A,A) ∀A ∈ Ob(C′).

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

7

8 1. PRELIMINARIES

Example 1.4. R-Mod and Mod-R are subcategories of Sets, but not fullsubcategories.

Definition 1.5. Let C be a category, let f : A→ B be a morphism in C.

(a) f is called a monomorphism (or monic) if ∀ X ∈ Ob(C), ∀g, g′ : X → Awith f g = f g′, we have g = g′.

(b) f is called an epimorphism (or epic) if ∀ Y ∈ Ob(C), ∀h, h′ : B → Y withh f = h′ f , we have h = h′.

(c) P ∈ Ob(C) is called initial if ∀ Y ∈ Ob(C), HomC(P, Y ) has only oneelement. Q ∈ Ob(C) is called terminal (or final) if ∀ X ∈ Ob(C),HomC(X,Q) has only one element.

Examples 2.

(1) Let C = R-Mod or Mod-R, f : A→ B in C. Then

f is monic ⇐⇒ f is injective.

f is epic ⇐⇒ f is surjective.

(2) If 0 is the zero module, then 0 is both initial and final.

Definition 1.6. Let C and D be categories. A covariant functor (resp. con-travariant functor) F : C→ D consists of:

(a) a map F : Ob(C)→ Ob(D)(b) ∀ A,B ∈ Ob(C) a map

F : HomC(A,B) −→ HomD(F (A), F (B))

(resp. F : HomC(A,B) −→ HomD(F (B), F (A)) )

such that ∀A ∈ Ob(C), ∀ f : B → C, g : A→ B in C

(i) F (idA) = idF (A)

(ii) F (f g) = F (f) F (g) (resp. F (f g) = F (g) F (f) )

Examples 3.

(1) Identity functor: IdC : C → C. For all C,C ′ ∈ Ob(C) and for all f : C →C ′ in C, IdC(C) = C and IdC(f) = f . This functor is covariant.

(2) Hom functors: Fix A ∈ Ob(C).(a) F = HomC(A,−) : C→ Sets is covariant. For all C,C ′ ∈ Ob(C) and

for all f : C → C ′ in C, F (C) = HomC(A,C) and

F (f) : HomC(A,C) −→ HomC(A,C ′),

g 7−→ f g.

Notation. F (f) = f∗.

(b) G = HomC(−, A) : C→ Sets is contravariant. For all C,C ′ ∈ Ob(C)and for all f : C → C ′ in C, G(C) = HomC(C,A) and

G(f) : HomC(C ′, A) −→ HomC(C,A),

h 7−→ h f.

Notation. G(f) = f∗.

2. NATURAL TRANSFORMATIONS 9

Definition 1.7. Let C be a category.

(a) The dual category (or opposite category) is defined to the the category Cop

where• Ob(Cop) = Ob(C), and• for all A,B ∈ Ob(C) = Ob(Cop)

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

ϕop ↔ ϕ

with composition law:

Cϕ−→B ψ−→ A in C, then

Cϕop

←−−B ψop

←−− A in Cop.

Then, we get a duality functor :

op : C −→ Cop

A 7−→ A

ϕ 7−→ ϕop.

Since (ψ ϕ)op = ϕop ψop, the duality functor is contravariant.

Note 1.8. A contravariant functor C→ D is the same as a covariantfunctor Cop → C.

(b) Let S be a statement in C. The dual statement of S is the statement Sop

in Cop obtained from S by applying the duality functor (i.e. by reversingarrows and compositions) and interpreting the resulting statement in C.

Examples 4.

(1) “monic” is dual to “epic”.(2) In module categories:

(a) “direct product” is dual to “direct sum”(b) “kernel” is dual to “cokernel”.(c) “exactness of sequences” is self-dual.

Remark 1.9. If R is a ring then (R-Mod)op is not a module category unlessR = 0. (see, e.g., Freyd’s book on abelian categories.)

2. Natural Transformations

Definition 2.1. Let C,D be categories.

(a) Let F,G : C → D be functors of the same variance (i.e., both covari-ant or both contravariant). A natural transformation τ : F → G is acollection τAA∈Ob(C) of morphisms in D, τA : FA → GA such that∀f ∈ HomC(A,A′), we get a commutative diagram:

F,G covariant:

FA FA′

GA GA′

Ff

τA τA′

Gf

F,G contravariant:

FA FA′

GA GA′

τA

FfτA′

Gf

10 1. PRELIMINARIES

If, additionally, τA is an isomorphism for all A ∈ Ob(C), we call τ anatural isomorphism.

(b) The category of covariant functors, denoted DC (resp. contravariant func-tors) has as objects covariant (resp. contravariant) functors C → D, andhas as morphisms natural transformations.

Note 2.2 (Issue with (b)). HomDC(F,G) may not be a set, but aproper class. So, we need to allow “bigger categories”. (See 3)

(c) If F,G : C → D are two functors of the same variance, we say they arenaturally isomorphic, written F ∼= G, if there exists a natural isomorphismτ : F → G.

(d) A functor F : C → D is called an equivalence of categories if there existsa functor F ′ : D→ C with F ′ F ∼= IdC and F F ′ ∼= IdD.

F ′ is called a quasi-inverse of F .

Note. F is called an isomorphism of categories if there exists a func-tor G : D→ C with G F = IdC and F G = IdD.

Not every equivalence of categories is an isomorphism.

Example 2.3. IdR-Mod∼= R⊗R − ∼= HomR(R,−)

Proof. For all C ∈ Ob(R-Mod), define

σC : C −→ R⊗R Cx 7−→ 1⊗ x

τC : C −→ HomR(R,C)

x 7−→ (αx : R→ C, r 7→ rx).

Check: σCC∈Ob(R-Mod) and τCC∈Ob(R-Mod) are natural isomorphisms.

3. Hom Functors and Tensor Product Functors

All rings are associative with 1. All modules are unital. The most importantfunctors are Hom functors and tensor product functors. Let R be a ring.

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

Definition 3.1. Let A = AR and B = RB, and L ∈ Ob(Z-Mod).

(a) A map ϕ : A×B → L is called R-balanced or R-biadditive if• ϕ(a+ a′, b) = ϕ(a, b) + ϕ(a′, b)• ϕ(a, b+ b′) = ϕ(a, b) + ϕ(a, b′)• ϕ(ar, b) = ϕ(a, rb)

∀ a, a′ ∈ A,∀ b, b′ ∈ B, ∀ r ∈ R.(b) A tensor product of A and B over R is a Z-module A ⊗R B together

with an R-balanced map h : A × B → A ⊗R B satisfying the followinguniversal mapping property: For every Z-module X with an R-balancedmap ϕ : A × B → X, there exists a unique Z-module homomorphismΦ : A⊗R B → X with Φ h = ϕ.

A×B A⊗R B

X

h

ϕ ∃!Φ

3. HOM FUNCTORS AND TENSOR PRODUCT FUNCTORS 11

Theorem 3.2. The tensor product (A ⊗R B, h) exists and is unique up toisomorphism.

Definition 3.3. Let R,S be rings. An additive abelian group M is called an(R,S)-bimodule, written RMS , if M = RM and M = MS and ∀ r ∈ R,∀ s ∈S,∀ m ∈M ,

(rm)s = r(ms).

Proposition 3.4.

(1) Let A = SAR, B = RB. Then A⊗R B is a left S-module with

s(a⊗ b) = (sa)⊗ b

(plus Z-linear extension).(2) Let A = AR, B = RBS. Then A⊗R B is a right S-module with

(a⊗ b)s = a⊗ (bs)

(plus Z-linear extension).

Corollary 3.5. Let A = SAR, B = RBS. Then

A⊗R − : R-Mod −→ S-Mod

and −⊗RB : Mod-R −→Mod-S

are additive covariant functors.

Proposition 3.6.

(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) = rf(a).

Corollary 3.7. Let A = RAS, B = RBS.

(1) HomR(A,−) : R-Mod → S-Mod and HomS(A,−) : Mod-S → Mod-Rare additive covariant functors.

(2) HomR(−, B) : R-Mod →Mod-S and HomS(−, B) : Mod-S → R-Modare additive covariant functors.

Let F be a covariant or contravariant functor between module categories.

Definition 3.8. Suppose F is covariant (resp. contravariant).

(a) F is left exact if

0→ Aα−→ B

β−→ C exact =⇒ 0→ FAFα−−→ FB

Fβ−−→ FC exact.

( resp. Aα−→ B

β−→ C → 0 exact =⇒ 0→ FCFβ−−→ FB

Fα−−→ FA exact.)

12 1. PRELIMINARIES

(b) F is right exact if

Aα−→ B

β−→ C → 0 exact =⇒ FAFα−−→ FB

Fβ−−→ FC → 0 exact.

( resp. 0→ Aα−→ B

β−→ C exact =⇒ FCFβ−−→ FB

Fα−−→ FA→ 0 exact.)

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

Note 3.9. If F,G : R-Mod→ S-Mod are naturally isomorphic then

F is left exact ⇐⇒ G is left exact.

F is right exact ⇐⇒ G is right exact.

Theorem 3.10. Let C = R-Mod or Mod-R, fix M ∈ Ob(C). Then HomC(M,−)and HomC(−,M) are left exact functors.

Example 3.11. 0→ Z ι−→ Q π−→ Q/Z→ 0 is exact in Z-Mod, but HomZ(−,Z/2)is not an exact functor since

HomZ(Q,Z/2)︸︷︷︸=0

ι∗−→ HomZ(Z,Z/2)︸︷︷︸∼=Z/2

→ 0

is not exact.

Theorem 3.12. If A = AR and B = RB then A ⊗R − and − ⊗R B are rightexact functors.

Example 3.13. Take the previous ses 0→ Z ι−→ Q π−→ Q/Z→ 0. Then Z/2⊗Z−is not an exact functor since

0→ Z/2⊗Z Z︸︷︷︸∼=Z/2

idZ/2⊗ι−−−−−→ Z/2⊗Z Q︸︷︷︸=0

is not exact.

Lemma 3.14. Let C = R-Mod or Mod-R.

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

β−→ C in C. If for all M ∈ Ob(C),the sequence

0→ HomR(M,A)α∗−−→ HomR(M,B)

β∗−→ HomR(M,C)

is exact in Ab, then (∗) is exact in C.

(b) Consider a sequence (∗∗) A α−→ Bβ−→ C → 0 in C. If for all N ∈ Ob(C),

the sequence

0→ HomR(C,N)β∗−→ HomR(M,B)

α∗−−→ HomR(M,A)

is exact in Ab, then (∗∗) is exact in C.

4. Products and Coproducts

Let C = R-Mod or Mod-R, let Λ be an arbitrary index set, let Aλλ∈Λ bea family of modules in C.

4. PRODUCTS AND COPRODUCTS 13

Theorem 4.1. The module∏λ∈ΛAλ in C with underlying set

(aλ)λ∈Λ | aλ ∈ Aλ ∀λ ∈ Λ

with componentwise addition and R-action together with projection homomorphisms

pµ :∏λ∈Λ

Aλ −→ Aµ, ∀µ ∈ Λ

(aλ)λ∈Λ 7−→ aµ

satisfies the universal mapping property: For all X ∈ Ob(C) with homomorphismsϕµ : X → Aµ ∀µ ∈ Λ, ∃!Φ : X →

∏λ∈ΛAλ in C such that pµ Φ = ϕµ ∀µ ∈ Λ.∏

λ∈Λ

Aλ Aµ

X

pµ

∃!Φ ϕµ

The pair(∏

λ∈ΛAλ, pµµ∈Λ

)is called the product (or direct product) of

Aλλ∈Λ. This is unique up to isomorphism.

Theorem 4.2. The module∐λ∈ΛAλ in C with underlying set

(aλ)λ∈Λ | aλ ∈ Aλ and aλ = 0 for all but finitely many λ ∈ Λ

with componentwise addition and R-action together with injection homomorphisms

ιµ : Aµ −→∐λ∈Λ

Aλ, ∀µ ∈ Λ

a 7−→ (aλ)λ∈Λ, where aλ =

a if λ = µ

0 if λ 6= µ

satisfies the universal mapping property: For all Y ∈ Ob(C) with homomorphismsψµ : Aµ → Y ∀µ ∈ Λ, ∃!Ψ :

∐λ∈ΛAλ → Y in C such that Ψ ιµ = ψµ ∀µ ∈ Λ.

Aµ∐λ∈Λ

Aλ

Y

ψµ

ιµ

∃!Ψ

The pair(∐

λ∈ΛAλ, ιµµ∈Λ

)is called the coproduct (or direct sum) of Aλλ∈Λ.

This is unique up to isomorphism.

Theorem 4.3. Let C = R-Mod or Mod-R. Let Aλλ∈Λ, Bλλ∈Λ be familiesof modules in C, let A,B ∈ Ob(C).

(a) The map

HomR

(A,∏λ∈Λ

Bλ

)−→

∏λ∈Λ

HomR(A,Bλ)

f 7−→ (pλ f)λ∈Λ

is a Z-module isomorphism.

14 1. PRELIMINARIES

(b) The map

HomR

( ∐λ∈Λ

Aλ, B)−→

∏λ∈Λ

HomR(Aλ, B)

g 7−→ (g ιλ)λ∈Λ

is a Z-module isomorphism.

Theorem 4.4. Let Aλλ∈Λ be a family of right R-modules, let Bλλ∈Λ be afamily of left R-modules, let A = AR, B = RB.

(a) The map

A⊗R( ∐λ∈Λ

Bλ

)−→

∐λ∈Λ

(A⊗R Bλ

)a⊗ (bλ)λ∈Λ 7−→ (a⊗ bλ)λ∈Λ

(plus Z-linear extension) is a Z-module isomorphism.(b) The map ( ∐

λ∈Λ

Aλ

)⊗R B −→

∐λ∈Λ

(Aλ ⊗R B

)(aλ)λ∈Λ ⊗ b 7−→ (aλ ⊗ b)λ∈Λ

(plus Z-linear extension) is a Z-module isomorphism.

5. Adjoints

Definition 5.1. Let C,D be categories, let F : C → D and G : D → C becovariant functors. Then (F,G) is called an adjoint pair , F is called a left adjointfunctor, and G is called a right adjoint functor if the following is true: ∀A ∈ Ob(C),∀C ∈ Ob(D), there exist a bijection

τA,C : HomD(FA,C) −→ HomC(A,GC)

such that τA,C is natural in each variable, i.e., ∀f : A → A′ in C and ∀g : C → C ′

in D, we have commutative diagrams:

HomD(FA,C) HomC(A,GC)

HomD(FA′, C) HomC(A′, GC)

τA,C

(Ff)∗

τA′,C

f∗

HomD(FA,C) HomC(A,GC)

HomD(FA,C ′) HomC(A,GC ′)

g∗

τA,C

(Gg)∗

τA,C′

Theorem 5.2. Let B = RBS

(a)(B ⊗S −,HomR(B,−)

)is an adjoint pair.

(b)(−⊗RB,HomS(B,−)

)is an adjoint pair.

Proof idea. (for (a). (b) is similar).For all A = SA, for all C = RC define

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

f 7−→ f,

5. ADJOINTS 15

where

f : A −→ HomR(B,C) and fa : B −→ C

a 7−→ fa, b 7−→ f(b⊗ a).

Then check

• fa is an R-module homomorphism.• f is an S-module homomorphism.• τA,C is an isomorphism of abelian groups and natural in each variable.

Theorem 5.3. Suppose F : R-Mod → S-Mod and G : S-Mod → R-Modare covariant functors such that (F,G) is an adjoint pair. Then F is right exactand G is left exact.

Proof. We prove the statement for F (the one for G is similar).

Let Aα−→ B

β−→ C → 0 be an exact sequence in R-Mod. Consider the sequence

(∗) FAFα−−→ FB

Fβ−−→ FC → 0

in S-Mod. Let N ∈ Ob(S-Mod) be arbitrary. We get a diagram in Ab

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

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

τC,N∼=

(Fβ)∗ (Fα)∗

τB,N∼= τA,N∼=β∗ α∗

Since HomR(−, GN) is left exact, the bottom row is exact. Since the verticalarrows are isomorphisms, and since the diagram commutes (because τ·,· is natural ineach variable), the top row is exact. Since this is true for all N = SN , Lemma 3.14shows that (∗) is exact.

Example 5.4. Let C = Sets, D = k-Vec = k-Mod (k a field). Let F : C→ D

and G : D→ C be as follows:

• F (S) = k 〈S〉 = k-vector space with basis S.• F (f) = k-linear map between k-vector spaces induced from the map on

the bases.• G : D → C forgetful functor, forgetting the k-vector space structure andk-linearity of maps.

Claim. (F,G) is an adjoint pair.

Proof idea. Let S ∈ Ob(C), let V ∈ Ob(k-Vec). Define

τS,V : Homk(FS, V ) −→ Maps(S,GV )

f 7−→ f |S

Show that τS,V is bijective and natural in each variable.

16 1. PRELIMINARIES

6. Direct and Inverse Limits

Definition 6.1. Let (I,≤) be a partially ordered set (or “poset”). (I can beuncountable). We say (I,≤) is directed if ∀i, j ∈ I, ∃k ∈ I with i ≤ k and j ≤ k.

Definition 6.2. Let (I,≤) be a poset. Let C be a category.

(a) A direct system in C with index set I consists of:• Fii∈I a collection of objects in C, and• ∀ i ≤ j in I, a morphism in C ϕij : Fi → Fj such that

• ϕii = idFi for all i ∈ I, and

• for all i ≤ j ≤ k in I, ϕik = ϕjkϕij .

Notation. Fi, ϕiji,j∈Ii≤j

or Fi, ϕij.

(b) An inverse system in C with index set I consists of:• Fii∈I a collection of objects in C, and

• ∀ i ≤ j in I, a morphism in C ψji : Fj → Fi such that• ψii = idFi for all i ∈ I, and

• for all i ≤ j ≤ k in I, ψki = ψjiψkj .

Notation. Fi, ψji i,j∈Ii≤j

or Fi, ψji .

(c) Define a category J by• Ob(J) = I• ∀i, j ∈ I, if i 6≤ j then HomJ(i, j) = ∅, and if i ≤ j then ∃! morphismi→ j in J.

Remark 6.3. Hence, a direct system (resp. inverse system) in C with indexset I is a covariant (resp. contravariant) functor J→ C.

Examples 5. Let C be any category, let (I,≤) be a poset.

(1) Let A ∈ Ob(C). For all i ∈ I, define Fi := A and for all i ≤ j in I, define

ϕij = idA = ψji . Then Fi, ϕij (resp. Fi, ψji ) is called the constantdirect system (resp. constant inverse system) associated to A, denoted by|A|.

(2) I = Z+ with usual (partial) order ≤. Then a direct system on Z+ is asequence

F1 → F2 → F3 → F4 → · · ·

in C, and an inverse system on Z+ is a sequence

F1 ← F2 ← F3 ← F4 ← · · ·

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

diagram in C:

F0 F1

F2

6. DIRECT AND INVERSE LIMITS 17

and an inverse system on I is a diagram in C:

F1

F2 F0

(4) Suppose I has trivial partial order, i.e., ∀i, j ∈ I, i ≤ j ⇐⇒ i = j. ThenA direct (resp. inverse) system on I is just a collection Fii∈I of objectsin C.

Definition 6.4. Let C be a category, let (I,≤) be a poset.

(a) Let Fi, ϕij be a direct system in C on I. A direct limit (or inductivelimit) of this system is an object lim−→Fi in C together with morphisms inC

αj : Fj → lim−→Fi, ∀j ∈ I with αj ϕij = αi ∀i ≤ j in I

satisfying the following universal mapping property:For all X ∈ Ob(C) with morphisms fj : Fj → X ∀j ∈ I in C with

fj ϕij = fi ∀i ≤ j in I, there exists a unique morphism Φ : lim−→Fi → Xin C so that fi = Φ αi for all i ∈ I, i.e., the following diagram commutesfor all i ≤ j in I:

X lim−→Fi

Fi

Fj

∃!Φ

fi αi

ϕij

fj αj

(b) Let Fi, ψji be an inverse system in C on I. An inverse limit (or projectivelimit) of this system is an object lim←−Fi in C together with morphisms inC

αj : lim←−Fi → Fj , ∀j ∈ I with ψji αj = αi ∀i ≤ j in I

satisfying the following universal mapping property:For all X ∈ Ob(C) with morphisms fj : X → Fj ∀j ∈ I in C with

ψji fj = fi ∀i ≤ j in I, there exists a unique morphism Φ : X → lim←−Fiin C so that fi = αi Φ for all i ∈ I, i.e., the following diagram commutesfor all i ≤ j in I:

X lim←−Fi

Fj

Fi

fj

fi

∃!Φ

αj

αiψji

18 1. PRELIMINARIES

Theorem 6.5. Let C = R-Mod or Mod-R, let (I,≤) be a poset.

(a) The direct limit of a direct system Fi, ϕij in C over I exists in C and isunique up to isomorphism.

(b) The inverse limit of an inverse system Fi, ψji in C over I exists in C

and is unique up to isomorphism.

Proof sketch.

(a) Let (∐i∈I Fi, ιjj∈I) be the coproduct (or direct sum) of Fii∈I in C.

Let S be the submodule of∐i∈I Fi generated by all the elements of the

formιj(ϕ

ij(ai))− ιi(ai) ∀ i ≤ j in I, ∀ ai ∈ Fi.

Define lim−→Fi := (∐i∈I Fi)/S and for all j ∈ I define

αj : Fj → lim−→Fi by αj = p ιjwhere p :

∐i∈I Fi → lim−→Fi is the natural projection.

Check: (lim−→Fi, αjj∈I) satisfies the universal mapping property of a

direct limit of Fi, ϕij.(b) Let (

∏i∈I Fi, pjj∈I) be the product (or direct product) of Fii∈I in C.

Define

lim←−Fi :=

(ai)i∈I ∈

∏i∈I

Fi

∣∣∣∣∣ ai = ψji (aj) ∀i ≤ j in I

.

Show that this is a submodule of∏i∈I Fi. Define

αj : lim←−Fi → Fj by αj = pj |lim←−Fi ∀j ∈ I.

Check: (lim←−Fi, αjj∈I) satisfies the universal mapping property of an

inverse limit of Fi, ψji

Examples 6. Let C = R-Mod or Mod-R, let (I,≤) be a poset.

(1) Let A ∈ Ob(C). Let |A| be the constant direct (resp. inverse) system in C

over I associated to A. If I is directed, the the direct (resp. inverse) limitof |A| is isomorphic to A.

X A

Fi = A

Fj = A

∃!Φ

fi idA

ϕijidAfj idA

If I is directed, i.e. ∀i, j ∈ I, there exists k ∈ I with i ≤ k and j ≤ k,then fi = fk = fj and so we get fi = fj ∀i, j ∈ I. So, let Φ = fi.

(2) Let R be a commutative ring, J ⊆ R and ideal. We get a sequence ofideals

R = J0 ⊇ J1 ⊇ J2 ⊇ · · ·If M is an R-module then we get a sequence

M = RM ⊇ J1M ⊇ J2M ⊇ · · ·


of submodules where

J iM =

n∑t=1

xtmt

∣∣∣∣∣ n ∈ Z+, xt ∈ Ji,mt ∈M

.

Let I = Z+ with the usual (partial) order ≤. If i ≤ j we have a projectionhomomorphism

ψji : M/JjM →M/J iM.

So we get an inverse system M/J iM,ψji . The inverse limit lim←−M/J iM

is called the J-adic completion of M and denoted by M .Let R = lim←−R/J

i. Then R is a commutative ring.

Note. R is the completion of R with respect to the topology of Rthat has as neighborhood basis of r ∈ R, r + J i | i ≥ 0.

Note. M is an R-module.

Example 6.6. Let R = Z and J = (p) for p a prime number. Then

R is called the ring of p-adic integers and denoted by Zp or Zp. ThenZp = lim←−n Z/p

nZ and Qp = Frac(Zp).

(3) I = 0, 1, 2 with 0 < 1, 0 < 2.(a) Direct System:

A B

C

f

g

The direct limit is called a pushout . One can show that the pushout isisomorphic to Q := (B⊕C)

/(f(a),−g(a)) | a ∈ A︸︷︷︸

=:K

with morphisms

αB : B → Q, b 7→ (b, 0) +K

αC : C → Q, c 7→ (0, c) +K.

(b) Inverse System:

B

C A

f ′

g′

The inverse limit is called a pullback . One can show that the pullbackis isomorphic to P := (b, c) ∈ B⊕C | f ′(b) = g′(c) with morphisms

αB : P → B, (b, c) 7→ b

αC : P → C, (b, c) 7→ c.

(c) Kernel and Cokernel are Dual:Let f : X → Y be an R-module homomorphism. Then

20 1. PRELIMINARIES

(i) The diagramX Y

0

f

0 has pushout

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

(ii) The diagramX

0 Y

f ′

0

has pullback

(x, 0) ∈ X ⊕ 0 | f(x) = 0 ∼= Ker f.

(4) If I has trivial partial order, then lim−→Fi =∐i∈I Fi and lim←−Fi =

∏i∈I Fi.

Definition 6.7. Let C be a category, let (I,≤) be a poset.

(a) The category of direct systems in C over I is denoted by Dir(I) or DirC(I),and has as objects all direct systems in C over I. A morphism

t : Fi, ϕij −→ Gi, λijbetween direct systems in C over I consists of a family of morphisms in C

ti : Fi −→ Gi

such that for all i ≤ j in I, the following diagram commutes:

Fi Gi

Fj Gj

ti

ϕij λij

tj

(b) The category of inverse systems in C over I is denoted by Inv(I) orInvC(I), and has as objects all inverse systems in C over I. A morphism

s : Fi, ψji −→ Gi, ρji

between inverse systems in C over I consists of a family of morphisms inC

si : Fi −→ Gi

such that for all i ≤ j in I, the following diagram commutes:

Fi Gi

Fj Gj

si

sj

ψji ρji

(c) If C has exact sequences, then a sequence

Fi, ϕijt−→ Gi, λij

u−→ Hi, µij

is an exact sequence in DirC(I) if for all i ∈ I, the sequence Fiti−→ Gi

ui−→Hi is exact in C.

Similarly, a sequence

Fi, ψji s−→ Gi, ρji

r−→ Hi, τji

is an exact sequence in InvC(I) if for all i ∈ I, the sequence Fisi−→ Gi

ri−→Hi is exact in C.


Definition 6.8. Let C = R-Mod or Mod-R, let (I,≤) be a poset. We havethe following covariant functors:

(a) | · | : C → DirC(I) (or | · | : C → InvC(I)) defined by: ∀A,A′ ∈ Ob(C),∀f ∈ HomC(A,A′),

A 7−→ |A| and f 7−→ |f | : |A| → |A′|, where |f |i = f ∀i ∈ I.

(b) lim−→ : DirC(I)→ C defined by:

∀ Ai, ϕij, Bi, λij ∈ Ob(DirC(I)) and ∀ t : Ai, ϕij → Bi, λij inDirC(I),

Ai, ϕij 7−→ lim−→Ai =(∐i∈I

Ai

)/S,

Bi, λij 7−→ lim−→Bi =(∐i∈I

Bi

)/T, (for certain S & T ),

and t 7−→→t : lim−→Ai → lim−→Bi, (ai)i∈I + S 7→ (ti(ai))i∈I + T.

Check:→t is a well defined R-module homomorphism.

(c) lim←− : InvC(I)→ C defined by:

∀ Ai, ψji , Bi, ρji ∈ Ob(InvC(I)) and ∀ s : Ai, ψji → Bi, ρ

ji in

InvC(I),

Ai, ψji 7−→ lim←−Ai ⊆∏i∈I

Ai

Bi, ρji 7−→ lim←−Bi ⊆∏i∈I

Bi

and s 7−→ ←s : lim←−Ai → lim←−Bi, (ai)i∈I 7→ (si(ai))i∈I .

Check:←s is a well defined R-module homomorphism.

Theorem 6.9. Let C = R-Mod or Mod-R, let (I,≤) be a poset. (lim−→, | · |) and

(| · |, lim←−) are adjoint pairs. In particular, lim−→ is right exact and lim←− is left exact.

Proof. We prove the statement for (lim−→, | · |).Let A := Ai, ϕij ∈ Ob(DirC(I) and C ∈ Ob(C). Define

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

(where (lim−→Ai, αii∈I) is the direct limit of Ai, ϕij). Then τA,C is bijective:

Let t = ti : Ai, ϕij → |C|. Then for all i ≤ j in I, we getAi C

Aj C

ti

ϕijtj

.

By the universal mapping property of direct limits, ∃!f ∈ HomC(lim−→Ai, C) withf αi = ti ∀i ∈ I. Check: τ·,· is natural in each variable.

Theorem 6.10. Let F : C → D, G : D → C be covariant functors such that(F,G) is an adjoint pair. If C and D admit direct limits and inverse limits, then Fpreserves direct limits and G preserves inverse limits.

22 1. PRELIMINARIES

Proof. We prove the statement for G.Let Ci, ψji ∈ Ob(InvD(I)). Let (lim←−Ci, αjj∈I) be its inverse limit in D.

Since G is a covariant functor, G(Ci), G(ψji ) ∈ Ob(InvC(I)). We check that(G(lim←−Ci), G(αj)j∈I) satisfies the universal mapping property of the inverse limit

lim←−G(Ci) in C. For a diagram in C for all i ≤ j in I,

G(lim←−Ci) X

G(Ci)

G(Cj)

G(αi)

G(αj)

∃!β

fi

fjG(ψji )

we want to show ∃!β : X → G(lim←−Ci) in C such that G(αi) β = fi for all i ∈ I.

Using that (F,G) is an adjoint pair, we get a commutative diagram ∀i ≤ j in I,

HomC(X,G(lim←−Ci)) HomD(FX, lim←−Ci)

HomC(X,G(Cj)) HomD(FX,Cj)

HomC(X,G(Ci)) HomD(FX,Ci)

τ∼=

G(αj)∗ (αj)∗

τj∼=

G(ψji )∗ (ψji )∗

τi∼=

We have a diagram in D:

lim←−Ci FX

Ci

Cj

αi

αj

τi(fi)

τj(fj)

∃!γ

ψji

Now ψji τj(fj) = τi(fi) since the bottom rectangle commutes in the previousdiagram. There exists a unique γ ∈ HomD(FX, lim←−Ci) such that αi γ = τi(fi)

for all i ∈ I. Define β := τ−1(γ). Then

G(αi) β = G(αi)∗(τ−1(γ)) = τ−1

i ((αi)∗(γ)) = τ−1i (αi γ) = fi.

Moreover, β is unique since γ is unique.

Corollary 6.11. If B = SBR then −⊗S B and B⊗R− preserve direct limitsand HomS(B,−) and HomR(B,−) preserve inverse limits.

Note 6.12. HomS(−, B) and HomR(−, B) send direct limits to inverse limits.For example,

HomS(lim−→Ai, B) ∼= lim←−HomS(Ai, B).


Corollary 6.13. Direct limits (possibly different index sets) commute, so doinverse limits. For example, there is a natural isomorphism of functors

lim−→i∈I

lim−→j∈J

∼= lim−→j∈J

lim−→i∈I

Let C = R-Mod or Mod-R, let (I,≤) be a poset. Let Ai, ϕij ∈ Ob(DirC(I)).Recall: The direct limit of this system is

lim−→Ai =(∐i∈I

Ai

)/S

where S is the submodule of∐i∈I Ai generated by all elements of the form

ιj(ϕij(ai))− ιi(ai) ∀ i ≤ j in I, ∀ ai ∈ Ai.

where

ιj : Aj −→∐i∈I

Ai, a 7−→ (ai)i∈I where ai =

a i = j

0 i 6= j.

Lemma 6.14. Suppose (I,≤) is a directed poset.

(a) lim−→Ai = ιi(bi) + S | i ∈ I, bi ∈ Ai.(b) ιi(bi) + S = 0 + S ⇐⇒ ∃k ∈ I, i ≤ k, ϕik(bi) = 0Ai .

Proof.

(a) Let (ai)i∈I +S ∈ lim−→Ai. Notice that (ai)i∈I =∑i∈I ιi(ai). Let I0 = i ∈

I | ai 6= 0. Then I0 ⊆ I is a finite subset. Since I is directed, there existsk ∈ I such that i0 ≤ k for all i0 ∈ I0. Then

(ai)i∈I + S =∑i0∈I0

ιi0(ai0) + S

=∑i0∈I0

ιi0(ai0) + ιk(ϕi0k (ai0))− ιi0(ai0) + S

= ιk

( ∑i0∈I0

ϕi0k (ai0))

+ S.

(b) (⇐) ιi(ai) + S = ιi(ai) + ιk(ϕik(ai))− ιi(ai) + S = (0) + S.(⇒) ιi(ai) + S = (0) + S means there exists I0 ⊆ I a finite subset and∀i0 ∈ I0, ∃j0 ∈ I with i0 ≤ j0, ∃bi0 ∈ Ai0 such that

ιi(ai) =∑i0∈I0

(ιj0(ϕi0j0(bi0))− ιi0(bi0)

).

24 1. PRELIMINARIES

Since I is directed, there exists k with i ≤ k and i0 ≤ j0 ≤ k for all i0 ∈ I0.Then

ιk(ϕik(ai)) = ιk(ϕik(ai))− ιi(ai) + ιi(ai)

= (ιk(ϕik(ai))− ιi(ai)

+∑i0∈I0

ιk(ϕj0k (ϕi0j0(−bi0))− ιj0(ϕi0j0(−bj0))

+∑i0∈I0

(ιk(ϕi0k )− ιi0(bi0)

)=∑t∈T

(ιk(ϕtk(ct))− ιt(ct)),

for certain ct ∈ At, where T = i ∪ I0 ∪ j0 | i0 ∈ I0.

Compare LHS and RHS:• If t ∈ T satisfies t < k (t 6= k) then

ιt(ct) = 0 =⇒ ct = 0 =⇒ ιk(ϕtk(ct))− ιt(ct) = (0).

• If t ∈ T with t = k then

ιk(ϕkk(ck))− ιk(ck) = (0).

So the RHS= (0) implies LHS= (0), and since ιk is injective, ϕik(ai) = 0.

Theorem 6.15. If (I,≤) is a directed poset then lim−→i∈I is exact.

Proof. We know lim−→ is always right exact. Suppose t : Ai, ϕij → Bi, λijis a morphism in DirC(I) with ti : Ai → Bi injective ∀i ∈ I. We need to show that→t : lim−→Ai → lim−→Bi is injective. Let(∐

i∈IAi, ιii∈I

)and

(∐i∈I

Bi, jii∈I)

be the coproducts of Aii∈I and Bii∈I respectively. Write

lim−→Ai =(∐i∈I

Ai

)/S and lim−→Bi =

(∐i∈I

Bi

)/T.

Using Lemma 6.14 (a), suppose

→t (ιi(ai) + S)︸︷︷︸=ji(ti(ai))+T

= (0) + T.

Then by Lemma 6.14 (b), there exists k ∈ I, i ≤ k such that

λik(ti(ai))︸︷︷︸=tk(ϕik(ai))

= 0Bk .

Since tk is injective, ϕik(ai) = 0, and so by Lemma 6.14 (b), ι(ai)+S = (0)+S.

Remark 6.16. It is not enough to assume (I,≤) is directed to get exactnessof lim←−i∈I .


Example 6.17. Let I = Z+ with the usual order, p ≥ 3 prime. Consider theinverse system Z/pnZ, ψnm where for m ≤ n

ψnm : Z/pnZ −→ Z/pmZa+ pnZ 7−→ a+ pmZ.

Consider t : |Z| → Z/pnZ, ψnm where

tn : Z −→ a+ pnZa 7−→ a+ pnZ.

However, the map←t : Z −→ lim←−Z/pnZ = Zp

a 7−→ (a+ pnZ)n∈Z+

is not surjective. For example,

x := (1 + p+ p2 + · · ·+ pn−1 + pnZ)n∈Z+

doe not lie in Im(←t ). Why? We have

1 + p+ p2 + · · ·+ pn−1 =pn − 1

p− 1

and −1p−1 6∈ Z, but x is the multiplicative inverse of −(p− 1).

Remark 6.18 (What goes wrong?). Given

Bi, ρji︸︷︷︸=|Z|

t−→ Ci, τ ji ︸︷︷︸Z/pnZ, ψnm

,

consider preimages: say for (ci)i∈I ∈ lim←−Ci, t−1i (ci), i ∈ I. These form an inverse

system of sets, but the inverse limit could be empty. Hence, we need additionalconditions.

Definition 6.19. Let I = Z+. An inverse system Ai, ψji ∈ Ob(InvC(I))satisfies the Mittag-Leffler (ML) condition if

∀i, ψji (Aj) | j ≥ istabilizes.

One can show that if

0→ Ai, ψji s−→ Bi, ρji

t−→ Ci, τ ji → 0

is a short exact sequence in InvC(I) and Ai, ψji satisfies the ML condition, then

0 lim←−Ai←s−→ lim←−Bi

←t−→ lim←−Ci → 0

is exact. (See Lang’s Algebra book).

CHAPTER 2

Module Categories

7. Projective Modules

Theorem 7.1. Let C = R-Mod or Mod-R. Let P ∈ Ob(C). TFAE:

(a) given a diagram in C

P

B C 0

∃γα

β

there exists γ ∈ HomR(P,B) with β γ = α.(b) HomR(P,−) is exact.(c) every short exact sequence 0→ X → Y → P → 0 splits.(d) P is a direct summand of a free module F , written P |F .

Proof. Standard.

Definition 7.2. A module P satisfying (a)–(d) of Theorem 7.1 is called aprojective module.

Example 7.3. Not every projective module is free. Let

R = Z× Z, P1 = (a, 0) | a ∈ Z, and P2 = (0, b) | b ∈ Z.

Then RR = P1 ⊕ P2, (inner direct sum), but P1 and P2 are not free since

Z(P1) ∼= Z ∼= Z(P2),

but for all n ∈ Z+, Z(Rn) ∼= Z2n, i.e. Rn has even rank as a free Z-module.

Proposition 7.4.

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

∐λ∈Λ Pλ is a projective

module.

Proof. Standard.

Let C = R-Mod or Mod-R.

Definition 7.5. Let M be an R-module in C. A projective resolution of M isan exact sequence in C

· · · → Pndn−→ Pn−1 → · · · → P1

d1−→ P0ε−→M → 0

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

Theorem 7.6. Every module M in C has a projective resolution.

27

28 2. MODULE CATEGORIES

Proof. Let F0 be a free module in C surjecting onto M . Then we get a shortexact sequence

0→ K0ι0−→ F0

ε−→M → 0

where K0 = Ker(ε) and ι0 is inclusion. Let F1 be a free module in C surjectingonto K0. Then we get a short exact sequence

0→ K1ι1−→ F1

p1−→ K0 → 0

where K1 = Ker(p1), ι1 is inclusion. Inductively, we get a short exact sequence forall n ∈ Z+

0→ Knιn−→ Fn

pn−→ Kn−1 → 0

Define dn := ιn−1 pn for all n ∈ Z≥1. Then

· · · → Fndn−→ Fn−1 → · · ·

d1−→ F0ε−→M → 0

is exact since for all n ∈ Z≥2:

Im(dn) = Im(ιn−1 pn) = Im(ιn−1),

Ker(dn−1) = Ker(ιn−2 pn−1) = Ker(pn−1) = Im(ιn−1),

and Im(d1) = Im(ι0 p1) = Im(ι0) = Ker(ε).

Definition 7.7. Let f : M ′ → M be an R-module homomorphism in C. Wesay f is an essential epimorphism if f(M ′) = M and for every submodule M ′′ (M ′

we have f(M ′′) 6= M .

Theorem 7.8. Let ε : P → M be a surjective R-module homomorphism in C

where P is a projective module. The following are equivalent:

(a) ε is an essential epimorphism(b) For all projective modules Q in C, for all surjective R-module homomor-

phisms ε′ : Q → M , and for all f : Q → P in C with ε f = ε′, we havethat f is surjective.

0

P M 0

Q

ε

fε′

Proof. (a)⇒(b): Let Q, ε′, f be as in (b). (Note: f exists since ε is surjective

and Q is a projective module). We have

ε(Im(f)) = ε(f(Q)) = ε′(Q) = M.

Since ε is an essential epimorphism, then Im(f) = P , i.e. f is surjective.

7. PROJECTIVE MODULES 29

(b)⇒(a): Let L ( P be a submodule. Suppose ε(L) = M . Let Q be a projectiveR-module with a surjective R-module homomorphism δ : Q→ L.

0

P M 0

L

Q

ε

ε|L

δ

ε′=ε|Lδf

Let f : Q→ P be δ followed by the inclusion L → P . Then f is not surjective, butε f = ε|L, contradicting (b).

Definition 7.9. A projective module P together with a module homomor-phism ε : P →M satisfying (a)⇐⇒ (b) of Theorem 7.8 is called a projective coverof M , denoted by (P, ε).

Theorem 7.10. If (P, ε) and (P ′, ε′) are projective covers of M then ∃f : P →P ′, an R-module isomorphism ε′ f = ε.

Proof.0

P M 0

P ′

ε

∃fε′

Since P is projective and ε′ is surjective, there exists f ∈ HomR(P, P ′) with ε′ f =ε. Since (P ′, ε′) is a projective cover, f is surjective. (See Theorem 7.8). Since P ′

is projective, the short exact sequence

0→ Ker(f) → Pf−→ P ′ → 0

splits. Then there exists a submodule P ′′ ⊆ P such that f |P ′′ : P ′′ → P ′ is anisomorphism and P = Ker(f)⊕ P ′′ (inner direct sum). Since

ε(P ′′) = ε′′(f(P ′′)) = ε(P ′) = M

and ε is an essential epimorphism, P ′′ = P . So Ker(f) = 0, and hence f is anisomorphism.

Remark 7.11. Projective covers do not always exists. For example, the Z-module Z/2 does not have a projective Z-module cover.

Theorem 7.12. If R is a left Artinian ring and M is a finitely generated R-module, then M has a projective R-module cover.


8. Injective Modules

Let C = R-Mod or Mod-R.

Theorem 8.1. Let E be an R-module in C. TFAE:

(a) For every diagram in C

E

0 A B (exact)

f

α

∃g

there exists g ∈ HomR(B,E) with g α = f .(b) HomR(−, E) is exact.(c) Every short exact sequence 0→ E → Y → Z → 0 in C splits.

Definition 8.2. An R-module E in C is called an injective module if E satisfies(a)–(c) of Theorem 8.1

Proposition 8.3.

(a) Every direct summand of an injective module is injective.(b) Suppose Eλλ∈Λ is a family of injective modules. Then

∏λ∈ΛEλ is an

injective module.

Note 8.4. The proof of Proposition 8.3 only uses the equivalence of (a) and(b) of Theorem 8.1.

Note 8.5. Theorem 8.1 is easy to prove (a) ⇐⇒ (b)⇒(c). To prove Theo-rem 8.1 (c)⇒(a), one proves the following: (using only Theorem 8.1 (a) ⇐⇒ (b)whenever dealing with injective modules)

Theorem 8.6 (Baer’s Criterion). If E is a left R-module, then E is injectiveif and only if for every left ideal I ⊆ R, and for every f ∈ HomR(I, E), f extendsto R.

Definition 8.7. Let M be a left R-module, let m ∈ M, r ∈ R. Then m isdivisible by r is there exists m′ ∈M with m = rm′.

M is a divisible left R-module if each m ∈M is divisible by each r ∈ R that isnot a right zero divisor (i.e. s ∈ R− 0 =⇒ sr 6= 0).

Theorem 8.8. Every injective left R-module is divisible. If R is a PID, thenthe converse is true.

Example 8.9. Let R = K[u, v], a polynomial ring over a field K in 2 variablesu, v. Let Q = Frac(R). Then Q/R is a divisible R module that is not an injectiveR module.

Theorem 8.10. Every Z-module M can be embedded into an injective Z-module.

Main Idea of Proof. There exists a free Z-module F =⊕

λ∈Λ Z together

with a surjective Z-module homomorphism Fπ−→M . Then

M ∼=(⊕λ∈Λ

Z)/Ker(π)

(⊕λ∈Λ

Q)/Ker(π),

the latter of which is a divisible Z-module, hence injective.

8. INJECTIVE MODULES 31

Theorem 8.11. If M is a left (resp. right) R-module, then M can be embeddedinto an injective left (resp. right) R-module.

Proof. Let D be an injective Z-module together with an injective Z-modulehomomorphism ι : M → D. Assume M is a left R-module.

Define E := HomZ(ZRR, ZD). So E is a left R-module. Define

f : M −→ E

m 7−→ (fm : R→ D, r 7→ ι(rm)).

Check: fm is a Z-module homomorphism; f is an injective R-module homomor-phism.

Claim 8.1. E is an injective left R-module.

Proof of Claim 8.1 We show that HomR(−, E) = HomR(−,HomZ(R,D)) is

exact. Let 0→ Aα−→ B be exact in R-Mod. We use (R ⊗R −,HomZ(R,−)) is an

adjoint pair.

HomR(B,HomZ(R,D)) HomR(A,HomZ(R,D)) 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 arrows areisomorphisms. The bottom square commutes because R⊗R − ∼= IdR-Mod (check).Since D is an injective Z-module, the bottom row is exact. Hence the top row isexact.

From now on, let C be R-Mod or Mod-R

Definition 8.12. Let M be an R-module in C. An injective resolution of Mis an exact sequence in C

0→Mι−→ E0 d0

−→ E1 → · · ·En−1 dn−1

−−−→ En → · · ·

where each Ei is an injective module.

Theorem 8.13. Every module in C has an injective resolution.

Proof. Follows from Theorem 8.11.

Definition 8.14. A hom. f : M → N in C is called an essential monomor-phism if f is a monomorphism and for all submodules 0 6= N ′ ⊆ N , N ′ ∩ f(M) 6=0.

Remark 8.15. “essential monomorphism” and “essential epimorphism” aredual.

Why? Let f : M → N be in C.


• f is an essential epimorphism if Coker(f) 6= 0 where f = f ι.

0

M ′′ 6= 0

M N 0 (exact)

M ′ N

0(exact)

f

ιf

• f is an essential monomorphism if Ker(f) 6= 0 where f = π f .

0

N ′ 6= 0

N M 0 (exact)

N ′′ M

0(exact)

π

f

f

Theorem 8.16. Let ι : M → E be a monomorphism in C, where E is injective.The following are equivalent:

(a) ι is an essential monomorphism.(b) ∀ Q ∈ Ob(C), for every monomorphism ι : M → Q, ∀ f : E → Q with

f ι = ι′, we have that f is injective.

0

0 M E

Q

ι

ι′

f

Definition 8.17. An injective module E in C together with an essential monomor-phism ι : M → E in C is called an injective hull of M .

Theorem 8.18. If (E, ι) and (E′, ι′) are injective hills of M then there existsan isomorphism f : E → E with f ι = ι′.

Theorem 8.19. Every R-module M in C has an injective hull.

Proof. We show this for C = R-Mod. Let D be an injective left R-moduletogether with a monomorphism f : M → D. If f is an essential monomorphism,we are done. Now, suppose f is not an essential monomorphism. Let

S = T ∈ Ob(C) | f(M) ⊆ T ⊆ D, fT : M → T,m 7→ f(m) is an ess. monom.Then S 6= ∅ since f(M) ∈ S. S is partially ordered by inclusion. Using Zorn’s

Lemma, S has a maximal element, say E.

9. GENERATORS AND NONGENERATORS 33

Claim 8.2. E is an injective left R-module.

Proof of Claim 8.2

M E DfE

f

ι

Since fE is an essential monomorphism and f is not, then ι is not an essen-tial monomorphism (since composition of essential monomorphisms is an essentialmonomorphism). So there exists 0 6= X, a submodule of D with E ∩ X = 0.By Zorn’s Lemma, there exists a maximal X satisfying these conditions. Let X besuch a maximal. Consider

0→ X → Dπ−→ D/X → 0

d 7→ d+X.

We will show that there exists an isomorphism j : D/X → E such that π j =idD/X . (Hence E is isomorphic to a direct summand of D, hence is itself injective.)

Let g = π|E : E → D/X. Then g is an essential monomorphism: Indeed, g isa monomorphism since E ∩X = 0. Now let X ( D′ ⊆ D, D′ a submodule. Bymaximality of X, D′ ∩ E 6= 0.

So there exists d′ ∈ D′ ∩ E, d 6∈ X. So d′ +X ∈ g(E) ∩D′/X.

D

0 E D/Xg

ι∃j

Since D is injective, there exists an R-module homomorphism j : D/X → Dwith j g = ι. Now j is a monomorphism: Ker(j) ∩ Im(g) = 0 + X. Since g isan essential monomorphism, Ker(j) = 0 +X.

Now, j : D/X → j(D/X) is an essential monomorphism, and so is

j g : E → j(D/X)

and also j g f : M → j(D/X) is an essential monomorphism. Notice that

j g f = fj(D/X).

By maximality of E in S, j(D/X) = E. Hence j : D/X → E is an isomorphism,and j g = idE . So g = j−1. Therefore, π j = g j = idD/X .

9. Generators and Nongenerators

Let C = R-Mod or Mod-R. We proved the following: (cf Theorem 8.11 andits proof)

Proposition 9.1. Let M be a module in C, let D be an injective Z-modulewith an injective Z-module homomorphism ι : M → D. Then E := HomZ(R,D) isan injective module in C and

f : M −→ E = HomZ(R,D)

m 7−→ (fm : R→ D)


wherefm : r 7−→ ι(mr) if C = Mod-R

ι(rm) if C = R-Mod.

Definition 9.2.

(a) A module P in C is called a generator for C if every module in C isisomorphic to a quotient module of a direct sum of copies of P .

(b) A module C in C is called a cogenerator for C if every module in C isisomorphic to a submodule of a direct product of copies of C.

Example. R (in fact Rn ∀n ∈ Z+) is a projective generator for both R-Modand Mod-R.

Lemma 9.3. Let C be a module in C. C is a cogenerator for C if and onlyif ∀ 0 6= M ∈ Ob(C),∀m ∈ M,m 6= 0,∃ fm : M → X a morphism in C withfm(m) 6= 0.

Proof. (⇒) Let 0 6= M ∈ Ob(C). Then there exists an index set Λ andan injective R-module homomorphism f : M →

∏λ∈Λ C.. For all µ ∈ Λ let

pµ :∏λ∈Λ C → C be the µth projection. If m ∈M,m 6= 0, there exists µ ∈ Λ with

pµ(f(m)) 6= 0. Let fm := pµ f .(⇐) Let 0 6= M ∈ Ob(C). For all ∈ M − 0, let fm : M → C be a morphism

in C with fm(m) 6= 0.By the universal mapping property for

∏, there exists a unique homomorphism

f : M →∏m∈M−0 C with pmf = fm for all m ∈M−0. Then f is an injective

morphism in C.

Corollary 9.4. Q/Z is an injective cogenerator for Z-Mod.

Proof. Q/Z is a quotient of a divisible Z-module, hence divisible, and there-fore an injective Z-module. Let 0 6= M ∈ Ob(Z-Mod). Let m ∈M −0. Let 〈m〉be the Z-submodule of M generated by m.

Define g : 〈m〉 → Q/Z by:

g(m) =

12 + Z if o(m) =∞1r + Z if o(m) = r <∞.

In all cases, g(m) 6= 0. Since Q is an injective Z-module, there exists fm makingthe diagram commute, and fm(m) 6= 0.

Q/Z

0 〈m〉 M

g∃fm

Definition 9.5. Let R∗ := HomZ(R,Q/Z). Then R∗ is an injective R-modulein C. A cofree module in C is defined to be a module that is isomorphism to a directproduct of copies of R∗.

Note. Cofree modules are injective R-modules.

Theorem 9.6. Let E be a module in C. Then E is an injective R-module in C

if and only if E is isomorphic to a direct summand of a cofree R-module in C.

10. FLAT MODULES 35

Proof. (⇐) Follows since direct summands of injective modules are injective.(⇒) Follows since R∗ is an injective cogenerator for C.

Remark 9.7 (Why is R∗ an injective cogenerator for C?). Let 0 6= M be amodule in C. Let Λ be an index set such that there exists an injective Z-modulehomomorphism ι : M →

∏λ∈Λ Q/Z.

By Proposition 9.1, there exists an injective R-module homomorphism

f : M −→ HomZ

(R,∏λ∈Λ

Q/Z)∼=∏λ∈Λ

HomZ(R,Q/Z) =∏λ∈Λ

R∗.

Remark 9.8. If M is a left (resp. right) R-module, then HomZ(M,Q/Z) is aright (resp. left) R-module, called the Pontryagin dual of M .

10. Flat Modules

Definition 10.1. A right (resp. left) R-module B = BR (resp. B = RB) iscalled flat (or R-flat) if B ⊗R − (resp. −⊗R B) is an exact functor.

Theorem 10.2. Let B = SBR and C = SC. If B is R-flat and C is an injectiveS-module, then HomS(B,C) is an injective left R-module.

Proof. The functor HomR(−,HomS(B,C)) is, by adjointness, naturally iso-morphic to the functor HomS(B ⊗R −, C) and

HomS(B ⊗R −, C) = HomS(−, C) (B ⊗R −).

Since both HomS(−, C) and B ⊗R − are exact by assumption, thenHomR(−,HomS(B,C)) is exact.

Theorem 10.3.

(a) R = RR is R-flat.(b) Let Bλλ∈Λ be a family of right R-modules. Then∐

λ∈Λ

is flat ⇐⇒ Bλ is flat ∀λ ∈ Λ.

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

Proof. (c) follows from (a) and (b).

(a) Follows since R⊗R − ∼= IdR-Mod.(b) Let fλ : Cλ → C ′λ be an S-module homomorphism ∀λ ∈ Λ. Then, there

exists a unique S module homomorphism∐λ∈Λ

fλ :∐λ∈Λ

Cλ −→∐λ∈Λ

C ′λ

(cλ)λ∈Λ 7−→ (fλ(cλ))λ∈Λ

and∐λ∈Λ fλ is monic ⇐⇒ fλ is monic ∀λ ∈ Λ.

Now consider an exact sequence of left R-modules 0→ Aα−→ A′. We

get a commutative diagram of Z-modules:

0(∐

λ∈ΛBλ

)⊗R A

(∐λ∈ΛBλ

)⊗R A′

∐λ∈Λ

(Bλ ⊗R A

) ∐λ∈Λ

(Bλ ⊗R A′

)(id∐Bλ ⊗α)

∼= ∼=∐λ∈Λ(idBλ ⊗α)


Hence

(id∐Bλ)⊗ α is monic ⇐⇒ the top row is exact

⇐⇒ bottomw row is exact

⇐⇒∐λ∈Λ

(idBλ ⊗ α

)is monic

⇐⇒ idBλ ⊗ α is monic ∀λ ∈ Λ.

This implies (b).

Definition 10.4. Let M be a right R-module. A flat resolution of M is anexact sequence

· · · → Fndn−→ Fn−1 → · · · → F1

d1−→ F0ε−→M → 0

in Mod-R, where each Fi is R-flat.

Note 10.5. Since every M has a projective resolution, Theorem 10.3 impliesevery M has a flat resolution.

Theorem 10.6. Let (I,≤) be a directed poset. Let Bi, ϕij be a direct systemin Mod-R over I. If Bi is flat for all i ∈ I, then lim−→Bi is flat. (The converse is

not true in general).

Proof. Let α : A→ A′ be a monic R-module homomorphism between left R-modules. By assumption, for all i ∈ I,

idBi ⊗ α : Bi ⊗R A −→ Bi ⊗R A′

is a monic Z-module homomorphism. Now we use that − ⊗R A and − ⊗R A′

commute with direct limits and that lim−→Iis exact since I is directed. We get a

commutative diagram

lim−→(Bi ⊗R A) lim−→(Bi ⊗R A′)

(lim−→Bi)⊗R A (lim−→Bi)⊗R A

−−−−→idBi ⊗ α

∼= ∼=idlim−→Bi

⊗ α

(Check that this commutes). Since lim−→ is exact (since I is directed), then−−−−−→idBi ⊗ α

is monic, and hence idlim−→Bi ⊗ α is monic.

Corollary 10.7. If R is an integral domain, then Q = Frac(R) is a flatR-module.

Proof. Let I := (R − 0)/ ∼, where r ∼ r′ for r, r′ ∈ R − 0 if and only ifthere exists u ∈ R× (i.e. u is a unit in R) with r = r′u.

Define [r] ≤ [r′] ⇐⇒ r|r′. Then (I,≤) is directed since ∀r, r′ ∈ R − 0,[r] ≤ [rr′] and [r′] ≤ [rr′]. Now for all r ∈ R − 0, define A[r] := 1

rR. ThenA[r]∼= R as R-modules, and hence A[r] is flat.

10. FLAT MODULES 37

For all r|r′ in R− 0, say r′ = ra, define

ϕ[r][r′] : A[r] −→ A[r′] = A[ra]

1

rx 7−→ 1

raax =

1

r′ax.

We get a direct system A[r], ϕ[r][r′].

Claim 10.1. Q ∼= lim−→A[r]

Proof of Claim 10.1For all r ∈ R− 0 define

α[r] : A[r] −→ Q

1

rx 7−→ x

r.

Let X be an R-module together with R-module homomorphisms f[r] : A[r] → X

such that ∀r, r′ ∈ R− 0 with r|r′, f[r′] ϕ[r][r′] = f[r].

Q X

A[r]

A[r′]

f

α[r] f[r]

ϕ[r]

[r′]

α[r′] f[r′]

Define f : Q→ X by a/r 7→ f[r](1ra) for all a, r ∈ R, r 6= 0.

Check:

• f is well-defined.• f is an R-module homomorphism• f is the unique R-module homomorphism making the diagram commute.

Examples 7. R = Z(1) Z is a flat and projective Z-module.(2) Q is a flat and injective Z module that is not projective.(3) Q/Z is an injective Z-module that is not flat. For example, look at the

inclusion morphism ι : Z → Q. Then

idQ/Z⊗ι : Q/Z⊗Z Z −→ Q/Z⊗Z Q(1

2+ Z

)⊗ 1 7−→

(1

2+ Z

)⊗ 1 = (1 + Z)⊗ 1

2= 0Q/Z⊗ZQ.

In fact Q/Z⊗Z Q = 0 since(rs

+ Z)⊗ p

q· ss

= (r + Z)⊗ p

qs= 0Q/Z⊗ZQ.


11. Character Modules

Definition 11.1. LetB = BR. Define (a leftR-module)B∗ = HomZ(ZBR,Q/Z),called the character module (or Pontryagin dual) of B.

Lemma 11.2. A sequence in Mod-R

Aα−→ B

β−→ C

is exact if and only if

C∗β∗−→ B∗

α∗−−→ A∗

is exact in R-Mod.

Proof. (⇒) Follows since HomZ(−,Q/Z) is exact.(⇐)Im(α) ⊆ Ker(β): Otherwise, ∃a ∈ A with β(α(a)) 6= 0. Using that Q/Z is a

cogenerator for Z-Mod, there exists f : C → Q/Z in Z-Mod, i.e. f ∈ C∗, withf(β(α(a))) 6= 0. Hence

(α∗ β∗)(f)(a) = (f β α)(a) 6= 0,

contradicting Im(β∗) = Ker(α∗).Ker(β) ⊆ Im(α): Otherwise, ∃b ∈ Ker(β) with b + Im(α) 6= 0 + Im(α) in

B/ Im(α). Since Q/Z is a cogenerator for Z-Mod, ∃g : B/ Im(α) → Q/Z inZ-Mod with g(b+ Im(α)) 6= 0.

Let π : B → B/ Im(α) be the natural projection. Then h := g π ∈ B∗ and

α∗(h) = h α = 0. Then h ∈ Ker(α∗) = Im(β∗), i.e. ∃h ∈ C∗ with h = β∗(h) =

h β. We have h(b) = g(b + Im(α)) 6= 0 but h(β(b)) = 0, since b ∈ Ker(β), acontradiction.

Theorem 11.3. Let B = BR. Then B is flat if and only if B∗ is injective.

Proof. (⇒) Follows since B = ZBR is R-flat and Q is an injective Z-module,and hence by Theorem 10.2, HomZ(B,Q/Z) is an injective left R-module.

(⇐) Let α : A→ A′ be an injective R-module homomorphism in R-Mod. Wewill show that idB ⊗ α is injective.

We have the following diagram:

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)∗ 0

α∗

∼= ∼=(idB ⊗R α)∗

(idB ⊗R α)∗

Since B∗ is an injective left R-module, the top row is exact. Moreover, the diagramcommutes since the vertical arrows are adjoint isomorphisms. So the bottom rowis exact, and by Lemma 11.2

0 −→ (B ⊗R A) (B ⊗R A′)(idB ⊗R α)

is exact.

11. CHARACTER MODULES 39

Theorem 11.4. Let B = BR. If, for each finitely generated left ideal I of Rwith inclusion ιI : I → R, idB ⊗ ιI : B ⊗R I → B ⊗R R is injective, then B is flat.

Proof. Step 1 : We first show that idB ⊗ ιJ is injective for every left ideal Jof R (whether or not J is finitely generated).

Let J be an arbitrary left ideal of R. Define

S = J ′ ⊆ J | J ′finitely generated left ideal of R.Now, S is partially ordered by inclusion, and S is directed, since if J, J ′′ ∈ S, thenJ ′ + J ′′ ∈ S.

Claim 11.1. J ∼= lim−→J′∈S J′, where ∀J ′ ∈ S, the morphism α′J : J ′ → J is

inclusion, and the direct system is J ′, ϕJ′J′′ where ϕJ′

J′′ : J ′ → J ′′ is inclusion∀J ′ ⊆ J ′′.

Proof of Claim 11.1 Given a diagram

J X

J ′

J ′′

f

αJ′ fJ′

ϕJ′J′′

αJ′′ fJ′′

define f : J → X by f(r) = fRr(r). Check: f is a well-defined R module homomor-phism and the unique R-module homomorphism making the diagram commute.

By assumption, ∀J ′ ∈ S,

idB ⊗ ιJ′ : B ⊗R J ′ → B ⊗R Ris injective. Now we use that S is a directed poset and hence lim−→J′∈S is exact to

see that−−−−→

idB ⊗ ιJ′ : lim−→J′∈S

(B ⊗R J ′) −→ lim−→J′∈S

(B ⊗R R)

is injective: Since B ⊗R − preserves lim−→, we have

lim−→J′∈S

(B ⊗R J ′) ∼= B ⊗R lim−→J′∈S

J ′ ∼= B ⊗R J.

So we get a diagram

lim−→J′∈S(B ⊗R J ′) lim−→J′∈S(B ⊗R R)

B ⊗R lim−→J′∈S

J ′

B ⊗R J B ⊗R R

∼=

−−−−→idB ⊗ ιJ′

∼=

∼=

idB ⊗ ιJ

Since the diagram commutes, it follows that for every left ideal J of R,idB ⊗ ιJ : B ⊗R J −→ B ⊗R R is injective.


Step 2 : To prove that BR is flat, we use Theorem 11.3 and prove instead thatB∗ is injective. By Baer’s criterion, we prove that if J ⊂ R is a left ideal and

ιJ : J → R is inclusion, then HomR(R,B∗)ι∗J−→ HomR(J,B∗) → 0 is exact (i.e.

∀f ∈ HomR(J,B∗), ∃f HomR(R,B∗) with f |J = f).Using adjointness, we get a 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 )∗

The bottom row is exact by Step 1 and by exactness of HomZ(−,Q/Z). Hence thetop row is exact, meaning B∗ is injective.

12. Schanuel’s Lemma

Theorem 12.1. Let C = R-Mod or Mod-R.

(a) (Schanuel’s Lemma) Given two short exact sequences in C

0→ K1 → P1f1−→ B → 0 and 0→ K2 → P2

f2−→ B → 0

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

(b) Dually: Given two short exact sequences in C

0→ Bg1−→ E1 → Q1 → 0 and 0→ B

g2−→ E2 → Q2 → 0

where E1, E2 are injective, we have E1 ⊕Q2∼= E2 ⊕Q1.

Proof of (a). (Proof for (b) is dual).Take the pullback of f1 and f2:

X P1

P2 B

π1

π2 f1

f2

where X = (a1, a2) ∈ P1 ⊕ P2 | f1(a1) = f2(a2) and πi : X → Pi, is given by(a1, a2) 7→ ai for i = 1, 2. We have (check this!):

• parallel arrows have isomorphic kernels• If fi is an epimorphism, then π3−i is an epimorphism for i = 1, 2.

Hence we get a short exact sequence for i = 1, 2:

0→ Ker(πi)︸︷︷︸∼=Ker(f3−i)=K3−i

→ Xπi−→ Pi → 0.

This splits since Pi is projective, and hence P1 ⊕K2∼= X ∼= P2 ⊕K1.

12. SCHANUEL’S LEMMA 41

Definition 12.2. 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 Theorem 12.1(a) (Schanuel’s Lemma), Ω(M) is unique up to “addingon or subtracting off” projective modules.

Let α : M1 → M2 be in C, let εi : Pi → Mi be an epimorphism in C

where Pi is projective for i = 1, 2. Since ε2 is epic and P1 is a projectivemodule, ∃α : P1 → P2 in C with ε2 α = α ε1. Define Ω(α) = α|Ω(M1).

0 Ω(M1) P1 M1 0

0 Ω(M2) P2 M2 0

Ω(α)

ε1

∃α α

ε1

Note. If x ∈ Ω(M1) then ε2(α(x)) = α(ε1(x)) = 0 and so Ω(α) maps

into Ω(M2).

Note. Ω(α) is unique up to adding morphisms factoring through aprojective module.

Why? If ˜α : P1 → P2 satisfies ε2 ˜α = α ε1, then

α− ˜α : P1 −→ Ω(M2)

since ε2 (α− ˜α) = 0. Hence

α|Ω(M1) − ˜α|Ω(M2)

factors through P1.

(b) Let ι : M → E be a monomorphism in C where E is an injective module.

Define Ω−1(M) := Coker(ι) = E/ Im(ι).

By Theorem 12.1(b), Ω−1(M) is unique up to “adding on or subtract-ing off” injective modules.

Let α : M1 →M2 be in C, let ιi : Mi → Ei be a monomorphism in C

where Ei is an injective module for i = 1, 2. Since ι1 is monic and E2 isan injective module, ∃α : E1 → E2 in C with α ι1 = ι2 α. Define

Ω−1(α) : Ω−1(M1) −→ Ω−1(M2)

x+ ι1(M1) 7−→ α(x) + ι1(M2).

This is well defined since ι1 α = α ι1.

0 M1 E1 Ω−1(M1) 0

0 M2 E2 Ω−1(M2) 0

α

ι1 ε1

∃α Ω−1(α)

ι2 ε2

Note. Ω−1(α) is unique up to adding morphisms factoring throughan injective module.

Remark. Our goal is to make Ω and Ω−1 into functors. We need “new” cate-gories R-Mod and R-Mod.


13. Stable Module Categories

Definition 13.1.

(a) The stable module category R-Mod has objects Ob(R-Mod) = Ob(R-Mod)and morphisms are as follows: For M,N ∈ Ob(R-Mod), define

P HomR(M,N) = α ∈ HomR(M,N) | α factors through a proj. module.

Define the morphisms from M to N in R-Mod to be

HomR-Mod(M,N) := HomR(M,N) := HomR(M,N)/P HomR(M,N).

Then Ω defines a functor

Ω : R-Mod −→ R-Mod .

(b) The category R-Mod has objects Ob(R-Mod) = Ob(R-Mod) and

HomR-Mod(M,N) := HomR(M,N) := HomR(M,N)/I HomR(M,N)

where

I HomR(M,N) = α ∈ HomR(M,N) | α factors through an inj. module..

Then Ω−1 defines a functor

Ω−1 : R-Mod −→ R-Mod .

Note. “α factors through a module X” means there exists a commutativediagram

M N

X

α

Definition 13.2. A ring R is called self-injective if RR is an injective leftR-module.

Theorem 13.3. The following are equivalent:

(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.

Definition 13.4. A ring R satisfying (a)-(d) of Theorem 13.3 is called quasi-Frobenius.

Suppose R is quasi-Frobenius. Then R-Mod = R-Mod and

Ω,Ω−1 : R-Mod −→ R-Mod

are quasi-inverse functors.A projective R-module P is injective since R is quasi-Frobenius, and so given

a sequence 0→ M → Pε−→M → 0, we have M ∼= Ω−1(Ω(M)) in R-Mod.

Similarly, an injective R-module E is projective since R is quasi-Frobenius, and

so given a sequence 0→Mι−→ E → Ω−1(M)→ 0, we have M ∼= Ω(Ω−1(M)).

Recall Note 2.2: For categories C and D, HomDC(F,G) may not be a set.MacLane says that most people base their category theory either on Godel-Bernays(GB) or Zermelo-Fraenkel (ZF) with a given universe.

14. MORITA THEORY 43

14. Morita Theory

Question: When are two module categories R-Mod and S-Mod equivalent?

Definition 14.1. A left R-module P is called a progenerator for R-Mod if

(a) P is projective.(b) P is finitely generated.(c) P is a generator for R-Mod.

Example 14.2. Rn is a progenerator for R-Mod for any n ∈ Z+.

Theorem 14.3. Let F : R-Mod → S-Mod be an additive covariant equiva-lence with additive quasi-inverse F ′ : S-Mod → R-Mod. Then P := F ′(S) is aprogenerator for R-Mod and EndR(P )

op ∼= S as rings.

Proof.

(a) P is projective:Applying F to a diagram in R-Mod

P

B C 0 (exact)

α

β

we obtain a diagram in S-Mod

F (P ) S

F (B) F (C) 0 (exact)

F (α)

ρ

∼=

∃γ

F (β)

where we find γ so that F (β) (γ ρ) = F (α) since S is projective. Defineγ : P → B to be the unique (since F is fully faithful) morphism such thatF (γ) = γ ρ. Then β γ = α.

(b) P is a generator:Let M be a left R-module. Then F (M) ∈ Ob(S-Mod). Since S is a

generator for S-Mod, there exists an index set Λ and an epimorphism

τ :∐λ∈Λ

S −→ F (M).

So we get an epimorphism

F ′(τ) : F ′( ∐λ∈Λ

S)−→ F ′(F (M)) ∼= M.

Now since (F ′, F ) is an adjoint pair, we have

F ′( ∐λ∈Λ

S)∼=∐λ∈Λ

F ′(S) =∐λ∈Λ

P

(c) P is finitely generated:


Use the same arguments as above to show F (R) is a projective gener-ator for S-Mod. Since S is a finitely generated left S-module, there exitsn ∈ Z+ and an epimorphism

κ : F (R)n −→ S.

Applying F ′, we get an epimorphism

F ′(κ) : F ′(F (R)n) −→ F ′(S) = P.

Since (F ′, F ) is an adjoint pair, F ′(F (R)n) ∼= F ′(F (R))n ∼= Rn.

Define

ξ : EndR(P )op

−→ S

F ′(β) 7−→ β(1)

for all β ∈ EndS(S).Note: P = F ′(S); since F ′ is fully faithful we can write each α ∈ EndR(P ) as

α = F ′(β) for unique β ∈ EndS(S).Check: ξ is a bijection and additive. ξ is a ring homomorphism since for all

β, γ ∈ EndS(S)

F ′(β)op

F ′(γ) = F ′(γ) F ′(β) = F ′(γ β) = (γ β)(1) = γ(β(1)) = β(1)γ(1).

Definition 14.4.

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

σ : P ⊗S Q −→ R, an (R,R)− bimodule hom.,

τ : Q⊗R P −→ S, an (S, S)− bimodule hom.,

which satisfy: ∀x, y ∈ P and ∀u, v ∈ Q,

σ(x⊗ u)y = xτ(u⊗ y)

and vσ(x⊗ u) = τ(v ⊗ x)u

(b) Two rings R and S are said to be Morita equivalent if there is a Moritacontext (R,S, P,Q, σ, τ) as in (a) where σ and τ are surjective.

Notation. R ∼M S.

Theorem 14.5 (Morita I). Let R ∼M S with a Morita context (R,S, P,Q, σ, τ).Then

(1) σ and τ are isomorphisms.(2) Q⊗R− : R-Mod→ S-Mod is an additive equivalence with quasi-inverse

P ⊗S−. Similarly, −⊗RP : Mod-R→Mod-S is an additive equivalencewith quasi-inverse −⊗S Q.

(3) P is a progenerator for R-Mod and Mod-S, and Q is a progenerator forS-Mod and Mod-R.

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

Proof. (3) and (4) follow from (2) by Theorem 14.3.


(1) By assumption σ and τ are surjective since R ∼M S. We show σ isinjective, (τ is treated similarly).

Since σ is surjective, there exists z =∑ni=1 xi ⊗ ui ∈ P ⊗S Q with

σ(z) = 1R. Let

t =

m∑j=1

yj ⊗ vj ∈ Ker(σ) ⊆ P ⊗S Q.

Then

t = tσ(1R) =∑i,j

yj ⊗ (vjσ(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 = 0Rz = 0P⊗SQ.

(2) Let F = Q ⊗R −, F ′ = P ⊗S −. We will show F ′ F ∼= IdR-Mod.(F F ′ ∼= IdS-Mod is similar, and so is showing that −⊗R P and −⊗S Qare quasi inverses of each other).

Define ρ : F ′F = P⊗SQ⊗R− −→ IdR-Mod by: ∀M ∈ Ob(R-Mod),

ρM : P ⊗S Q⊗RM −→M,

ρM = αM(σ ⊗ idM ),

where αM : R⊗RM −→M

r ⊗M 7−→ rm.

We know α = αMM∈Ob(R-Mod) is a natural isomorphism and by (1), σis an isomorphism. It remains to show that ρ is natural with respect toR-module homomorphisms f : M → N .

P ⊗S Q⊗RM M

P ⊗S Q⊗R N N

ρM

idP⊗SQ⊗f f

ρN

We have

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

and

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

Proposition 14.6. Let R be a ring, let P be a progenerator for R-Mod, andlet S = EndR(P )op. Then R ∼M S.

Proof. S acts on the right of P by: ∀s ∈ S, ∀x ∈ P , x.s := s(x). Check: Pis an (R,S)-bimodule.

Let Q := HomR(RPS ,RRR), where ∀r ∈ R, ∀s ∈ S, ∀u ∈ Q, ∀x ∈ P ,

(sur)(x) = u(xs)r.


Check: Q is an (S,R)-bimodule.Define

σ : P ×Q −→ R

(x, u) 7−→ u(x).

Check: σ is S-balanced. So there exists a unique Z-module homomorphism σ :P⊗SQ→ R with σ(x⊗u) = u(x). Check: σ is an (R,R)-bimodule homomorphism.

We now show that σ is surjective. Since P is a generator for R-Mod and

RR is a finitely generated left R-module, there exists n ∈ Z+ and an epimorphismπ : Pn → R. For all 1 ≤ j ≤ n, let ιj : P → Pn be the jth injection. Let r ∈ R.Then there exists x := (x1, . . . , xn) ∈ Pn with π(x) = r. Then

σ

(n∑i=1

xi ⊗ π ιi

)=

n∑i=1

π(ιi(xi)) = π

(n∑i=1

ιi(xi)

)= π(x) = r.

Define τ : Q× P → S by: ∀u ∈ Q, ∀x, y ∈ P ,

τ(u, x)(y) = u(y)x.

Check: τ is R-balanced. So there exists a unique Z-module homomorphism τ :Q⊗R P → S with τ(u⊗ x)(y) = u(y)x. Check: τ is an (S, S)-bimodule.

We now show that τ is surjective. Since P is finitely generated an projective,there exists m ∈ Z+ such that P |F = Rm. So there exists R-module homomor-phisms π : F P : ι where π is surjective and π ι = idP . For all 1 ≤ j ≤ m, letιJ : R → F = Rm be the jth injection homomorphism, and let pj : F → R be thejth projection homomorphism. Let s ∈ S. Then for all y ∈ P ,

τ

m∑j=1

(pj ι)⊗ (s π ιj)(1R)

(y) =

m∑j=1

pj(ι(y))s(π(ιj(1R)))

= s

π m∑j=1

ιj(pj(ι(y)))

= s(π(ι(y))) = s(y).

Let x, y ∈ P and u, v ∈ Q. Then

σ(x⊗ u)y = u(x)y,

xτ(u⊗ y) = τ(u⊗ y)(x) = u(x)y,

and (vσ(x⊗ u))(y) = v(y)σ(x⊗ u)

= v(y)u(x),

(τ(v ⊗ x)u)(y) = u(yτ(v ⊗ x))

= u(τ(v ⊗ x)(y))

= u(v(y)x)

= v(y)u(x).

Corollary 14.7. Let R,S be rings. The following are equivalent:

(1) There exists an additive covariant equivalence R-Mod→ S-Mod.(2) R ∼M S.


(3) There exists a progenerator P for R-Mod with S ∼= EndR(P )op.

Proof. (1) =⇒ (3) Follows from Theorem 14.3(2) =⇒ (1) Follows from Theorem 14.5(3) =⇒ (2) Follows from Proposition 14.6

Example 14.8. Let R be a ring, n ∈ Z+, and S = Matn(R). Then R ∼M S.

Proof. We will show that there exists a progenerator P for R-Mod withEndR(P )op ∼= S. Let P = Rn with standard R-basis e1, . . . , en. Define

ϕ : Matn(R) = S −→ EndR(P )op

(aij) = A 7−→ ϕA,

where ϕA(ei) =

n∑j=1

aijej

Check: ϕ is additive and bijective.

ϕAB(ei) =

n∑j=1

n∑k=1

aikbkjej =

n∑k=1

aik

n∑j=1

bkjej =

n∑k=1

aikϕB(ek)

= ϕB

(n∑k=1

aikek

)= ϕB(ϕA(ei)).

Hence ϕAB = ϕB ϕA.

Theorem 14.9 (Morita II). Let F : R-Mod → S-Mod and F ′ : S-Mod →R-Mod be additive covariant functors that are quasi-inverses of each other. LetP = F ′(S) and Q = F (R). Then

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

Proof sketch. Let η : F ′ F → IdR-Mod be a natural isomorphism. Defineρ : F → HomR(P,−) by: ∀M ∈ Ob(R-Mod),

ρM : F (M) −→ HomR(P,M)

X 7−→ ηM F ′(ϕX),

where ϕX : S → F (M)

1S 7→ x.

Then ρM is an isomorphism since

F (M) HomS(S, F (M)) HomR(F ′(S), F ′(F (M))) HomR(P,M).∼= F ′

∼=(ηM )∗∼=

Check ρM is natural. So F ∼= HomR(P,−). Similarly, one shows F ′ ∼= HomS(Q,−).Since (F, F ′) and (F ′, F ) are adjoint pairs, then (F,HomS(Q,−)) and (F ′,HomR(P,−))

are adjoint pairs. Since (Q⊗R−,HomS(Q,−)) and (P ⊗S −,HomR(P,−)) are alsoadjoint pairs, it follows that

F ∼= Q⊗R − and F ′ ∼= P ⊗S −.

CHAPTER 3

More on Categories

15. Godel-Bernays System

The following axioms are taken from P.J. Cohen, Set theory and the continuumhypothesis. W. A. Benjamin, Inc., New York, 1966.

Upper case letters denote class variables; lower case letters denote set variables.Axioms for Sets:

(1) Y ∈ X → Y is a set.(2) X = Y ↔ ∀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 Information:

(9) ∃X ∀a (a ∈ X ↔ ∃b ∃c (a = 〈b, c〉 ∧ b ∈ c))(10) ∀X ∀Y ∃Z ∀u (u ∈ Z ↔ u ∈ X ∧ u ∈ Y )(11) ∀X ∃Y ∀u (u ∈ Y ↔ ¬u ∈ X)(12) ∀X ∃Y ∀u (u ∈ Y ↔ ∃v (〈v, u〉 ∈ X))(13) ∀X ∃Y ∀u (u ∈ Y ↔ ∃r ∃s (u = 〈r, s〉 ∧ s ∈ X))(14) ∀X ∃Y ∀a (a ∈ Y ↔ ∃b, c (〈b, c〉 = a ∧ 〈c, b〉 ∈ X))(15) ∀X ∃Y ∀u (u ∈ Y ↔ ∃a, b, c(〈a, b, c〉 = u ∧ 〈c, a, b〉 ∈ X))(16) ∀X ∃Y ∀u (u ∈ Y ↔ ∃a, b, c(〈a, b, c〉 = u ∧ 〈a, c, b〉 ∈ X))(17) Axiom of Choice:

∃X ∀a (a 6= ∅→ ∃!u (u ∈ a ∧ 〈a, u〉 ∈ X))(18) Axiom of Regularity:

∀X(X 6= ∅→ ∃u (u ∈ X ∧ ∀y (y ∈ X → ¬y ∈ u)))

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

For the above axioms, we have:Axioms for Sets:

(1) A class has only sets as members.(2) is called “extensionality of classes”(3) Empty set axiom: x = ∅(4) Pair set axiom: z = x, y(5) Union set axiom: y =

⋃x

49

50 3. MORE ON CATEGORIES

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

(7) Power set axiom: y = power set of x(8) ∀X ((∀u ∃!v 〈u, v〉 ∈ X)→ ∀u ∃v (v is the range of the function defined by X on u)),

where 〈u, v〉 = u, u, vAxioms for Class Information:

(9) means: X is the collection of all 〈b, c〉 with b ∈ c.(10) intersection of classes(11) complement of a class(12) means: Y picks out second entry in all ordered pairs in X(13) means: Y picks out all ordered pairs with second entry in X(14) means Y picks out the reversed ordered pairs of all ordered pairs in X.(15) about ordered triples(16) about ordered triples(17) means: there is a class which picks out one member from each non-empty

set(18) means: each class X has a member u such that no element of X is a

member of u.

Definition 15.1.

(a) A proper class is a class that is not a set. (In other words, the word “class”means either a set or a proper class.)

(b) 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.

(c) A category C is said to be large if Ob(C) is a class and ∀X,Y ∈ Ob(C),HomC(X,Y ) is a class.

So far, our categories were always assumed to be large with small Hom sets.

16. Functor Categories

Let C and D be categories and consider the functor category DC. If both C,Dare small, then DC is small. If C is small and D is large, then DC is large. Ifmoreover, D has small Hom sets, then DC has small Hom sets.

Example 16.1. Suppose D is a category where Ob(D) is a proper class. DefineC by Ob(C) = ∗ with HomC(∗, ∗) = id∗. Then, any functor F : C → D

is uniquely determined by F (∗). Hence Ob(DC) = Ob(D) and HomDC(F,G) =HomD(F (∗), G(∗)).

Example 16.2. Let R be the Russell class

R = x | (x is a set) ∧ (¬x ∈ x).Recall that R is a proper class, since if R were a set, R ∈ R ⇐⇒ ¬R ∈ R.

Define categories C and D: Ob(C) = R, only morphisms are identity morphisms;Ob(D) = 0, 1, only morphisms are identity morphisms.

Any functor F : C → D is uniquely determined by UF = x ∈ R | F (x) = 0.Then Ob(DC) = U | (x ∈ U)→ (x ∈ R), and so R ∈ Ob(DC), and hence Ob(DC)is not a class by the GB axiom (1). So DC is “larger than large”.

17. EQUIVALENCES OF CATEGORIES 51

17. Equivalences of Categories

Definition 17.1. Let F : C→ D be a covariant functor.

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

ΦXY : HomC(X,Y ) −→ HomD(FX,FY )

α 7−→ Fα

has a two-sided inverse.(b) F is dense if there exists a map T : Ob(D) → Ob(C) such that ∀X ′ ∈

Ob(D), we have an isomorphism ρX′ : X ′ → F (TX ′) in D.

Theorem 17.2. A covariant functor F : C→ D is an equivalence of categoriesif and only if F is fully faithful and dense.

Proof. (⇒) There exists a quasi-inverse functor F ′ : D→ C and there existsnatural isomorphisms

τ : IdC −→ F ′ F and

σ : IdD −→ F F ′.

Let X,Y ∈ Ob(C) and α ∈ HomC(X,Y ). We get a commutative diagram

F ′(FX) X

F ′(FY ) Y

F ′(Fα)

τX

α

τY

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

Fα = Fβ, then α = β.Similarly, using F ′ instead of F , we see that ∀X ′, Y ′ ∈ Ob(D), ∀α′β′ ∈

HomD(X ′, Y ′), we have F ′(α′) = F ′(β′) implies α′ = β′.Define

ΨXY : HomD(FX,FY ) −→ HomC(X,Y )

α′ 7−→ τ−1Y (F ′α′) τX .

Then

ΨXY (ΦXY (α)) = τ−1Y (F ′(Fα)) τX = α

and

ΦXY (ΨXY (α′)) = F (τ−1Y (F ′α′) τX).

Define α := τ−1Y (F ′α′) τX . We will show that F (α) = α′. Applying F ′ F to

α, we get

F ′(F (α)) = τY α τ−1X = F ′(α′).

Hence F (α) = α′. Therefore, F is fully faithful. To see that F is dense, define

T : Ob(D) −→ Ob(C)

X ′ 7−→ F ′(X ′)

and define ρX′ : X ′ → F (F ′(X ′)) to be σX′ .


(⇐) Suppose F is fully faithful and dense. Define a functor F ′ : D → C asfollows: ∀X ′ ∈ Ob(D), define F ′(X ′) := T (X ′) and ∀α′ ∈ HomD(X ′, Y ′) considerthe diagram

F (F ′X ′) X ′

F (F ′Y ′) Y ′

∼=ρX′

α′

∼=ρY ′

Since F is fully faithful, we can define F ′(α′) : F ′(X ′) → F ′(Y ′) to be the uniquemorphism in C such that

F (F ′(α′)) = ρY ′ α′ ρ−1X′ .

Check: F ′ is a functor ( use that F is fully faithful).Now, ρ = ρX′X′∈Ob(D) gives a natural isomorphism ρ : IdD → F F ′,

(naturality follows from the definition of F ′ on morphisms).Define τ : IdC → F ′ F by: ∀X ∈ Ob(C), τX : X → F ′(FX) is the unique

morphism in C such that F (τX) = ρFX . Check: τ is a natural isomorphism, (usesthat ρ is a natural isomorphism and F is fully faithful).

18. Yoneda’s Lemma

Let C be a large category with small Hom sets.

Definition 18.1. Let F : C→ Sets be a functor.

(a) If F is covariant, we say F is representable if ∃X ∈ Ob(C) with F ∼=HomC(X,−).

(b) If F is contravariant, we say F is representable if ∃Y ∈ Ob(C) with F ∼=HomC(−, Y ).

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

Example 18.2. Let C = R-Mod, let A ∈ Ob(Z-Mod) be fixed, and letF : R-Mod→ Ab be F = HomZ(−, A).

Claim 18.1. F is representable.

Proof of Claim 18.1. We need to show ∃Y ∈ Ob(R-Mod) such that F ∼=HomR(−, Y ). Define Y := HomZ(ZRR, ZA), a leftR-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.

18. YONEDA’S LEMMA 53

Lemma 18.3 (Yoneda’s Lemma).

(a) Let C be a large category with (small) Hom sets, let F : C → Sets be acovariant functor, let X ∈ Ob(C). Then

ΦX : HomSetsC(HomC(X,−), F ) −→ F (X)

τ 7−→ τX(idX)

has a two-sided inverse. In particular, HomSetsC(HomC(X,−), F ) is a set.(b) Let F : C→ Sets be a contravariant functor, let Y ∈ Ob(C). Then

ΨY : Hom ˇSetsC(HomC(−, Y ), F ) −→ F (Y )

σ 7−→ σY (idY )

has a two-sided inverse. In particular, Hom ˇSetsC(HomC(−, Y ), F ) is a set.

Proof. We prove (a). Define

ΦX : F (X) −→ HomSetsC(HomC(X,−), F )

s 7−→ τs

where, ∀Y ∈ Ob(C)

(τs)Y HomC(X,Y ) −→ F (Y )

f 7−→ F (f)(s).

Check: τs is a natural transformation. We have

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

and for all Y ∈ Ob(C), ∀f ∈ HomC(X,Y ), using the commutative diagram

HomC(X,X) F (X)

HomC(X,Y ) F (Y )

τX

f∗ F (f)

τY

we have

ΦX(ΦX(τ))Y (f) = ΦX(τX(idX))Y (f)

= F (f)(τX(idX))

= τY (f∗(idX))

= τY (f).

hence ΦX(ΦX(τ)) = τ .


Corollary 18.4 (Yoneda’s Embedding). Let C be an in Yoneda’s Lemma 18.3.

(a) Define H : C→ SetsC by: ∀X,Y ∈ Ob(C) and ∀α ∈ HomC(X,Y ),

H(X) := HomC(X,−) and

H(α) := HomC(Y,−) −→ HomC(X,−)

defined by

H(α)Z : HomC(Y, Z) −→ HomC(X,Z)

f 7−→ α∗(f),

i.e. H(α)Z = α∗ ∀Z ∈ Ob(C). Then H is a contravariant “functor” thatis fully faithful.

(b) Define H : C→ ˇSetsC by: ∀Y,Z ∈ Ob(C) and ∀β ∈ HomC(Y,Z),

H(Y ) := HomC(−, Y ) and

H(β) := HomC(−, Y ) −→ HomC(−, Z)

defined by

H(β)X : HomC(X,Y ) −→ HomC(X,Z)

g 7−→ β∗(g),

i.e. H(β)X = β∗ ∀X ∈ Ob(C). Then H is a covariant “functor” that isfully faithful.

Proof. We prove (a). Check: H is contravariant. For showing H is fullyfaithful, we use Yoneda’s Lemma 18.3(a) for F = H(X) = HomC(X,−) and Y asthe object in C. So we get a bijection

HomSetsC(H(Y ), H(X)) −→ F (Y ) = HomC(X,Y )

τ 7−→ τY (idY )

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

(σα)Z : HomC(Y, Z) −→ HomC(Y,X)

f 7−→ F (f)(α) = f∗(α) = α∗(f).

Hence (σα)Z = α∗, meaning σα = H(α). Therefore, H is fully faithful.

CHAPTER 4

Classical Homological Algebra – Tor and Ext

19. Basic Definitions

Definition 19.1. Let C be a module (or, more generally, an abelian category).

(a) Chain complex: C· = Cn, ∂nn∈Z,

· · · → Cn+1∂n+1−−−→ Cn

∂n−→ Cn−1 → · · ·∂n is the differential and for all n, ∂n ∂n+1 = 0. x ∈ Cn is in degree n.

(b) Cochain complex: C· = Cn, δnn∈Z,

· · · → Cn−1 δn−1

−−−→ Cnδn−→ Cn+1 → · · ·

δn is the differential and for all n, δn δn−1 = 0. x ∈ Cn is in degree n.

(c) C· (resp. C·) is bounded below if Cn = 0 (resp. Cn = 0) ∀n << 0.

C· (resp C·) is bounded above if Cn = 0 (resp. Cn = 0) ∀n >> 0.

A complex is bounded if it is bounded above and below.(d) Homology: C· = Cn, ∂nn∈Z,

Zn(C·) := Ker(∂n), n-cycles

Bn(C·) := Im(∂n+1), n-boundaries

Hn(C·) := Zn(C·)/Bn(C·), nth homology group

(e) Cohomology: C· = Cn, δnn∈Z,

Zn(C·) := Ker(δn), n-cocycles

Bn(C·) := Im(δn−1), n-coboundaries

Hn(C·) := Zn(C·)/Bn(C·), nth cohomology group

(f) Chain map: f : C· → D· consists of fn : Cn → Dn in C for all n ∈ Zwith ∂n,D fn = fn−1 ∂n,C .

Cochain map: f : C· → D· consists of fn : Cn → Dn in C for all n ∈ Zwith δnD fn = fn+1 δnC .

Note 19.2. A chain map f : C· → D· induces homomorphisms of abelian

groups on homology:

Hn(f) : Hn(C·) −→ Hn(D·)[x] 7−→ [fn(x)]

Similarly for cochain maps and cohomology.

55

56 4. CLASSICAL HOMOLOGICAL ALGEBRA – Tor AND Ext

Let C be a module category (or, more generally, an abelian category).

Remark 19.3.

(1) Let C· = Cn, ∂n be a chain complex over C. We can define a cochain

complex C· = Cn, δn by Cn := C−n, δn := ∂−n

(2) Homology groups (resp. cohomology groups) of chain (resp. cochain)complexes are objects in C

Definition 19.4. Let f, f : C· → D· be chain maps. We say f, f ′ are chain

homotopic, written f ∼ f ′ if for all n ∈ Z, there exists hn : Cn → Dn+1 such thatfn − f ′n = ∂n+1,D hn + hn−1 ∂n,C for all n ∈ Z.

Cn+1 Cn Cn−1

Dn+1 Dn Dn−1

fn f ′nhn

∂n,C

hn−1

∂n+1,D

We say C· and D· are chain homotopy equivalent if there exists chain mapsf :

C· → D· and g : D· → C· with g f ∼ idC and f g ∼ idD.

Notation. C· ∼ D·Similarly, we can define cochain homotopic cochain maps and cochain homotopy

equivalent cochain complexes.

Theorem 19.5. Let f, f ′ : C· → D· be chain homotopic chain maps. Then

for all n ∈ Z, Hn(f) = Hn(f ′) : Hn(C·) → Hn(D·). In particular, if C· ∼ D·then Hn(C·) ∼= Hn(D·). Similarly for cochain homotopic cochain maps.

Proof. Let f, f ′ : C· → D· be cochain maps with f ∼ f ′. Then for all n ∈ Z,there exists kn : Cn → Dn such that

fn − f ′n = δn−1D kn + kn+1 δnC .

For all x ∈ Zn(C·), let [x] = x+Bn(C·) ∈ Hn(C·). Then

(Hn(f)−Hn(f ′))([x]) = [fn(x)− f ′n(x)]

= [δn−1D (kn(x)) + kn+1(δnC(x))]

= [0] ∈ Hn(D·)

20. Snake Lemma & Long Exact Homology/Cohomology Sequence

Theorem 20.1 (Snake Lemma). Consider the following diagram in C withexact rows and commuting squares

M ′ M M ′′ 0

0 N ′ N N ′′

f1

α′

g1

α α′′

f2 g2

20. SNAKE LEMMA & LONG EXACT HOMOLOGY/COHOMOLOGY SEQUENCE 57

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

∂ : Ker(α′′) −→ Coker(α′)

m′′ 7−→ f−12 (α(m)) + Im(α′) ∀m ∈ g−1

1 (m′′).

where f−12 : Im(f2)→ N ′ is the inverse morphism of f2 : N ′ → Im(f2). Moreover,

we get an exact sequence in C

0 0 0

Ker(α′) Ker(α) Ker(α′′)

M ′ M M ′′ 0

0 N ′ N N ′′

Coker(α′) Coker(α) Coker(α′′)

0 0 0

f1 g1

∂f1

α′

g1

α α′′

f2 g2

f2 g2

where f1, g1 are the restrictions of f1, g1 and f2, g2 are induced from f2, g2.

Note 20.2. g2(α(m)) = α′′(g1(m)) = α′′(m′′) = 0, so α(m) ∈ Ker(g2) =Im(f2).

Proof sketch. ∂ is well-defined:Suppose m, m ∈ g−1

1 (m′′) where m′′ ∈ Ker(α′′). Then m − m ∈ Ker(g1) =Im(f1) and so there exists m′ ∈M ′ with m− m = f1(m′). Then

f−12 (α(m− m)) = f−1

2 (α(f1(m′))) = f−12 (f2(α′(m′))) = α′(m′) ∈ Im(α′.

Since the original diagram commutes and has exact rows, we get

Im(f1) = Ker(g1) and Im(f2) = Ker(g2).

We need to show Im(g1) = Ker(∂) and Im(∂) = Ker(f2). We will show the first:

∂(g1(m)) = f−12 (α(m)) + Im(α′) = 0 + Im(α′)

Let m′′ ∈ Ker(∂), i.e., α′′(m′′) = 0 and f−12 (α(m)) ∈ Im(α′) for all m ∈ g−1

1 (m′′).So there exists m′ ∈ M ′ with f−1

2 (α(m)) = α′(m′), and so α(m) = f2(α′(m′)) =α(f1(m′)). So m− f1(m′) ∈ Ker(α) and

g1(m− f1(m′)) = g1(m− f1(m′)) = g1(m) = m′′,

and hence m′′ ∈ Im(g1).


Theorem 20.3 (Long Exact Homology Sequence). Let 0 → C ′· f−→ C· g−→C ′′· → 0 be a short exact sequence of chain complexes. Then we get an exact

sequence in C:

· · · → Hn(C ′·) Hn(C·) Hn(C ′′·)

Hn−1(C ′·) Hn−1(C·) Hn−1(C ′′·)→ · · ·

Hn(f) Hn(g)

∂n

Hn−1(f) Hn−1(g)

Proof. Use the Snake Lemma (20.1): For all n ∈ Z, we have a well-definedmorphism in C

ϕn : Cn/ Im(∂n+1) −→ Ker(∂n−1)

x+ Im(∂n+1) 7−→ ∂n(x).

We haveKer(ϕn) = Ker(∂n)/ Im(∂n+1) = Hn(C·)

andCoker(ϕn) = Ker(∂n−1)/ Im(∂n) = Hn−1(C·).

So, we get a snake diagram:

0 0 0

Hn(C ′·) Hn(C·) Hn(C ′′·)

C ′n/ Im(∂′n+1) Cn/ Im(∂n+1) C ′′n/ Im(∂′′n+1) 0

0 Ker(∂′n−1) Ker(∂n−1) Ker(∂′′n−1)

Hn−1(C ′·) Hn−1(C·) Hn−1(C ′′·)

0 0 0

Hn(f) Hn(g)

∂n

fn

ϕ′n

gn

ϕn ϕ′′n

fn−1 gn−1

Hn−1(f) Hn−1(g)

21. LEFT DERIVED FUNCTORS 59

Theorem 20.4 (Long Exact Cohomology Sequence). Let 0 → C′· f−→ C· g−→

C′′· → 0 be a short exact sequence of cochain complexes. Then we get an exact

sequence in C:

· · · → Hn(C′·) Hn(C·) Hn(C

′′·)

Hn+1(C′·) Hn+1(C·) Hn+1(C

′′·)→ · · ·

Hn(f) Hn(g)

δn

Hn+1(f) Hn+1(g)

21. Left Derived Functors

Let C be a module category.

Theorem 21.1 (Comparison Theorem). Given a diagram in C where the rowsare chain complexes

X· : · · · X2 X1 X0 M 0

X ′· : · · · X ′2 X ′1 X ′0 M ′ 0

∂2 ∂1 ε

α

∂′2 ∂′1 ε′

If the Xn are projective for n ≥ 0 and the bottom row is exact, then there exists achain map f : X· → X ′· with f−1 = α. If f ′ : X· → X ′· is another such chain

map with f ′−1 = α, then f ∼ f ′.

Proof.Let X−1 = M , ∂0 = ε, Xi = 0 ∀i ≤ −2, ∂i = 0 ∀i ≤ −1.Let X ′−1 = M ′, ∂′0 = ε′, Xi = 0 ∀i ≤ −2, ∂′i = 0 ∀i ≤ −1.Define fiXi → X ′i to be fi = 0 ∀i ≤ −2 and f−1 = α.Suppose by induction, we have constructed fk : Xk → X ′k ∀k ≤ n such that

∂′k fk = fk−1 ∂k.

· · ·Xn+1 Xn Xn−1 · · ·

· · ·X ′n+1 X ′n X ′n−1 · · ·

fn+1

∂n+1 ∂n

fn fn−1

∂′n+1 ∂′n

We have ∂′n fn ∂n+1 = fn−1 ∂n ∂n+1 = 0, and since Ker(∂′n) = Im(∂′n+1), weget a diagram

Xn+1

X ′n+1 Im(∂′n+1) 0

fn∂n+1

∂′n+1

Since Xn+1 is projective, there exists fn+1 : Xn+1 → X ′n+1 in C with ∂′n+1 fn+1 =fn ∂n+1.

Now suppose we have another chain map f ′ : X· → X ′· with f ′−1 = α. Define

hi : Xi → X ′i as follows: hi = 0 ∀i ≤ −2 and h−1 = 0. This works since

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


Assume by induction we have found hk : Xk → X ′k+1 ∀k ≤ n such that

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+2 f ′n+2

∂n+1

hn+1

fn+1 f ′n+1

∂n

hn

fn f ′n

hn−1

fn−1 f ′n−1

∂′n+2 ∂′n+1 ∂′n

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

∂′n+1 g = ∂′n+1fn+1 − ∂′n+1f′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

= 0.

Since Ker(∂′n+1) = Im(∂′n+2), we get a diagram

Xn+1

X ′n+2 Im(∂′n+2) 0

fn+1−f ′n+1−hn∂n+1

∂′n+2

Since Xn+1 is projective, there exists hn+1 : Xn+1 → X ′n+1 in C with

∂′n+1 hn+1 = fn+1 − f ′n+1 − hn ∂n+1.

Definition 21.2. Let T : C → Ab be an additive covariant functor. For allM ∈ Ob(C), choose a projective resolution

· · · → Pn∂n−→ Pn−1 → · · · → P1

∂1−→ P0ε−→M → 0.

Let (PM )· be the truncated chain complex

(PM )· : · · · → Pn∂n−→ Pn−1 → · · · → P1

∂1−→ P0∂0−→ 0.

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

(LnT )M := Hn(T (PM )·).For all α : M →M ′ in C, let f : (PM )· → (PM ′)· be a chain map associated to α

by the Comparison Theorem, and define (LnT )α := Hn(Tf).

Theorem 21.3. Let T : C → Ab be as in the definition. For all n ≥ 0,LnT : C→ Ab is an additive covariant functor, called the nth left derived functorof T .

Proof Sketch. By the Comparison Theorem (21.1), it follows that (LnT )αis well-defined since if f ′ : (PM )· → (PM ′)· is another chain map associated to

α then f ∼ f ′. So Tf ∼ Tf ′ and hence Hn(Tf) = Hn(Tf ′). Check: LnT is afunctor. LnT is additive since T is additive.

22. TOR 61

Theorem 21.4. Let T and LnT be as above for n ≥ 0. Suppose LnT : C→ Abis another nth left derived functor of T obtained by choosing different projectiveresolutions for all M ∈ Ob(C). Then LnT and LnT are naturally isomorphic for

all n ≥ 0. In particular, (LnT )M ∼= (LnT )M for all M ∈ Ob(C) and for all n ≥ 0.

Proof. For all M ∈ Ob(C), let (PM )· (resp. (PM )·) be a truncated com-

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

(LnT )M). By the Comparison Theorem (21.1), there exists chain maps g : (PM )· →(PM )· and h : (PM )· → (PM )· associated to idM .

Define σM : (LnT )M → (LnT )M by σM = Hn(Tg) and τM : (TnT )M →(LnT )M by τM = Hn(Th).

Note: By the Comparison Theorem (21.1), hg ∼ id(PM )· and gh ∼ id(PM )· .Hence

τM σM = Hn(T (h g)) = Hn(T (id(PM )·)) = Hn((idT (PM )·)) = id(LnT )M ,

and σM τM = id(LnT )M .

We now show that σ is natural. Let α : M →M ′ in C. Let f : (PM )· → (P ′M )·(resp. f : (PM )· → (P ′M )·) be a chain map associated to α by the Comparison

Theorem (21.1).

(LnT )M (LnT )M

(LnT )M ′ (LnT )M ′

σM

Hn(Tf) Hn(T f)

σM′

To show this diagram commutes, we need

Hn(T (f g)) = Hn(T (g′ f)),

and hence it suffices to show f g ∼ g′ f . Now f g is associated to α idM = αand g′ f is associated to idM ′ α = α. By the Comparison Theorem (21.1), we

get f g ∼ g′ f .

22. Tor

Definition 22.1. Let M ′ = M ′R. Define

TorRn (M ′,−) = Ln(M ′ ⊗R −) : R-Mod −→ Ab .

Let M = RM . Choose a projective resolution

· · · → P2∂2−→ P1

∂1−→ P0ε−→M → 0.

Applying M ′ ⊗R − to the truncated chain complex (PM )·, we get

(M ′ ⊗R (PM )·) : · · · idM′ ⊗∂2−−−−−−→M ′ ⊗R P1idM′ ⊗∂1−−−−−−→M ′ ⊗R P0 → 0.

SoTorRn (M ′,M) = Ker(idM ′ ⊗∂n)

/Im(idM ′ ⊗∂n+1)


Note 22.2.

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

TorR0 (M ′,M) = M ′ ⊗R P0

/Im(idM ′ ⊗∂1)

= M ′ ⊗R P0

/Ker(idM ′ ⊗ε)

∼= Im(idM ′ ⊗ε)= M ′ ⊗RM.

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

resolution of M , and so

TorRn (M ′,M) = 0 ∀n ≥ 1 and TorR0 (M ′,M) = M ′ ⊗RM.

23. Right Derived Functors

Theorem 23.1 (Dual Comparison Theorem). Given a diagram in C where therows are cochain complexes

X· : · · · M X0 X1 X2 0

X′· : · · · M ′ X ′0 X ′1 X ′2 0

ι δ0 δ1

α′

ι′ δ′0 δ′1

If the Xn are injective for n ≥ 0 and the bottom row is exact, then there exists a

cochain map f : X′· → X· with f−1 = α′. If f ′ : X

′· → X· is another suchcochain map with f ′−1 = α′, then f ∼ f ′.

23.0.1. Covariant Version.

Definition 23.2. Let T : C → Ab be an additive covariant functor. For allM ∈ Ob(C) choose an injective resolution

0→Mι−→ E0 δ0

−→ E1 → · · · → Enδn−→ En+1 → · · ·

Let E·M be the truncated cochain complex

E·M : 0→ E0 δ0

−→ E1 → · · · → Enδn−→ En+1 → · · ·

For all n ≥ 0, define RNT : C→ Ab as follows: For all M ∈ Ob(C) define

(RnT )M := Hn(TE·M ).

For all α : M → M ′ in C, let f : E·M → E·M ′ be a cochain map associated to α bythe Dual Comparison Theorem, and define (RnT )α := Hn(Tf).

Theorem 23.3. Let T : C → Ab be as above. Then for all n ≥ 0, RnT : C →Ab is an additive covariant functor, called the nth right derived functor of T . Upto natural isomorphism, RnT is independent of the choices of injective resolutionsfor all M ∈ Ob(C) and for all n ≥ 0.

Proof. Dual to LnT , using the Dual Comparison Theorem (23.1).

24. EXT 63

23.0.2. Contravariant Version.

Definition 23.4. Let T : C→ Ab be a contravariant additive functor. For allM ∈ Ob(C) choose a projective resolution with associated truncated chain complex(PM )·. For all n ≥ 0, define

(RnT )M := Hn(T (PM )·).So

T (PM )· : 0→ TP0T∂1−−→ TP1

T∂2−−→ TP2 → · · ·and

(RnT )M = Ker(T∂n+1)/

Im(T∂n).

For all α : M → M ′ in C let f : (PM )· → (PM ′) be an associated chain complex

by the Comparison Theorem. Define

(RnT )α := Hn(Tf).

Theorem 23.5. Let T : C → Ab be an additive contravariant functor. Thenfor all n ≥ 0, RnT : C → Ab is an additive contravariant functor, called the nthright derived functor of T . Up to natural isomorphism RnT does not depend on thechoices of projective resolutions for all M ∈ Ob(C).

24. Ext

Definition 24.1. Let C = R-Mod or Mod-R. Fix M ′ ∈ Ob(C). For alln ≥ 0, define

ExtnR(−,M ′) = Rn HomR(−,M ′).If M ∈ Ob(C), choose a projective resolution

· · · → P3∂2−→ P1

∂1−→ P0ε−→M → 0.

Applying HomR(−,M ′) to the truncated chain complex (PM )·, we get

HomR((Pm)·,M ′) : 0→ HomR(P0,M′)

∂∗1−→ HomR(P1,M′)

∂∗2−→ · · ·

SoExtnR(M,M ′) = Ker(∂∗n+1)

/Im(∂∗n).

Note 24.2.

(a) Since HomR(−,M ′) is left exact, we get

Ext0R(M,M ′) = Ker(∂∗1 ) = Im(ε∗) ∼= HomR(M,M ′).

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

resolution of M , and so

ExtnR(M,M ′) = 0 ∀n ≥ 1 and Ext0R(M,M ′) = HomR(M,M ′).


25. Examples of Tor and Ext in Z-Mod

Example 25.1. Let R = Z, M0 ∈ Ob(Z-Mod), and M = Z/mZ. Then aprojective resolution of M is

0→ Z m−→ Z π−→ Z/mZ→ 0

So we have a truncated complex

(PM )· : 0→ Zdeg. 1

m−→ Zdeg. 0

→ 0.

(1) TorZn(M0,M):

M0 ⊗Z (PM )· : 0→M0 ⊗Z Z︸︷︷︸deg. 1

idM0⊗m

−−−−−−→M0 ⊗Z Z︸︷︷︸deg. 0

→ 0

We have a commuting diagram

0 M0 ⊗Z Z M0 ⊗Z Z 0

0 M0 M0 0

∼=

idM0⊗m

∼=

m

So

TorZ0 (M0,M) ∼= M0/mM0

TorZ1 (M0,M) ∼= x ∈M0|mx = 0

TorZn = 0 ∀n ≥ 2.

(2) ExtnZ(M,M0) :

HomZ((PM )·,M0) : 0→ HomZ(Z,M0)m∗−−→ HomZ(Z,M0)→ 0

We have a commuting diagram

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

0 M0 M0 0

m∗

∼= ∼=

m

So

Ext0Z(M,M0) = x ∈M0|mx = 0

Ext1Z(M,M0) = M0/mM0

ExtnZ(M,M0) = 0 ∀n ≥ 2.

26. HORSESHOE LEMMA 65

26. Horseshoe Lemma


Lemma 26.1 (Horseshoe Lemma). Consider a diagram in C where the columnsare projective resolutions of M ′ and M ′′ with corresponding truncated chain com-plexes (PM ′)· and (PM ′′)· and the bottom row is 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 truncated complex (PM )· and there exists chain maps f, g associated to α, β,

respectively, such that we get a short exact sequence of chain complexes

0→ (PM ′)· f−→ (PM )· g−→ (PM ′′)· → 0.

Proof. It suffices to show we can start with the diagram in C where the leftand right columns and the bottom row are exact and P ′ and P ′′ are projective, andthen fill in the bold part such that the middle column and top and middle roware exact, P is projective, and all the squares commute.

0 0 0

0 K ′ K K ′′ 0

0 P ′ P P ′′ 0

0 L′ L L′′ 0

0 0 0

λ0|K′ µ0|K

λ0 µ0

λ µ

Define P = P ′ ⊕ P ′′ and λ0 : x 7→ (x, 0), µ0 : (x, y) 7→ y. Then the middle rowis split exact. Since µ is surjective, and P ′′ is projective, there exists τ : P ′′ → Lin C with µ τ = ∂′′. Define ∂ : P → L, (x, y) 7→ λ(∂′(x)) + τ(y). Check: ∂ is amorphism in C. Define K := Ker(∂). We have

∂(λ0(x)) = ∂(x, 0) = λ(∂′(x))


and µ(∂(x, y)) = µ(λ(∂′(x)) + τ(y)) = µ(τ(y)) = ∂′′(y) = ∂′′(µ0(x, y)). Check: toprow is exact and top squares commute.

Why is this sufficient?: Use induction on n. Start with

L′ = M ′, L = M, L′′ = M ′′,

λ = α, µ = β, P ′ = P ′0,

P ′′ = P ′′0 , K ′ = Ker(ε′), K ′′ = Ker(ε′′).

Then use

(PM ′)· : · · ·P ′n+2 P ′n+1 P ′n P ′n−1 · · ·

Ker(∂′n+1)

=

Im(∂′n+2)

Ker(∂′n)

=

Im(∂′n+1)

Ker(∂′n−1)

=

Im(∂′n)

∂′n+2 ∂′n+1 ∂′n

to get

0 0

Ker(∂′n) Ker(∂′′n)

P ′n P ′′n

0 Ker(∂′n−1) Ker(∂n−1) Ker(∂′′n−1) 0

0 0

∂′n ∂′′n

Lemma 26.2 (Dual Horseshoe Lemma). Consider a diagram in C where thecolumns are injective resolutions of M ′ and M ′′ with corresponding truncated cochain

complexes (EM ′)· and (EM ′′)· and the bottom row is exact

...(E′)1

...(E′′)1

(P ′)0 (P ′′)0

0 M ′ M M ′′ 0

0 0

(δ′)0 (δ′)0

ι′

α β

ι′′

Then there exists an an injective resolution of M

0→Mι−→ E0 δ0

−→ E1 → · · ·

27. LONG EXACT SEQUENCES FOR LEFT AND RIGHT DERIVED FUNCTORS 67

with truncated cochain complex (EM )· and there exists cochain maps f, g associatedto α, β, respectively, such that we get a short exact sequence of cochain complexes

0→ (EM ′)· f−→ (EM )· g−→ (EM ′′)· → 0.

27. Long Exact Sequences for Left and Right Derived Functors


Theorem 27.1. Let 0 → M ′α−→ M

β−→ M ′′ → 0 be a short exact sequence inC. Let T : C → Ab be an additive covariant functor. Then we have a long exactsequence in Ab

· · · −→ (LnT )M ′ (LnT )M (LnT )M ′′

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

(L0T )M ′ (L0T )M (L0T )M ′′ −→ 0

(LnT )α (LnT )β

∂n

(Ln−1T )α (Ln−1T )β

∂1

(L0T )α (L0T )β

In particular, L0T : C→ Ab is right exact.

Remark 27.2. If C = R-Mod and M0 = (M0)R, we get a long exact Torsequence in Ab:

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

′′)

TorRn−1(M0,M′) TorRn−1(M0,M) TorRn−1(M0,M

′′) · · ·

M0 ⊗RM ′ M0 ⊗RM M0 ⊗RM ′′ −→ 0

∂n

∂1

idM0⊗α idM0

⊗β

Proof. Let (PM ′)· and (PM ′′)· be truncated chain complexes associated to

projective resolutions of M ′ and M ′′. By the Horseshoe Lemma (26.1), there existsa projective resolution of M with associated truncated chain complex (PM )· and

there exists chain maps f, g associated to α, β, respectively, such that we have a

short exact sequence of chain complexes 0 → (PM ′)· f−→ (PM )· g−→ (PM ′′)· → 0.

For all n ≥ 0, we have a split short exact sequence:

0 PM ′,n PM,n PM ′′,n 0

0 PM ′,n PM ′,n ⊕ PM ′′,n PM ′′,n 0

fn

∃γ ∼=

gn

ι′ π′′


Applying T to this diagram and adding a third row:

0 TPM ′,n TPM,n TPM ′′,n 0

0 TPM ′,n T (PM ′,n ⊕ PM ′′,n) TPM ′′,n 0

0 TPM ′,n TPM ′,n ⊕ TPM ′′,n TPM ′′,n 0

Tfn

Tγ ∼=

Tgn

Tι′

(Tπ′

Tπ′′

)Tπ′′

( 10 ) (1 0)

wherePM ′,n ⊕ PM ′′,n

PM ′,n PM ′′,n

π′ π′′

The lower squares commute since(Tπ′

Tπ′′

) Tι′ =

(T (π′ ι′)T (π′′ ι′)

)=

(10

)and (

0 1)(Tπ′

Tπ′′

)= Tπ′′.

By the Short 5 Lemma,(Tπ′

Tπ′′

)is an isomorphism. So

0→ TPM ′,nTfn−−→ TPM,n

Tgn−−→ TPM ′′,n → 0

is exact for all n, and so

0→ T (PM ′)· Tfn−−→ T (PM,n)· Tgn−−→ T (PM ′′,n)· → 0

is a short exact sequence of chain complexes. Apply the Long Exact HomologySequence (20.3) to get the long exact sequence of the LnT . Now L0T is right exactsince PM ′,−1 = 0 and hence (L−1T )M ′ = 0.

Corollary 27.3. Let M0 = (M0)R. 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. (c) =⇒ (a) follows from the long exact Tor sequence:

Suppose 0→M ′α−→M

β−→M ′′ → 0 is exact in R-Mod. Then

0 = TorR1 (M0,M′′)

∂1−→M0 ⊗RM ′idM0

⊗α−−−−−→M0 ⊗RM

idM0⊗β

−−−−−→M0 ⊗RM ′′ → 0

is exact.(b) =⇒ (c) is obvious.

(a) =⇒ (b): Let M = RM with projective resolution · · · → P1∂1−→ P0

ε−→M →0 and truncates complex (PM )· : · · · →→ P1

∂1−→ P0 → 0. Since M0 is R-flat,

M0 ⊗R (M0)· : · · · →M)9⊗R P1

idM0⊗∂1−−−−−−→M0 ⊗R P0 → 0

has zero homology, i.e., TorRn (M0,M) = Hn(M0 ⊗ (PM )·) = 0 for all n > 0.

27. LONG EXACT SEQUENCES FOR LEFT AND RIGHT DERIVED FUNCTORS 69

Theorem 27.4. Let 0→M ′α−→M

β−→M ′′ → 0 be a short exact sequence in C.

(a) Let T : C → Ab be an additive covariant functor. Then we get a longexact sequence in Ab:

0 −→ (R0T )M ′ (R0T )M (R0T )M ′′

(R1T )M ′ (R1T )M (R1T )M ′′ · · ·

(RnT )M ′ (RnT )M (RnT )M ′′ −→ · · ·

(R0T )α (R0T )β

δ0

(R1T )α (R1T )β

δn−1

(RnT )α (RnT )β

In particular, R0T : C→ Ab is left exact.(b) Let T : C→ Ab is an additive contravariant functor. Then we get a long

exact sequence in Ab:

0 −→ (R0T )M ′′ (R0T )M (R0T )M ′

(R1T )M ′′ (R1T )M (R1T )M ′ · · ·

(RnT )M ′′ (RnT )M (RnT )M ′ −→ · · ·

(R0T )β (R0T )α

δ1

(R1T )β (R1T )α

δn

(RnT )β (RnT )α

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) Ext1

R(M,M0) Ext1R(M ′,M0) · · ·

ExtnR(M ′′,M0) ExtnR(M,M0) ExtnR(M ′,M0) −→ · · ·

β∗ α∗

δ1

δn

Corollary 27.5. Let M0 ∈ Ob(C) where C = R-Mod or Mod-R. Thefollowing are equivalent:

(a) M0 is injective.(b) ExtnR(M,M0) = 0 ∀n ≥ 1, ∀M ∈ Ob(C)(c) Ext1

R(M,M0) = 0 ∀M ∈ Ob(C)

Proof. (c) =⇒ (a) follows from Theorem 27.4(b) =⇒ (c) is obvious.(a) =⇒ (b) follows since Hom(R(−,M) is exact if M0 is injective.


We also get long exact Tor and Ext sequences in the “other variable”:

Proposition 27.6.

(a) Let 0 → M0α−→ M1

β−→ M2 → 0 be a short exact sequence in Mod-R.Then for all M = RM , we have a long exact sequence in Ab:

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

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

M0 ⊗RM M1 ⊗RM M2 ⊗RM −→ 0

∂n

∂1

α⊗idM β⊗idM

(b) Let 0 → M0α−→ M1

β−→ M2 → 0 be a short exact sequence in Mod-R.Then for all M = RM , we have a long exact sequence in Ab:

0 −→ HomR(M,M0) HomR(M,M1) HomR(M,M2)

Ext1R(M,M0) Ext1

R(M,M1) Ext1R(M,M2) · · ·

ExtnR(M,M0) ExtnR(M,M1) ExtnR(M,M2) −→ · · ·

α∗ β∗

δ0

δn−1

Proof. Let M = RM . Let (PM )· be the truncated complex associated to a

projective resolution of M .

(a) Since PM,n is projective, hence flat, we get a short exact sequence of chaincomplexes for all n:

0→M0 ⊗R (PM )·α⊗id(PM )·−−−−−−−→M1 ⊗ (PM )·

β⊗(PM )·−−−−−−→M2 ⊗R (PM )· → 0.

Now use the Long Exact Homology Sequence (20.3).(b) Since PM,n is projective, we get a short exact sequence of cochain com-

plexes

0→ HomR((PM )·,M0)α∗−−→ HomR((PM )·,M1)

β∗−→ HomR((PM )·,M2)→ 0.

Now use the Long Exact Cohomology Sequence (20.4).

28. PROPERTIES OF Tor AND Ext 71

28. Properties of Tor and Ext

Theorem 28.1.

(a) If M ′ = M ′R,M = RM , then for all n ≥ 0,

Ln(M ′ ⊗R −) ∼= Ln(−⊗RM)M ′

(b) If C = R-Mod or Mod-R, and M,M ′ ∈ Ob(C), then for all n ≥ 0,

Rn(HomR(−M ′))M ∼= Rn(HomR(M,−))M ′

Proof. We prove (b). We use the method of dimension shifting. 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 variables. Considera short exact sequence in C

0→Miι−→ P

π−→M → 0

where P is projective. Then for n ≥ 2, we get 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 ′).

Since P is projective, Extn−1R (P,M ′) = ExtnR(P,M ′) = 0. Since HomR(P,−) is

exact, extn−1R (P,M ′) = extnR(P,M ′) = 0. Hence for all n ≥ 2, we have

ExtnR(M,M ′) ∼= Extn−1R (M1,M

′) and extnR(M,M ′) ∼= extn−1R (M1,M

′).

By induction, we only have to consider the case n = 1. (The case n = 0 follows sinceR0 in both cases is just the original functor.) We get exact sequences of abeliangroups:

0 HomR(M,M ′) HomR(P,M ′) HomR(M1,M′) Ext1

R(M,M ′) 0

0 HomR(M,M ′) HomR(P,M ′) HomR(M1,M′) ext1

R(M,M ′) 0

π∗ ι∗ α

π∗ ι∗ β

Hence

Ext1R(M,M ′) ∼= HomR(M1,M

′)/Ker(α) = HomR(M1,M′)/ Im(ι∗)

= HomR(M1,M′)/Ker(β)

∼= ext1R(M,M ′).

Corollary 28.2. If M ′ or M is flat, then TorRn (M ′,M) = 0 for all n ≥ 1. IfM ′ is injective or M is projective, then ExtnR(M,M ′) = 0 for all n ≥ 1.


Theorem 28.3. For all n ≥ 0,

(a) TorRn

(∐k∈K

Ak, B)∼=∐k∈K

TorRn (Ak, B) and TorRn

(A,∐k∈K

Bk

)∼=∐k∈K

TorRn (A,Bk).

(b) If Ak, ϕk` is a direct system of right R-modules over a directed index setK, then

TorRn (lim−→Ak, B) ∼= lim−→TorRn (Ak, B).

Similarly, if Bk, ϕk` is a direct system of left R-modules over a directedindex set K, then

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

Proof.

(a) For the case n = 0, tensor products preserve direct limits. For n = 1, weshow the first isomorphism. For all k ∈ K, choose a short exact sequence

0→Mk → Pk → Ak → 0

where Pk is projective. Then we get a short exact sequence

0→∐k∈K

Mk →∐k∈K

Pk →∐k∈K

Ak → 0,

and∐k∈K Pk is projective. So we get a commutative diagram with exact

rows:

0 TorR1

(∐k∈K

Pk, B)

TorR1

(∐k∈K

Ak, B) (∐

k∈K

Mk

)⊗R B

(∐k∈K

Pk

)⊗R B

0∐k∈K

TorR1 (Pk, B)∐k∈K

TorR1 (Ak, B)∐k∈K

(Mk ⊗R B)∐k∈K

(Pk ⊗R B)

∂

γ

f

∃α∼= ∃β∼=∐∂k

∐fk

We have(∐k

fk) α ∂ = β f ∂ = 0.

Since the left two terms in each row are zero, this means there exists awell-defined γ that makes the diagram commute. Buy the 5-Lemma, itfollows that γ is an isomorphism.

For n > 1, use dimension shifting and induction since

TorRn (M ′, B) ∼= TorRn−1(M ′, B1)

when 0→ B1 → P → B → 0 is a short exact sequence with P projective.(b) Same idea: For n = 0, tensor products preserve direct limits. For n = 1,

we do the first isomorphism. For all k ∈ K, choose a short exact sequence

0→Mk → Pk → Ak → 0

where Pk is projective (hence flat). For all k ≤ l in K, there existsψk` : Pk → P` since Pk is projective. Let αk` = ψk`

∣∣Mk

.

0 Mk Pk Ak 0

0 M` P` A` 0

αk` ψk` ϕk`

28. PROPERTIES OF Tor AND Ext 73

Then we get a short exact sequence of direct system and hence of directlimits since K is directed:

0→ lim−→Mk → lim−→Pk → lim−→Ak → 0,

where lim−→Pk is flat since Pk is flat for all k ∈ K. So we get a commutativediagram 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)

∂

γ

f

∃α∼= ∃β∼=→∂k

→fk

We have →fk α ∂ = β f ∂ = 0.

Since the left two terms in each row are zero, this means there exists awell-defined γ that makes the diagram commute. By the 5-Lemma, itfollows that γ is an isomorphism. For n > 1, use dimension shifting andinduction since

TorRn (M ′, B) ∼= TorRn−1(M ′, B1)

when 0→ B1 → P → B → 0 is a short exact sequence with P projective.

Theorem 28.4. For all n ≥ 0, we have

ExtnR(∐k∈K

Ak, B) ∼=∏k∈K

ExtnR(Ak, B)

andExtnR(A,

∏k∈k

Bk) ∼=∏k∈K

ExtnR(A,Bk).

Proof. We show the first isomorphism. For n = 0, the result follows sinceHom functors have this property. For n > 1, the result follows by dimensionshifting. For n = 1, choose, for all k ∈ Ka short exact sequence

0→Mk → Pk → Ak → 0

where Pk is projective. Then we get a short exact sequence

0→∐

Mk →∐

Pk →∐

Ak → 0,

where∐Pk is projective. So we get a commutative diagram with exact rows:

HomR

(∐k∈K

Pk, B)

HomR

(∐k∈K

Mk, B)

Ext1R

(∐k∈K

Ak, B)

Ext1R

(∐k∈K

Pk, B)

0

∐k∈K

HomR

(Pk, B

) ∐k∈K

HomR

(Mk, B

) ∐k∈K

Ext1R

(Ak, B

) ∐k∈K

Ext1R

(Pk, B

)0

∃α∼= ∃β∼= γ

Argue dually to before to get γ that makes the diagram commute, then use the5-Lemma.


29. Tor for abelian groups

Recall the following example:

Example 29.1. Let m ∈ Z+. Then we have a projective resolution of Z/mZ:

0→ Z m−→ Z→ Z/mZ→ 0

with truncated complex

P· : 0→ Z m−→ Z→ 0.

For all Z-modules A,

A⊗Z P· ∼= 0→ Am−→ A→ 0.

So

TorZ0 (A,Z/mZ) ∼= A/mA,

TorZ1 (A,Z/mZ) ∼= x ∈ A|mx = 0

TorZn(A,Z/mZ) = 0 ∀n ≥ 2.

Recall the following definition:

Definition 29.2. If M is a Z-module then

t(M) = x ∈M | ∃r ∈ Z− 0 with rx = 0is a submodule of M , called the torsion submodule of M . We say M is a torsionabelian group if M = t(M). We say M is torsion free if t(M) = 0.

Proposition 29.3. For all Z-modules A,B, TorZ1 (A,B) is a torsion abelian

group, and TorZn(A,B) for all n ≥ 2.

Proof. Recall: Let Bα be the collection of all finitely generated submodulesof B. Note this is a directed index set under inclusion, and B ∼= lim−→Bα.

For all n ≥ 0, we have

TorZn(A,B) ∼= lim−→TorZn(A,Bα).

Since lim−→ of torsion abelian groups is torsion, and since lim−→ 0 = 0, we can restrictto the case when B is a finitely generated Z-module. So

B ∼= Zr ⊕ Z/m1Z⊕ · · ·Z/mkZand hence

TorZn(A,B) ∼= TorZn(A,Zr)︸︷︷︸

=

0for n≥1 since

Zr is a flat Z-module

⊕k⊕i=1

TorZn(A,Z/miZ)︸︷︷︸

=x ∈ A|mix = 0 if n = 1

0 if n ≥ 2

30. Ext FOR ABELIAN GROUPS 75

Proposition 29.4.

(a) If B is a torsion abelian group, then

TorZ1 (Q/Z) ∼= B

(b) Let A be a Z-module. Then A is torsion free(1)⇐⇒ TorZ1 (A,−) = 0 if and

only if(2)⇐⇒ TorZ1 (−, A) = 0.

Proof.

(a) Consider the short exact sequence 0 → Z → Z → Q/Z → 0. We get anexact sequence

TorZ1 (Q, B)︸︷︷︸

=

0since Q is flat

→ TorZ1 (Q/Z, B)→ Z⊗Z B︸︷︷︸∼=B

→ Z⊗Z B︸︷︷︸

=

0since B is torsion.

(b) ((1)⇒) We have A ∼= lim−→Aα where Aα is the collection of finitely generated

submodules of A. Since each Aα is a finitely generated torsion free Z-module, hence a free Z-module. Hence

TorZ1 (A,−) ∼= lim−→TorZ1 (Aα,−) = 0

(Similarly, TorZ1 (−, A) = 0.)

((1)⇐) In particular, ∀m ∈ Z+,

x ∈ A|mx = 0 = TorZ1 (A,Z/m) = 0.

Hence A is torsion free.(Similarly, one shows TorZ1 (Z/mZ, A) = x ∈ A|mx = 0 to see A is

torsion free when TorZ1 (−, A).)

30. Ext for abelian groups

Proposition 30.1. Let A,B be Z-modules. Then ExtnZ(A,B) = 0 for all n ≥ 2.

Proof. Let E be an injective Z-module such that there is a short exact se-quence

0→ Bι−→ E

π−→ Q→ 0.

Now, E is an injective Z-module hence divisible, and since Q/ι(B), Q is a quotientof a divisible Z-module, hence divisible. So Q is an injective Z-module. So theshort exact sequence is an injective resolution of B with truncated complex

E· : 0→ Eπ−→ Q→ 0.

HenceHomZ(A,E·) : 0→ HomZ(A,E)

π∗−→ HomZ(A,Q)→ 0

and so ExtnZ(A,B) = 0 for all n ≥ 2.


Example 30.2. If A is a torsion abelian group then Ext1Z(A,Z) ∼= A∗ =

HomZ(A,Q/Z). This is because from the injective resolution

0→ Z→ Q→ Q/Z→ 0

of Z, we get an exact sequence:

HomZ(A,Q)︸︷︷︸=0

since A is torsion

→ HomZ(A,Q/Z)→ Ext1Z(A,Z)→ Ext1

Z(A,Q)︸︷︷︸=0

since Q is aninjective Z-module

31. Ext and extensions

Let R be a ring. Let A,C be left R-modules. A extension of C by A is a shortexact sequence in R-Mod:

ξ : 0→ Aι−→ E

π−→ C → 0.

Two extensions ξ, ξ′ of C by A are equivalent if there is a commutative diagram inR-Mod:

ξ : 0 A E C 0

ξ′ : 0 A E′ C 0

ι

f

π

ι′ π′

(By the 5-Lemma, f is an isomorphism).Define e(C,A) to be the set of equivalence classes of extension 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∂2−→ P1

∂1−→ P0∂0−→ C → 0.

By the Comparison Theorem (21.1), there exists ψξ and φξ making the diagramcommute:

· · · P2 P1 P0 C 0

ξ : · · · 0 A E C 0

∂2

0

∂1

φξ

∂0

ψξ

ι π

We get φξ ∂2 = 0, hence φξ ∈ Ker(∂∗2), so

[φξ] ∈ Ext1R(C,A) = ker(∂∗2 )/ Im(∂∗1 ).

So we get a map

H : e(C,A) −→ Ext1R(C,A)

[ξ] 7−→ [φξ]

Check: h is well-defined.

Note 31.1. 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.

CHAPTER 5

Abelian Categories and Homotopy Categories

32. Abelian Categories

Definition 32.1. Let C,D be categories (large with (small) Hom sets.)

(a) C is called an Ab-category if for all X,Y ∈ Ob(C), HomC(X,Y ) is anadditive abelian group such that composition distributes over addition.

(b) Let C,D be Ab-categories, let F : C→ D be a covariant functor. Then Fis called an additive functor if for all X,Y ∈ Ob(C), the map

HomC(X,Y ) −→ HomD(FX,FY )

α 7−→ Fα

is a group homomorphism under addition.(c) C is called an additive category if:

• C is an Ab-category,• C has a zero object (i.e. an object that is both initial and terminal),

and• For all A,B ∈ Ob(C), there exists a product (A×B, πA, πB) in C.

Examples 8.

(1) R-Mod and Mod-R are additive categories.(2) Let D = R-Mod or Mod-R. Let C(D) be the category of cochain

complexes and cochain maps.Then C(D) is an Ab-category, the zero object is the zero complex, and

we have arbitrary products and coproducts: If Cnα , δnαα∈Λ = C·α α∈Λ

is a family of cochain complexes in C(D), then∏α∈Λ

C·α : · · · →∏α∈Λ

Cnα

∏δnα−−−→

∏α∈Λ

Cn+1α → · · ·

∐α∈Λ

C·α : · · · →∐α∈Λ

Cnα

∐δnα−−−→

∐α∈Λ

Cn+1α → · · ·

So C(D) is an additive category.

Definition 32.2. Let C be an additive category, let f : A→ B be a morphismin C.

(a) A kernel if f is a pair (K, ι) where K ∈ Ob(C) and ι : K → A in C suchthat f ι = 0 and (K, ι) is universal with this property, i.e. if (X, g) isanother pair with X ∈ Ob(C) and g : X → A in C such that f g = 0,

77

78 5. ABELIAN CATEGORIES AND HOMOTOPY CATEGORIES

then there exists a unique α : X → K in C with g = ι α.

K A B

X

ι f

g ∃!α

Note. ι is monic.

(b) A cokernel if f is a pair (Q, π) where Q ∈ Ob(C) and π : B → Q in C

such that π f = 0 and (Q, π) is universal with this property, i.e. if (Y, h)is another pair with Y ∈ Ob(C) and h : B → Y in C such that h f = 0,then there exists a unique β : Q→ Y in C with h = β π.

A B Q

Y

f π

h ∃!β

Note. π is epic.

Lemma 32.3. Let C be an additive category, let f : A → B be a morphism inC.

(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 a kernel of f . Then

(β f) ι = β (f ι) = 0.

If (X, g) is such that g : X → A is in C with (β f) g = 0 = β 0,then since β is monic, this means f g = 0. Hence there exists a uniqueα : X → K in C with g = ι α.

(b) dual.

Definition 32.4. Let D = R-Mod or Mod-R, let C = C(D) be the category

of cochain complexes. Let C· = Cn, δnC ∈ Ob(C).

(a) A subcomplex of C· is a cochain complex A· = An, δnA where An ⊆ Cnis a submodule and δnA = δnC |An for all n ∈ Z.

(b) Given a subcomplex A· of C· we can build the quotient complex Q· =Qn, δnQ where Qn = Cn/An and

δnQ : Cn/An −→ Cn+1/An+1

x+An 7−→ δnC(x) +An+1.

(This is well-defined!)

Example 32.5. Let f : C· → D· be a cochain map.

(1) Ker(fn)n∈Z forms a subcomplex of C·, denoted by Ker(f).

Note. The inclusion homomorphisms ιn : Ker(fn) → Cn form a

cochain map ι : Ker(f)→ C·. Then (Ker(f), ι) is a kernel of f . (Check).

32. ABELIAN CATEGORIES 79

(2) Coker(fn)n∈Z forms a quotient complex of D·, denoted by Coker(f).

Note. The projection homomorphisms πn : Dn Coker(fn) form a

cochain map π : D· → Coker(f). Then (Coker(f), π) is a cokernel of f .(Check).

Definition 32.6. A category A is called an abelian category if the followingare satisfied:

(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 32.7. Part (c) of Definition 32.6 means that if f : A → B is amonomorphism in A, (Q, π) = Coker(f) and (K, ι) = Ker(π), then there exists aunique isomorphism ω : A→ K with f = ι ω.

A B Q

K

f

∃!ω

π

ι .

Note 32.8. The statements (a) - (d) in Definition 32.6 imply:

(e) Every morphism f : A→ B in A can be factored as f = m e where e isepic and m is monic.

A I Be

f

m

Lemma 32.9. Let A be an abelian category.

(a) Let f : A → B be in A, let (K,m) = Ker(Coker(f)) and (C, e) =Coker(Ker(f)). Then there exists a unique α : A → K with f = m αand there exists a unique β : C → B with f = β e.

(b) If m : Y → Z is monic, λ : X → Y is such that Coker(m) = Coker(mλ)then λ is epic.If e : X → Y is epic, µ : Y → Z is such that Ker(e) = Ker(µ e) then µis monic.

Proof.

(a) Let (X, ι) = Ker(f) and (Y, π) = Coker(f).

K

X A B Y

C

m

ι f

e

π

Since π f = 0, then by the universal property of (K,m), there existsa unique α : A → K in A with f = m α. Since f ι = 0, then by theuniversal property of (C, e), there exists β : C → B in A with f = β e.


(b) By property (e) of abelian categories, we can factor λ as

X I Yε

λ

µ

where ε is epic and µ is monic.

Coker(m µ) = Coker(m µ ε) = Coker(m),

where the first equality follows since ε is epic and the last equality is byassumption. Since m and m µ are monic, they are the kernels of theircokernels.

So (Y,m) = (I,m µ). The equality here means: There exists aunique isomorphism ω : I → Y with m ω = m µ. Since m is monic,ω = µ, so µ is an isomorphism. Then λ = µ ε is epic since ε is epic.

Theorem 32.10. Let A be an abelian category, let f : A → B be in A. Let(X, ι) = Ker(f), (Y, π) = Coker(f), (K,m) = Ker(π), (C, e) = Coker(ι). Let α, βbe Lemma 32.9(a). Then

X A K B Yι α

f

m π

where α is epic and (K,α) = Coker(ι) = (C, e), and

X A C B Yι e β π

where β is monic and (C, β) = Ker(π) = (K,m). Moreover, there exists an iso-morphism λ : C → K with m λ = β and λ e = α.

Proof. We have

Coker(m α) = Coker(f) = (Y, π) = Coker(Ker(π)) = Coker(m).

By Lemma 32.9(b), α is epic. So

(K,α)(∗)= Coker(Ker(α))

(∗∗)= Coker(Ker(m α))

Coker(Ker(f))

Coker(ι)

= (C, e).

where (∗) follows since α is epic, and (∗∗) follows since m is monic.Dually, β is monic and (C, β) = Ker(π) = (K,m). In particular, there exists

unique isomorphisms λ, ν : C → K with

m λ = β and ν e = α

by the universal properties of kernel and cokernel. Hence

m λ e = β e = f = m α = m ν e,and since m is monic and e is epic, λ = ν.

33. CATEGORY OF COCHAIN COMPLEXES, C(A) 81

Definition 32.11. Let A be an abelian category.

(a) Let f : A→ B be a morphism in A. Define the image of f to be

Im(f) = Ker(Coker(f)) =: (I,m).

(b) A sequence Aα−→ B

β−→ C of morphisms in A is said to be exact (at B) ifIm(α) = Ker(β).

A B C

I

α β

m

(Ker(β) = (I,m) = Im(α)).(c) A subcategory B of A is called an abelian subcategory if B is abelian and

exact sequences in B are also exact when viewed as sequences in A.

Example 32.12. Let A = R-Mod, R 6= 0 ring. Let X ∈ Ob(A), X 6= 0.Define B by: Ob(B) = X and HomB(X,X) = idX. Then X is the zero objectin B and

XidX−−→ X

idX−−→ X

is exact in B but not in A. Hence B is abelian but not an abelian subcategory ofA.

Theorem 32.13 (Freyd-Mitchell Embedding Theorem (1964)). Let A be asmall category, (i.e. Ob(A) is a set.) The there exists a ring R (could be thezero ring) and an exact fully faithful additive functor F : A → R-Mod which em-beds A into a full subcategory of R-Mod. (i.e. HomR(A,B) ∼= HomR(FA,FB) asabelian groups for all A,B ∈ Ob(A))

33. Category of Cochain Complexes, C(A)

Theorem 33.1. Let A be an abelian category. Then C(A) = category of cochaincomplexes and cochain maps, is an abelian category.

Proof sketch.

• Check: C(A) is an additive category. (Uses that A is additive; do every-thing “dimension by dimension”.)

• Let f : C· → D· be a cochain map in C(A). Let Ker(fn) = (Kn, ιn).We have

fn+1 δnC ιn = δnD fn ιn = 0.

By the universal property of (Kn+1, ιn+1), there exists a unique δnK :Kn → Kn+1 making the diagram commute.

Kn Kn+1

Cn Cn+1

Dn Dn+1

ιn

δnK

ιn+1

fn

δnC

fn+1

δnD


Then

ιn+1 δnK δn+1K = δnC ιn δn−1

K = δnC δn−1C ιn−1 = 0.

Since ιn+1 is monic, δnK δn−1K = 0.

Check: (K·, ι) is a kernel of f . Similarly, show f has a cokernel basedon Coker(fn) = (Qn, πn) for all n ∈ Z. Now

f is monic ⇐⇒ ∀α, β : X· → C· with f α = f β, we have α = β.

⇐⇒ ∀αn, βn : Xn → Cn with fn αn = fn βn,we have αn = βn ∀n.

⇐⇒ fn is monic ∀n.By the construction of kernel and cokernel in C(A), this means if f

is monic, then f is the kernel of its cokernel. Similarly, we get the dualstatement for epic f .

Note 33.2. For all n ∈ Z we get an additive functor Hn : C(A)→ A as follows:

C· = Cn, δnC in C(A)

(Zn(C·), ιn) = Ker(δnC)

(Bn(C·), jn) = Im(δn−1C ) = Ker(Coker(δn−1

C )).

Let (Qn−1, ρn−1) = Coker(δn−1C ).

Bn(C·) Qn−1

· · · Cn−1 Cn Cn+1 · · ·

Zn(C·)

jn

δn−1C δnC

ρn−1

ιn

Since δnC δn−1 = 0, then by the universal property of Coker(δn−1C ) there exists

a unique morphism α : Qn−1 → Cn+1 such that δnC = α ρn−1. Then

δnC jn = α ρn−1 jn = α 0 = 0.

By the universal property of Ker(δnC), there exists a unique morphism Bn(C·) →Zn(C·).

Bn(C·) Qn−1

· · · Cn−1 Cn Cn+1 · · ·

Zn(C·) Hn(C·)

jn α

δn−1C δnC

ρn−1

ιn

π

Let (Hn(C·), π) = Coker(Bn(C·)→ Zn(C·)).

34. BICOMPLEXES AND TOTAL COMPLEXES 83

34. Bicomplexes and Total Complexes

Let A be an abelian category and let C(A) be the (abelian) category of cochaincomplexes in A.

Definition 34.1.

(a) A bicomplex (or double complex ) in A is a collection of objects Cp,qp,q∈Zin 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,qu = 0 and dp,qh d

p,q−1h = 0 and

dp+1,qh dp,qu + dp,q+1

u dp,qh = 0,

i.e., each square “anti-commutes”:

......

· · · Cp,q Cp+1,q · · ·

· · · Cp,q+1 Cp+1,q+1 · · ·

......

dh dh

dv

dh

dv

dh dh dh

(b) C·,· = Cp,q, dp,qh , dp,qv is said to be bounded if along each diagonalp+q = n, there are only finitely many nonzero objects Cp,q. (For example,

if Cp,q = 0 for all p < 0 and for all q < 0, then C·,· is bounded.)

Remark 34.2. d·,qV : C·,q → C·,q+1is not a cochain complex because of the

anticommutativity. So, we introduce a “sign trick”: Define

f·,q : C·,q → C·,q+1, fp,q = (−1)pdp,qv .

Then f·,q is a cochain map.Hence, we can identify the category of bicomplexes in A with the category

C(C(A)), i.e. the category of cochain complexes of cochain complexes in A.

Definition 34.3. Let C·,· be a bicomplex in A. We can define total complexes

TotΠ(C)· and Tot⊕(C)· by

TotΠ(C)n =∏

p+q=n

Cp,q and Tor⊕(C)n =∐

+p+ q = nCp,q

and

dn : TotΠ(C)n −→ TotΠ(C)n+1

(xp,q)p+q=n 7−→(dr−1,sh xr−1,s + dr,s−1

v xr,s−1)r+s=n+1

.

Similarly for Tot⊕(C)·.


Remark 34.4. If C·,· is not bounded, then TotΠ(C)· or Tot⊕(C)· may notexists. (For example, if A is the category of finite abelian groups, this does notexist.)

Definition 34.5. If A is an abelian category such that arbitrary productsexist, then A is called complete. If arbitrary coproduct exists, then A is calledcocomplete.

35. Shift/Translation Functor

Definition 35.1. Let p ∈ Z. For all C· ∈ Ob(C(A)), define C[p]· by

C[p]n = Cp+n and dnC[p] = (−1)pdp+nC .

So we “shift C· p degrees to the left”:

C[p]· : · · · → Cp−1︸︷︷︸deg. −1

(−1)pdp−1C−−−−−−−→ Cp︸︷︷︸

deg. 0

(−1)pdpC−−−−−→ Cp+1︸︷︷︸deg. 1

→ · · ·

If f : C· → D· is in C(A) define f [p] : C[p]· → D[p]· by f [p]n = fp+n.Then [p] : C(A)→ C(A) is an additive functor called the shift (or translation)

functor of degree p.

Note 35.2. Hn(C[p]·) = Hn+p(C·.Remark 35.3. 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 snake diagram. Then C is a small category. By Freyd-Mitchell (32.13),C can be embedded into R-Mod for some ring R as a full abelian subcategory.Since the snake lemma is true in R-Mod (20.1), then it is also true in C and hencein A.

As a consequence, if 0 → C′· f−→ C· g−→ C

′′· → 0 is a short exact sequence inC(A), then we have a long exact cohomology sequence in A:

· · · → Hn(C′·) Hn(C·) Hn(C

′′·)

Hn+1(C′·) Hn+1(C·) Hn+1(C

′′·)→ · · ·

Hn(f) Hn(g)

δn

Hn+1(f) Hn+1(g)

Note 35.4. We can identify A with a full abelian subcategory of C(A) byidentifying C ∈ Ob(A) with

C· = 0→ C︸︷︷︸deg. 0

→ 0

in C(A), and Cf−→ D in A with

C· · · · 0 C 0 · · ·

D· · · · 0 D 0 · · ·

f· 0 f 0

36. HOMOTOPY CATEGORIES 85

36. Homotopy Categories

Definition 36.1. Let A be an abelian category. Then f : C· → D· inC(A) is said to be homotopic to zero, written f ∼ 0 if for all n ∈ Z, there existskn : Cn → Dn−1 in A such that

fn = dn−1D kn + kn+1 dnC .

DefineHt(C·, D·) = f ∈ HomC(A)(C·, D·) | f ∼ 0.

Then Ht(C·, D·) ≤ HomC(A)(C·, D·) under addition (check).

Moreover, if f : C· → D·, g : D·, E·, h : B·, C· are cochain maps withf ∼ 0, then

(36.1) g f ∼ 0 and f h ∼ 0.

To see that g f ∼ 0, suppose fn = dn−1D kn + kn+1 dnC . Then

gn fn = gn dn−1D kn + gn kn+1 dnC

= dn−1E (gn−1 kn) + (gn kn+1) dnC .

Define a new category K(A) by Ob(K(A)) = Ob(C(A)) and for all C·, D· inK(A),

HomK(A)(C·, D·) := HomC(A)(C·, D·)/Ht(C·, D·).Note. We need 36.1 to get well-defined compositions of morphisms.

Remark 36.2. K(A) is an additive category (check). We have an additivefunctor

F : C(A) −→ K(A)

C· 7−→ C·(f : C· → D·) 7−→ f + Ht(C·, D·)

Remark 36.3 (“Decorations on C(A)”). Let A be an abelian category. ThenCb(A)

C+(A)

C−(A)

is the full subcategoryof C(A) consisting of

bounded complexes

bounded below complexes

bounded above complexes

.

Lemma 36.4. The cohomologies of cochain complexes in A define for all n ∈ Za well-defined additive functor Hn : K(A)→ A.

Proof. For A = R-Mod, we already proved that f ∼ 0 =⇒ Hn(f) = 0. Forarbitrary A, use Freyd-Mitchell Embedding Theorem (32.13).

Definition 36.5. A cochain map f : X· → Y · in C(A) is called a quasi - iso-

morphism (or qis) if for all n ∈ Z, Hn(f) : Hn(X·)→ Hn(Y ·) is an isomorphismin A. A morphism f in K(A) is called a qis if Hn(f) is an isomorphism in A forall n ∈ Z.

Note 36.6. If f = f + Ht(X,Y ), then f is a qis in K(A) ⇐⇒ f is a qis inC(A).


Remark 36.7 (What’s the big picture?). Let A = R-Mod or Mod-R. LetX,Y ∈ Ob(A). We know that HomA(X,−) and HomA(−, Y ) are left exact, butusually not right exact. Our “remedy” was to construct right derived functors

ExtnA(X,−) and ExtnA(−, Y ). Then is 0 → Aα−→ B

β−→ C → 0 is a short exactsequence in A , then we get a long exact Ext sequence:1

0 −→ HomA(X,A) HomA(X,B) HomA(X,C)

Ext1A(X,A) Ext1

A(X,B) Ext1A(X,C) · · ·

ExtnA(X,A) ExtnA(X,B) ExtnA(X,C) −→ · · ·

α∗ β∗

δ0

δn−1

Recall that in order to define Extn and Torn, we used projective and injectiveresolutions, i.e. we used complexes. So the idea is now to define a new category ofcomplexes, D(A), called the derived category , such that

ExtnA(X,A)def= Hn HomA(X, I·A) = HomD(A)(X·, IA[n]·)

where I·A is a truncated injective resolution of A, X· = 0→ X → 0, and IA[n]· isthe shifted complex. The construction of D(A) will be as follows:

cochain complexes C(A) (abelian)

homotopy category K(A) (additive & triangulated)

derived category D(A) = K(A)Qis (additive & triangulated)

where K(A)Qis is the localization of K(A) at quasi-isomorphisms. In D(A), quasi-isomorphisms become isomorphisms.

37. Mapping Cones

Definition 37.1. Let f : X· → Y · be in C(A). Define the mapping cone of

f to be the cochain complex M(f)· where

M(f)n = Xn+1 ⊕ Y n and dnM(f) =

(−dn+1

X 0

−fn+1 dnY

).

Then, we get a short exact sequence in C(A):

(37.1) 0 Y · M(f)· X[1]· 0α(f)0

1

β(f)(0 1)

1Hence, the sense in which these right derived functors serve as a “remedy” for the Hom

functors not being exact is that while the Hom functors may not send exact sequences to exactsequences, the Ext functors send exact sequences to long exact sequences. Similarly for ⊗R and

TorRn .

37. MAPPING CONES 87

Note. (We use column vectors!)

Lemma 37.2. The long exact cohomology sequence associated to (37.1) is

· · · → Hn−1(Y ·) Hn−1(M(f)·) Hn(X·)

Hn(Y ·) Hn(M(f)·) Hn(X·)→ · · ·

Hn−1(α(f)) Hn−1(β(f))

δn−1=Hn(f)

Hn(α(f)) Hn(β(f))

Proof. We prove this for A = R-Mod and then use the Freyd-Mitchell Em-bedding Theorem (32.13). We only need to show δn−1 = Hn(f).

Let z ∈ Zn(X·), i.e. z ∈ Xn 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)).

Hence

δn−1([z]) = [fn(z)] = Hn(f)([z]).

Corollary 37.3. f : X· → Y · in C(A) is a qis if and only if M(f)· is

acyclic (i.e. Hn(M(f)·) = 0 for all n ∈ Z).

Lemma 37.4. There exists φ : X[1]· → M(α(f))· in C(A) such that thefollowing diagram commutes in K(A) (so, up to homotopies):

Y · M(f)· X[1]· Y [1]·

Y · M(f)· M(α(f))· Y [1]·

α(f) β(f) −f [1]

φ

α(f) α(α(f))(0 01 00 1

) β(α(f))

(−1 0 0 )

Proof. We have M(α(f))n = Y n+1 ⊕Xn+1 ⊕ Y n with

dnM(α(f)) =

−dn+1Y 0 0

0 −dn+1X 0

−1 −fn+1 dnY

.

Define φ : X[1]· →M(α(f))· by

φn : Xn+1

(fn+1

−10

)−−−−−−→M(α(f))n,

and ψ : M(α(f))· → X[1]· by

ψn : M(α(f))n

(0 − 1 0

)−−−−−−−−→ Xn+1.

Then

ψn φn =(0 − 1 0

)fn+1

−10

= idXn+1 ,


and

φn ψn =

0 −fn+1 00 1 00 0 0

.

Define

sn : M(α(f))n M(α(f))n−1

Y n+1 ⊕Xn+1 ⊕ Y n Y n ⊕Xn ⊕ Y n−1

(0 0 −10 0 00 0 0

)

Then

idM(α(f))n −φn ψn =

1 fn+1 00 0 00 0 1

,

and

dn−1M(α(f)) s

n + sn+1 dnM(α(f)) =

0 0 dn+1Y

0 0 00 0 1

+

1 fn+1 −dn+1Y

0 0 00 0 0

=

1 fn+1 00 0 00 0 1

.

Hence ψn φn ∼ idXn+1 and φn ψn ∼ idnM(α(f)) for all n, meaning φ ∼= ψ in K(A).Moreover, we have

ψn α(α(f))n =(0 −1 0

)0 01 00 1

=(−1 0

)= β(f)n

and

β(α(f))nφn =(−1 0 0

)−fn+1

−10

= −fn+1 = −f [1]n.

38. PROJECTIVE AND INJECTIVE OBJECTS 89

38. Projective and Injective Objects


(a) P ∈ Ob(A) is projective if HomA(P,−) is exact, i.e. for every diagram

P

X Y 0 (exact)

α

f

in A, there exists β ∈ HomA(P,X) with f β = α.(b) E ∈ Ob(A) is injective if HomA(−, E) is exact, i.e. for every diagram

E

0 X Y (exact)g

α

in A, there exists β ∈ HomA(Y,E) with β g = α.


(a) Let f : P· → X· be in C(A) such that X· is acyclic and P· is a boundedabove complex of projective objects. Then f ∼ 0.

(b) Let : X· → E· be in C(A) such that X· is acyclic and E· is a boundedbelow complex of injective objects. Then f ∼ 0.

Proof. We prove (a). Without loss of generality, assume Pn = 0 for all n > 0.Define kn : Pn → Xn−1 to be zero for all n > 0. Suppose we have constructed kn

for all n ≥ n0.

· · · Pn0−2 Pn0−1 Pn0 Pn0+1 · · ·

· · · Xn0−2 Xn0−1 Xn0 Xn0+1 · · ·

fn0−2

kn0−1

fn0−1

kn0

fn0

kn0+1

fn0+1

We have

dn0−1X (fn0−1 − kn0 dn0−1

P ) = dn0−1X fn0−1 − dn0−1

X kn0 dn0−1P

= fn0 dn0−1P − dn0−1

X kn0 dn0−1P

= (dn0−1X kn0 + kn0+1 dn0

P ) dn0−1P − dn0−1

X kn0 dn0−1P

= 0.

So we get

Pn0−1

Xn0−2 Im(dn0−2X ) 0 (exact)

fn0−1−kn0dn0−1

P

dn0−2

X

Since Pn0−1 is projective, there exists kn0−1 : Pn0−1 → Xn0−2 with

dn0−2X kn0−1 = fn0−1 − kn0 dn0−1

P .

By induction we obtain that f ∼ 0.



(a) Let s : X· → P· be a qis and P· a bounded above complex of projective

objects. Then s has a right homotopy inverse, i.e. there exists t : P· →X· with s t ∼ idP · . If X· is also a bounded above complex of projectiveobjects then we also get t s ∼ idX· .

(b) Let s : E· → X· be a qis and E· a bounded below complex of injective

objects. Then s has a left homotopy inverse, i.e. there exists t : X· → E·with ts ∼ idE· . If X· is also a bounded below complex of injective objectsthen we also get s t ∼ idX· .

Proof. We prove (a). (We use the notation for mapping cones as in modulecategories.)

Since s is a qis, M(s)· is acyclic. So we get a cochain map

P·0

1

−−−→M(s)·

In degree n:

Pn

01

−−−→M(s)n = Xn+1 ⊕ Pn.

Recall that dnM(s) =

(−dn+1

X 0−sn+1 dnP

).

By Lemma 38.2,

(01

): P· → M(s)· is homotopic to zero. Let the homotopy

be (−tnun

): Pn −→M(s)· = Xn+1 ⊕ Pn−1

such that (01

)= dn−1

M(s) (−tnun

)+

(−tn+1

un+1 dnP

).

Then0 = dnX tn − tn+1 dnP

so t : P· → X· is a cochain map, and

idPn = sn tn un + un+1 dnP ,i.e., s t ∼ idP · (where un : Pn → Pn−1).

Now suppose X· is a bounded above complex of projective objects. Since s, stare qis, then t : P· → X· is a qis. By what we have just proved, t has a righthomotopy inverse, say u. Then

u ∼ (s t) u = s (t u) ∼ sand so t s ∼ t u ∼ idX· .

CHAPTER 6

Triangulated Categories

39. Triangles and Distinguished Triangles in K(A)

Definition 39.1.

(1) A triangle in K(A) is a sequence of morphisms in K(A):

X· u−→ Y · v−→ Z· w−→ X[1]·.The shorthand notation for such a sequence is a diagram:

Z·

X· Y ·+1

w

u

v

A morphism of triangles is a commutative diagram in K(A) (i.e. thediagram commutes up to homotopy):

X· Y · Z· X[1]·

X′· Y

′· Z′· X ′[1]·

u

φ

v

µ

w

ν φ[1]

u′ v′ w′

This is called an isomorphism of triangles if φ, µ, ν are isomorphisms inK(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]·Lemma 39.2. If X· u−→ Y · v−→ Z· w−→ X[1]· is a distinguished triangle in

K(A), then we have a long exact cohomology sequence in A:

· · · Hn(X·) Hn(Y ·) Hn(Z·)

Hn+1(X·) Hn+1(Y ·) Hn(Z·) · · ·

Hn−1(w) Hn(u) Hn(v)

Hn(w)

Hn+1(u) Hn+1(v) Hn+1(w)

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.In particular, it sends isomorphisms to isomorphisms.

91

92 6. TRIANGULATED CATEGORIES

40. The Triangle Axioms for K(A)

Let A be an abelian category. Then the collection of distinguished trianglesin K(A) satisfies the triangle axioms (TR0) - (TR5):

(TR0) A triangle isomorphic to a distinguished triangle is distinguished.

(TR1) For any X· ∈ Ob(K(A)), X· idX·−−−→ X· → 0 → X[1]· is a distinguished

triangle.

(TR2) And f : X· → Y · in K(A) can be embedded into a distinguished triangle

X· 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 commutative diagram in K(A):

X· Y ·

X′· Y

′·

u

φ µ

u′

, there exists a morphism of triangles:

X· Y · Z· X[1]·

X′· Y

′· Z′· Z ′[1]·

u

φ µ

v

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):

Y′·

Z′· X

′·

X· Z·

Y ·

g

+1

f

+1

h

+1

u

vu

+1

v

41. TRIANGLE AXIOMS SATISFIED IN K(A) 93

41. Triangle Axioms satisfied in K(A)

Theorem 41.1. The distinguished triangles in K(A) satisfy the triangle axioms(TR0)-(TR5).

Proof. (TR0) and (TR2) follows from the definition of distinguished triangles.

(TR3) is Lemma 37.4. For (TR1), if 0 : 0· → X· is the zero map, then M(0)· =

X·, and we have a distinguished triangle

0· 0−→ X· α(0)=idX·−−−−−−−→M(0)· β(0)=0−−−−→ 0[1]·,

giving a distinguished triangle X· idX·−−−→ X· → 0· → X[1]·.. For (TR4), assume

Z· = M(u)·, Z ′· = M(u′)· have µ u = u′ φ in K(A). So there exists ahomotopy kn : Xn → Y ′n−1 such that

µn un − u′n φn = dn−1Y ′ k

n + kn+1 dnX .Define

νn : Zn = Xn+1 ⊕ Y n

φn+1 0−kn+1 µn

−−−−−−−−−−−→ Z ′n = X ′n+1 ⊕ Y ′n

Check: ν defines a cochain map and makes the diagram of triangles commute inK(A).

For (TR5), let Z′· = M(u)·, X ′· = M(v)·, Y ′· = M(v u)·, and define

fn : Z ′n = Xn+1 ⊕ Y n

1 00 vn

−−−−−−−→ Y ′n = Xn+1 ⊕ Zn

and

gn : y′n = Xn+1 ⊕ Zn

un+1 00 1

−−−−−−−−−→ X ′n = Y n+1 ⊕ Zn.

Define h = α(u)[1] β(v). Then

hn : X ′n = Y n+1 ⊕ Zn

0 0−1 0

−−−−−−−→ Z ′[1]· = Xn+2 ⊕ Y n+1

Check: This makes the big diagram commute in K(A).Check: There exists φ, ψ that make the diagram commute in K(A) such that

φ ψ ∼ idX′· and ψ φ ∼ idM(f)· .

Z′· Y

′· X′· Z ′[1]·

Z′· Y

′· M(f)· Z ′[1]·

f g h

ψ

f α(f) β(f)

φ

Try

φ =

(1 1 u[1] 00 0 0 1

)and ψ =

0 01 00 00 1

.


42. Triangulated Categories

Definition 42.1. A triangulated category is an additive category C togetherwith

(1) an additive automorphism T : C → C of categories, called shift or trans-lation functor. (sometimes T is denoted by [1], and T k is denoted by[k]).

(2) a collection of triangles in C called distinguished triangles such that thetriangle axioms are satisfied.

Notation. We sometimes write a triangle Xu−→ Y

v−→ Zw−→ T (X) as

Z

X Y

+1

Example 42.2. If A is an abelian category, then K(A) is a triangulated cate-gory.

Definition 42.3.

(a) Let (C, T ) and (C′, T ′) be triangulated categories. An additive covariantfunctor F : C → C′ is called a covariant ∂-functor (or a functor of trian-gulated categories) if F T = T ′ F and F sends distinguished trianglesin C to distinguished triangles in C′.

(b) Let (C, T ) be a triangulated category, let A be an abelian category. Anadditive covariant (resp. contravariant) functor H : C→ A is called a co-variant (resp. contravariant) cohomological functor if for all distinguished

triangles Xu−→ Y

v−→ Zw−→ T (X), there is a long exact sequence in A:

· · · −→ H(T i(X)) H(T i(Y )) H(T i(Z))

H(T i+1(X)) H(T i+1(Y )) H(T i+1(Z)) −→ · · ·

H(T iu) H(T iv)

H(T iw)

H(T i+1u) H(T i+1v)

(resp.

· · · −→ H(T i(Z)) H(T i(Y )) H(T i(X))

H(T i−1(Z)) H(T i−1(Y )) H(T i−1(X)) −→ · · ·

H(T iv) H(T iu)

H(T i−1w)

H(T i+1v) H(T i−1u)

)

Example 42.4. Let A be an abelian category. Then H0 : K(A) → A is a

covariant cohomological functor since H0(X[i]·) = Hi(X·) for all i ∈ Z and for

all X· ∈ Ob(K(A)).

Remark 42.5. Let (C, T ) be a triangulated category, let A be an abelian cat-egory, let H : C → A be an additive covariant (resp. contravariant) functor.Then H is a cohomological functor if and only if for every distinguished trian-

gle Xu−→ Y

v−→ Zw−→ T (X), the sequence H(X)

H(u)−−−→ H(Y )H(v)−−−→ H(Z) (resp.

H(Z)H(v)−−−→ H(Y )

H(u)−−−→ H(X)) is exact in A.

42. TRIANGULATED CATEGORIES 95

Proposition 42.6. Let (C, T ) be a triangulated category.

(a) If 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. contravariant) cohomological functor C→ Ab.

Proof.

(a) By (TR3), it suffices to show v u = 0. The top triangle (top row) isdistinguished by (TR1):

X X 0 T (X)

X Y Z T (X)

idX

u ∃ν

u v

By (TR4), there exists ν : 0 → Z in C making the diagram commute inC. Hence v u = ν 0 = 0.

(b) We show this for HomC(M,−). Let Xu−→ Y

v−→ Zw−→ T (X) be a distin-

guished triangle. To show

HomC(M,X)u∗−→ HomC(M,Y )

v∗−→ HomC(M,Z)

is exact in Ab. We have v∗ u∗ = (v u)∗ = 0, where the last equalityfollows from part (a). Hence Im(u∗) ⊆ Ker(v∗). Let f : M → Y withv∗(f) = 0, i.e. v f = 0. The top triangle (top row) is distinguished by(TR1):

M M 0 T (M)

X Y Z T (X)

∃g

idM

f 0 T (g)

u v w

By (TR3) and (TR4), there exists g : M → X in C making the diagramcommute in C. Hence f = u g = u∗(g), i.e. f ∈ Im(u∗).

Remark 42.7. Let A be an abelian category, let f : X· → Y · be in C(A).We have a distinguished triangle

X· Y · M(f)· X[1]·f α(f)0

1

β(f)(−1 0

)

Now, α(f) f =

(0f

)6=(

00

)in C(A). Show:

(0f

)∼(

00

)Lemma 42.8. Let (C, T ) be a triangulated category. Given a morphism between

distinguished triangles in C:

(42.1)

X Y Z T (X)

X ′ Y ′ Z ′ T (X ′)

u

φ

v

µ

w

ν T (φ)

u′ v′ w′


If φ and µ are isomorphisms in C, then so is ν. 1

Proof. Apply HomC(Z ′,−) to the diagram (42.1) to get a commutative dia-gram inAb with exact rows.

HomC(Z ′, X) HomC(Z ′, Y ) HomC(Z ′, Z) HomC(Z ′, T (X)) HomC(Z ′, T (Y ))

HomC(Z ′, X ′) HomC(Z ′, Y ′) HomC(Z ′, Z ′) HomC(Z ′, T (X ′)) HomC(Z ′, T (Y ′))

u∗

φ∗

v∗

µ∗

w∗

ν∗

T (u)∗

T (φ)∗ T (µ)∗

u′∗ v′∗ w′∗ T (u′)∗

Since φ, µ are isomorphisms in C, then φ∗, µ∗, T (φ)∗, T (µ)∗ are isomorphisms inAb. By the 5-Lemma, ν∗ is an isomorphism. Hence there exists ζ : Z ′ → Z in C

with ν ζ = ν∗(ζ) = idZ′ .Now we apply HomC(−, Z) to the diagram (42.1) to obtain that ν∗ is an iso-

morphism in Ab. Hence there exists ξ : Z ′ → Z in C with ξ ν = ν∗(ξ) = idZ . Wehave

ξ = ξ (ν ζ) = (ξ ν) ζ = ζ.

43. Localization of Categories

Definition 43.1. Let C be a category, let S be a collection of morphisms inC. Then S is called a multiplicative system if (S1)-(S4) are satisfied:

(S1) idX ∈ S for all X ∈ Ob(C)(S2) If f, g ∈ S and g f is defined, then g f ∈ S.(S3) Given a diagram in C

Z

Z Y

f

s

with s ∈ S, we can complete this to a commutative diagram in C

W Z

Z Y

t

g f

s

with t ∈ S. Ditto for the diagrams with all arrows reversed.(S4) Given f, g : X → Y in C, the following are equivalent:

(i) There exists s : X ′ → X in S with f s = g s.(ii) There exists t : Y → Y ′ in S with t f = t g.

Definition 43.2. Let C be a category, let S be a multiplicative system ofmorphisms in C. The localization of C with respect to S is a category CS togetherwith a functor Q : C→ CS , called the localization functor , such that

(i) Q(s) is an isomorphism in CS for all s ∈ S.(ii) For every category D, and for every functor F : C → D such that for

all s ∈ S, F (s) is an isomorphism in D, there exists a unique functorG : CS → D with G Q = F .

1In fact, if any two vertical arrows in the diagram are isomorphisms, then the third verticalarrow is an isomorphism, using (TR3).

43. LOCALIZATION OF CATEGORIES 97

Moreover, if C is an additive category, then CS is an additive category.

Proposition 43.3. Let C be a category, let S be a multiplicative system ofmorphisms in C. Then the localization CS can be obtained as follows: Ob(CS) =Ob(C), and for all X,Y ∈ Ob(CS),

HomCS (X,Y ) =

X ′

X Y

s f

∣∣∣∣∣ X ′ ∈ Ob(CS),s, f morphisms in C,

s ∈ S.

/∼

where

X ′

X Y

s f ∼ X ′′

X Y

t g

if there exists X ′′′ ∈ Ob(C) and morphisms u : X ′′′ → X ′ in S and h : X ′′′ → X ′′

in C such that the following diagram commutes:

X ′′′

X ′ X ′′

X Y

u h

fs

tg

i.e., s u = t h and f u = g h.The composition of two morphisms

X ′

X Y

s f andY ′

Y Z

t g

in CS (so s, t ∈ S) is defined as follows: By (S3), there exists a commutativediagram in C:

Z ′

X ′ Y ′

X Y Z

u h

s f t g

where u ∈ S. Define Y ′

Y Z

t g

X ′

X Y

s f

=Z ′

X Z

su gh

Proof sketch. Check: CS is a category (in particular, show that compositionof morphisms is well-defined.) Note that the identity of X ∈ Ob(CS) in CS is

represented byX

X X

idX idX . Define Q : C→ CS by Q(X) = X for all X ∈ Ob(C)

and for all f ∈ HomC(X,Y ), define

Q(f) =X

X Y

idX f

Check: Q is a functor. We will write an equal sign for the identity morphism.


Let s : X → Y be in S. ThenQ(s) =X

X Y

s andQ(s)−1 =X

Y X

s

sinceX

X X

X Y X

s s=

X

X X

andX

X X

Y X Ys s

=X

Y Y

s s ∼ Y

Y Y

Let F : C → D be a functor such that F (s) is an isomorphism in D for alls ∈ S. Then define G : CS → D by G(X) = F (X) for all X ∈ Ob(C) and

G

X ′

X Y

s f

= F

X ′

X ′ Y

f

X ′

X X

s

−1 = F (f)F (s)−1.

Check: G is a functor and G is unique with G Q = F . Last thing to check: C

additive =⇒ CS additive.Check that Q(0) is the zero object in CS . Also check that if X,Y ∈ Ob(C) and

(X∏Y, pX , pY ) is the product of X and Y in C, then (Q(X

∏Y ), Q(pX), Q(pY ))

is the product of X and Y in CS .Finally, check the details of the following, showing that CS is an Ab-category:

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

where the first equality is obtained using the following commutative diagram, which

exists by (S3):

X ′′

X ′ X ′′

X

u v

s t

where u ∈ S, and the second equality follows since

su = tv.

44. LOCALIZATION OF TRIANGULATED CATEGORIES 99

Proposition 43.4. Let C be a category, let S be a multiplicative system ofmorphisms in C. Let C′ be a full subcategory of C. Let S′ = S ∩C′. Suppose S′ is amultiplicative system of C′ such that one of the following two conditions is satisfied:

(i) For all s : X → X ′ in S with X ′ ∈ Ob(C′), there exists f : X ′′ → X in C

such that X ′′ ∈ Ob(C′) and s f ∈ S.(ii) For all t : X ′ → X in S with X ′ ∈ Ob(C′, there exists g : X → X ′′ in C

such that X ′′ ∈ Ob(C′) and g t ∈ S.

Then C′S′ is a full subcategory of CS, in the sense that the natural functor C′S′ → CSis fully faithful.

Proof. Exercise.

44. Localization of Triangulated Categories

Definition 44.1. Let (C, T ) be a triangulated category, let S be a multiplica-tive system of morphisms. We say S is compatible with the triangulation if S satisfies(S5) and (S6):

(S5) T (s) ∈ S for all s ∈ S.

(S6) Given two distinguished triangles Xu−→ Y

v−→ Zw−→ T (X) and X ′

u′−→Y ′

v′−→ Z ′w′−→ T (X ′) in C together with morphisms f : X → X ′ and

g : Y → Y ′ in S such that g u = u′ f , there exists h : Z → Z ′ in S suchthat we have a morphism of triangles:

X Y Z T (X)

X ′ Y ′ Z T (X ′)

f

u

g

v

h

w

T (f)

u′ v′ w′

Proposition 44.2. Let (C, T ) be a triangulated category, let S be a multiplica-tive system of morphisms in C that is compatible with triangulation. Then CS hasa unique structure of a triangulated category such that

• Q : C→ CS us a ∂-functor, and• For every ∂-functor F : C → D of triangulated categories such that F (s)

is an isomorphism in D for all s ∈ S, there exists a unique functor G :CS → D such that G Q = F .

Proof sketch. Define the distinguished triangles in CS to be those that areisomorphic to images of distinguished triangles in C under Q. Check (TR0)-(TR5).Check that Q is a ∂-functor by (S5), (S6). Check the universal property withrespect to ∂-functors.

Theorem 44.3. Let A be an abelian category, let (C, T ) be a triangulated cat-egory, and let H : C→ A be a cohomological (covariant) functor. Define

S =s morphism in C | H(T i(s)) is an isomorphism for all i ∈ Z

.

Then S is a multiplicative system that is compatible with triangulation.

Proof. We check (S1)-(S6).


(S1) Let X ∈ Ob(C). Then XidX−−→ X

0−→ 00−→ T (x) is a distinguished triangle.

Since H is a cohomological functor, we get an exact sequence

· · · −→ H(T iX)H(T i idX)−−−−−−→ H(T iX) −→ 0 · · · .

Hence H(T i idX) is an isomorphism for all i ∈ Z, and so idX ∈ S.(S2) Let f, g ∈ S such that g f is defined. Then for all i ∈ Z,

H(T i(g f)) = H(T i(g)) H(T i(f))

is an isomorphism (since both H(T i(g)) and H(T i(f)) are isomorphisms).So g f ∈ S

(S3) Given a diagram in C, with s ∈ S,

Z

X Y

s

u

the triangle axiom (TR2) gives a distinguished triangle

Zs−→ Y

f−→ Ug−→ T (Z).

By (TR2) and (TR3), there exists a distinguished triangle

Wt−→ X

fu−−→ Uv−→ T (W ).

Note that since s ∈ S, H(T i(s)) is an isomorphism for all i ∈ Z. Sincewe have the exact sequence

· · · → H(T i−1U → H(T iZ)H(T i(s))−−−−−−→ H(T iY )→ H(T iU)→ · · · ,

then H(T iU) = 0 for all i ∈ Z. And so, since we also have the exactsequence

· · · → H(T i−1U)︸︷︷︸=0

→ H(T iW )H(T i(t))−−−−−−→ H(T iX)→ H(T iU)︸︷︷︸

=0

→ · · · ,

we get that H(T i(t)) is an isomorphism for all i ∈ Z, and so t ∈ S.We get a commutative diagram

W X U T (W )

Z Y U T (Z)

t fu

u

v

s f g

By (TR3) and (TR4), there exists w : W → Z in C with sw = ut.(S4) Since C is additive, it suffices to shoe the following: Given : X → Y in C,

TFAE:(i) There exists s : Y → Y ′ in S with s f = 0.

(ii) There exists t : X ′ → X in S with f t = 0.We prove (i) =⇒ (ii). ( (ii) =⇒ (i) ) is similar. We have

Xf−→ Y

s−→ Y ′

44. LOCALIZATION OF TRIANGULATED CATEGORIES 101

By (TR2) and (TR3), there exists a distinguished triangle Zu−→ Y

s−→Y ′

v−→ T (Z). Apply the cohomological functor HomC(X,−) to get anexact sequence in Ab:

HomC(X,Z)u∗−→ HomC(X,Y )

s∗−→ HomC(X,Y ′).

Since 0 = sf = s∗(f), there exists g : X → Z in C with u∗(g) = ug = f .By (TR2) and (TR3), there exists a distinguished triangle

X ′t−→ X

g−→ Z → T (X ′)

Since s ∈ S, we have H(T i(s)) for all i ∈ Z. (Compare to our argumentsfor (S3)). Hence H(T i(t)) is an isomorphism for all i ∈ Z, and so t ∈ S.Apply HomC(−, Y ) to the distinguished triangle above to get an exactsequence in Ab:

HomC(Z, Y )g∗−→ HomC(X,Y )

t∗−→ HomC(X ′, Y ).

Hencef t = t∗(f) = t∗(u g) = t∗(g∗(u)) = 0.

(S5)

s ∈ S ⇐⇒ H(T i(s)) is an isomorphism for all i ∈ Z

⇐⇒ H(T i(T (s))) is an isomorphism for all i ∈ Z⇐⇒ T (s) ∈ S.

(S6) By (TR4), it is enough to show: Given a morphism of triangles

X Y Z T (X)

X ′ Y ′ Z ′ T (X ′)

u

f

v

g

w

h T (f)

u′ v′ w′

with f, g ∈ S, we need to show h ∈ S.Since H is a cohomological functor, we get a commutative diagram in

A for all i ∈ Z with exact rows:

H(T iX) H(T iY ) H(T iZ) H(T i+1X) H(T i+1Y )

H(T iX ′) H(T iY ′) H(T iZ ′) H(T i+1X ′) H(T i+1Y ′)

H(T if) ∼= H(T ig) ∼= H(T ih) H(T i+1f) ∼= H(T i+1g) ∼=

The vertical isomorphisms come from our assumption that f, g ∈ S. Bythe 5-Lemma, H(T ih) is an isomorphism. This is true for all i ∈ Z, andhence h ∈ S.

Corollary 44.4. Let A be an abelian category, let K(A) be the homotopy cate-gory of C(A). Let Qis be the collection of quasi-isomorphisms in K(A). Then Qis isa multiplicative system of morphisms in K(A) that is compatible with triangulation.

CHAPTER 7

Derived Categories

45. Derived Categories

Definition 45.1. Let A be an abelian category. The derived category of A,denoted by D(A), is defined as

D(A) = K(A)Qis.

The additive functor Hn : K(A) → A (for all n ∈ Z) satisfies Hn(s) is an isomor-phism for all s ∈ Qis. By the universal property of Q, Hn factors through D(A).We denote the resulting additive functor again by Hn : D(A)→ A.

LetKb(A)

K+(A)

K−(A)

be the full subcategoryof K(A) consisting of

bounded complexes

bounded below complexes

bounded above complexes

.

Define

D−(A) = K−(A)Qis, D+(A) = K+(A)Qis, Db(A) = Kb(A)Qis.

Proposition 45.2. 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 for all n >> 0(resp. for all n << 0, resp. for all |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 with

Hn(X·) = 0 for all n 6= 0.

Proof. This follows from our earlier result about localizing full subcategories.(Proposition 43.4)

Remark 45.3. Let Q : K(A)→ D(A) be the localization functor.

(a) Let X· ∈ Ob(K(A)). Then Q(X·) = 0 in D(A) if and only if X· isquasi-isomorphic to 0 in K(A).

(b) Let f : X· → Y · be in C(A) with corresponding morphism f = f +

Ht(X·, Y ·) in K(A). Then

Q(f) = 0 in D(A) ⇐⇒ there exists a qis s : X′· → X· in C(A) with f s ∼ 0

⇐⇒ there exists a qis t : Y · → Y′· in C(A) with t f ∼ 0.

Proof.

103

104 7. DERIVED CATEGORIES

(a) (⇐) Clear, since quasi-isomorphisms in K(A) become isomorphisms inD(A).

(⇒) Q(X·) = 0 in D(A) means that there existsX′·

X· 0·s f

with inverseX′′·

0· X·t g . So there exists Z·, a qis u : Z· → X

′·,and a morphism h : Z· → X

′′· such that f u = t h in K(A).

Z·X′· X

′′·X· 0· X·

u h

s f t g

Moreover, by the definition idX· in D(A), s u = g h in K(A). Thenfor all n ∈ Z,

Hn(Z·) Hn(X·)

Hn(X′′·)︸︷︷︸

=0

since t:X′′·→0

is a qis

Hn(su)

∼=

Hn(h) Hn(g)

Hence Hn(h) = 0 = Hn(g), and so Hn(su) = 0 giving Hn(X·) = 0 for

all n ∈ Z. Therefore, 0 : X· → 0· is a qis. (Being quasi-isomorphic to 0

just means X· is acyclic, i.e., all the cohomology groups vanish.)(b) The second equivalence follows from (S4). For the first equivalence (⇐) ,

since s is a quasi-isomorphism in C(A), then Q(s) is an isomorphism, so

0 = Q(f s) = Q(f) Q(s) =⇒ 0 = 0 Q(s)−1 = Q(f).

(⇒) Q(f) = 0 in D(A) means

X′·

X· Y ·f ∼ X

′′·X· Y ·s′

0

where s′ is a qis in K(A). So there exists X′·, s, g in K(A) with f s =

0 g = 0, and hence f s ∼ 0.

X′·

X· X′′·

X· Y ·

s g

f s′

0

45. DERIVED CATEGORIES 105

Remark 45.4. If f, g : X· → Y · are in C(A) and f ∼ g, then Q(f) = Q(g)in D(A). However, the converse is false in general.

Example 45.5. Let

X· : · · · → 0→ 0→ Z ·2−→ Z π−→ Z/2→ 0→ 0→ · · ·

Then X· is acyclic, i.e. Hn(X·) = 0 for all n ∈ Z. Let f = idX and

t : X· → 0· be the zero map. Then t is a qis, so t f = 0, and hence Q(f) = 0 inD(A). But f 6∼ 0 since such a homotopy would require a nonzero map Z/2→ Z.

Theorem 45.6. Let A be an abelian category. Suppose there exists a full ad-ditive subcategory I such that for all X ∈ Ob(A), there exists E ∈ Ob(I) and amonomorphism f : X → E. Then

(a) For all X· ∈ Ob(K+(A)), there exists E· ∈ Ob(K+(I)) and a qis f :

X· → E·.(b) The natural functor D+(I)→ D+(A) is an equivalence of categories.

Proof.

(a) Without loss of generality, assume Xn = 0 for all n < 0. Define En = 0,fn = 0 for all n < 0.

X· · · · 0 0 X0 X1 X2 · · ·

E· · · · 0 0 E0

f· 0 0 f0

d0X d1

X

By our assumption, there exists E0 ∈ Ob(I) and a monomorphism f0 :X0 → E0.

Suppose we have constructed En, fn, dn−1E . Consider the induced

morphisms

dnX : Coker(dn−1X )→ 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

dnX

fn kn+1

`n

Again by assumption, there exists En+1 ∈ Ob(I) and a monomorphismgn+1 : Zn+1 → En+1. Define fn+1 := gn+1kn+1 and dnE := gn+1`nπn.

So dnE dn+1E = 0.

Coker(dn−1X ) Xn+1

En Coker(dn−1E ) Zn+1 En+1

dnX

fn kn+1

fn+1

dnE

πn `n gn+1

We prove by induction:• fn is a monomorphism.


• fn dn−1X = dn−1

E fn−1.

• Hn−1(f) : Hn−1(X·)→ Hn−1(E·) is an isomorphism.We prove this in R-Mod and then apply Freyd-Mitchell (32.13). We havea commutative diagram

0 Hn(X·) Xn/

Im(dn−1X ) Xn+1

0 Ker(`n) En/

Im(dn−1E ) Zn+1

0 Hn(E·) En/

Im(dn−1E ) En+1

dnX

fn

fn+1

kn+1

`n

gn+1

dnE

where

dnE = gn+1 `n and dnE = dnE πn

We have

Zn+1 =En/

Im(dn−1E )⊕Xn+1

(fn(x) + Im(dn−1

E ),−dnX(x))| x ∈ Xn

• fn+1 is a monomorphism:Let x ∈ Xn+1 with fn+1(x) Im(dnE) i.e. there exists x0 ∈ En with

gn+1(kn+1(x)) = fn+1(x) = gn+1(`n(x0 + Im(dn−1E )))

Since gn+1 is monic, kn+1(x) = `n(x0 + Im(dn−1E )). There exists

y ∈ Xn with

(x0 + Im(dn−1E ),−x) = (fn(y) + Im()dn−1

E ,−dnX(y)).

Hence x = dnX(y) and so x+Im(dnX) = 0+Im(dnX). So fn+1 is monic.

• fn+1 dnX = dnE fn follows since fn+1 dnX = dnE fn by thecommutativity of the diagram.

• Hn(f) is an isomorphism:Since fn is monic, Hn(f) is monic. To show it is epic, let z ∈ Enwith dnE(z) = 0. Hence gn+1(`n(z + Im(dn−1

E ))) = 0. Since gn+1 is

monic, `n(z + Im(dn−1E )) = 0. So there exists a ∈ Xn with

(z + Im(dn−1E ), 0) = (fn(a) + Im(dn−1

E ),−dX(a)).

Then dnX(a) = 0 so [a] ∈ Hn(X·), and z+Im(dn−1E ) = fn(a) Im(dn−1

E )so [z] = Hn(f)([a]).

(b) Note that K+(I) is a triangulated category and Qis∩K+(I) forms a mul-tiplicative system that is compatible with triangulation. To show thenatural functor D+(I) → D+(A) is fully faithful, we show the following(which comes from an earlier result on how to localize full subcategories):

If s : X′· → X· is a qis in K+(A) and X

′· ∈ Ob(K+(I) then there

exists f : X· → X′′· in K+(I) such that X

′′· ∈ Ob(K+(I)) and f s isa qis.

But the above statement follows directly from (a). To show that

D+(I) → D+(A) is dense, let X· ∈ Ob(D+(A)) = Ob(K(A)). By (a),

47. DERIVED FUNCTORS 107

there exists E· ∈ Ob(K+(I) = Ob(D+(I) and a qis f : X· → E·. Butf becomes an isomorphism when viewed as a morphism in D+(A).

We also have the dual to Theorem 45.6:

Theorem 45.7. Let A be an abelian category. Suppose there exists a full ad-ditive subcategory P such that for all X ∈ Ob(A), there exists P ∈ Ob(P) and aepimorphism f : P → X. Then

(a) For all X· ∈ Ob(K−(A)), there exists P· ∈ Ob(K−(P)) and a qis f :

P· → X·.(b) The natural functor D−(P)→ D−(A) is an equivalence of categories.

46. Enough Injectives or Projectives


(a) We say A has enough injectives if for all X ∈ Ob(A) there exists aninjective object E in A and a monomorphism f : X → E in A.

(b) We say A has enough projectives if for all X ∈ Ob(A), there exists aprojective object P in A and an epimorphism f : P → X.

Theorem 46.2. Let A be an abelian category.

(a) If A has enough injectives, let I be the full subcategory of injective objects.Then the natural functor K+(I)→ D+(A) is an equivalence of categories.

(b) If A has enough projectives, let P be the full subcategory of projectiveobjects. Then the natural functor K−(P) → D−(A) is an equivalence ofcategories.

Proof. We prove (a).If suffices to show that each qis in K+(I) is an isomorphism in K(I), since

then K+(I) = D+(I). By Lemma 38.3, every qis s : E· → E′· in C+(I) has a

2-sided homotopy inverse, i.e., s defines an isomorphism in K+(I).

Definition 46.3. Let A be an abelian category. Let K∗(A) be a full triangu-lated subcategory of K(A). We call K(A) a localizing subcategory of K(A) if thenatural functor

K∗(A)K∗(A)∩Qis −→ D(A)

is fully faithful. In this case, we define

D∗(A) := K∗(A)K∗(A)∩Qis.

Example 46.4. K+(A),K−(A),Kb(A) are localizing subcategories of K(A).Intersections of localizing subcategories are localizing.

47. Derived Functors

Motivation: let A and B be abelian categories. Let F : A→ B be an additivefunctor. Then F induces an additive functor F : C(A) → C(B). Since F sendshomotopies 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 map-ping cones, it also sends distinguished triangles in K(A) to distinguished trianglesin K(B). Hence F : K(A)→ F (B) is a (covariant) ∂-functor. Our goal is to extendF in some way to a ∂-functor D(A)→ D(B).


Example 47.1. F sends qis in K(A) to qis in K(B) (e.g. if F is an exactfunctor). Hence QB F sends qis in K(A) to isomorphisms in D(B), so by the

universal property of localization functors, there exists a unique functor F : D(A)→D(B) that makes the diagram commute:

K(A) K(B)

D(A) D(B)

F

QA QB

∃!F

Definition 47.2. Let A,B be abelian categories. Let K∗(A) be a localizingsubcategory 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)

Q

F

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 Qsatisfying the following universal property:

For every covariant ∂-functor G : D∗(A) → D(B) with a naturaltransformation ζ : Q F → F Q, there exists a unique natural transfor-mation 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)

together with a natural transformation

ξ : L∗F Q −→ Q Fsatisfying the following universal property:

For every covariant ∂-functor G : D∗(A) → D(B) with a naturaltransformation ζ : G Q→ Q F , there exists a unique natural transfor-mation η : G→ L∗F such that ξ (η Q) = ζ.

Remark. If R∗F (resp. L∗F ) exists, it is unique up to natural isomorphism.

48. EXISTENCE THEOREM FOR DERIVED FUNCTORS 109

Example 47.3. Let A be an abelian category. Let I be the full subcategoryof injective objects. Let F : K+(I)→ K(B) be a covariant ∂-functor.

We know Q : K+(I) → d+(I) is an isomorphism of categories. We can formQFQ−1:

D+(I) K+(I) K(B) D(B)∼=Q−1

QFQ−1

F Q

Then (QFQ−1) Q = Q F , and hence QFQ−1 is both the left and right derivedfunctor of F . (check the details).

Dually, if P is the full subcategory of A of projective objects and F : K−(P)→K(B) is a covariant ∂-functor, thenQFQ−1 is both the left and right derived functorof F , where

D−(P) K−(P) K(B) D(B)∼=Q−1

QFQ−1

F Q

Notation. If there is no confusion, write RF for R∗F and LF for L∗F .

Definition 47.4. We have the following hyper derived functors:

RiF := Hi(RF ) and LiF := H−i(LF ).

48. Existence Theorem for Derived Functors

Theorem 48.1. Let A,B be abelian categories.

(a) Let F : K+(A) → K(B) be a covariant ∂-functor. If A has enoughinjectives, then the right derived functor R+F : D+(A) → D(B) exists.

Moreover, if E· is a bounded below 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, then the left derived functor L−F : D−(A) → D(B) exists.

Moreover, if P· is a bounded above 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)

(where α = Q∣∣K+(I)

for Q : K+(A)→ D+(A) the localizing functor) is an equiva-

lence of categories. Let β : D+(A)→ K+(I) be a quasi-inverse, let τ : IdD+(A) →α β be a natural isomorphism.

Define R+F as the following composition:

D+(A) K+(I) K(B) D(B)β

R+F

F Q

Claim 48.1.

HomK+(A)(X·, βQX·) ∼=−→ HomD+(A)(QX·, αβX·)f 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)). By the claim, there exits a unique fX· ∈HomK+(A)(X·, βQX·) such that Q(fX·) = τ

QX· . Define ξX· := (Q F )(fX·).Since τ is a natural transformation, then ξ is a natural transformation (check).

We check the universal property: Let G : D+(A) → D(B) be a covariant ∂-functor with a natural transformation ζ : Q F → F Q. We need to define

η : R+F = Q F β → G. Let X· ∈ Ob(D+(A)). Then define

Q(F (βX·)) G(Q(βX·)) = G(Q(β(Q(X·)))) G(QX·) = GX·ζβX·

ηX·

G(Q(fX· )−1)

(Note: Q(fX·)−1 = τ−1QX·). Since ζ and τ are natural transformations, also η is a

natural transformation.Check: ηQX· ηX· = ξX· and η is unique with this property. Hence (R+F, ξ)

is the right derived functor of F .

Let E· be a bounded below complex of injectives. Then

(R+F )(E·) = Q(F (βE·))= Q(F ((β α)(E·)))∼= Q(F (E·))

where the last isomorphism follows since β α ∼= IdK+(I). We now prove the claim

Proof of Claim 48.1. We can define morphisms in D+(A) also as

HomD+(A)(U·, V ·) =

U· V ·W·f t

∣∣∣∣∣ f, t,W· in K+(A),t a qis.

/∼

where

U· V ·W·f t

∼ U· V ·W′·f u

if there exists Z· ∈ Ob(K+(A)) and morphisms v : W′· → Z·, h : W· → Z· in

K+(A), v a qis, such that the following diagram commutes:

U· V ·

W· W′·

Z·

f

g t

u

h v

i.e., v g = h f and h g = v u.

49. EXT AND RHOM: HYPEREXT 111

Now Q(f) = Q(g) in the claim means

X· βQX·

βQX· βQX·

Z·

f

g

t t

Hence t f = t g in K+(A). Since t is a qis and βQX· is in K+(I), then t has aleft homotopy inverse, s. So in K+(A)

f = s t f = s t g = g.

Let

X· βQX·W·f t

be an arbitrary element of HomD+(A)(QX·, αβQX·). Since βQX· is in K+(I),t has a left homotopy inverse, s. Hence

X· βQX·W·f t

∼X· βQX·

βQX·sf= Q(s f).

49. Ext and RHom: Hyperext

Definition 49.1. Let A be an abelian category. Let X·, Y · ∈ Ob(D(A)).

For all n ∈ Z, define the nth hyperext of X· and Y · to be

Extn(X·, Y ·) := ExtD(A)(X·, Y ·) := HomD(A)(X·, Y [n]·)Remark 49.2.

(1) Recall: D+(A) is a full subcategory of D(A). So if X·, Y · ∈ Ob(D+(A)),then

Extn(X·, Y ·) = HomD+(A)(X·, Y [n]·).(2) Since A can be embedded as a full subcategory into D(A), we get a def-

inition of Extn(X,Y ) for all X,Y ∈ Ob(A). One of our goals is to showthat this coincides with the classical definition of ExtnA(X,Y ) using aninjective resolution of Y (provided A has enough injectives.)

Lemma 49.3. Let A be an abelian category. Let 0→ X· f−→ Y · g−→ Z· → 0 be

a short exact sequence in C(A). Then there exists 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· Y · M(f)· X[1]·f α(f)0

1

β(f)(−1 0

)

where

M(f)n = Xn+1 ⊕ Y n and dnM(f) =

(−dn+1

X 0

−fn+1 dnY

).

(we write as in R-Mod, using Freyd-Mitchell (32.13)). Define s : M(f)· → Z· by

sn : M(f)n = Xn+1 ⊕ Y n(0 gn

)−−−−−−→ Zn.

Now, s is a cochain map: Since gn+1 fn+1 = 0,

sn+1 dnM(f) =(0 gn+1 dnY

),

and

dnX sn =(0 dnZ gn

),

and gn+1 dnY = dnZ gn since g is a cochain map.

Claim 49.1. s is a qis.

Proof of Claim 49.1. Consider the long exact cohomology sequences asso-

ciated to 0 → X· f−→ Y · g−→ Z· → 0 and to 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((0 g))=Hn(s)

Hn(β(f))

−1

Hn+1(f)

−1

Hn(f) Hn(α(f)) ∂n Hn+1(f)

The first and fourth squares commute trivially. The second square commutessince (0 g) α(f) = g. For the third square, let z = ( z1z2 ) ∈ M(f)n = Xn+1 ⊕ Y nand dnM(f)(z) = 0, i.e.

dn+1X (z1) = 0 and dnY (z2) = fn+1(z1).

Now,

∂n((Hn(0 g))([z])) = ∂n([gn(z2)]).

Recall how ∂n is defined:

z2∈Y n Zn3gn(z2)

z1∈Xn+1fn+1(z1)∈Y n+1

dnY

gn

fn+1

Hence

∂n((Hn(0 g))([z])) = [z1] = −Hn(β(f))([z]),

51. TOTAL HOM COMPLEX 113

where the last equality follows since β(f) = (−1 0). So the third square commutes.By the 5-Lemma, Hn(s) = Hn((0 g)) is an isomorphism for all n ∈ Z. Hence s isa qis, proving the claim.

Consider the following diagram, the top row of which is a distinguished trianglein K(A) and D(A):

X· Y · M(f) X[1]·

X· Y · Z· X[1]·

f α(f) β(f)

s=(0 g)

f g β(f)s−1

The morphism β(f) s−1 exists in D(A) since s is an isomorphism in D(A).So the diagram commutes in D(A), which means the bottom row is a distinguishedtriangle in D(A). Define h := β(f) s−1.

50. Long exact hyperext sequences

Proposition 50.1. Let A be an abelian category. Let 0→ X· f−→ Y · g−→ Z· →0 be a short exact sequence in C(A). Then for all V · ∈ Ob(D(A)), we get longexact hyperext sequences

· · · −→ Exti(V ·, X·) Exti(V ·, Y ·) Exti(V ·, Z·)

Exti+1(V ·, X·) Exti+1(V ·, Y ·) Exti+1(V ·, Z·) −→ · · ·and

· · · −→ Exti(Z·, V ·) Exti(Y ·, V ·) Exti(X·, V ·)

Exti+1(Z·, V ·) Exti+1(Y ·, V ·) Exti+1(X·, V ·) −→ · · ·Proof. Use the distinguished triangle 0 → X· f−→ Y · g−→ Z· h−→ Z[1]· in

D(A) from Lemma 49.3, and that HomD(A)(V ·,−) and HomD(A)(−, V ·) are co-homological functors.

51. Total Hom complex

Let A be an abelian category, let X·, Y · ∈ C(A). Define a bicomplex C·,· asfollows:

Cp,q = HomA(X−p, Y q),

with

dp,qh : HomA(X−p, Y q) −→ HomA(X−p−1, Y q)

f 7−→ f d−p−1X


and

dp,qv : HomA(X−p, Y q) −→ HomA(X−p, Y q+1)

f 7−→ (−1)p+q+1dqY f

Check:

• dh dh = 0• dv dv = 0• dv dh + dh dv = 0

Define

Hom·(X·, Y ·) := TotΠ(C·,·),i.e.,

Homn(X·, Y ·) =∏

p+q=n

HomA(X−p, Y q)

=∏i∈Z

HomA(Xi, Y i+n).

with differential

dn : Homn(X·, Y ·) −→ Homn+1(X·, Y ·)(f−p,q) 7−→ (f−r+1,s d−rX + (−1)r+sds−1

Y f−r,s−1)

(p+ q = n) (r + s = n+ 1)

where each f−p,q is in HomA(X−p, Y q), i.e.

dn : (f i)i∈Z 7−→ (f i+1 diX + (−1)n+1di+nY f i)i∈Z

where f i ∈ HomA(Xi, Y i+n) for each i ∈ Z. Then our n-cocycles are

Zn Hom·(X·, Y ·) = Ker(dn)

= (f i)i∈Z | f i ∈ HomA(Xi, Y i+n) and f i+1 diX = (−1)ndi+nY f i= (f i)i∈Z | f i ∈ HomA(Xi, Y i+n) and f i+1 diX = diY [n] f

i

= HomC(A)(X·, Y [n]·).And we have n-coboundaries

Bn Hom·(X·, Y ·) = Im(dn−1)

= (f i+1 diX + (−1)ndi−1+nY f i)i∈Z | f i ∈ HomA(Xi, Y i−1+n)

= (f i+1 diX + di−1Y [n] f

i)i∈Z | f i ∈ HomA(Xi, Y i−1+n)

= Ht(X·, Y [n]·)Remark 51.1. The work above then gives

Hn Hom·(X·, Y ·) = HomK(A)(X·, Y [n]·)Note 51.2. Hom· : K(A)op ×K(A)→ K(Ab) is a bi-∂-functor.

52. RIGHT DERIVED BIFUNCTOR OF TOTAL HOM COMPLEX 115

52. Right Derived Bifunctor of Total Hom Complex

Definition 52.1. Let C1,C2,D be categories. A bifunctor F : C1×C2 → Ob(D)consists of

(a) a map F : Ob(C1)×Ob(C2)→ Ob(D)(b) for all Xi, Yi ∈ Ob(Ci) (i = 1, 2), a map

F : HomC1(X1, Y1)×HomC2

(X2, Y2) −→ HomD(F (X1, X2), F (Y1, Y2))

such that for all Xi, Yi ∈ Ob(Ci) and for all fi ∈ HomCi(Xi, Yi), (i = 1, 2), both

F (X1,−) : C2 −→ D

and

F (−, X2) : C1 −→ D

are covariant functors, and

F (Y1, f2) F (f1, X2) = F (f1, Y2) F (X1, f2).

A bifunctor is called additive, or exact, or a ∂-functor if this is the case with respectto each variable.

Suppose A is an abelian category having enough injectives. Let I be the full

subcategory of A of injective objects. Fix X· ∈ Ob(K(A)). Then

Hom·(X·,−) : K+(A) −→ K(Ab)

is a ∂-functor. So we can build a right derived functor

R+II Hom·(X·,−) : D+(A) −→ D(Ab)

as follows:

D+(A) K+(I) K(Ab) D(Ab)β

equiv.of cat.

R+II Hom·(X·,−)

Hom·(X·,−) Q

Fix Y · ∈ Ob(D+(A)). Then

R+II Hom·(−, Y ·) = Q Hom·(−, βY ·) : K(A)op −→ D(Ab)

is a ∂-functor. We get:

R+II Hom·(−,−) : K(A)op ×D+(A) −→ D(Ab)

is a bi-∂-functor (check).

Claim 52.1. Fix again Y · ∈ Ob(D+(A)). R+II Hom·(−, Y ·) sends qis to qis

(i.e. isomorphisms in D(Ab))


Proof of 52.1. Since s : A· → B· is a qis if and only if M(s)· is acyclic,

it suffices to show that R+II Hom·(−, Y ·) sends acyclic complexes to acyclic com-

plexes.

Let X· ∈ Ob(K(A)) be acyclic. Then for all n ∈ Z

Hn(R+II Hom·(X·, Y ·)) = Hn(QHom·(X·, βY ·))

= Hn(Hom·(X·, βY ·))= HomK(A)(X·, βY [n]·)= 0,

where the last equality follows from Claim 48.1.

So R+II Hom·(−, Y ·) passes to a functor, (namely, the right derived functor)

RIR+II Hom·(−, Y ·) : D(A)op −→ D(Ab).

Then we get a right derived bifunctor

R+ Hom· = RIR+II Hom· : D(A)op ×D+(A) −→ D(Ab).

Note 52.2.

• We have

R+ Hom·(X·, Y ·) = RIQHom·(X·, βY ·) ∼= Hom·(X·, βY ·).• If Y · ∈ Ob(D+(I)), then βY · ∼= βαY · ∼= Y ·.

Similarly, if A has enough projective objects, let P be the full subcategory ofprojective objects. Then we get a right derived bifunctor:

R−Hom· = RIIR−I Hom· : D−(A)op ×D(A) −→ D(Ab).

53. Yoneda’s Result on HiRHom

Theorem 53.1 (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 for all i ∈ Z

Hi(R+ Hom·(X·, Y ·)) ∼= ExtiD(A)(X·, Y ·) def= HomD(A)(X·, Y [i]·)

(b) If A has enough projectives, let X· ∈ Ob(D−(A)) and let Y · ∈ Ob(D(A)).Then for all i ∈ Z

Hi(R−Hom·(X·, Y ·)) ∼= ExtiD(A)(X·, Y ·) def= HomD(A)(X·, Y [i]·)

Proof. We prove (a).

Hi(R+ Hom·(X·, Y ·)) ∼= Hi(Hom·(X·, βY ·))∼= HomK(A)(X·, βY [i]·)∼= HomD(A)(QX·, αβY [i]·)∼= HomD(A)(X·, Y [i]·))= HomD(A)(X·, Y [i]·),

53. YONEDA’S RESULT ON HiRHom 117

where the second isomorphism follows from Remark 51.1, and we prove the thirdisomorphism below.

We will show that the following map is an isomorphism:

HomK(A)(X·, βY [i]·) −→ HomD(A)(QX·, αβY [i]·)f 7−→ Qf,

where α = Q|K+(I). Recall that

HomD(A)(QX·, αβY [i]·) =

X· βY [i]·W·f t

∣∣∣∣∣ f, t,W· in K(A),t a qis.

/∼

where

X· βY [i]·W·f t

∼X· βY [i]·

W′·f ′ t′

if there exists Z· ∈ Ob(K(A)) and morphisms u : W′· → Z·, g : W· → Z· in

K(A), u a qis, such that the following diagram commutes:

X· βY [i]·

W· W′·

Z·

f

gt

t′

g u

i.e., u t′ = g t and g f = u f ′.For injectivity, suppose Q(f) = Q(g). Then there exists a t making the diagram

commute:

X· βY [i]·

βY [i]· βY [i]·

Z·

f

g

t t

So tf = tg in K(A). Since βY [i]· ∈ Ob(K+(I)), t has a left homotopy inverse,and hence f = g in K(A).

Let

X· βY [i]·W·f t


be in HomD(A)(QX·, αβY [i]·). Again, since βY [i]· ∈ Ob(K+(I)), t has a lefthomotopy inverse, say s. Then

X· βY [i]·W·f t

∼X· βY [i]·

βY [i]·sf= Q(f).

Corollary 53.2. Let A be an abelian category. Let X,Y ∈ Ob(A). Let X·(resp. Y ·) be the corresponding complex in D(A), i.e.

X· : · · · 0 X 0 · · ·

(resp. Y · : · · · 0 Y 0 · · ·

)with X (resp. Y ) in the 0 degree.

If A has enough injectives (resp. enough projective) then for all i ≥ 0

ExtiD(A)(X·, Y ·) ∼= ExtiA(X,Y )

where ExtiA(X·, Y ·) is the classical Exti defined by taking an injective resolutionof Y (resp. a projective resolution of X).

Proof. We prove this when A has enough injectives. Let

0→ Yι−→ E0

deg. 0

d0E−−→ E1

textdeg.1

d1E−−→ E2

deg. 2

d2E−−→ · · ·

be an injective resolution of Y with corresponding truncated complex E·Y .

Then Y · ι−→ E·Y is a qis and E·Y ∼= βE·Y since E·Y ∈ Ob(K+I)). Now,

ExtiD(A)(X·, Y ·) ∼= ExtiD(A)(X·, E·Y )

∼= Hi(R+ Hom·(X·, E·Y ))

∼= Hi(Hom·(X·, βE·Y ))

∼= Hi(Hom·(X·, E·Y ))

= Hi(HomA(X,E·Y ))

= ExtiA(X,Y ).

where the second isomorphism follows from Theorem 53.1.

55. LEFT DERIVED BIFUNCTOR OF TOTAL TENSOR PRODUCT 119

54. Total Tensor Product Complex

Let R be a ring (with 1). Let X· ∈ Ob(C(Mod-R)), Y · ∈ Ob(C(R-Mod)).

Define a bicomplex 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)p idXp ⊗dqY : Xp ⊗R Y q −→ Xp ⊗R Y q+1

Define

X· ⊗R Y · := Tot⊕(C·,·),i.e.,

(X· ⊗R Y ·)n =⊕p,q

p+q=n

Xp ⊗R Y q,

and

dn : (X· ⊗R Y ·)n −→ (X· ⊗R Y ·)n+1

(zp,q)p+q=n 7−→(

(dr−1X ⊗ idY s)(zr−1,s) + ((−1)r idXr ⊗ds−1

Y )(zr,s−1))r+s=n+1

.

We get a bi-∂-functor

−⊗R − : K(Mod-R)×K(R-Mod) −→ K(Ab).

We want to construct a left derived bifunctor

−⊗LR − : D−(Mod-R)×D−(R-Mod) −→ D(Ab)

55. Left Derived Bifunctor of Total Tensor Product

We know R-Mod has enough projectives. Let RP be the full subcategory ofR-Mod of projective modules. We have seen earlier that the functor

α = Q|K−(RP) : K−(RP) −→ D−(R-Mod),

where Q : K−(R-Mod) → D−(R-Mod) is the localization functor, is an equiva-lence 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 derived functor

L−II(X· ⊗R −) : D−(R-Mod) −→ D(Ab)

defined as follows:

D−(R-Mod) K−(RP) K(Ab) D(Ab)γ

equiv.of cat.

L−II(X·⊗R−)

X·⊗R− Q

Then

L−II(X· ⊗R −) : K(Mod-R)×D−(R-Mod) −→ D(Ab)

is a bi-∂-functor (check).


Fix Y · ∈ Ob(D−(R-Mod)). Then

L−II(−⊗R Y ·) : K−(Mod-R) −→ D(Ab)

is a ∂-functor. One can show: L−II(− ⊗R Y ·) sends acyclic complexes to acycliccomplexes. (This is more involved than for RHom; it uses spectral sequences).

Hence L−II(−⊗R Y ·) sends qis to isomorphisms, and so it passes to a functor (theleft derived functor)

L−I (L−II(−⊗R Y ·)) : D−(Mod-R) −→ D(Ab).

Then we get a left derived bifunctor

−⊗LR − := L−I (L−II(−⊗R −)) : D−(Mod-R)×D−(R-Mod) −→ D(Ab).

Note 55.1. L−I (L−II(−⊗R −)) is naturally isomorphic to L−II(L−I (−⊗R −)).

How to compute − ⊗LR −:

LetX· ∈ Ob(D−(Mod-R), Y · ∈ Ob(D−(R-Mod)). Let P· ∈ Ob(K−(Mod-R))

(resp. Q· ∈ Ob(K−(R-Mod))) be quasi-isomorphic to X· (resp. Y ·). Then

X· ⊗LR Y · ∼= P· ⊗R Y · ∼= X· ⊗R Q·.56. Connection with Classical Tor

Recall. Hyperderived functors: Li(−⊗R −) := H−i(−⊗LR −), i ∈ Z.Proposition 56.1. Let A ∈ Ob(Mod-R), B ∈ Ob(R-Mod)) with associated

one term complexes A·, B·), respectively. Then for all i ≥ 0, we have

TorRi (A,B) ∼= H−i(A· ⊗LR B·)Proof. Consider a projective resolution of B is R-Mod:

· · · dP,3−−−→ P2dP,2−−−→ P1

dP,1−−−→ P0ε−→ B → 0

with truncated chain complex PB,· : · · · → P2dP,2−−−→ P1

dP,1−−−→ P0 → 0. We have the

associated cochain complex

S· : · · · → S

=

P2

−2 → S

=

P1

−1 → S

=

P0

0 → 0

with d−iS := dP,i. Then ε induces a qis,

S· = PB,· · · · P1 P0 0 · · ·

B· · · · 0 B 0 · · ·

ε 0 ε 0

Hence

H−i(A· ⊗LR B·) ∼= H−i(A· ⊗R S·)= H−i(A⊗R S·)= Hi(A⊗R PB,·)def= TorRi (A,B).

Index

Ab-category, 77abelian category, 79

cocomplete, 84complete, 84

abelian subcategory, 81acyclic, 87additive category, 77

Baer’s Criterion, 30bicomplex, 83

bounded, 83bifunctor

additive, 115∂-functor, 115exact, 115

bifunctors, 115bimodule, 11

category, 7localization of, 96of covariant functors, 10derived (D(A)), 86derived (D(A)), 103equivalence, 10, 51homotopy (K(A)), 85large, 50localization

of full subcategory, 99of triangulated category, 99

of abelian groups (Ab), 7of cochain complexes, 77, 81of direct systems, 20of functors, 10, 50of inverse systems, 20of modules (R-Mod or Mod-R),

7of sets (Sets), 7opposite, 9

small, 50triangulated, 94

chain complex, 55bounded below, 55bounded, 55bounded above, 55differential, 55

chain map, 55homotopy, 56

equivalent, 56character module, 38class

proper, 50axioms, 49, 50

cochain complex, 55bounded below, 55bounded, 55bounded above, 55quotient complex, 78subcomplex, 78

cochain map, 55homotopy, 56

equivalent, 56cogenerator, 34cohomology, 55cokernel, 78comparison theorem, 59

dual, 62

direct limit, 17of flat modules, 36

direct product, 13direct sum, 13direct system, 16

constant, 16dual statement, 9

enough injectives, 107

121

122 INDEX

enough projectives, 107epimorphism (epic), 8

essential, 28exact sequence, 81

of direct systems, 20of inverse systems, 20

Ext, 63example of, 64for abelian groups, 74

flatmodule, 38resolution, 36

Freyd-Mitchell embedding theorem,81

functorcontravariant, 8covariant, 8opposite (duality), 9additive, 77adjoint pair, 14, 15cohomological, 94dense, 51derived

existence theorem, 109direct limit, 20exact, 12flat, 35fully faithful, 51Hom, 8

in a module category, 11hyper derived, 109identity, 8left adjoint, 14left derived, 59, 108

bifunctor, of total tensorproduct, 119

left exact, 11localization, 96of triangulated categories (∂), 94representable, 52right adjoint, 14right derived, 62, 108

bifunctor, of total Homcomplex, 115

right exact, 12shift (translation), 84, 94tensor product, 11

Godel-Bernays system, 49generator, 34

homological algebra, 55homology, 55homotopic to zero, 85horseshoe lemma, 65

dual, 66hyperext, 111

injectivehull, 32module, 30object, 89resolution, 31

inverse limit, 17, 20inverse system, 16

constant, 16isomorphism, 7

natural, 10of categories, 10

J-adic completion, 19

kernel, 77

long exact cohomology sequence, 59long exact homology sequence, 58

mapping cone, 86Mittag-Leffler (ML) condition, 25module

cofree, 34monomorphism (monic), 8

essential, 31, 32Morita

context, 44equivalent, 44theory, 43

morphismR-balanced, 10identity, 7image of, 81in a category, 7natural, 14

multiplicative system, 96compatible with triangulation, 99

natural transformation, 9

object

INDEX 123

in a category, 7initial, 8terminal (final), 8

p-adic integers, 19Pontryagin dual, 35, 38poset

directed, 16with trivial ordering, 17

progenerator, 43projective

cover, 29generator, 34module, 27object, 89resolution, 27

pullback, 19pushout, 19

quasi-inverse, 10quasi-isomorphism (qis), 85

Schanuel’s Lemma, 40self-injective (ring), 42set

axioms, 49snake lemma, 56stable module category, 42subcategory, 7

full, 7localizing, 107

tensor product, 10Tor, 61

example of, 64connection with left derived

bifunctor of total tensorproduct, 120

for abelian groups, 74torsion

abelian group, 74free, 74submodule, 74

total complex, 83total hom complex, 113

cohomology groups, 114triangle(s), 91

morphism of, 91axioms, 92

in homotopy category, 93distinguished, 91, 94isomorphism of, 91

YonedaEmbedding, 54HiRHom, 116Lemma, 53

homological algebra - nicholas camacho · 2019-09-23 · chapter 4. classical homological algebra {...

Documents