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Homological Algebra

Instructor: Professor Claudia Miller & Typist: Caleb McWhorter

Spring 2015

Contents

1 Review/Disclaimer 3

1.1 Products and Coproducts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.6 Free Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.14 Exactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.18 Projective and Injective Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.32 Tensor Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.61 Flat Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.72 A Few Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2 Categories 28

2.1 Categories and Morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.31 Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.44 Natural Transformations between Functors . . . . . . . . . . . . . . . . . . . . . . . 33

2.49 Kernel and Cokernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3 Categorical Limits & Constructions 36

3.1 Additive Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.16 Adjoint Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.22 Colimit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.45 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.58 Exactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.63 Directed/Filtered Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.73 Flatness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4 Homology 52

4.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.2 Chain Complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4.20 Long Exact Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.23 Homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4.29 Bicomplexes and Tot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4.34 Mapping Cone & Mapping Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

1

Homological Algebra 2

5 Resolutions 625.1 Projective Resolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625.9 Injective Resolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 685.13 Left Derived Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 685.22 Balancing Tor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 725.28 Right Derived Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 745.32 Contravariant Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755.34 Balancing Ext . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755.44 Kunneth Formula and the Universal Coefficient Theorem . . . . . . . . . . . . . . . 785.51 Tor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 795.52 Ext and Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805.58 Baer Sum (1934) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 825.60 Yoneda Extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

6 The Derived Category 836.1 Derived Categories & Localization of Categories . . . . . . . . . . . . . . . . . . . . . 836.3 Localization of a Category . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 836.7 Homotopy Category, K(A) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 876.13 Distinguished Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 906.17 Rotating Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 926.19 Ext & D(A) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

7 Triangulated Categories 957.1 Distinguished Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 957.6 Cohomological Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 967.11 The Cone in Triangulated Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

8 Spectral Sequences 1018.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1018.2 Homology Spectral Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1038.6 Cohomology Spectral Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1048.10 Filtrations and Bounded Spectral Sequences . . . . . . . . . . . . . . . . . . . . . . . 105

3 C. McWhorter

1 Review/Disclaimer

For this section, indeed in much of the course, we assume all M,N are [left] R-modules. All mapsare assumed to be R-module homomorphisms. These following few sections (up to “A Few Notes”)are taken from the author’s notes from MAT 731: Rings & Modules taught by Professor DanZacharia with minor alteration of ordering as a “brief” review before the 732: Homological Algebramaterial. Any errors in this text should be attributed to the typist – Caleb McWhorter – and notthe instructor or any referenced text. Any credit as to the organization of the topics, choice ofexamples, et cetera should be attributed to the referenced texts as well as the instructor.

1.1 Products and Coproducts

First, note that I will denote any indexed set.

Definition 1.2 (Product). Let C be a category and {Ai | i ∈ I} be a family of objects of C. Aproduct for the family {Ai | i ∈ I} is an object P of C together with a family of morphisms.

{πi : P → Ai | i ∈ I}

such that for any object B and any family of morphisms

{ϕi : B → Ai | i ∈ I}

there is a unique morphism ϕ : B → P such that πi ◦ ϕ = ϕi for all i ∈ I. That is, there is aUniversal Mapping Property.

B P

Ai

ϕ

ϕi πi

ϕ(B) =∏i∈I

ϕi(B)

b 7→∏i∈I

ϕi(b)

A product P of {Ai | i ∈ I} is usually denoted∏i∈I Ai.

It is usually most helpful to describe products in terms of their commutative diagrams. Aproduct for {A1, A2} is a diagram of objects and morphisms A1

π1← Pπ2→ A2 such that for any other

diagram of the form A1ϕ1← B

ϕ2→ A2, there is a unique morphism ϕ : B → P such that the followingdiagram commutes:

P

A1 B A2

π1 π2ϕ

ϕ1 ϕ2

It is important to note that a product need not exist for a category. However, this will not bea problem for the category we will be working with – abelian groups & sets. For example in thecategory of sets, the Cartesian product

∏i∈I Ai is a product of the family {Ai | i ∈ I}.


Theorem 1.3. If (P, {πi}) and (Q, {ϕi}) are both products of the family {Ai | i ∈ I} objects of acategory C, then P and Q are equivalent.

Proof: Since P,Q are both products, they each have their family of morphisms to the Ai’s. Weobtain the following commutative diagrams:

P Q

Ai

f

πi ϕi

Q P

Ai

g

ϕi πi

Then g ◦ f : P → P , i.e

P P

Ai

g◦f

πi πi

Q Q

Ai

f◦g

ϕi ϕi

But by definition, such a morphism is unique. We have the map P1P−→ P . By uniqueness, we

know that g ◦ f = 1P . Similarly, we know that f ◦ g = 1Q. But then f, g are isomorphisms.

We also obtain the dual definition and theorem by reversing arrows in the definition and theoremabove.

Definition 1.4 (Coproduct). A coproduct (or sum) for the family {Ai | i ∈ I} of objects in acategory C is an object S of C together with a family of morphisms {τi : Ai → S | i ∈ I} such thatfor any object S and any family of morphisms {τi : Ai → S | i ∈ I} there is a unique morphismϕ : S → B such that

ϕ ◦ τi = ϕi

A coproduct S of {Ai | i ∈ I} is denoted⊕

i∈I Ai,∑

i∈I Ai, or sometimes qi∈IAi.

Notice again these do not assure existence, just uniqueness.

Theorem 1.5. If (S, {τi}) and (S′, {λi}) are both coproducts for the family {Ai | iI} of objects ofa category C, then S and S′ are equivalent.

Proof: Simply use the dual of the argument of Theorem 1.3.

Consider the following diagram:

Ai B

A1 ×A2 × · · · ⊃ S

ϕi

τiϕ

5 C. McWhorter

For ai ∈ Ai, let τi(ai) = 0 everywhere except for the ith factor, where it is ai. To make thediagram commute, we want ϕ to be ϕi in the ith factor. Then set

ϕ(a1, a2, · · · )def=∑i∈I

ϕi(ai)

Then S is the submodule of∏i∈I Ai consisting of all elements that are nonzero in at most finitely

many spots. This construction works for abelian groups and R-modules. It does not work fornonabelian groups. Together, this gives the following diagram

B P

S Aα

ϕ

ϕαπiϕ

τi

ϕi

1.6 Free Modules

Definition 1.7 (Linearly Independent). A subset X of an R-module M is said to be linearlyindependent if given distinct x1, x2, · · · , xn ∈ X and ri ∈ R, if r1x1 + r2x2 + · · ·+ rnxn = 0 then itmust be that ri = 0 for all i. Any subset of X which is not linearly independent is said to be linearlydependent.

If the subset M of X is generated as an R-module by a set A, then we say that A spans M .Furthermore, if R has identity (which we assume it does) and M is unitary, then A spans M if andonly if every element of M can be written as a linear combination r1a1 + r2a2 + · · ·+ rnan, whereri ∈ R and ai ∈ A. If A spans M and A is linearly independent, we say that A is a basis for M .This is a generalization of vector spaces.

Theorem 1.8. Let R be a ring with identity. The following conditions on a unitary R-module Fare equivalent:

(i) F has a nonempty basis

(ii) F is the internal direct sum of a family of cyclic R-modules, each of which is isomorphic as a[left] R-module to R

(iii) F is isomorphic as an R-module to a direct sum of copies of the [left] R-module R

(iv) There is a nonempty set X and some function ϕ : X → A such that given any unitaryR-module A and function f : X → A, there is a unique R-module homomorphism f : F → Asuch that fϕ = f . That is, F is a free object in the category of unitary R-modules.

Definition 1.9 (Free R-module). A unitary module F over a ring R with identity satisfying theconditions of Theorem 1.8 is called a free R-module on the set X.

By the Theorem, observe that F is a free object in the category of unitary [left] R-modules. ButF is not a free object in the category of all [left] R-modules. By definition, the zero module is thefree module on the empty set. One could define free modules in the category of all [left] R-modulesover any ring R. Such a free module is not isomorphic to a direct sum of copies of R, even when Rhas identity. However when we say “free module”, we mean a unitary free module.


Corollary 1.10. Every (unitary) module A over a ring R (with identity) is the homomorphic imageof a free R-module. If A is finitely generated, then F may be chosen to be finitely generated.

Lemma 1.11. A maximal linearly independent subset X of a vector space V over a division ringD is a basis of V .

Theorem 1.12. Every vector space V over a division ring D has a basis and is therefore a freeD-module. More generally, every linearly independent subset of V is contained in a basis of V .

Theorem 1.13. If V is a vector space over a division ring D and X is a subset that spans V , thenX contains a basis of V .

1.14 Exactness

Definition 1.15 (Exactness). A pair of module homomorphisms Af−→ B, B

g−→ C is said to beexact at B provided im f = ker g.

Af−→ B

g−→ C

For longer sequences (finite),

A0f1−→ A1

f2−→ A2f3−→ · · · fn−→ An

is exact provided im fi = ker fi+1 for 1 ≤ i ≤ n− 1. For infinite sequences, we say the sequence isexact if and only if im fi = ker fi+1 for all i.

Example 1.15.1. Let A be a submodule of B

0 −→ Af−→ B

g−→ B/A −→ 0

is exact. .

An exact sequence of the form

0 −→ Af−→ B

g−→ C −→ 0

is called a short exact sequence.

Lemma 1.16 (The Short 5 Lemma). Let R be a ring and let

0 A B C 0

0 A′ B′ C ′ 0

f

α

g

β γ

f ′ g′

be a commutative diagram of R-modules and R-module homomorphisms such that each row is ashort exact sequence, then

(i) If α, γ are monomorphisms then β is a monomorphism.

(ii) If α, γ are epimorphisms then β is an epimorphism.

7 C. McWhorter

(iii) If α, γ are isomorphisms then β is an isomorphism.

Proof: Notice that (iii) follows from the first two propositions, so it suffices to show them. Ourproof is by diagram chase.

(i) Let b ∈ B such that β(b) = 0, we want b = 0 (so we are going to show that the kernel istrivial).

B C

B′ C ′

g

β γ

g′

We look at γ(g(b)) and using the fact that the diagram commutes find

γ(g(b)) = g′(β(b)) = g′(0) = 0

But γ is a monomorphism so that g(b) = 0. Therefore as the rows are exact, g ∈ ker g = im f .Then we have b = f(a) for some a ∈ A. We then use our other commuting diagram.

A B

A′ B′

α

f β

f ′

Using our initial assumption that β(b) = 0, we have

f ′(α(a)) = β(f(a)) = β(b) = 0

Then f ′ is injective so that it must be α(a) = 0. We have α injective by assumption so thatit must be that a = 0. Finally, we know that b = f(a) = f(0) = 0 so that β must be amonomorphism.


(ii) Let b′ ∈ B′. We want to show that there is a b ∈ B such that β(b) = b′. We know that g, g′,and α are surjective. We proceed by going about the following diagram clockwise, “adjustingour target” so that we hit our goal “the long way” [about the diagram].

B C

B′ C ′

g

β γ

g′

We know that g′(b′) ∈ C. As γ is an epimorphism, there is some c ∈ C such that g′(b′) = γ(c).But g is an epimorphism so that c = g(b) for some b ∈ B. Now we use the commutativity ofthe diagram to find

g′(β(b)) = γ(g(b)) = γ(c) = g′(b′)

which is only helpful in that we now know g′(β(b))− g′(b′) = 0; that is, that g′(β(b)− b′

)= 0.

What we is for β(b) − b′ = 0. However, this is not necessarily so. But we do know thatβ(b)− b′ ∈ ker g′ = im f . So say f ′(a′) = β(b)− b′ for some a′ ∈ A′. The β(b)− b′ is like an“error”. Using the fact that α is an epimorphism, a′ = α(a) for some a ∈ A. Now considerb− f(a) ∈ B (this is our adjusting the error). We have β

(b− f(a)

)= β(b)− β(f(a)).

Now using the commutativity of the diagram, we have

β(f(a)) = f ′(α(a)) = f ′(a′) = β(b)− b′

where the last equality follows from the last fact mentioned in the preceding paragraph. Butthen

f ′(a′) = β(b)− b′ = β(b)− (β(b)− b′) = b′

Now as f(a) ∈ B, there exists a b0 ∈ B such that b0 = f(a) so that β(b0) = b′. But then β isan epimorphism.

0 A B C 0

0 A′ B′ C 0

f g hf−1 g−1h−1

An isomorphism between these two exact sequences will be maps f, g, h such that f, g, h areisomorphisms and they make the diagram commute. We need the diagram to commute with f, g, h

9 C. McWhorter

and f−1, g−1, h−1. This follows from the commutativity of the diagram with f, g, h. Consider thefollowing diagram of short exact sequences that commutes with f, g, h:

0 A B C 0

0 A′ B′ C 0

f

r

g

s

hf−1

r′

g−1

s′

h−1

We want to check if rf−1(a′) = g−1r′(a′) for all a′ ∈ A′. We have f−1(a′) = a ∈ A for some a.That is, f(a) = a′. Then we have r′(a′) = r′f(a). The commutativity of the diagram in f, g givesr′f(a) = gr(a). Then

r′f(a) = r′(a′)

gr(a) = r′(a′)

g−1gr(a) = g−1r′(a′)

r(a) = g−1r′(a′)

rf−1(a′) = g−1r′(a′)

as desired. We need now check commutativity of the right square. That is, we want to show thatsg−1(b′) = h−1s′(b′) for all b′ ∈ B′. We have g−1(b′) = b ∈ B for some b ∈ B so that g(b) = b′. Thecommutativity of the diagram in g, h gives s′g(b) = hs(b). Then

s′(b′) = s′g(b)

s′(b′) = hs(b)

h−1s′(b′) = h−1hs(b)

h−1s′(b′) = s(b)

h−1s′(b′) = sg−1(b′)

as desired. One can easily verify that isomorphisms of short exact sequences form an equivalencerelation.

Theorem 1.17. Let R be a ring and

0 −→ A1f−→ B

g−→ A2 −→ 0

a short exact sequence of R-module homomorphisms. Then the following conditions are stillequivalent:

(i) There is an R-module homomorphism h : A2 → B with g ◦h = 1A2 . That is, h has a one-sidedinverse.

(ii) There is an R-module homomorphism k : B → A1 with k ◦ f = 1A1.

(iii) The given sequence is isomorphic with the identity maps on A1 and A2 to the direct sum exactsequence

0 −→ A1i1−→ A1 ⊕A2

π2−→ A2 −→ 0

and up to isomorphism there is only one such sequence. In particular, B ∼= A1 ⊕A2.

A short exact sequence satisfying these conditions is said to be split or a split exact sequence. Themaps h, k are sometimes called splittings.


1.18 Projective and Injective Modules

Definition 1.19 (Projective Module). A module RP is projective if for all homomorphisms Bg−→

C −→ 0 and map f : P → C, then f can be lifted to B; that is, if there exists a homomorphismh : B → B with gh = f .

B C 0

P

g

fh

Lemma 1.20. Every free module is projective.

Proof: Assume that

B C 0

F

g

sh

on a basis B = {bi}. We look at {f(bi)} ⊆ C. As g is onto, we have f(bi) = g(bi) for some bi ∈ B.Let h : F → B be the unique homomorphism with h(bi) = bi. So we have

h

(∑i

ribi

)=∑

ribi

But then gh = f .

Proposition 1.21. The following are equivalent for a module P :

(i) P is a projective module.

(ii) Hom(P,−) is exact. That is, if

0 −→ Af−→ B

g−→ C −→ 0

is any exact sequence, then

0 −→ Hom(P,A)f∗−→ Hom(P,B)

g∗−→ Hom(P,C) −→ 0

is exact.

(iii) P is isomorphic to a direct summand of a free module.

(iv) For all Xg−→ P with g onto, the mapping splits. That is, there exists L : P → X with

gh = 1P .

11 C. McWhorter

Proof: 1→ 2:

X P 0

P

g

id∃h

and gh = 1.

2 → 3: There exists a free module F and an onto homomorphism Fπ−→ P −→ 0. But

by assumption, there exists a map i : P → f such that πi = 1P . We look at e : F → F , wheree = iπ. Observe that e2 = e. Then F = im e⊕ker e. One of im e, ker e is isomorphic to P . (Exercise)

3→ 1: Let F be free on a basis S. Assume that F decomposes: P = P ⊕Q. To show that P isprojective, observe

B C 0

P

F = P ⊕Q

g

f

πP

h

Now πP iP = 1P . Now F is projective by the previous lemma so that there exists a map h : F → Bwith gh = fπP . Let h : P → B be hiP . Then

gh = ghiP = f

Remark 1.22. Observe that in the proof of the previous proposition we actually showed that asummand of projective modules is projective.

Remark 1.23. If R is a local principal ideal domain, then R is free as RR.

Example 1.24. Not all projective modules are free. Let

R =

Q 0Q QQ Q Q

That is, let R be the set of lower triangular matrices. We know dimQR = 6. So if F is a finitedimensional free module, then dimF is a multiple of 6.

P =

0 0 00 Q 00 Q 0

⊆ RAs P is a submodule (in fact a left ideal of R), then as a left module dimR P = 2 so that P is notfree.


Proposition 1.25. Let {Pi}i∈I be a family of modules. Then⊕

i∈I Pi is projective if and only ifPi is projective.

Proof: The forward direction was shown in Proposition 1.21. Now consider

B C 0

⊕i∈I Pi

Pi

g

hf

ki

hi

where ki is the canonical injection. Then there exists hi : Pi → B with ghi = fki. Then there existsh :⊕

i∈I Pi → B with gh = f . One need only show that these maps “stitch together” correctly.(Exercise)

Proposition 1.26. Let RM be a module. Then M is a quotient of some projective module. If Mis finitely generated, we can also choose this projective module to be finitely generated.

Proof: Assume we have the following commutative diagram with exact rows.

A B C 0

A′ B′ C ′ 0

f

α

g

β h

f ′ g′

Then there exists a unique h : C → C ′ such that hg = g′β. We show this first. Let x ∈ C. As g isonto, there is a b ∈ B such that x = g(b).

b x = g(b)

β(b) g′β(b)

g

β

We “try” the map h(x)def= g′β(b). We need check that this map is well defined. Let b1 ∈ B such

that g(b1) = g(b) = x. We have

b1

b x = g(b)

β(b) g′β(b)

g

β

13 C. McWhorter

We need check that g′(β(b)) = g′(β(b; )). Note that g(b1 − b) = 0 so that b1 − b = f(a) for somea ∈ A. Then β(b1 − b) = β(f(a)) = f ′(α(a)). Moreover, g′(β(b1 − b)) = g′(f ′(α(a))) = 0, because ofexactness. So we know that g′(β(b1)) = g′(β(b)) so that the map is well defined. By construction, itcommutes the diagram.

However, there is an easier way of demonstrating this.

A B C 0

A′ B′ C ′ 0

f

α

g

β h

f ′ g′

Note we have Bg−→ C and look at the cokernel of f . We have g′βf = g′f ′α = 0. Then simply use

the Universal Property of the Cokernel.

A B C

C ′

f

g′β

g

∃!h

We have yet to show that h is unique and a homomorphism. We show this by showing that for anycommutative diagram with exact rows, as below,

0 A B C

0 A′ B′ C ′

f

h

g

β γ

f ′ g′

there exists a unique h : A → A′ with f ′h = βf . The idea is the same as above but we use theUniversal Property of the Kernel. We have f ′ : A′ → B′ is the kernel of g′.

A

0 A′ B′ C ′h

βf

f ′ g′

We know also that g′βf = 0. But then the Universal Property of the Kernel says there exists aunique h : A→ A′ with f ′h = βf .

Let f : A→ B be a homomorphism and let M be another R-module. We have an induced mapof abelian groups

HomR(B,M) −→HomR(f,M)=fM∗HomR(A,M)

given by fM∗ (β) = βf .

Proposition 1.27. Let 0 −→ Af−→ B

g−→ C −→ 0 be a short exact sequence. Then the sequence

0 −→ Hom(C,M)gM∗−→ Hom(B,M)

fM∗−→ Hom(A,M)

is exact.


Proof: Exercise

Note that fM∗ is onto if for any h : A→M , there is a p : B →M with pf = h.

0 A B

M

f

h∃p

Furthermore, we know that if 0 −→ Af−→ B

g−→ C −→ 0 is a split exact sequence then for all Mthe sequence

0 −→ HomR(C,M) −→ HomR(B,M) −→ HomR(A,M) −→ 0

is exact.

Definition 1.28 (Injective Module). An R-module I is injective if whenever we have

0 A B

I

f

hg

h can be “extended” to B. That is, there exists g : B → I with gf = h.

First note that RI is injective if and only if for all short exact sequences

0 −→ Af−→ B

g−→ C −→ 0

the following sequence is exact

0 −→ Hom(C,M)gM∗−→ Hom(B,M)

fM∗−→ Hom(A,M) −→ 0

Proposition 1.29. RI is injective if and only if for all monomorphisms 0 −→ If−→ X splits, i.e.

there exists p : X → I with pf = idI . So I is isomorphic to a direct summand of X.

Theorem 1.30 (Baer’s Criterion). Let R be a ring and E a left R-module. Then RE is injective ifand only if for all left ideals RI of R and

0 I R

E

incl

fs

then f can be extended to R, i.e. there exists a homomorphism s : R→ E such that s|I = f .

15 C. McWhorter

Proof: The forward direction is trivial as when E is injective the result is trivial. Now assumethe converse. Let the following the homomorphisms of R-modules

0 A B

E

f

Without loss of generality, assume that Aincl−→ B. (Why?) Then we have

0 A B

E

f

Let S = {(A′, f ′) | A ⊆ A′ ⊆ B, f ′ : A′ → E and extends f}.

A A′ B

E

ff ′

We know that S 6= ∅ as the pair (A, f) ∈ S. We put an ordering on S given by (A′, f ′) ≤ (A′′, f ′′) if

A ⊆ A′ ⊆ A′′ ⊆ B

E

ff ′ f ′′

This is an ordering. (Why?) We claim that S has a maximal element. Pick a chain {(Ai, fi)}i in S.We look at

(⋃Ai, f

), where f :

⋃Ai → E defined by f(ai) = fi(ai). This is an upper bound for

the chain. Then Zorn’s Lemma says that there exists (A∗, f∗) which is a maximal element of S.

0 A A∗ B

E

ff∗

We want to show that A∗ = B. Suppose that this is not the case. Then there is a b ∈ B \A∗. Then

0 A A∗ A∗ + 〈b〉 = A

E

f∗


Let I = {r ∈ R | rb ∈ A∗}. We know that 0 ∈ I so that I 6= ∅. It is trivial to show that I is a leftideal of R. Then

0 I R

E

j∃j

where j(r)def= f∗(rb). The map j is a R-module homomorphism where j|I = j. We look at j(1) ∈ E.

We construct f : A→ B so that f |A∗ = f∗. Then we have f |A = f . So we have (A, f) > (A∗, f∗)as A ) A∗. This contradicts our ordering from above; that is, this contradicts the maximality of(A∗, f∗).

Define f(a∗ + rb) = f∗(a∗) + rj(1). We claim that f is well defined and is a map of R-modules.We first show the map is well defined. Let a∗1 + r1b = a∗2 + r2b. Then we have a∗1 − a∗2 = (r2 − r1)b.Then r2 − r1 ∈ I.

f∗(a∗1 − a∗2) = f∗((r2 − r1)b) = j(r2 − r1) = j(r2 − r1) = j(r2)− j(r1)

But f∗(a∗1) − f∗(a∗2) = f∗(a∗1 − a∗2) so that f∗(a∗1) + j(r1) = f∗(a∗2) + j(r2) = f(a∗2 + r2b). Butf(a∗1 + r1b) = f∗(a∗1) + j(r1). Therefore, the map is well defined. It remains to show that f is aR-module homomorphism. (Exercise)

Now recall that if {Pi}i∈I is a family of modules, then each Pi is projective if and only if⊕

i∈I Piis projective. We have a similar result for injective modules.

Proposition 1.31. Let {Ei}i∈I be a family of R-modules. Then each Ei is injective if and only if∏i∈I Ei is injective.

Proof: Let each Ei be injective.

0 A B

∏Ei

Ei

f

j

∃g

gi

πi

We need find a map g : B →∏Ei. As the Ei are injective, there is a map gi : B → Ei. Let g =

∏gi.

That is, let g(b) =(gi(b)

)i∈I . This map works. (Why?)

Now assume∏Ei is injective.

0 A B

Ei

∏Ei

f

j

∃gi

∃f

17 C. McWhorter

Let ki(x) = (0, 0, · · · , x, 0, 0, · · · , 0), where the x occurs in the ith position. As the∏Ei is injec-

tive, there exists a module homomorphism f : B →∏Ei. Note that there is also πi :

∏Ei → Ei,

where πi((xi)i∈I

) def= xi and πiki = 1Ei . Let gi = πif . Then one easily checks πifj = πiki︸︷︷︸

id

fi = fi.

It is important to take note of a few things:

1. A finite direct sum of injective modules is injective because a finite direct sum is equal to a finitedirect product.

2. You can have infinitely many injective modules and have⊕Ei not injective.

3. You can have an infinite family of projective modules {Pi} but∏Pi not projective.

4. Each summand of an injective module is injective.

1.32 Tensor Product

Definition 1.33 (Biadditve R-function). Let R be a ring. Let AR and RB be right and left R-modules, respectively. Let G be a Z-module. Then a biadditive R-function is a function A×B → Gwith

1.

f(a1 + a2, b) = f(a1, b) + f(a2, b)

f(a, b1 + b2) = f(a, b1) + f(a, b2)

2. f(ar, b) = f(a, rb)

for all a, a1, a2 ∈ A and b, b1, b2 ∈ B. If R is commutative, then the function f is just a bilinearfunction.

Definition 1.34 (Tensor Product). A tensor product over R of AR and RB is an abelian group

A ⊗R B together with a biadditive function A× B f−→ A ⊗R B satisfying the following universalproperty: for all g : A×B → G, a biadditive homomorphism of abelian groups, there is a uniquehomomorphism h : A⊗R B → G commuting the following diagram

A×B A⊗R B

G

f

g∃!h

Theorem 1.35. Given AR,RB, their tensor product exists and is unique up to isomorphism.

Proof: First, we prove the existence of such an object. Let F be the free abelian group withbasis A×B. So the elements of F are finite linear combinations of ordered pairs in A×B togetherwith integer coefficients. Let S be the subgroup of F generated by elements of the form

(i) (a1 + a2, b)− (a1, b)− (a2, b)


(ii) (a, b1 + b2)− (a, b1)− (a, b2)

(iii) (ar, b)− (a, rb)

Define A⊗RBdef= F/S. Furthermore, define a⊗b to be the coset of (a, b), i.e. (a, b)+S. Observe

that for all a, a1, a2 ∈ A, b, b1, b2 ∈ B, and r ∈ R, we have

(i) (a1 + a2)⊗ b = a1 ⊗ b+ a2 ⊗ b

(ii) a⊗ (b1 + b2) = a⊗ b1 + a⊗ b2

(iii) ar ⊗ b = a⊗ rb

We have a map A×B f−→ A⊗R B. It is simple to show that f is biadditive. We also have anexact sequence of Z-modules:

0 −→ Sincl↪−→ F

f−→ A⊗R B −→ 0

We need show that the universal property of tensor products is satisfied.

A×B

0 S F A⊗B 0

G

f

g ∃!g∃!h

Now g is a biadditive map and G is an abelian group. Now F is free on A×B so that there existsa unique map g : F → G such that g|A×B = g. Now as g is biadditive, we have S ⊆ ker g. Usingthe universal property of the cokernel, there exists a unique map h : A⊗B → G with hf = g. Nowcomparing this with A×B ↪→ F , we get hf = g.

Now we demonstrate uniqueness of the tensor product. Assume that G1, G2 are tensor productsof A,B. Then

A×B G1

G2

f1

f2 ∃!h1∃!h2

Now there exists a unique map h1 since G1 is a tensor product with h1f1 = f2. But as G2 is atensor product, there exists a unique map h2 with h2f2 = f1. But then (h2h1)f1 = f1. Note thatthe map 1G1f1 = f1 works and must be unique. But then h2h1 = 1G1 . Similarly, h1h2 = 1G2 sothat we must have G1

∼= G2.

19 C. McWhorter

Remark 1.36. The elements of A ⊗R B are of the form∑

i∈I ai ⊗ bi, where ai ∈ A and bi ∈ B.These elements are usually not of the form a⊗ b.

Remark 1.37. Assume R is a commutative ring. Let A,B be two R-modules. Form the tensorproduct A⊗R B with F and S as given before. Then A⊗R B is an R-module.

r

(∑i∈I

ai ⊗ bi

)=∑i∈I

rai ⊗ bi =∑i∈I

ai ⊗ rbi

Later, we will prove the following:

Theorem 1.38. Let R be a commutative ring. Let A be a free module with basis {ei}i∈I and B afree module with basis {fj}j∈J . Then A⊗R B is a free R-module with basis {ei ⊗ fj}i∈I,j∈J .

Corollary 1.39. Let F be a field and V a vector space with basis {ei}i∈I and W a vector space withbasis {fj}j∈J , then V ⊗FW is a vector space with basis {ei⊗fj}i∈I,j∈J . If we have dimV = n <∞,dimW = m <∞, then dimV ⊗F W = nm.

Example 1.40. Let F be a field. Let V be a 2-dimensional vector space with basis {v1, v2}. We lookat V⊗FV . We know dimV⊗FV = 4 and a basis for this vector space is {v1⊗v1, v1⊗v2, v2⊗v1, v2⊗v2}.We look at v1 ⊗ v1 + v2 ⊗ v2 ∈ V ⊗F V . We claim we cannot write v1 ⊗ v1 + v2 ⊗ v2 as u⊗ w forsome u ∈ V , w ∈ V .

Assume to the contrary that this is possible. Let u = a1v1 + a2v2 and w = b1v1 + b2v2. Then

v1 ⊗ v1 + v2 ⊗ v2 = (a1v1 + a2v2)⊗ (b1v1 + b2v2)

= a1b1v1 ⊗ v1 + a2b2v2 ⊗ v2 + a1b2v1 ⊗ v2 + a2b1v2 ⊗ v1

But then using the chosen basis for the tensor product, we must have

a1b1 = 1

a2b2 = 1

a1b2 = 0

a2b1 = 0

So ai 6= 0 6= bj for all i, j. This method works for any dimension, not just dimension 2. Only obtainsonly u⊗ v’s alone in dimension 1.

Example 1.41. Q⊗Z Z/nZ for n > 1. The tensor product Q⊗Z Z/nZ is generated (not equal tobut generated by) elements of the form q ⊗ a, where q ∈ Q and a ∈ Z/nZ. We claim that q ⊗ a = 0for all q ∈ Q and a ∈ Z/nZ. Let 1 = α/β, then

q ⊗Z a =α

n⊗Z a =

nα

nβ⊗Z a =

α

nβ⊗Z na =

α

nβ⊗Z 0 = 0

However, one can’t always “move things over”. Take A⊗R B. Suppose a⊗R b and b = 0 for somer ∈ R and a = a′r, where a′ is not necessarily in A. Then one cannot write

a⊗R b 6= a′ ⊗R rb = 0

One is reminded of ideals, for a = a′r with a′ /∈ A.


Proposition 1.42. Let M be an R-module. Then R⊗RM ∼= M .

Proof: We know R⊗RM is a left R-module via

r

(∑i∈I

ri ⊗mi

)=∑i∈I

rri ⊗mi

Let∑ri ⊗mi ∈ R⊗RM . As ri = 1 · ri and ri ⊗mi = 1⊗ rimi, we have∑

1⊗ rimi = 1⊗∑

rimi

But∑rimi ∈ M as M is an R-module. Define g(r,m)

def= rm. Now g is biadditive (as an M

module). We have f(r,m) = r ⊗m.

R×M M

R⊗RM

g

f∃!h

Then there exists a unique h : R ⊗RM → M such that the above diagram commutes. We haveh(r⊗m) = rm. Now h is a left homomorphism of modules as sh(r⊗m) = s(rm) = (sr)m = h(sr⊗m).We need an inverse homomorphism h. Let h : M → R⊗RM be h(m) = 1⊗m. Then

h(rm) = 1⊗ rm= r ⊗m= r(1⊗m)

= rh(m)

so that h is a homomorphism. Observe also that hh = hh = 1.

Proposition 1.43. If MR is an R-module, then there exists an isomorphism of right R-modules,M ⊗R R

∼→M .

Proposition 1.44. If R is a commutative ring and A,B are R-modules then there is an isomorphismof R-modules A⊗R B

∼→ B ⊗R A.

Proposition 1.45. Let I be a two-sided ideal of R (I C R). Let RM be a left R-module, thenR/I ⊗M is a left R-module. In fact, it is a left R/I-module via

R :r(s⊗m)def= rs⊗m

R/I :r(s⊗m)def= rs⊗m

Theorem 1.46. If I C R and RM , then there exists an isomorphism of R-modules (in fact ofR/i-modules) from R/I ⊗RM

∼→M/IM , where

IM = {∑

αimi | αi ∈ I,mi ∈M}

21 C. McWhorter

Proof: An element in R/I ⊗RM is of the form∑ri ⊗R mi, where ri ∈ R/I. But∑

ri ⊗R mi =∑

(1 · ri)⊗R mi

=∑

1⊗R rimi

= 1⊗m

for some m ∈ M . Now let R/I ⊗R Mh−→ M/IM be given by h(1 ⊗R m)

def= m + IM . We

will denote m + IM as m. It is clear that h is a homomorphism of left modules. (Why?) Leth : M/IM → R/I ⊗RM be given by h(m+ IM) = 1⊗R m. It is clear that h is a homomorphism.(Why?) Then hh = hh = 1.

Example 1.47. Z/mZ⊗Z Z/nZ. We can take R = Z, I = mZ, and M = Z/nZ.

Suppose we have ARf−→ BR, RA

′ f ′−→R B′ be homomorphisms. We claim that there exists an

induced homomorphism of abelian groups

A⊗R A′f⊗f ′−→ B ⊗R B′

such that (f ⊗ f ′)(a⊗ a′) = f(a)⊗ f ′(a′). We can say that

A×A′ B ⊗B′

A⊗A′

g

fh

where g(a, a′) = f(a)⊗ f ′(a′) is a biadditive function. Then the Universal Property of the Kernelsays that there exists a unique homomorphism h : A⊗A′ −→ B ⊗B′ with the “right” property.

Example 1.48. If MR is a right R-module and we have a homomorphism of left R-modules

RAf−→R B, then we have a homomorphism of abelian groups

M ⊗R A1M⊗f−→ M ⊗R B

given by m⊗ a m−→ ⊗f(a).

Now suppose that we have a short exact sequence

0 −→R Af−→R B

g−→R C −→ 0

Given a right R-module MR, we can ask about the sequence

0 −→M ⊗A 1M⊗f−→ M ⊗B 1M⊗g−→ M ⊗ C −→ 0

This sequence is not usually exact.


Theorem 1.49. If 0 −→R Af−→R B

g−→R C −→ 0 is exact and MR is a right R-module, then the

sequence M ⊗ A 1M⊗f−→ M ⊗ B 1M⊗g−→ M ⊗ C −→ 0 of abelian groups is exact. Furthermore, if the

sequence 0 −→ Af−→ B

g−→ C −→ 0 is split exact, then 0 −→M⊗A 1M⊗f−→ M⊗B 1M⊗g−→ M⊗C −→ 0is split exact.

Proof: Let x ∈M ⊗R C. Now as g is onto

x =∑

mi ⊗ ci

=∑

mi ⊗ g(bi)

=∑

1M (mi)⊗ g(bi)

=∑

(1M ⊗ g)(mi ⊗ bi)

= (1M ⊗ g)∑

mi ⊗ bi

so that 1M ⊗ g is onto. Now

(1M ⊗ g)(1M ⊗ f)(m⊗ a) = (1M ⊗ g)(m⊗ f(a))

= m⊗ g(f(a))︸︷︷︸=0

= 0

as gf = 0. Therefore, im(1M ⊗ f) ⊆ ker(1M ⊗ g). To prove that ker(1M ⊗ g) ⊆ im(1M ⊗ f) ismuch harder and will not be treated here (see page 210 Hungerford Algebra or Dummit & Foote).

Let R be a ring. Let RM be a left R-module. Then HomR(R,M) is a left R-module via

(rf)(s)def= f(sr), where r, s ∈ R and f ∈ HomR(R,M). We claim that rf is in HomR(R,M).

Proposition 1.50. There is an isomorphism of left R-modules HomR(R,M)ψ−→M .

Proof (Sketch): We have ψ : f 7→ f(1) a homomorphism. Then let Mϕ−→ HomR(R,M) be given

by ϕ(m)def= fM , where fM (r) = rm. Now fM is a homomorphism so that ϕ is a homomorphism.

Furthermore, ϕ and ψ are inverses of each other.

Theorem 1.51. Let R be a ring. There is a functorial isomorphism R⊗M ϕM−→M for every leftmodule M .

Functorial means for all modules RMg−→R N , the following diagram commutes

R⊗RM M

R⊗R N N

ϕM

1R⊗g g

ϕN

where ϕM , ϕN are isomorphisms. (Show this) We also have functorial isomorphisms M ⊗RR −→M .Now let AR and RB be R-modules with A′R,RB

′ submodules. Let a ∈ A′ and b ∈ B′. Thena ⊗ b ∈ A′ ⊗R B′ and a ⊗ b ∈ A ⊗R B. But it can be the case that a ⊗ b = 0 in A ⊗R B but benonzero in A′ ⊗R B′ so that A′ ⊗R B′ is not a subgroup of A⊗R B.

23 C. McWhorter

Example 1.52. Let R = Z, A = Z, B = Z/3Z, A′ = 3Z, and B′ = Z/3Z. Let 0 6= b ∈ B, say b = 1.We look at 3⊗ 1. We have 3⊗ 1 ∈ A′ ⊗B′ nonzero but inside A⊗B, A⊗R B = Z⊗Z Z/3Z.

3⊗Z 1 = 1⊗Z 3 = 1⊗ 0 = 0

Example 1.53. We look at 0 −→ Z f−→ Z −→ Z/3Z −→ 0, where f(m) = 3m. Let M = Z/3Z.Then f ⊗ 1M is not injective. We have f ⊗ 1M : Z⊗Z Z/3Z −→ Z⊗Z Z/3Z.

(f ⊗ 1M )(a⊗ b) = f(a)⊗ b = 3a⊗Z b = a⊗Z 3b = a⊗ 0 = 0

so f ⊗ 1M is the zero map. This shows that the tensor product of injective maps need not beinjective.

Theorem 1.54 (Adjoint Isomorphism Theorem). Let R,S be rings. Let AR,RBS , CS. Then thereis an isomorphism of abelian groups

HomS(A⊗R B,C)∼−→ HomR(A,HomS(B,C))

which is functorial in A,B, and C.

We explain what functorality in A,B, and C means. To be functorial in C, we have if g ishomomorphism CS

g−→ C ′S , then we have

HomS(A⊗B,C) HomR(A,HomS(B,C))

HomS(A⊗B,C ′) HomR(A,HomS(B,C ′))

ϕA,B,C

1R⊗g g

ϕA,B,C′

where the map on the left is Hom(A⊗B, g) and the map on the right is Hom(A,Hom(B, g)).

Functorality in A means given a homomorphism Ag−→ A′, then A⊗B g⊗1B−→ A′ ⊗B −→ C.

Hom(A′ ⊗B,C) Hom(A′,Hom(B,C))

Hom(A⊗B,C) HomR(A,HomS(B,C))

ϕA′,B,C

1R⊗g g

ϕA,B,C

where the map on the left is Hom(g ⊗ 1B, C) and the map on the right is Hom(g,HomS(B,C)) andϕA′,B,C , ϕA,B,C are isomorphisms.

Now we show that it means to be functorial in B. (Exercise)

Theorem 1.55 (Adjoint Isomorphism Theorem). Let R,S be rings. Let AR,RBS , CS. Then thereis an isomorphism of abelian groups

HomS(A⊗R B,C)∼−→ HomR(A,HomS(B,C))

which is functorial in A,B, and C.


Proof: We need a map ϕ : HomS(A⊗RB,C) −→ HomR(A,HomS(B,C)). Let f : A⊗RB −→ C.

We have ϕ(f) : A −→ HomS(B,C) and ϕ(f)(a) : B −→ C. Now let ϕ(f)(a)(b)def= f(a⊗b). We need

to show that ϕ(f)(a) is a S-homorphism, ϕ(f) is a R-homomorphism, and ϕ is a homomorphismof abelian groups. We need an inverse ψ : HomR(A,HomS(B,C)) −→ HomS(A ⊗R B,C). Letg : A −→ HomS(B,C) be a R-module homomorphism. Then ψg is a unique homomorphismA⊗RB −→ C such that ψg(a⊗ b) = g(a)(b). To demonstrate this, we apply the Universal Propertyof Tensor Products

A×B

A⊗B C∃!ψg

It remains to show that ψg is a homomorphism of abelian groups and show that ψ,ϕ are inverses ofeach other. (Exercise)

If RM , then there exists a left projective R-module P with P −→M −→ 0. If RM , then thereexists an injective R-module E and a monomorphism M −→ E. Furthermore, recall that over Z, amodule is injective if and only if it is divisible. So Q is an injective Z-module. Then we obtain thefollowing result:

Proposition 1.56. Every Z-module can be embedded in an injective module.

Proof: Let ZM =⊕

Z/L. Observe⊕

Z is free. As every quotient of a free module is free, ZMis free. Now Z ( Q is a submodule so

⊕Z (

⊕Q. So we have

⊕Q/L. This is a quotient of a

divisible module so that it is divisible so that it is injective.

Proposition 1.57. Let R be any ring and let ZD be a divisible group. Then the R moduleE = HomZ(ZRR,ZD) is a left injective R-module, where E is a left R-module via (rf)(s) = f(sr).

Proof: Let f : A→ E. We need to show f can be extended to B.

0 A B

E

f

i

That is, we have to prove that HomR(B,E)HomR(i,E)−→ HomR(A,E) −→ 0 or showing HomR(B,HomZ(R,D)) −→

HomR(A,HomZ(R,D)) −→ 0. To show this, one need only follow the diagram arrows and getisomorphism compositions with onto maps to get the middle map onto and then perform the same

25 C. McWhorter

to the top row. (Exercise)

HomR(B,HomZ(R,D)) HomR(A,HomZ(R,D)) 0

HomZ(R⊗R B,D) HomR(R⊗R A,D) 0

HomZ(B,D) HomZ(A,D) 0

Example 1.58. If Mh−→ N is an isomorphism, then Hom(M,X)

∼−→ Hom(N,X). (Exercise)

Example 1.59. If M ∼= N and A is another R-module, then

Hom(M,A) ∼= Hom(N,A)

andHom(A,M) ∼= Hom(A,N)

It is also useful to take note of the following theorem,

Theorem 1.60. Let M1f−→ M2

g−→ M3 be a sequence of R-modules. Then the following areequivalent:

(i) M1f−→M2

g−→M3 −→ 0 is exact.

(ii) g ◦ f = 0 and for all h : M → X with hf = 0, there exists a unique h′ : M3 → X with h′g = h(g is the cokernel of f so this gives us exactness).

(iii) For all R-modules X, we have an induced exact sequence of abelian groups

0 −→ Hom(M3, X)Hom(g,X)−→ Hom(M2, X)

Hom(f,X)−→ Hom(M1, X)

1.61 Flat Modules

Theorem 1.62. (i) Let {Ai}i∈I be a family of right R-modules and let B be a left R-module.Then there exists an isomorphism of abelian groups⊕

i∈IAi ⊗R B

∼−→⊕i∈I

(Ai ⊗R B

)

(ii) Let A be a right R-module and let {Bi}i∈I be a family of left R-modules. Then there exists anisomorphism of abelian groups

A⊗R

(⊕i∈I

Bi

)∼−→⊕i∈I

(A⊗R Bi)

Moreover if R is commutative, these are R-module isomorphisms.


Assume that we have SAR,RB then A⊗R B is also a left S-module via s(a⊗ b) def= sa⊗ b. If

AR,RBS → A⊗R B is a right S-module. Recall that if 0 −→ Af−→ B

g−→ C −→ 0 is a short exactsequence of left R-modules then for all MR, we have the exact sequence

M ⊗A idM⊗f−→ M ⊗B idM⊗g−→ M ⊗ C −→ 0

Example 1.63. If 0 −→ Af−→ B

g−→ C −→ 0 splits then 1M ⊗ f is also injective.

Definition 1.64 (Flat Module). A right R-module M is called flat if for every short exact sequenceof left modules

0 −→ Af−→ B

g−→ C −→ 0

the induced sequence

0 −→M ⊗A 1M⊗f−→ M ⊗B 1M⊗g−→ M ⊗ C −→ 0

is exact. That is, tonsuring with flat modules preserves exactness.

Example 1.65. RR is flat. To see this let Af−→ B be a monomorphism. Then

R⊗R A R⊗R B

A B

ϕA

1R⊗f

ϕB

f

where the left map is given by r ⊗ a 7→ ra and the right map is given by r ⊗ b 7→ rb. Going aroundthe left and bottom of the diagram is a homomorphism and going around the other away is amonomorphism so that 1R ⊗ f is a monomorphism.

Proposition 1.66. Let {Mi}i∈I be a family of right R-modules, then⊕

i∈IMi is flat if and onlyif Mi is flat for i ∈ I.

Proof (Sketch): The forward direction is an exercise. For the reverse direction, assume that each

Mi is flat. We look at 0 −→R Af−→R B. But then Mi ⊗R A

1M⊗f−→ Mi ⊗R B is a monomorphism

for all i ∈ I. Then we use the fact that Xihi−→ Yi is a monomorphism for all i if and only if⊕

Xi⊕hi−→

⊕Yi is a monomorphism: h((xi))→ (hi(xi))i.⊕

Xi⊕Yi

Xi Yi

0 0

kXi

hi

kYi

27 C. McWhorter

Corollary 1.67. M = L⊕K is flat if and only if L and K are flat.

Corollary 1.68. (i) Every free module is flat.

(ii) Every projective module if flat.

Proof:

(i) We know that a module is free if and only if it is isomorphic to a direct sum of copies of R.

(ii) We know that if P is projective then there exists a free module F such that F = P ⊕Q forsome Q. Then P is flat using the previous part.

It is important to note

Free Modules −→ Projective Modules −→ Flat Modules

We also have the following:

Theorem 1.69. If R is a PID and P is projective then P is free.

Theorem 1.70. If R is Noetherian and M is finitely generated then M is flat if and only if M isprojective.

Example 1.71. ZQ is flat but not projective.

1.72 A Few Notes

Theorem 1.73. HomR(K,−) and HomR(−,K) are left exact while −⊗RK and K ⊗R− are rightexact.

We know that HomR(RM,RN) is not necessarily an R-module but is an abelian group. However,there are cases where this has more structure. The group HomR(RM,RNS) is a right S-module.Furthermore, HomR(RMS ,RN) is a left S-module.

An important language piece to note is that covariant modules do not “change direction” whenapplying things such as Hom to them while contravariant modules reverse arrow direction whenapplying things such as Hom. The first component of Hom is the “bad” component because it iscontravariant. However, the second component of Hom is “nice” in that it is covariant. That is,Hom(−,M) is contravariant while Hom(M,−) is covariant.

Recall that if M,N are R-modules with Mf−→ N , we define the cokernel to be cokerϕ = N/ imϕ.

This gives us the exact sequence

0 −→ kerϕ −→Mϕ−→ N −→ cokerϕ −→ 0


2 Categories

2.1 Categories and Morphisms

We generalize the world of modules to an arbitrary world with the same features.

Definition 2.2 (Category). A category C consists of a class of objects, a set of morphisms, and acomposition for these morphisms. The objects of C are denoted obj C and need not even be a set.These objects are called small if the objects of C are a set. The morphisms are maps between objects.If A,B are objects, then we write HomC(A,B) for the set of morphisms f : A → B (more often

written Af−→ B). This HomC(A,B) has the following properties:

(i) There is an identity morphism, denoted 1A or idA, such that 1A : A→ A for all A ∈ obj C.

(ii) There is a composition operation

HomC(A,B)×HomC(B,C) −→ HomC(A,C)

written g ◦ f - or gf when no confusion will arise - so that (f, g) 7→ gf .

This is the composition for the category C and it must satisfy the following properties:

(i) Associativity: For all morphisms Af−→ B

g−→ Ch−→ D, we have (hg)f = h(gf).

(ii) Unit: For each object A, there is an identity morphism 1A ∈ HomC(A,A) such that f1A = f

and 1Bf = f for all Af−→ B.

Remark 2.3. Note that one will often write A ∈ C when what one really means is A ∈ obj C. Thisis not entirely unreasonable given there is a one-to-one correspondence between objects A and theiridentity morphism 1A so that a category could be considered to be a class of maps alone.

Example 2.4. Sets: The objects are sets, the morphisms are functions, and the composition isordinary function composition.

Example 2.5. Ab: The objects are abelian groups, the morphisms are group homomorphisms,and the composition is ordinary function composition.

Example 2.6. Veck: The objects are [finite dimensional] vector spaces over k, the morphisms arelinear transformations, and the composition is ordinary function composition.

Example 2.7. Groups: The objects are groups, the morphisms here are homomorphisms, andthe composition is ordinary function composition.

Example 2.8. Ring: The objects are rings, the morphisms here are ring homomorphisms, andthe composition is ordinary function composition.

Example 2.9. Top: The objects are topological spaces, the morphisms are continuous maps, andthe composition is ordinary function composition.

Example 2.10. Mfd: The objects are smooth manifolds, the morphisms are smooth maps, andthe composition is ordinary function composition.

29 C. McWhorter

Example 2.11. R-mod: The objects are [left] R-modules, the morphisms are R-module homo-morphisms, and the composition is ordinary function composition. Here, R is some fixed ring.

The objects and morphisms in a category C need not be actual sets and functions between them.

Example 2.12. Take a poset (a partially ordered set) P and form a category by making the objectsthe set of points, or elements, of P and the morphisms be defined by if p ≤ q then there is amorphism called p ≤ q, or denoted p→ q. That is, if p ≤ q, then HomC(p, q) = {p→ q}. If p 6≤ q,then HomC(p, q) = ∅. Moreover, HomC(p, p) = {p → p}. We can visualize this in a diagram. Seethe example below:

d

e f

b c

a

as a ≤ b, there is a morphism from a to b. However as e 6≤ f , there is no morphism from e to f .The identity morphism exists as a ≤ a. Note at each note there is really an arrow going to itself;however, we recognize that this arrow is there and do not write it. Similarly, there should be anarrow from a to e in the diagram as there is an arrow from a to b and b to e, but be recognize thisand simply to not draw it to simplify the diagram.

The properties of morphisms generalize the properties of R-module homomorphisms. However,not all the properties of R-module homomorphisms hold for any category.

Definition 2.13 (Monic). Let Bf−→ C be a morphism. We say that f is monic if for all e1, e2

morphisms in C

A Be1e2

with fe1 = fe2 then e1 = e2. That is, they are the same element in HomC(A,C). Equivalently, ife1 6= e2 then fe1 6= fe2.

Definition 2.14 (Epi). Let Bf−→ C be a morphism. We say that f is epi if for all morphisms

g1, g2

A Be1e2

with g1f = g2f , then g1 = g2. Equivalently, if g1 6= g2 then g1f 6= g2f . We also call f epic if it hasthis property.


Definition 2.15 (Isomorphism). Given a morphism Bf−→ C, we say that f is an isomorphism if

it has an inverse. That is, there exists a morphism Cg−→ B such that gf = 1B and fg = 1C . This

morphism g is denoted f−1.

Example 2.16. In R-mod, epi is the same as surjective and monic is the same as injective.

Remark 2.17. In R-mod, a morphism is epi and monic if and only if it is an isomorphism.However, this is not true in all categories.

Example 2.18.

Category Isomorphisms

Set BijectionR-mod R-module Isomorphism

Top HomeomorphismMfd Diffeomorphism

Definition 2.19 (Subcategory). A subcategory B of a category C is a collection of some of theobjects of C and some of the morphisms such that B is category. That is, we need 1B to be in thecategory B for all objects B ∈ B and we need the morphisms to be closed under composition. Thesubcategory B is called full if it contains all possible morphisms between the objects of B. That is forall B,B′ ∈ B, we have HomB(B,B′) = HomC(B,B

′).

Example 2.20. Ab ⊆ Groups

Definition 2.21 (Initial Object). An object A in a category C is called an initial object if for allobjects X in C, there is a unique morphism A→ X.

Example 2.22. The emptiest is an initial object in Sets.

Example 2.23. The zero module is an initial object in R-mod.

Example 2.24. 0 is the initial object of the natural numbers N, when viewed as a poset.

Definition 2.25 (Terminal Object). An object A in a category C is called a terminal object if forall objects C ∈ C, there is a unique morphism X → C, where X is a set.

Example 2.26. Every one-point set is a terminal object in Sets. Note that the empty set is not aterminal object in Sets as HomSets(X, ∅) = ∅ for every nonempty set X.

Example 2.27. The zero module is a terminal object in R-mod.

Example 2.28. Viewed as a poset, N has no terminal object.

Definition 2.29 (Zero Object). In a category C, a zero object is an object that is both an initialand a terminal object.

Definition 2.30 (Opposite Category). If C is a category, define the opposite category, denotedCtextop, to be the category with the objects of Cop to be the objects of C with morphisms HomCop(A,B) =HomC(B,A). One might write the morphisms of Cop as fop, where f is a morphism of C. The

composition rule in Cop is defined by gopfopdef= (fg)op.

31 C. McWhorter

2.31 Functors

We now look at maps between categories.

Definition 2.32 (Functor). If C and D are categories, then a function T : C → D is a functionsuch that

(i) if A ∈ obj C, then T (A) ∈ obj D

(ii) if Af−→ A′, where A,A′ ∈ C, then T (A)

T (f)−→ T (A′) in D

(iii) If Af−→ A′

g−→ A′′ in C, then T (A)T (f)−→ T (A′)

T (g)−→ T (A′′) in D and T (gf) = T (g)T (f)

(iv) T (1A) = 1T (A) for every A ∈ obj C

That is, functors preserves the identity and preserves compositions. One could say instead, thatfunctors preserve commutative diagrams. Why?

· ·

·

gf

f g

As functors preserve triangles, they must preserve all commutative diagrams as all commutativediagrams can be triangulated and vice versa.

Example 2.33. The identity functor id : C → C is given by id(C)def= C for all objects C ∈ C and

id(f)def= f for all morphisms f ∈ C.

Example 2.34. If C,D, E are categories and C F−→ D, D G−→ E are functors, then GF is a functor

C GF−→ E in the obvious way.

Example 2.35. If M is a right R-module, then M ⊗R − is a functor R-modM⊗R−−→ Ab given

by RNM−→ ⊗RN on morphisms and N

h−→ N ′ 7→ F (N)1N⊗h−→ F (N ′). If M is a left R-module,

then HomR(M,−) is a functor R-mod → Ab given by N 7→ HomR(M,N) and Nh−→ N ′ 7→

HomR(M,N)h∗−→ HomR(M,N ′).

Example 2.36. The forgetful functor is a functor that simply “forgets” some of the structure onthe objects of a category C. This is easiest to define in an example. The objects in R-mod areR-modules, hence abelian groups. Therefore, we have a functor R-mod −→ Ab where the objectC ∈ obj R-mod is simply identified with its corresponding abelian group under addition. Simplyput, the forgetful functor has forgotten the R-module structure of the set and only remembered itsabelian group structure. We also have a forgetful functor from Ab to Sets, where the functor onlyremembers the set structure on Ab.


Example 2.37. Take P the poset given by the diagram below

·

· ·

·

b

d

ac

Since this poset is a category, there is a unique map from the bottom vertex to the top one. That is,ba = dc. We have a functor F : R→ R-mod where · 7→ R-mod and →7→ R-mod homomorphism.We have F (ba) = F (dc) and by the properties of a functor, F (b)F (a) = F (d)F (c). That is, we have

· N

· · M L

· T

F

But then a functor F : P → R-mod is just a single commutative square of R-modules.

Definition 2.38 (Covariant/Contravariant Functor). A contravariant functor F : C → D is a

functor that reverses directions of arrows; that is, if obj C F−→ obj D then HomC(C1, C2) −→HomD(F (C2), F (C1)) such that F (id) = id and F (gf) = F (f)F (g). A covariant functor preservesdirections of arrows and are defined exactly as functor previously. So a functor which is notcontravariant is covariant and vice versa.

Functors can have further properties which makes them “nicer.”

Definition 2.39 (Faithful). A functor F : C → D is called faithful if for all A,B ∈ obj C, thefunctions HomC(A,B)→ HomD(TA, TB) given by f 7→ Ff are injections.

Definition 2.40 (Full). A functor F : C → D is called faithful if for all A,B ∈ obj C, the functionsHomC(A,B)→ HomD(TA, TB) given by f 7→ Ff are surjections.

Definition 2.41 (Fully Faithful). A functor which is both faithful and full is called fully faithful. Thatis, a functor F : C → D is called fully faithful if for all A,B ∈ obj C, the functions HomC(A,B)→HomD(TA, TB) given by f 7→ Ff are bijections.

Remark 2.42. Observe that the definitions above refer only to the maps in Hom and not to theobjects in the category.

Example 2.43. Let C be a category. Let skeleton C is the set of isomorphism classes of the objectsof C. The functor F : C → skeleton C is fully faithful.

33 C. McWhorter

2.44 Natural Transformations between Functors

In different sets, maps compare the objects. For example, homomorphisms compare algebraic objects,continuous maps compare topological objects, etc. Natural transformations are what comparefunctors.

Definition 2.45 (Natural Transformations). Let C F−→ D, C G−→ D be (covariant) functors. Anatural transformation η : F → G is a one-parameter family of morphisms in D

η = (ηC : F (C)→ G(C))C∈obj C

such that the following diagram commutes for all f : C → C ′ in obj C:

F (C) G(C)

F (C ′) F (′C)

ηC

F (f) G(f)

ηC′

A natural transformation between contravariant functors is defined mutatis mutandis by changing Cby Cop. A natural isomorphism is a natural transformation η for which each τC is an isomorphism.

We write η : F∼=−→ G. Sometimes a natural isomorphism is called functorial.

Informally, we say that “the morphism(s) ηC : F (C) → G(C) is natural in C. That is, themorphism is compatible with the change/morphism in C. If η is a natural isomorphism, there existnatural transformation ε : G→ F such that ηε = 1 : G→ G and εη = 1 : F → F .

Example 2.46. The Adjoint Isomorphism Theorem is an isomorphism of abelian groups

HomS(M ⊗R N,L)∼−→ HomR(M,HomS(N,L))

Recall that if M −→M ′, then

HomS(M ′ ⊗N,L) ∼= HomR(M ′, Hom(M,L))

and we have the commutative diagram

HomS(M ′ ⊗N,L) Hom(M,Hom(N,L))

Hom(M ⊗N,L) Hom(M ′,Hom(N,L))

That is, there is a natural isomorphism of functors the functors F = Hom(− ⊗ N,L) and G =Hom(−,Hom(N,L)).

F (M) G(M)

F (M ′) G(M ′)

We say that this isomorphism is “natural in M”. Thinking of this as a functor of L, then it is“natural in L”.


Example 2.47. Let Vec be finite dimensional vector spaces over a field k. Then ( )∗ = HomR(−,K)is a (contravariant) functor. This is dualizing the vector space. We know that Hom(kn,k) =Hom(k,k)n = k

n. Even though V ∼= V ∗, the isomorphism between them is not natural in V . Thatis, there is not a natural transformation id

∼−→ ( )∗. This is easy to see as one is covariant and oneis contravariant. However, V → V ∗∗ is natural in V ! That is, there exists a natural isomorphismid→ ( )∗∗ making the diagram commute.

Definition 2.48 (Equivalence). A functor F : C → D is called an equivalence of categories if thereexists a functor which undoes it. That is, if there exists a functor G : D → C such that GF

∼−→ idCand FG

∼−→ idD, where the maps are natural transformations.

2.49 Kernel and Cokernel

Let C be a category.

Definition 2.50 (Initial Object). An initial object I is one such that for all C ∈ objC, there existsa unique morphism I → C.

Example 2.51. The zero module only maps to any other module in one way. Therefore, the zeromodule is an initial object in R-mod.

Definition 2.52 (Terminal Object). A terminal object T is one such that for all C ∈ obj C, thereis a unique morphism C → T .

Definition 2.53 (Zero Object). A zero object is an object which is both initial and terminal. Thezero object is written as 0.

Remark 2.54. It is clear that initial objects, terminal objects, and zero objects, when they exist,are unique up to isomorphism.

If there is a zero object, then the zero map on any object C is as follows

C −→ 0 −→ C

Then map above HomC(C,C) is the zero map (morphism).

Example 2.55. In Sets, ∅ is initial. Singleton sets in Sets are terminal. Therefore, Sets have nozero object and hence have no zero map.

Example 2.56. In R-mod, 0 is the zero object and the the zero map is exactly what one wouldexpect.

Definition 2.57 (Kernel). Let Bf−→ C be a morphism in C. A kernel of f is a morphism A

i−→ Bsuch that

(i) fi = 0

(ii) For all A′i′−→ B′ with fi′ = 0, there is a unique θ with i′ = iθ. That is, the following diagram

commutes:

A B C

A′

i f

∃!θ i′

35 C. McWhorter

The kernel of a map f is written ker f .

In the definition above, the first requirement essentially is for existence while the second conditionforces the ker f to be the “largest” such object.

Definition 2.58 (Cokernel). Let Bf−→ C be a morphism in C. A cokernel of f is a morphism

Cp−→ D such that

(i) pf = 0

(ii) For all Cp′−→ D′ with p′f = 0, there is a unique θ with p′ = θp. That is, the following diagram

commutes:

B C D

D′

f p

p′∃!θ

The cokernel of a map f is written coker f .

Again, the first requirement is for existence and the second is for maximality. Furthermore, wehave the following proposition

Proposition 2.59. Whenever the kernel and cokernel of a morphism exist, they satisfy the followingproperties:

(i) Any kernel is monic. However, the converse is not true.

(ii) Any cokernel is epi(c).

(iii) In Ab and R-mod, kernels and cokernels agree with the usual definition. That is, given

Mf−→ N , ker f = {m ∈M | f(m) = 0} and coker f = N/ im f .


3 Categorical Limits & Constructions

3.1 Additive Categories

Definition 3.2 (Additive Category). A category C is additive if

(i) HomC(A,B) is an additive abelian group for every A,B ∈ obj C.

(ii) the distributive laws hold: given morphisms

X A B Yafg b

where X,Y ∈ obj C, then

b(f + g) = bf + bg and (f + g)a = fa+ ga

(iii) C has a zero object.

(iv) C has finite products and coproducts.

Definition 3.3 (Additive Functor). If C and D are additive categories, a functor F : C −→ D isadditive if for all A,B and all f, g ∈ Hom(A,B), we have

F (f + g) = F (f) + F (g)

That is, the function HomC(A,B) −→ HomD(F (A), F (B)) given by f 7→ Ff is a homomorphism ofabelian groups.

Recall the definitions of kernels and cokernels:

Definition 3.4 (Kernel). Let Bf−→ C be a morphism in C. A kernel of f is a morphism A

i−→ Bsuch that

(i) fi = 0

(ii) For all A′i′−→ B′ with fi′ = 0, there is a unique θ with i′ = iθ. That is, the following diagram

commutes:

A B C

A′

i f

∃!θ i′

The kernel of a map f is written ker f .

In the definition above, the first requirement essentially is for existence while the second conditionforces the ker f to be the “largest” such object.

37 C. McWhorter

Definition 3.5 (Cokernel). Let Bf−→ C be a morphism in C. A cokernel of f is a morphism

Cp−→ D such that

(i) pf = 0

(ii) For all Cp′−→ D′ with p′f = 0, there is a unique θ with p′ = θp. That is, the following diagram

commutes:

B C D

D′

f p

p′∃!θ

The cokernel of a map f is written coker f .

However, note that these definitions require the existence of the kernel and cokernel. Not everymorphism in every category will have such nice properties. Even if a morphism has these propertiesin a particular category, there is no reason that any other morphism in the category should havethese properties. An abelian category is a category in which every morphism has these “nice”properties. In fact, it is strong enough for us to define a notion of exactness.

Definition 3.6 (Abelian Category). A category C is an abelian category if it is an additive categorysuch that

(i) every morphism has a kernel and cokernel

(ii) every monomorphism is a kernel and every epimorphism is a cokernel

One immediate consequence of the existence of finite direct sums, kernels, and cokernels in anabelian category is the existence of finite direct and inverse limits.

Remark 3.7. In a general category, if a monic is a kernel, then it is a kernel of its cokernel.Similarly, if a epi(c) is a cokernel then it is a cokernel of its kernel.

Furthermore, in an abelian category every map Bf−→ C factors as

Be−→ im f

m−→ C

with e epi(c) and m monic. We can now define a notion of exactness.

Definition 3.8 (Exact). A sequence of maps

Af−→ B

g−→ C

is exact (at B) if im f ∼= im ker g. Longer sequence are exact if they are exact at each object exceptthose at the ends (this is not a problem if the sequence is infinite). A short exact sequence is asequence of the form

0 −→ Af−→ B

g−→ C −→ 0


Proposition 3.9. R-mod and Ab are abelian categories with ker f, coker f, im f satisfying thetraditional definitions.

Definition 3.10 (Exact Functor). Let F : A → B be an additive (covariant) functor between abeliancategories. We say that F is left exact (mutatis mutandis right exact) if for all short exact sequences

0 −→ A −→ B −→ C −→ 0

in A, the sequence

0 −→ F (A) −→ F (B) −→ F (C)

is exact (we say that this sequence is left exact). The functor F is called exact if it is both leftand right exact. A contravariant functor is left exact (mutatis mutandis right exact) if the functorF ′ : Aop → B is left exact.

Proposition 3.11. F is exact if and only if it preserves exact sequences.

Definition 3.12 (Functor Category). Given two categories A,B, the functor category BA isa category whose objects are all covariant functors A → B and whose morphisms are naturaltransformations.

Proposition 3.13. The functor category is a category. Furthermore, if B is an abelian categorythen so is BA.

Proposition 3.14 (Yoneda Embedding). The Yoneda embedding of a small category A is thefunctor A → SetsA

opthat is injective on objets and whose image is a full subcategory of SetsA.

Proposition 3.15. If A is a small abelian category, then there is a ring R and an exact, fullyfaithful functor from A into R-mod, which embeds A as a full subcategory in the sense thatHomA(M,N) ∼= HomR(M,N).

Proof: See Weibel 1.6.1 pg 25. (1st ed, 1995).

3.16 Adjoint Functors

Definition 3.17 (Adjoint Pair). Let A,B be categories. A pair of (covariant) functors A BF

G

are an adjoint pair if for all A ∈ objA and B ∈ objB, there is a bijection

HomB(F (A), B)τA,B−→ HomA(A,G(B))

that are natural transformations in A and B. That is, the following two diagrams commute for allf : A′ → A in A and g : B → B′ in B:

HomB(F (A), B) HomB(F (A′), B)

HomA(A,G(B)) HomA(A′, G(B))

(Ff)∗

τA,B τA′,B

f∗

39 C. McWhorter

HomB(F (A), B) HomB(F (A), B′)

HomA(A,G(B)) HomA(A,G(B′))

g∗

τA,B τA,B′

(Gg)∗

We say that F is the left adjoint of G and G is the right adjoint of F and that (F,G) form anadjoint pair.

Example 3.18. If R,S are rings and SMR is a bimodule, then (−⊗SM,HomR(M,−)) is an adjointpair. Similarly, if RMS is a bimodule, (M ⊗S −,HomR(M,−)) is an adjoint pair.

Example 3.19. Consider Sets VecF

G

, where Vec is all vector spaces over a field k, where

F (X) = the vector space generated by the basis X over k and G(V ) is just V as a set (that is theforgetful functor on Vec). We know that HomVec(F (X), V )

∼−→ HomSets(X,G(V )) so that (F,G)is an adjoint pair.

Theorem 3.20. Let A BF

G

be (covariant) functors. Then (F,G) is an adjoint pair if and

only if there are natural transformations 1Aη−→ GF and FG

ε−→ 1B so that the compositions

FFη−→ FGF

εF−→ F are the identity (natural transformation).

Proof: Exercise. Hint: Given ηε, we get Hom(F (A), B)τ−→ Hom(A,G(B)) where f 7→ GfηA in

one direction. ηA : A→ GF (A)→ B.

Adjoint pairs have nice exactness properties.

Theorem 3.21. Let A,B be abelian categories and let (F,G) be an adjoint pair. Then F is rightexact and G is left exact.

Proof: (For F , G is homework). Let 0 −→ A′ −→ A −→ A′′ −→ 0 be an exact sequence in A.We know that Hom is left exact for any abelian category (the proof of such is similar to that of theproof for R-modules). So for all B ∈ objB, we have the exact sequence

0 −→ HomA(A′′, G(B)) −→ HomA(A,G(B)) −→ HomA(A′, G(B))

Now using the adjoint isomorphism on the adjoint pair (F,G), we have the exact sequence

0 −→ HomB(F (A′′), B) −→ Hom(F (A), B) −→ Hom(F (A′), B)

The squares in the diagram below commute by the naturality of the adjoint isomorphism so thatthe above sequence must also be exact.

0 HomA(A′′, G(B)) HomA(A,G(B)) HomA(A′, G(B))

0 HomB(F (A′′), B) HomB(F (A), B) HomB(F (A′), B)


By a lemma of Yoneda, Hom reflect exact sequences so that

F (A′) −→ F (A) −→ F (A′′) −→ 0

is exact. But then F is right exact.

3.22 Colimit

Here is an overview of what will be to come:

Covariant Contravariant

Colimit (Direct Limit): colimMi, lim−→Mi Limit (Inverse Limit): limMi, lim←−Mi

Coproduct (Direct Sum):∐Mi,

⊕Mi Product:

∏Mi

Cokernels Kernels

Unions of Sets Intersection of Sets

Pushout:

L N

M ∗

Pullback:

∗ N

M L

Recall given a collection of {Mi} in C, the direct sum ⊕Mi (that is the coproduct,∐Mi) is

an object in C with maps (called “inclusions”) Miηi−→ ⊕Mi and the Universal Mapping Property:

for any N and collection of maps {Mifi−→ N} in C, there is a unique map ⊕Mi

f−→ N such thatfi = fηi. We have the following diagram

...

...

Mj

...⊕Mi N

Mk

...

...

ηi

∀fj

∃!f

ηk

∀fk

That is, this object is the “closest thing to all the Mi’s so that all maps from all the Mi’s are forcedto go through it.” This is the idea of the colimit. Before we can rigorously define it, we need somepreliminary definitions.

41 C. McWhorter

Definition 3.23 (Direct System). A (direct) system of objects in a category C is a functor F : I → C,where I is a poset. That is, given a partially ordered set I and a category C, a direct system inC is an ordered pair

((Mi)i∈I , (ϕ

ij)i≤j

), often abbreviated {Mi, ϕ

ij}, where ϕii = 1Mi such that the

following diagram commutes for all i ≤ j ≤ k:

Mi Mk

Mj

ϕik

ϕij ϕjk

The partial ordered set I, when viewed as a category in its own right, has objects being theelements of I and morphisms the unique morphisms κij whenever i ≤ j. Then a direct system in Cover I are the covariant functors M : I → C, M(i) = Mi and M(κij) = ϕij .

Remark 3.24. One should be clear when one is talking about a direct system and when one istalking about a directed system. We define a directed system later.

Example 3.25. The diagram on the left is a poset while the diagram on the right are R-modules.The (direct) system is a functor which takes the objects in the poset to the corresponding moduleson the right and takes the arrows to their corresponding arrows.

· M1

· M3

· · −→ M2 M4

· M5

Again, recall that there is a loop at each vertex (the identity morphism) which we omit.

Example 3.26. If I = {1, 2, 3} is the partial ordered set in which 1 ≤ 2 and 1 ≤ 3, then a directsystem over I is a diagram of the form

A B

C

f

g

Definition 3.27 (Colimit Limit). Let I be a partially ordered set, let C be a category, and let{Mi, ϕ

ij} be a direct system in C over I. The colimit (or direct limit, injective limit, or inductive

limit) is an object lim−→Mi and insertion morphisms (αi : Mi → lim−→Mi)i∈I such that

(i) αjϕij = αi for i ≤ j


(ii) If X ∈ obj C and fi : Mi → X be a morphism with fjϕij = fi for all i ≤ j. There exists a

unique morphism θ : lim−→Mi → X making the following diagram commute

lim−→Mi X

Mi

Mj

θ

αi fi

ϕijαj

fj

Proposition 3.28. The colimit (direct limit) of a system is unique up to isomorphism (wheneverit exists).

Proof: Exercise

Example 3.29. Consider the (direct) system

M1

M2

ϕ′2=ϕ

What is the colimit (if it exists) of this system. Let’s assume that the colimit exists and call it L.Then we have the following diagram

M1

L T

M2

f1

i1

ϕ∃!f

i2

f2

where the diagram is such that f1 = f2ϕ. So f1 is determined by f2. Observe also that i2ϕ = i1.

We need an object L with a maps M1i1−→ L, M2

i2−→ L, and Lf−→ T . We guess L = M2, with

i1 = ϕ, i2 = 1M2 and f = f2. It is simple to check that these definitions satisfy the requirements.As the colimit is unique, we have found the colimit.

Example 3.30. In Sets, consider a chain of subsets of a set X

U1 ⊆ U2 ⊆ U3 ⊆ · · ·

43 C. McWhorter

where ⊆ is really the inclusion map. What is the colimit of this system? We want maps to an objectL that are compatible with the system, i.e. compatible with inclusion. As the Ui’s do not necessarilycover X, it is necessary that we choose L =

⋃Ui. Take the map Ui → L to be f(l) = fi(l) if l ∈ Ui,

the normal inclusion map. One need show that this is well defined, i.e. fi+1

∣∣Ui

= fi. But then it is

clear that lim−→(U1 ⊆ U2 ⊆ · · · ) =⋃Ui.

Example 3.31 (Pushout). The colimit of the diagram/system of the form

L N

M

ϕ

ψ is called

the pushout. Specifically, given two morphisms ϕ : L → N and ψ : L → M in a category C (orfibered sum) is a triple (D,α, β) with βg = αf that is a solution to the universal mapping problem.That is, for every triple (Y, α′, β′) with β′g = α′g, there is a unique θ : D → Y making the diagramcommute. The pushout is denoted M ∪L N .

L N

M D

Y

ϕ

ψ β

β′α

α′θ

Example 3.32. If B,C are submodules of some left R-module U , there are inclusions f : B∩C → Band g : B ∩ C → C. It is simple to show that the pushout exists and that it is B + C.

Example 3.33. If B,C are subsets of some set U , there are inclusions f : B ∩ C → B andg : B ∩ C → C. The pushout in Sets is B ∪ C.

Example 3.34. The pushout exists for Groups. The pushout for two injective homomorphismsis called the free product with amalgamation. Observe this is closely related to the van KampenTheorem!

Proposition 3.35. In R-mod, the pushout of two maps ϕ : L→ N and ψ : L→M exists.

Proof: Let S = {(ψ(l),−ϕ(l)) ∈ M ⊕ N | l ∈ L}. Define D = (M ⊕ N)/S, α : M → D bym 7→ (m, 0) + S, and β : N → D by n 7→ (0, n) + S. Then we have the diagram

L N

M D

ϕ

ψ β

α


It is simple to check that this diagram commutes. Given another triple (X,α′, β′), simply defineθ : D → X by θ : (m,n) + S 7→ α′(m) + β′(n).

Proposition 3.36. In an abelian category, a pushout

L N

M

ϕ

ψ is coker(L→M ×N), where

the map is given by the universal mapping property for A×B.

Proof: This is similar to the previous proof.

Example 3.37. Let k,A,B be commutative rings. Then the pushout of

k A

B

, i.e. A,B

are k-algebras, is A⊗k B.

Corollary 3.38. The coproduct in the category of commutative k-algebras is A⊗k B.

Remark 3.39. Taking k = Z (any ring A has a map Z→ A given by 1 7→ 1 so that n 7→ n1A), weobtain the coproduct in commutative rings A⊗Z B.

Note that in Rings, pushouts behave oddly because this category is not abelian. For the nextproof, it will be beneficial to consider the following examples of pushouts:

Example 3.40. The colimit (pushout) of the system

0 N

M

ϕ

ψ is simple M⊕N/ im θ = M⊕N .

Example 3.41. The pushout of

M N

0

ϕ

ψ is simply coker f .

Proposition 3.42. Let A be an abelian category. Then colimits exist (over posets) if and only ifcoproducts exist (over an indexed set).

Proof: The forward direction is trivial as a coproduct is just a special case of a colimit wherethere are no maps in the system. For the reverse direction, let I be a poset and I → A given byi 7→Mi be a system in R-mod. By a similar proof to the existence of the pushout in R-mod, onecan show that colimi∈IMi is

coker

⊕ϕ:i→j

Miθ−→⊕i∈I

Mi

45 C. McWhorter

where ϕ : i→ j gives a map Mi →Mj and θ is the map constructed from the universal mappingproperty for the direct sum, i.e. from Mi →

⊕i∈IMi with maps ϕ : i→ j given by mi 7→ ϕ(mi)−mi.

In any abelian category A, the same proof without elements applies. Simply build θ from

Miθϕ−→⊕i∈I

Mi

where ϕ : Mi →Mj and θϕdef= ijϕ− ii1.

Remark 3.43. In R-mod, the colimit is just the quotient of⊕

i∈IMi identifying elements alongthe ϕij ’s.

Proposition 3.44. The colimit (direct limit) of any direct system {Mi, ϕij} of left R-modules over

a partially ordered index set I exists.

Proof: For each i ∈ I, let λi be the morphism of Mi into the direct sum ⊕iMi. Define

D =

(⊕i

Mi

)/S

where S is the submodule of ⊕iMi generated by all the elements λjϕijmi − λimi, where mi ∈Mi

and i ≤ j. Now define the insertion morphisms αi : Mi → D by

αi : mi 7→ λi(mi) + S

It is simple to check that D and the maps αi satisfy the Universal Mapping Property so thatD ∼= lim−→.

3.45 Limits

Limits work the same as colimits except that the direction of the arrows are reversed. Hence, thenotions for limits are dual to those for colimits, i.e. epi goes to monic, quotients go to subobjects,and direct sums to to direct products.

Definition 3.46 (Inverse System). Given a partially ordered sets I and a category C, an inversesystem in C is an ordered pair

((Mi)i∈I , (ψ

ji )j≥i

), abbreviated {Mi, ψ

ji }, where (Mi)i∈I is an indexed

family of objects in C and (ψji : Mj →Mi)j≥i is an indexed family of morphisms for which ψji = 1Mi

for all i and such that the following diagram commutes for whenever k ≥ j ≥ i:

Mk Mi

Mj

ψki

ψkj ψji


Example 3.47. In N, then an inverse system is a diagram

M0 ←M1 ←M2 ← · · ·

Note that we have again excluded identity morphisms and the composites of the arrows above.Observe also that this is just a system over −N.

Definition 3.48 (Limit). Let I be a partially ordered sets and C. Let {Mi, ψji } be an inverse system

in C over I. The limit (or projective limit or inverse limit), written limMi or lim←−Mi, is an objectlim←−Mi and a family of projections (αi : lim←−Mi →Mi)i∈I such that

(i) ψjiαj = αi for i ≤ j

(ii) for all X ∈ obj C and morphisms fi : X →Mi satisfying ψji fj = fi for all i ≤ j, there exists aunique morphism θ : X → lim←−Mi making the diagram commute

lim←−Mi X

Mi

Mj

αi

αj

θ

fi

fj

ϕji

Remark 3.49. The limit is really just the colimit of F op : Iop → Cop.

Example 3.50. For a system {Mi}i∈I with no maps between the Mi’s, the limit of the system (ifit exists) is

∏i∈IMi.

Example 3.51. In Sets, an inverse system of a chain of subsets of a fixed set X U1 ⊇ U2 ⊇ U3 ⊇ · · ·over the poset N with the maps the inclusion of subsets is L =

⋂∞i=1 Ui.

Example 3.52. Given two morphisms f : B → A and g : C → A in a category C, the pullback (orfibered sum) is a triple (D,α, β) with gα = fβ that is a solution to the universal mapping property.That is, for all (X,α′, β′) with gα′ = fβ′, there exists a unique morphism θ : X → D making thediagram commute. The pullback is denoted by B uA C

X

D C

B A

α′

θ

β′α

β g

f

Example 3.53. Pullbacks exists in Groups: they are subgroups of a direct product.

47 C. McWhorter

Remark 3.54. The kernel is a type of pullback. Specifically, if f : B → A is a morphism inR-mod, then the pullback formed by A,B, and f is (ker f, i), where i : ker f → B is inclusion.

Proposition 3.55. In an abelian category, the pullback is just ker(A×B (f,g)−→ C). In particular, inR-mod, it is the submodule of A×B {(a, b) | f(a) = g(b)} ⊆ A×B.

So in some sense, limits are products with extra identifications.

Proposition 3.56. Let A be an abelian category. Then limits exist (over posets) if and only ifproducts (over index sets) exist.

Corollary 3.57. In R-mod, all limits exist over posets.

3.58 Exactness

Definition 3.59 (Exactness of Systems). Suppose that {Mi}, {Ni} are systems over a poset I in acategory C.

(i) A map from Mi → Ni is just a natural transformation of the functors I → C defining each.Equivalently, a map from Mi → Ni are maps such that for all i ≤ j, the following squarecommutes

Mi Ni

Mj Nj

(ii) A sequence {Mi} → {Ni} → {Wi} is exact at {Ni} if it is at each i.

Proposition 3.60. In an abelian category A (so that exactness is defined), the following hold:

(i) The direct limit, colim (that is, lim−→) is right exact if it exists.

(ii) The inverse limit, lim, (that is, lim←−) is left exact if it exists.

Proof: We want to show that the colimit is left adjoint to some functor (so that it is right exact)and that the limit is right adjoint to some functor (so that it is left exact). We will only show thatthe colimit is left adjoint as the proof for the limit runs similarly.

Fix a poset I. We know that AI → A is a functor. Define the diagonal functor ∆ : A → AI byA 7→ the constant functor with value A. That is, ∆(A)(i) = A, a diagram/system with A at eachpoint and all the arrows are 1A. It is easy enough to check that (lim−→,∆) is an adjoint pair

HomA(colim({Mi}), B) ∼= HomAI ({Mi},∆(B))

The above isomorphism follows from the universal mapping property for the colimit. The otherdirection follows similarly.

Remark 3.61. One can easily remember whether the direct limit, lim−→, or inverse limit, lim←−, isright/left exact by the direction of their arrows.


We now investigate the behavior of lim−→ and lim←− with respect to tensoring and homming.

Proposition 3.62. Let W ∈ R-mod, then

(i) −⊗RW preserves the colimit, lim−→(ii) HomR(W,−) preserves limits, lim←−

(iii) HomR(−,W ) converts colimits to limits.

Proof: We shall only prove one of these. We know that −⊗RW preserves coproducts

⊕Mi ⊗RW∼−→⊕

(Mi ⊗RW )

Note that in general,∏Mi ⊗ W 6∼=

∏(Mi ⊗ W ). We know also that S−1M = M ⊗A S−1A.

Furthermore, −⊗RW preserves cokernels:

coker((Mf−→ N)⊗W ) = coker(M

f−→ N)⊗W

To see the proof, take the exact sequence

Mf−→ N −→ coker −→ 0

Now as the tensor product is right exact, the following sequence is exact

M ⊗RW −→ N ⊗RW −→ coker⊗RW −→ 0

Now as coker⊗RW is coker(M ⊗RWf⊗1W−→ N ⊗RW ). So −⊗RW preserves colim by the proof of

the previous result. We know that −⊗RW preserves the colimit

colim{Mi} ⊗RW ∼= colim{Mi} ⊗RW

The proof of the other two are similar. We know that HomR(W,∏Mi) ∼=

∏HomR(W,Mi) us-

ing the Universal Mapping Properties for the Product. Furthermore, we show Hom(⊕Mi,W ) ∼=∏Hom(Mi,W ) using the Universal Mapping Property for the Direct Sum. Use the fact that Hom

is left exact.

Note that a stronger result holds: if (F,G) is an adjoint pair then F preserves colimits and Gpreserves limits. Note also that adjoints are unique.

3.63 Directed/Filtered Limits

Definition 3.64 (Directed Set). A directed set is a partially ordered set I such that for everyi, j ∈ I, there is k ∈ I with i ≤ k and j ≤ k. That is, if there are arrows (morphisms) from i, j to k.

Example 3.65. The following system is not directed

·

·

·

49 C. McWhorter

However, the following system is directed

·

· ·

·

Notice that for a finite poset, we could draw the diagram so that there is a “top most” point if thesystem is a directed system.

Example 3.66. Another very common directed system is the poset

· −→ · −→ · −→ · · · · · ·

More generally, a category I is called filtered if this holds and for all i mapping to j, there is a mapl such that γα = γβ

i j lα

β

γ

Definition 3.67. A colimit (direct limit) over a directed set is called a

(i) colimit over a directed set

(ii) colimit over a filtered set

(iii) directed limit

(iv) directed colimit

(v) filtered limit

(vi) filtered colimit

(vii) direct limit

However, a colimit is written in the obvious way: colimMi, lim−→Mi

Now for the main reason that a limit over a directed system is “nice”

Lemma 3.68. Let {Mi} be a directed system in R-mod. Then

(i) For every m ∈ lim−→Mi comes from some Mi; that is, m = ιi(mi) for some mi ∈Mi

(ii) For all i, ker(Mi → lim−→Mi) = ∪j≥i ker(Mi →Mj)

Corollary 3.69.lim−→Mi = tiMi/ ∼

where mi ∼ mj – that is mi −mj ∼ 0 – if ϕil(mi) = ϕjl (mj) for some l ≥ i, j.


Theorem 3.70. Directed limits (filtered limits) are exact in R-mod.

Proof: We know that lim−→ is right exact. Suppose

0 −→ {Mi}fi−→ {Ni} −→ {Li} −→ 0

is a short exact sequence. Then

0 −→ lim−→Mif−→ lim−→Ni

is exact.

Mi lim−→Mi

Ni lim−→Ni

fi f

Suppose f(m) = 0. Then by the lemma, m = ιi(mi) for some mi ∈ Mi. Then ιi(fi(mi)) = 0in lim−→Ni. By the lemma, there is i → j such that ϕij(f(mi)) = 0. But f(ϕij(mi)) = ϕij(f(mi)).

However, f is monic so that ϕij(mi) = 0. Then by the lemma, m = ιi(mi) = 0.

Corollary 3.71. In R-mod, a directed limit commutes with homology:

Hn(lim−→Mi) = lim−→Hn(Mi)

Proof: lim−→ is an exact functor.

Example 3.72. (i) If f ∈ R, then Rf = lim−→(Rf−→ R

f−→ · · · )

(ii) The p-adic numbers.

(iii) Let X be a space. Let I be a poset of open subsets of X ordered by reverse inclusion. That is,U ⊆ V then V → U , where U, V are open. This is called the sheaf of rings, OX , the structure

sheaf on X. For each U , let Γ(U) = Γ(U,OX) be the continuous functions Uf−→ C. Of course,

we could have chosen polynomials or holomorphic functions instead of continuous functions.The smaller the set, the easier this is to write explicitly.

3.73 Flatness

Definition 3.74. A left R-module M is flat if −⊗RM is exact.

Example 3.75. If M is free, then it is also projective and therefore flat.

Example 3.76. The localization S−1R is flat as an R-module. To see this, given an exact sequence

0 −→M −→ N −→ L −→ 0

Then we know that we have the exact sequence

0 −→M ⊗R S−1R −→ N ⊗R S−1R −→ L⊗R S−1R −→ 0

51 C. McWhorter

But this is precisely the exact sequence

0 −→ S−1M −→ S−1N −→ S−1L −→ 0

Using this, we can see that Q is a flat Z-module.

Example 3.77. We know that Z/nZ is not flat for n ≥ 2.

Proposition 3.78. We know that we have the following implications:

Free −→ Projective −→ Flat

The converse holds in some cases: finitely generated modules over commutative local rings

Proposition 3.79. If {Mi}i∈I is a directed system of R-modules and if each Mi is flat, then lim−→Mi

is flat.

Theorem 3.80 (Lazard). If M is a left R-module, then M is flat if and only if M is a directedlimit of finitely generated free R-modules.


4 Homology

4.1 Overview

The motivation for homology begins in Algebraic Topology. Take a space and triangulate, that is“simplicialize” the space. Take the following examples of the circle (on the left) and the disk (on theright):

a

b

c

u v

w a

b

c

u v

w

t

where u = [a b], v = [b c], w = [a c], and t = [a b c]; that is, u, v, w, t are convex linearcombinations of vertices. We then define a sequence of maps of free Z-modules on the simplices.

· · · −→ 0︸︷︷︸3-dimensional

−→ 0︸︷︷︸2-dimensional

∂2−→ Zu⊕ Zv ⊕ Zw︸︷︷︸1-dimensional

∂1−→ Za⊕ Zb⊕ Zc︸︷︷︸0-dimensional

∂0−→ 0︸︷︷︸(-1)-dimensional

We call ∂ the boundary map; it takes the “edge” of a simplex. In our above examples we have∂2 = ∂0 = 0 and ∂1 is given by

u 7→ [b]− [a]

v 7→ [c]− [b]

w 7→ [c]− [a]

We define the nth homology group (that is at spot n) to be Hn = ker ∂n/ im ∂n+1. But does thehomology group have a geometric meaning? Observe that ker ∂1 = Z(u+ v − w) and im ∂2 = 0 sothat H1 = Z(u+ v − w)/0 ∼= Z. Notice that the boundary ∂(u+ v − w) = ∂(u)− ∂(v)− ∂(w) = 0.Furthermore, ker ∂1 are all the loops (cycles) in the space.

Now for the disk, we have

· · · −→ 0︸︷︷︸3-dimensional

−→ Zt︸︷︷︸2-dimensional

∂2−→ Zu⊕ Zv ⊕ Zw︸︷︷︸1-dimensional

∂1−→ Za⊕ Zb⊕ Zc︸︷︷︸0-dimensional

∂0−→ 0︸︷︷︸(-1)-dimensional

so that H1 = ker ∂1/ im ∂2 = Z(u+ v − w)/Z(v − w + u) = 0; that is, the sequence is exact in thatposition. We know that im ∂2 is the “boundary”. What this calculation shows is that the “loop”has been filled in so that there are no 1-dimensional “holes”. Whereas in calculation of H1 for thecircle, we found a 1-dimensional “hole” (the single copy of Z). So geometrically, the nth homologygroup, Hn, counts the number of n-dimensional holes. However, this notion works not just in theconfines of the idea of a simplex but rather in the more general situation of an abelian category.

We define the nth homology group as HN = Zn/Bn = ker dN/ im dn+1. Note that Hn measureshow far from being exact a sequence is. Then we know that Hn = 0 if and only if the sequence isexact at n. Of course, as we discussed above, this has geometric meaning in many topological cases.We have now set the stage for the general theory.

53 C. McWhorter

4.2 Chain Complexes

Definition 4.3 (Chain Complex). A chain complex C· is a sequence of R-modules and maps (calleddifferentials) dn such that the composition dndn+1 = 0, often written d2 = 0.

· · · −→ Cn+1dn+1−→ Cn

dn−→ Cn−1dn−1−→ · · ·

Definition 4.4 (Cycles, Boundaries, Homology Groups). Given a chain complex C· of R-modules,

· · · −→ Cn+1dn+1−→ Cn

dn−→ Cn−1dn−1−→ · · ·

The nth cycle is Zndef= ker dn. The nth boundary is Bn

def= im dn+1. Then the nth homology group is

Hndef= Zn/Bn.

Remark 4.5. Note that the indices are often defined differently than here. Often, the indices aredropped or ignored except when necessary. Often, especially in an introductory text such as this,keeping track of indices can become tedious. The reader is warned that on occasion, indices ordifferences in differentials will be dropped when avoiding them would bring more clarity to theproof. In general, the reader should be wary of the validity of the indices in this text.

Example 4.6. In a short exact sequence, we know that d2 = 0 ad im f = ker g so that gf = 0.

Example 4.7.

· · · −→ Z/8Z 4−→ Z/8Z 4−→ · · ·Notice that d2 = 16 = 0.

Definition 4.8. A morphism (map) of complexes f : C· → D· is a family of maps fn : C· → D·such that the squares commute

· · · Cn+1 Cn Cn−1 · · ·

· · · Dn+1 Dn Dn−1 · · ·

dn+1

fn+1

dn

fn

dn−1

fn−1

dn+1 dn dn−1

Note that the above differentials for C· and D· are denoted d, but they are not the samedifferential. This is an example where we will not differentiate between the two for “clarity” sake.This will be done throughout where confusion will not arise. The readers attention is again broughtto this fact.

Proposition 4.9. If f : C· → D· is a map of complexes sends cycles to cycles and boundaries toboundaries. That is, fn(Zn(C)) ⊆ Zn(D) and fn(Bn) ⊆ Bn(D). Hence, we obtain an induced map

Hn(C·)Hn(f)−→ Hn(D·).

Definition 4.10. For a map of complexes f : C· → D·, define ker f to be the complex

· · · −→ ker fndCn | ker−→ ker fn−1 −→ · · ·

and coker f to be the complex

· · · −→ coker fndCn | coker−→ coker fn−1 −→ · · ·


Definition 4.11 (Subcomplex). B· is a subcomplex of a complex C· if Bn ⊂ Cn for all n anddBn = dCn |Bn. If so, then we also get a quotient complex, written C/B or C·/B·,

· · · d−→ Cn+1/Bn+1d−→ Cn/Bn

d−→ · · ·

given by d(x) = d(x). Furthermore, B → C → C/B are chain maps (maps of complexes).

Definition 4.12. Given a chain map f : C· → D·, then the individual modules ker fn, coker fn, andim fn and maps induced by the differentials of C or D form complexes called ker f ⊆ C, coker f =D/ im f, im f ⊆ D.

Remark 4.13. In fact, these really are the categorical ker, coker, im of f in the category Ch(R−mod),the category of chain complexes (the objects) and chain maps (the morphisms).

It is easy to see that then Ch(R) is abelian. In fact, if A is abelian then Ch(A) is abelian. Infact, 0· → A· → B· → C· → 0· is an exact sequence of chain complexes if and only if each one (ineach degree) is exact. That is, if 0→ An → Bn → Cn → 0 is exact.

Definition 4.14. A cochain complex C · is a sequence of R-modules and maps

· · · −→ Cn+1 dn+1

−→ Cndn−→ Cn−1

dn−1

−→ · · ·

with d2 = 0.

Note that a cochain complex is just a chain complex with increasing indices. We can convertone to the other by reindexing Cn = C−n for all n ∈ Z.

Definition 4.15 (Cocycles, Coboundary, Cohomology). Given a cochain complex C ·, we define thenth cocycle as Zn = ker dn, the nth coboundary as Bn = im dn−1, and the nth cohomology group tobe Hn(Cn) = Zn/Bn.

Definition 4.16 (Quasi-Isomorphism). A chain map f : C· → D· is called a quasi-iso (quiz, q-iso,

quism, or quasi-isomorphism) if it induces an isomorphism on homology, Hn(C·)Hn(f)−→ Hn(D·), for

all n.

Remark 4.17. (i) It is clear that C· is exact if and only if Hn(C) = 0 for all n. Furthermore,0· → C· is a quasi-isomorphism if and only if C· → 0· is a quasi-isomorphism.

(ii) Not every complex is quasi-isomorphic to its homology as there need not necessarily exista map between them. Therefore, it is possible to have two complexes C,D with the samehomology groups but not be quasi-isomorphic.

Example 4.18. Consider the sequence, say C·

· · · 4−→ Z/12Z 3−→ Z/12Z 4−→ Z/12Z 3−→ Z/12Z→ 0

We have H0(C) = ker d0/ im d1 = (Z/12Z)/(Z/12Z) ∼= Z/3Z (the cokernel of the last map),H1(C) = (4Z/12Z)/(4Z/12Z) ∼= 0, and H2(C) = (3Z/12Z)/(3Z/12Z) ∼= 0. Therefore, we have

Hn(C) =

{Z/3Z, if n = 0

0, otherwise

55 C. McWhorter

So this sequence is exact at all n with the exception of n = 0. Now is the chain map

· · · Z/12Z Z/12Z Z/12Z 0

· · · 0 0 Z/3Z 0

4 3

π

It is clear that the squares commute (only the last square is a concern but π3 = 0 in Z/3Z). Sothis is indeed a chain map. Furthermore, the induced map is clearly an isomorphism of homologygroups.

Example 4.19. Take the following two complexes

C : 0 −→ Z/4Z 2−→ Z/4Z −→ 0

D : 0 −→ Z/2Z 0−→ Z/2Z −→ 0

These complexes have isomorphic homology groups but are not quasi-isomorphic (the induced mapson homology are the zero maps but these both have nonzero homology groups).

4.20 Long Exact Sequences

To understand how homologies behave, we now need to develop the most useful tool for any goodhomology theory - long exact sequences.

Theorem 4.21. Let 0· −→ A·f−→ B·

g−→ C· −→ 0 be a short exact sequence of chain complexes.Then for all n, there are natural maps δ : Hn(C)→ Hn−1(A), called connecting homomorphisms,such that the sequence

· · · → Hn+1(A) −→ Hn+1(B) −→ Hn+1(C)δn+1−→ Hn(A)

Hn(f)−→ Hn(B)Hn(g)−→ Hn(C)

δn−→ Hn−1(A) −→ · · ·

is exact. Similarly, if instead we have a short exact sequence of cochain complexes, then for all nthere are natural maps δ : Hn(C)→ Hn+1(A) such that the sequence

· · · → Hn+1(A) −→ Hn+1(B) −→ Hn+1(C)δn+1−→ Hn(A)

Hn(f)−→ Hn(B)Hn(g)−→ Hn(C)

δn−→ Hn−1(A) −→ · · ·

is exact.

Proof: This is just a simple application of the Snake Lemma. Given a short exact sequence ofcomplexes 0 −→ A· −→ B· −→ C· −→ 0, apply the Snake Lemma to

0 An Bn Cn 0

0 An−1 Bn−1 Cn−1 0

dAn dBn dCn

in order to get for each n an exact sequence

0 −→ ZAn −→ ZBn −→ ZCnδ−→ An−1/ im dAn −→ Bn−1/ im dBn −→ Cn−1/ im dCn −→ 0


We rearrange to a commutative diagram with exact rows

An/ im dnBn/ im d Cn/ im d 0

0 ZAn−1 ZBn−1 ZCn−1

Applying the Snake Lemma we get an exact sequence

Hn(A) −→ Hn(B) −→ Hn(C)δ−→ Hn−1(A) −→ Hn−1(B) −→ Hn−1(C)

Note that we often write long exact sequences of homology groups as follows: given an exactsequence of complexes 0 −→ A· −→ B· −→ C· −→ 0, we write the exact sequence of homologygroups as

H∗(A) H∗(B)

H∗(C)

δ

Proposition 4.22 (Naturality of Long Exact Sequence). The map δ is natural; that is, given acommutative diagram of complexes

0 A· B· C· 0

0 A′· B′· C ′· 0

α β γ

the following diagram commutes

· · · Hn(B)Hn(β) Hn(C) Hn−1(A) · · ·

· · · Hn(B) Hn(C) Hn−1(A) · · ·

δ

Hn(γ) Hn(α)

δ

Proof: The squares with only Hn(−) clearly commute as Hn is a functor: Hn(fg) = Hn(f)Hn(g)for all composable f, g, as is easy to check. For the squares with the δ’s, note that δ is constructedin the same way as they are in the Snake Lemma. If one lifts c ∈ Hn(C) to a b ∈ Bn and d(b) is theimage of a ∈ An−1,

0 An Bn Cn 0

An−1 Bn−1 Cn−1

57 C. McWhorter

then β(b) lifts γ(c) = γ(c). As β is a chain map, we know d(β(b)) = β(d(b)) is the image of α(a).So δ(γ(c)) = α(a). So the square commutes

δ(γ(c)) = α(a) = α(δ(c))

4.23 Homotopy

Definition 4.24. A complex is split exact if it is exact and for all n

0 −→ Zn⊆−→ Cn

d−→ Bn−1 −→ 0

is split exact.

Note that you actually get Cn = Zn ⊕ s(Bn−1) and C is the following complex Xn = s(Bn−1)

Zn+1 Zn Zn−1

⊕ ⊕ ⊕ · · ·

Xn+1 Xn

∼ ∼

where d(s(Bn−1)) = 1Bn−1 . Furthermore, any exact complex C over a field (or semisimple ring) issplit exact. However, less “drastic” is the notion of homotopy.

Definition 4.25 (Nulltomotopy). A chain map f : C· → D· is nullhomotopic f ' 0 if there aremaps sn : Cn → Dn+1 such that f = ds′ + sd. The map s is called a (null)homotopy.

· · · Cn+1 Cn Cn−1 · · ·

· · · Dn+1 Dn Dn−1 · · ·

d

fs

d

s fs s

d′ d′ d′

Two chain maps f, g : C· → D· are homotopic f ' g if f − g is nullhomotopic, i.e. f − g ' 0.Finally, f : C· → D· is a homotopy equivalence if it has an inverse up to homotopy; that is, there isg : D → C such that fg ' 1D and gf ' 1C .

Proposition 4.26. If f ' 0 then H(f) = 0. Therefore, f ' g then H(f) = H(g) as maps.

Proof: Given a chain map f : C· → D·. If f ' 0, then there is s : f = sd+ d′s.

Cn Cn−1

Dn+1 Dn

d

fs

d′


If x ∈ Hn(C) = Zn/Bn, then

f(x) = (sd+ d′s)(x)

= (sd)(x) + (d′s)(x)

= s(d(x)) + 0

= s(0) + 0

= 0 + 0

= 0

Note that d(s(x)) ∈ Bn so (d′s)(x), which is 0 in Hn(D). Then f ' g so that f − g ' 0 by theabove. So Hn(f − g) = 0 so that Hn(f) = Hn(g).

Proposition 4.27. C is split exact if and only if 1C ' 0.

Example 4.28. Take R = Z[x, y]/(xy).

· · · x−→ Rx−→ R

x−→ Rx−→ R −→ 0

and multiplication by x on C·. We claim that x ' 0.

· · · R R R · · ·

· · · R R R · · ·

x

xs

x

xs

xs s

x x x

Create s by taking the zero map or the identity in the appropriate places. This also shows thathomotopies are not necessarily unique as we could have chosen s = x everywhere.

4.29 Bicomplexes and Tot

Definition 4.30 (Bicomplex). A double complex or bicomplex is a diagram of R-modules and maps

......

· · · Cp−1,q Cp,q · · ·

· · · Cp−1,q−1 Cp,q−1 · · ·

......

d′

d

d′

d′

d

d′

where d2 = 0, d′2 = 0, and that the diagram commutes, i.e. dd′ = d′d.

59 C. McWhorter

1. We can think that a bicomplex as a complex of complexes (vertical or horizontal).

2. The index is the position in (x, y) in Cartesian coordinates. The arrows are usually drawn to allflow towards the origin or all to flow away from the origin.

3. Some authors have all squares in the diagram anticommute: dd′ = −d′d.

4. C·· is bounded if along each diagonal p + q = n (fixed) there are only finitely many nonzeromodules. For example, a first quadrant bicomplex is bounded.

Definition 4.31 (Total Complex). Given a bicomplex C, form the total complex Tot(C) as(Tot

⊕C)n

def=

⊕p+q=n

Cp,q(Tot

∏C)n

def=

∏p+q=n

Cp,q

with differentials dTotdef= d+ (−1)pd′.

Remark 4.32. We place the minus sign on odd columns to make the diagram anticommute. If C··is bounded, then Tot⊕ = Tot

∏. One can check that this diagram does indeed anticommute and

(dTot)2 = 0 so that this indeed a complex.

Example 4.33. Given two complexes

P· : · · · −→ P1 −→ P0 −→ 0

Q· : · · · −→ Q1 −→ Q0 −→ 0

Then the tensor product P· ⊗R Q· is a bicomplex

......

P0 ⊗R Q2 P1 ⊗R Q2

P0 ⊗R Q1 P1 ⊗R Q1. . .

P0 ⊗R Q0 P1 ⊗R Q0 P2 ⊗R Q0 P3 ⊗R Q0d⊗1 d⊗1 d⊗1

or its Tot(P· ⊗R Q·)n =

⊕p+q=n

Pp ⊗R Qq

Similarly, we can define a bicomplex from HomR(P·, Q·).

Furthermore, if 0 −→ A·· −→ B·· −→ C·· −→ 0 is an exact sequence of bicomplexes and arebounded, then there is a short exact sequence. To see this, note that the direct sum of finitelymany short exact sequences is exact and that if A → B is a map of bicomplexes then the mapTot(A)→ Tot(B) is a map.


4.34 Mapping Cone & Mapping Cylinder

Definition 4.35 (Mapping Cone). Let f : B· → C· be a chain map, then the mapping cone of f is

C(f)def= Tot(C

←−f )

C3 B3

C2 B2

C1 B1

C0 B0

d −d′

d −d′

d −d′

f

(C(f))n = Cn ⊕Bn−1Cn Cn−1

⊕ ⊕

Bn−1 Bn−2

d

f

−d

with differential

[d f0 −d

].

Proposition 4.36. 1. There is a short exact sequence of complexes

0 −→ Ci1−→ C(f)

π2−→ B[−1] −→ 0

where B[−1] is the complex B shifted one unit to the left, i.e. (B[−1])n = Bn−1.

2. There is an associated long exact sequence of homology

· · · δ−→ Hn(C) −→ Hn(C(f)) −→ Hn(B[−1])δ−→ Hn−1(C) −→ Hn−1(C(f)) −→ · · ·

with connecting map δ = Hn(f).

3. The map f is a quasi-isomorphism if and only if Hn(C(f)) = 0.

Definition 4.37 (Mapping Cylinder). Let f : B· → C· be a chain map. The mapping cylinder of f

is Cyl(f)def= C(C(f)[1]

π2−→ B). One could think of this as “Tot” of (Cf←− B −→ B). However,

the arrows go the other way but this is a tricomplex. Also, (Cyl(f))ndef= Cn ⊕Bn−1 ⊕Bn with the

obvious differentials.

Essentially, we have subtracted off the complex B from the cone up to homology. To verify thisinstinct, we prove the following lemma.

61 C. McWhorter

Lemma 4.38. There is an exact sequence

0 −→ Ci−→ Cyl(f) −→ C(1B)[−1] −→ 0

But C(1B) is always split exact. Hence by the long exact sequence, H(i) is an isomorphism so thati is a quasi-isomorphism. In fact, i is a homotopy equivalence.

Proof: Look at Cyl(f)j−→ C given by (C, b, b′) 7→ c+ f(b). One need check the following details:

• This is a chain map

• The composition ji = 1C

• c 7→ (c, 0, 0) 7→ c+ f(0) = c

• ij ' 1Cyl(f)

Lemma 4.39. There is a commutative diagram

B C C(f)

0 B Cyl(f) C(f) 0

f

∼= j

i

∼=

i3

i

Remark 4.40. The above lemma is useful for triangulated and derived categories.


5 Resolutions

5.1 Projective Resolutions

We now develop a homology theory that gives “correction” terms for a functor, such as − ⊗RN,Hom(−, N), · · · , not being left/right exact. For instance, the tensor functor is right exact butdoes not preserve all short exact sequences

· · · −→ A⊗R N −→ B ⊗R N −→ C ⊗R N −→ 0

What terms exist on the left to make this sequence entirely exact? The key to filling in these terms is“approximating” modules by projective or injective modules. Recall a presentation of an R-moduleM is an expression of it as M ∼= F/K, where F is free and K is a kernel. We do this by mapping afree module onto M :

0 −→ K⊆−→ F

π−→M −→ 0

We can then continue this process by finding a free module to map into K, which would then give amap from the new F ′ to F

F ′

0 K F M 0

d1

⊆π

But then we have an exact sequence

F1 −→ F0 −→M −→ 0

Continuing this process again and again gives a “full” exact sequence. We need only use projectivemodules to complete this process.

Definition 5.2 (Projective Resolution). A projective resolution of R-modules of M is a complex P·of projective modules together with a map to M , P0

ε−→M , so that the augmented complex

· · · −→ P1 −→ P0 −→M −→ 0

is exact. Equivalently, there is a quasi-isomorphism

· · · P2 P1 P0 0

· · · 0 0 M 0

ε

Proposition 5.3 (Existence). In R-modules or any category in which all objects have projectivemapping onto them, any module M has a projective resolution.

63 C. McWhorter

Proof: Take a free module F0 mapping onto M . Let K0 = ker ε. Take F1 to be a free modulemapping onto K0 by π1 and let d1 = iπ1. Proceed by induction taking Kn = ker dn = kerπn. Thenthe sequence is exact by construction.

F1 F0 M 0

K0

0 0

d1

π1

ε

i

Example 5.4 (Koszul complex). Take R = k[x, y] and let M = R/(x, y) ∼= k, where k is a field.

0 R R2 R M 0

K (x, y)

0

f=

y

−x

[x y] π

f

Where e1 7→ x and e2 7→ y. This is a famous resolution called the Koszul complex.

Example 5.5. Take R = k[x, y]/(xy) and M to be the module R/(x).

· · · R R R R R M 0

(y) (x)

y x π

Theorem 5.6 (Comparison Theorem). Given a map f ′ : M → N and complexes

· · · P2 P1 P0 M

· · · Q2 Q1 Q0 N 0

ε

f ′

η

with each Pi projective and the bottom row exact, then the map f ′ can be lifted, i.e. ηf = f ′ε to a

chain map P·f−→ Q· and any two lifts are homotopic.


Proof: We do this by induction. Assume we have a diagram

0 Z0 P0 M

0 Z ′0 Q0 N

f0 f0

ε

f ′

η

As P0 is projective and η is surjective, there is a f0 such that the diagram

P0

Q0 N

f0f ′ε

η

We have ηf0 = f ′ε so that the last square commutes. Therefore, we have a map Z0 → Z ′0 using aprevious lemma. Then f0(Z0) ⊆ Z ′0 so f0 restricts to a map Z0 → Z ′0.

Now if you have constructed the maps for all I ≤ n, then we have a commutative diagram

Pn+1 Pn Pn−1

Zn

Qn+1 Qn Qn−1

Z ′n

d

d

d

fn fn−1

d

d′

d

As the right square commutes, fn restricts to give a map Zn = ker d → Z ′n = ker d′. Note thatim d : Pn+1 → Pn ⊆ Zn as d2 = 0. Likewise in the bottom row, im d′ ⊆ Z ′n as Q· is exact: im d′ = Z ′nso that the map is surjective onto Z ′n. This yields a diagram

Pn+1

Qn+1 Z ′n

fn+1

fnd

d′

Since Pn+1 is projective, we get a map fn+1 : Pn+1 → Qn+1 such that the triangle commutes –forcing commutativity of the square. Therefore, we obtain the result via induction.

We need only see that this map is unique up to homotopy. We do this by induction as well.Suppose g : P· → Q· is another lift of f ′. We want to show that given another lift g, we have f ' g,

65 C. McWhorter

i.e. f − g ' 0. Let h = f − g. Note that h is a difference so it is a lift M → N , f ′ − f ′ = 0 so it is alift of 0.

0

M

· · · P1 P0 0 0

· · · Q1 Q0 0 0

N

0

h1

ε

h0 0s−1 s−2

d′

For n < 0, we know that sn = 0. We need find a map s0 : P0· → Q0·. We need h0 : sd︸︷︷︸ 0 + ds. Let

Z ′0 = ker η. As Q· → N → 0 is exact, we know that im d′ = ker η = Z ′0.

0

M

P1 P0 0

Q1 Q0 0

Z0 N

0

d

h1

ε

s0h0

d′

d′ η

Now ηh0 = 0. So imh0 ⊆ ker η = Z ′0. Then we have a diagram

P0

Q1 Z ′0

h0

d′


As P1 is projective, there is a map s0 : P0 → Q1 such that the triangle commutes. This completes thebase step. We now proceed with the induction argument. Given si with i ≤ n and hn = d′sn+sn−1d.

Pn+2 Pn+1 Pn Pn−1

Qn+2 Qn+1 Qn Qn−1

hn+2 hn+1sn+1hn

d

hn−1sn−1

d′

We want to find an sn+1 such that hn+1 = d′sn+1 + snd. That is, hn+1 − snd = d′sn+1. Then

Z ′n+1 = ker d′ ⊆ Qn+1. By exactness, im d′ = Z ′n+1 so that d′ restricts to a surjection Qn+2d′−→ Z ′n+1.

d′(hn+1 − snd) = d′hn+1 − d′snd= d′hn+1 − (hn − sn−1d)d

= d′hn+1 − hnd+ sn−1d2

= 0

so that im(hn+1 − snd) ⊆ Z ′n+1. Then we have a diagram

Pn+1

Qn+2 Z ′n+1

sn+1hn+1−snd

d′

Corollary 5.7. A projective resolution of a module is unique up to homotopy, i.e. any two projectiveresolutions are homotopy equivalent.

Proof: Let

P0 M

Q0 M

1M

be two projective resolutions of M . By the Comparison Theorem, there is a lift P0·f−→ Q·, Q·

g−→ P·of the identity map. Note that both fg : Q· → Q· and 1Q· are both lifts of 1M . So by uniqueness,fg ' 1Q· and similarly gf ' 1P· .

The following lemma is useful for building resolutions for extensions.

Lemma 5.8 (Horseshoe Lemma). Given a short exact sequence of R-modules

0 −→M ′ −→M −→M ′′ −→ 0

67 C. McWhorter

and projective resolutions P ′· →M ′, P ′′· →M ′′, then one can build a projective resolution P· →Mwith Pn = P ′n ⊕ P ′′n and differential d of the form

P ′n

[d′ ∗0 d′′

]P ′n−1

⊕ −→ ⊕P ′′n P ′′n−1

so as to get an exact sequence

0 −→ P ′·i1−→ P·

π2−→ P ′′· −→ 0

Proof:

0 P ′ P ′1 ⊕ P ′′1 P ′′1

0 P ′0 P ′0 ⊕ P ′′0 P ′′0 0

0 M ′ M M ′′ 0

0 0

ε′

i1 π2

ε′′θ

i π

Since P ′′0 is projective and π is onto, there is a map θ : P ′′0 → M such that πθ = ε′′. This mapθ with iε′ : P ′0 → M gives a map ε : P ′0 ⊕ P ′′0 → M given by (a, b) 7→ iε(a) + θ(b). By the SnakeLemma, coker ε = 0 so that ε is onto. Then the sequence

0 −→ ker ε′ −→ ker ε −→ ker ε′′δ−→ 0

is exact. Repeat the argument to

0 P ′1 P ′1 ⊕ P ′′1 P ′′1 0

0 ker ε′ ker ε ker ε′′ 0

0 0

i1

d′

π2

d′′

Since P ′ε′−→ M ′ and P ′′

ε′′−→ M ′′ are resolutions, we can repeat the argument above to find thatthe squares commute and

d =

[d′ ∗0 d′′

]


5.9 Injective Resolutions

Really that any R-module M can be indebted into an injective R-module.

M ↪→ E(M)

where E(M) is the smallest injective R-module containing M , i.e. the injective hull/envelope.

Definition 5.10 (Injective Resolution). Given an R-module M , an injective resolution of M is anexact sequence

0 −→M −→ I0 −→ I1 −→ I2 −→ · · ·

where each Ii is an injective R-module.

We need justify the existence of an injective resolution for any R-module M . We can embed Minto an injective R-module, take the cokernel of the map and embed it into an injective R-moduleI ′. Then we can take the cokernel of this map and continue this process:

0 M I0 I ′

coker ε

ε d0=iπ

π i

Theorem 5.11 (Comparison Theorem). Given a map Mf ′−→ N and complexes

0 M J0 J1 J2 · · ·

0 N I0 I1 I2 · · ·

f ′

with each In an injective R-module and 0 → M → J· exact, f ′ can be lifted to a cochain mapf : J · → I · uniquely up to homotopy.

Proof: This follows mutatis mutandis from the proof of the Comparison Theorem for ProjectiveResolutions.

Remark 5.12. From now on, functors will be assumed to be additive!

We now generalize the concept of injective and projective resolutions to a type of functor: derivedfunctors. These were first developed by, among others, Grothendieck and made widely accessible inthe book by Cartan and Eilenberg.

5.13 Left Derived Functors

Definition 5.14 (Left Derived Functor). Let F : A → B be a right exact functor between abeliancategories. Given a short exact sequence,

0 −→ A −→ B −→ C −→ 0

69 C. McWhorter

we obtain an exact sequence

? −→ F (A) −→ F (B) −→ F (C) −→ 0

If A has projective resolutions, define the left derived functor LnF of F for n ≥ 0 by the following:

if M ∈ A, choose a projective resolution P· →M and defined LnF (M) = Hn(F (P·)) = Hn(· · · F (d)−→F (P1)

F (d)−→ F (P0) −→ 0).

There is one one particular type of left functor of great enough importance to warrant its ownname and separate definition:

Definition 5.15 (Tor). The function F (−) = − ⊗R N is a right exact functor from R-mod to

Z-mod. Its left derived functors are called TorRn (−, N)def= LnF (−), i.e. for any M

TorRn (M,N) = LnF (M) = Hn(F (P·))

where P· →M is a projective resolution; that is, the homology groups of

· · · d⊗1−→ P1 ⊗Nd⊗1−→ P0 ⊗N −→ 0

Example 5.16. Let R be a ring and f a nonzero divisor in R. Let M = R/(f2) and N = R/(f).We compute TorRn (M,N). Take the projective resolution of M

· · · 0 0 R′ R′ M 0

(f2)

f2 π

Note that if xf2 = 0, then x = 0 as f is a nonzerodivisor. Now apply the functor −⊗R N :

0 R⊗R N R⊗R N 0

N N

f2⊗1

∼ ∼

f2=0

So we have

TorR0 (M,N) = H0(complex) = R/(f)/0 = R/(f)

TorR1 (M,N) = H1(complex) = R/(f)/0 = R/(f)

TorRn (M,N) = 0 n ≥ 2

However, there is a problem with the above example: we did not justify why the Tor groupswould not be different had we chosen any other projective resolution.

Proposition 5.17. (i) LnF is well defined; that is, LnF is independent of the choice of projectiveresolution.


(ii) LnF is a functor and given f : M → K, we get Ln(F ) : LnF (M)→ LnF (K)

(iii) L0F = F

(iv) For any short exact sequence 0→M ′ →M →M ′′ → 0, there exist connecting maps δn and along exact sequence

· · · Ln+1F (M ′) Ln+1F (M) Ln+1F (M ′′)

LnF (M ′) LnF (M) LnF (M ′′)

Ln−1F (M ′) Ln−1F (M) · · ·

L0F (M ′) L0F (M) L0F (M ′′) 0

δn+1

δn

δ1

such that the δn are natural; that is, if for any map of short exact sequences

0 M ′ M M ′′ 0

0 K ′ K K ′′ 0

the “obvious” ladder diagram of the two long exact sequences and maps between them commutes, i.e.for all n the following diagram commutes:

LnF (M ′′) Ln−1F (M ′)

LnF (K ′′) Ln−1F (K ′)

δ0

δn

Proof:

(i) If P· →M and P ′· →M are two projective resolutions of M , then by the Comparison Theorem,they are homotopy equivalent. So there exist lifts of 1M

P· P ′·

f

g

such that fg ' 1P ′· and gf ' 1P· . Applying the function F (−), we obtain F (fg) ' F (1) viathe homotopy F (s), where s is the homotopy for fg ' 1. Then we have F (f)F (g) ' 1P ′·and likewise F (g)F (f) ' 1P· . So F applied to the complexes gives a homotopy equivalence.Then this must induce the same map on homotopy so that HnF (f)HnF (g) = Hn(1) = 1.Therefore, we have HnF (g)HnF (f) = 1 so that HnF (f) is an isomorphism. But thenHnF (f) : HnF (P·) = LnF (M)→ HnF (P ′· ) = LnF (M) is an isomorphism.

71 C. McWhorter

(ii) This part follows similarly to (i). Let Mf−→ K. Let P· → M and Q· → K be projective

resolutions of M and K, respectively. By the Comparison Theorem, we then get a lift of

P·′f ′−→ Q·. Applying the function F , we get a chain map F (P·)

F (f)−→ F (Q·). Hence, we get aninduced map on homology: LnF (M)→ LnF (K).

(iii) Simply compute L0F (M) given that P· →M is a projective resolution.

L0F (M) = H0(F (P·)) = coker(F (P1)→ F (P0)) = F (M)

since F (P1)→ F (P0)→M → 0 is exact.

(iv) Let0 −→M ′ −→M −→M ′′ −→ 0

be a short exact sequence. By the Horseshoe Lemma, there is a projective resolution fitting ina diagram

0 P ′· P· P ′′· 0

0 M ′ M M ′′ 0

ε′ ε ε′′

with exact rows (in fact, this diagram is even split). Applying F preserves exactness in eachrow (check this, use the “splitness”). Then we have

0 FP ′· FP· FP ′′· 0

0 FM ′ FM FM ′′ 0

Now we use the long exact sequence on homology and we are done. [Naturality has yet to beproven but for this see Weibel.]

Proposition 5.18. F is a right exact additive functor if and only if LnF = 0 for all n > 0 if andonly if L1F = 0.

Proof: The reverse direction follows from the definition and the forward direction follows easilyfrom the long exact sequence.

Definition 5.19 (δ-functor). A δ-functor is a collection {Tn}n≥0 of additive functors such that (iv)holds. That is for all exact

0 −→ A −→ B −→ C −→ 0

there is a natural δn’s such that

· · · −→ T1A −→ T1B −→ T1Cδ−→ T0A −→ T0B −→ T0C −→ 0


is a long exact sequence. The functor δ is called universal if given any other δ-functor, {δn} andnatural transformation S0 → T0, there is a unique map of δ-functors from

{Sn}{fn}−→ {Tn}

that commute with the δ’s.

Corollary 5.20. Any two universal δ-functors, {Tn}, {T ′n} with isomorphic T0, T0; are isomorphic,i.e. Tn ∼= T ′n and the isomorphisms commute with the δ’s.

Theorem 5.21. Given a right exact F , {LnF}n≥0 are a universal δ-functor (so they are unique).

Recall torRn (M,N) = LnF of functor F (−) = −⊗R N .

5.22 Balancing Tor

There are two derived functors we can associate to

M,N −→M ⊗R N

Namely, F (−) = − ⊗r N and G(−) = M ⊗R −. We have F (M) = G(N). What about LnF (M)and LnG(N)? These behave as expected; however, the proof needs many details. We attend to theproof first then fill in the necessary lemmas afterwards.

Theorem 5.23. If P·ε−→M and Q·

η−→ N are projective resolutions, then

HN (P· ⊗N) ∼= Hn(Tot(P· ⊗Q0)) ∼= Hn(M ⊗Q·) = torRn (M,N)

(note that the Tot is often dropped).

Proof: Let P· →M be a projective resolution. That is

· · · −→ P2 −→ P1 −→ P0ε−→M

is exact or equivalently

· · · P2 P1 P0 0

· · · 0 0 M M

ε

=

is a quasi-isomorphism. Hence, we have a map of bicomplexes:We want to show that Tot(1 ⊗ η) is a quasi-isomorphism, likewise for Tot(ε ⊗ 1). However,

the latter follows mutatis mutandis. By Lemma 5.24, it is enough to show that 1 ⊗ η is a quasi-isomorphism on each row. By Lemma 5.26, it will follow since each Pn is projective, hence flat. SoPn ⊗− is an exact functor so it preserves quasi-isomorphisms (show this).

Lemma 5.24. If A··f−→ B·· is a map of first quadrant bicomplexes (with arrows towards the origin

or away from the origin) such that the map on each row/column is a quasi-isomorphism, then

Tot(A··Tot(f)−→ Tot(B··) is a quasi-isomorphism.

73 C. McWhorter

Proof (Sketch): By taking cones of f on the rows, one obtains a bicomplex, say Crow/col(f). Note, fis a quasi-isomorphism on the rows so that Crow/col(f) is exact in each row/column. Also,

0 −→ B··i

↪−→ Crow/col(f)π−→ A··[−1] −→ 0

is exact. Then0 −→ Tot(B) −→ Tot(Crow/col) −→ Tot(A··)[−1] −→ 0

is exact. But the middle term is precisely C(Tot(f)). By Lemma 5.25, the sequence is exact soTot(f) is a quasi-isomorphism.

Lemma 5.25. If C· is a first quadrant bicomplex with exact rows (or columns) then Tot(C··) isexact.

Proof: Let X = (Xij) ∈ ⊕i+j=nCij be a cycle in Tot(C), i.e. for each i, (−1)id(Xi,n−i) +

d′(Xi−1,n−(i−1)) = 0, where (−1)id are the downwards maps in the complex and d′ are the leftarrows in the complex. Start at the upper left. Let Y0,n+1 = 0. Since d′(X0,n) = 0 and the rows areexact, there is a Y1,n such that d′(Y1,n) = X0,n. Check that X1,n−1 − (−d(Y1,n)) ∈ ker d′. Since therows are exact, there is a Y2,n−1 ∈ d′(Y2,n−1) = X1,n−1 + d(Y1,n). Then “d” of the Y ’s so far areX1,n−1. Continue this process by induction. This induction must end as the complex is 1st quadrant


(at the most when it hits the axis). So “d” of the Y ’s is (Xij)i+j=n, so (Xij) is a boundary.

Lemma 5.26. Let A·f−→ B· be a chain map (map of complexes). If f is a quasi-isomorphism and

P is flat, then P ⊗A·1⊗f−→ P ⊗B· is a quasi-isomorphism.

Proof: Suppose that Af−→ B is a quasi-isomorphism. Then C(f) is exact so that as P is flat,

P ⊗ C(f)· is exact. But this is isomorphic to C(1⊗ f). Then 1⊗ f is a quasi-isomorphism.

The same proof works for F (−) instead of P ⊗− if F is an exact functor. Suppose

0 −→ A −→ B −→ C −→ 0

is an exact sequence. Then we get the exact sequence

A⊗N −→ B ⊗N −→ C ⊗N −→ 0

We get a similar sequence for M ⊗−.

Corollary 5.27. We have two long exact sequences for

tor(M,N) = Hn(P· ⊗N) = Hn(M ⊗Q·)

5.28 Right Derived Functors

Definition 5.29. Let F : A → B be a left exact functor, where A and B are abelian and F isadditive. If A has injective resolutions, then we can define a right derived functor by taking N ∈ Aand take injective resolution N → I ·:

RnF (N)def= Hn(F (I ·))

This is the same construction as before applied to F op : Aop → Bop. That is, F is left exact ifand only if F op is right exact. Similarly, we have injective resolutions in A if and only if we haveprojective resolutions in Aop. So we have an automatic proof of:

Theorem 5.30. (i) RnF : A → B is well-defined (independent of the choice of injective resolu-tion) and is a functor for all n

(ii) R0F = F

(iii) if 0→ A→ B → C → 0 is exact, then there are a functorial, i.e. natural, maps δn such thatwe get a long exact sequence

0 −→ R0F (A) −→ R0F (B) −→ R0F (C)δ−→ R1F (A) −→ R1F (B) −→ R1F (C)

δ−→ · · ·

The major example will be the Ext functor. Let M,N ∈ R−mod. Let F (−)def= HomR(M,−).

Then F is covariant and left exact. These right derived functors are called the Ext functors.

ExtnR(M,N) = RnF (N)

i.e. Hn(Hom(M, I·)).

75 C. McWhorter

Example 5.31. ExtnZ(Z/kZ,Z) for k ≥ 2. Recall over a PID, injective is equivalent to divisible.So we have an injective resolution of Z

0 −→ Z ↪−→ Q −→ Q/Z −→ 0

where I0 = Q and I1 = Q/Z, which are divisible so they are injective Z-modules. Then we have

0 −→ HomZ(Z/kZ,Q) = 0 −→ HomZ(Z/kZ,Q/Z) −→ 0

So then we have

Ext0Z(Z/kZ,Z) = H0(above) = 0/0 = 0

Ext1Z(Z/kZ,Z) = H1(above) = HomZ(Z/kZ,Q/Z)/0

where the maps in Ext1Z are given by 1 7→ p/q with (p, q) = 1 such that kp/q = 0 ∈ Z so that q | kwhich is {p/q ∈ Q/Z | kp/q ∈ Z}.

Note that Ext0R(M,N) = HomR(M,N).

5.32 Contravariant Functors

If F : A → B is contravariant and left exact. We get F 1 : Aop → B is contravariant. So this is stillleft exact so we can use the construction from before. So F ′ has right derived functors which are

RnF ′(M)def= Hn(F ′(injective resolution of M in A))

which is Hn(F (projective resolution of M in A)) = RnF (M).

Example 5.33. F (−) = HomR(−, N) is left exact. This gives RnF (M) = Hn(Hom(P·, N)), whereP· is a projective resolution of M .

5.34 Balancing Ext

We have two derived functors to associate to M,N → HomR(M,N). Namely, Hom(M,−) andHomR(−, N).

Theorem 5.35. If Nη−→ I · is an injective resolution and O·

ε−→M is a projective resolution

Ext0R(M,N) = Hn(Hom(M, I ·)) ∼= Hn(Tot(Hom(P·, I·))) ∼= Hn(Hom(P·, N))

Example 5.36. We redo the example from before. Take ExtnZ(Z/kZ,Z). So instead of taking aprojective resolution of Z/kZ, we take a projective resolution of Z/kZ

0 −→ Z k−→ Z −→ Z/kZ −→ 0

where P1 = Z and P0 = Z. Then we have Extn = Hn(HomZ(P·,Z)) so we have

0 −→ HomZ(Z,Z)k∗−→ HomZ(Z,Z) −→ 0

which is0 −→ Z k−→ Z −→ 0

So that Ext0Z(Z/kZ,Z) = 0 and Ext1Z(Z/kZ,Z) = Z/kZ, which is the subgroup 〈1/k〉 of Q/Z.


Corollary 5.37. If

0 −→ A −→ B −→ C −→ 0

is exact. There are two types of long exact sequences

0→ Hom(M,A)→ Hom(M,B)→ Hom(M,C)δ−→ Ext1(M,A)→ Ext1(M,B)→ Ext1(M,C)

δ−→ · · ·

and the long exact sequence

0→ Hom(C,M)→ Hom(B,M)→ Hom(A,M)δ−→ Ext1(C,M)→ Ext1(C,M)→ Ext1(A,M)

δ−→ · · ·

Note that since Tor and Ext are balanced, there is another way to get the same long exactsequence (avoiding the Horseshoe Lemma). For example for Ext, if we have the exact sequence

0 −→ A −→ B −→ C −→ 0

and an R-module M , we take a projective resolution P· →M of M , then

0→ Hom(P·, A)→ Hom(P·, B)→ Hom(P·, C)→ 0

is an exact sequence of complexes since each Pn is projective (hence preserving exactness). Takingthe long exact sequence on homology.

Note also that if R is commutative with M,N being R-modules, if Mr−→ N is multiplication

by r, then the induced maps on M ⊗N , Hom(M,N), and Hom(N,M) are just multiplication by r.So too are the maps on Tor(N,M),Tor(M,N),Extn(M,n), and Extn(M,N).

Corollary 5.38 (Tor Balanced). We have

TorRn (M,N) ∼= TorRop

n (N,M)

Proof (Sketch): This is simple as M ⊗R N ∼= N ⊗Rop M , which is easily demonstrated.

Corollary 5.39. The following are equivalent for an R-module F :

(i) F is flat; that is, F ⊗R − is exact.

(ii) TorRn (F,N) = 0 for all n > 0.

(iii) Tor1(F,N) = 0

Proof:

(i)→(ii): Suppose that F is flat. Take a projective resolution P· → N of N , then we have

TorRn (F,N)def= Hn(F ⊗R P·) = F ⊗R Hn(P·) = 0

since F ⊗R − is an exact functor.

(ii)→(ii): This is obvious.

77 C. McWhorter

(iii)→(i): Start with an exact sequence

0 −→ A −→ B −→ C −→ 0

Tensoring with F preserves right exactness

Tor1(F,C) = 0 −→ F ⊗R A −→ F ⊗R B −→ F ⊗R C −→ 0

as Tor1(F,C) = 0 by assumption.

Corollary 5.40. One can use flat resolutions instead of projective resolutions to compute Tor.

Proof: This follows easily by the proceeding corollary and dimension shifting. One can also easilyshow this using the fact that Tor is balanced.

Proposition 5.41. The following are equivalent for an R-module P :

(i) P is projective

(ii) ExtnR(P,−) = 0 for n > 0

(iii) Ext1R(P,−) = 0

Proof: This is left as an exercise but it follow quickly using the fact that Hom(P,−) is exact.

Proposition 5.42. The following are equivalent for any R-module I:

(i) I is injective

(ii) ExtnR(−, I) = 0 for all n > 0

(iii) Ext1R(−, I) = 0

Proof: L.T.R.

Proposition 5.43 (Localization). Let R be a commutative ring and S be a multiplicatively closedsubset of R.

(i) S−1 TorRn (M,N) ∼= TorS−Rn (S−1M,S−1N)

(ii) If M is finitely presented, we have ExtnR(M,N) ∼= ExtS−RR (S−1M,S−1N).


5.44 Kunneth Formula and the Universal Coefficient Theorem

Theorem 5.45 (Kunneth Formula). Let P· be a chain complex of right R-modules such that

(i) each Pn is flat

(ii) each Bn−1 = d(Pn) is flat

then for each n and each left R-module M , there is an exact sequence

0 −→ Hn(P·)⊗M −→ Hn(P· ⊗M) −→ TorR1 (Hn−1(P ),M) −→ 0

Proof: Consider the short exact sequence

0 −→ Zn ↪−→ Pndn−→ Bn−1 −→ 0

where Pn, Bn−1 are flat. By the long exact sequence for Tor along with the fact that Pn and Bn−1are flat, we know that Zn are flat. Tensoring with M yields

· · · → 0 = Tor1(Bn−1,M)→ Zn ⊗M → Pn ⊗M → Bn−1 ⊗M → 0

where the left is 0 as Bn−1 is flat. Then we have

0 −→ Zn ⊗M −→ Pn ⊗M −→ Bn−1 ⊗M −→ 0

Putting these together for all n, we get a short exact sequence of complexes

0 −→ Z· ⊗M −→ P· ⊗Md⊗1−→ (B· ⊗M)[−1] −→ 0

with differentials on Z and B inherited from d on P·, so 0. That is, dZ = 0 and dB = 0. Then thelong exact sequence on homology is

· · · −→ Hn+1((B ⊗M)[−1])δ−→ Hn(Z ⊗M) = Zn ⊗M −→ Hn(P ⊗M) −→

Hn((B ⊗M)[−1])δ−→ Hn−1(Z ⊗M) −→ · · ·

We have Zn ⊗M/(i⊗ 1)(Bn ⊗M) = Zn/Bn ⊗M . But Hn(P·) = Zn/Bn. Since ⊗ is right exact, wehave

0 −→ Bn−1 −→ Zn−1 −→ Hn−1 −→ 0

This is a flat resolution of Hn−1(P ). So ker δ = ker(i⊗ 1) = Tor1(Hn−1(P ),M) by the resolutionabove.

Note that the term of the Kunneth Formula is the middle term. However, when does this evenhold? Particularly, when does (ii) hold? The major case when P· is projective (a fortiori flat) and Ris hereditary.

Definition 5.46 (Hereditary Ring). We call R a hereditary ring if submodules of projective modulesare projective. Equivalently, every module has projective dimension at most 1.

Example 5.47. Some easy examples of hereditary rings are R = Z, PIDs, and Dedekind domains.

79 C. McWhorter

In fact, we get a stronger result.

Theorem 5.48 (Universal Coefficient Theorem). Let P· be a chain complex of right R-modulessuch that

(i) each Pn are projective

(ii) each Bn−1 = d(Pn)

then the Kunneth Formula holds and the sequence (non-canonically) splits.

Hn(P· ⊗M) ∼= (Hn(P·)⊗M)⊕ TorR1 (Hn−1(P·),M)

We also have similar formulas for complexes:

Theorem 5.49 (Kunneth Formula for Complexes). If P·, Q· are complexes of left/right R-moduleswith each Pn flat and each Bn−1 = d(Pn) flat, then for all n there is an exact sequence

0 −→ ⊕p+q=nHn(P )⊗Hq(Q) −→ Hn(Tot(P· ⊗Q·)) −→ ⊕p+q=n−1 torR1 (Hp(P ), Hq(Q)) −→ 0

If each Pn, Bn−1 are projective, then this sequence splits.

Notice this result is useful for products of spaces!

Theorem 5.50 (Universal Coefficient Theorem for Cohomology). Let M be a left R-module and P·be a chain complex such that

(i) each Pn is projective

(ii) each Bn−1 = d(Pn) are projective

then for all n we have a (non-canonically) split exact sequence

0 −→ Ext1R(Hn−1(P ),M) −→ Hn(Hom(P·,M)) −→ Hom(Hn(P·,M)) −→ 0

5.51 Tor

Historically, Tor is related to

(i) to torsion (see 7.13 of Rotman’s Introduction to Homological Algebra)

(ii) Serre used Tor to give an algebraic foundation for intersection multiplicity:

χ(R/I,R/J)def=

d∑i=0

(−1)il(TorRi (R/I,R/J))

for varieties V (I), V (J) is a smooth space given by R, SpecR.


5.52 Ext and Extensions

For this section, we will be in R-mod.

Definition 5.53 (Extension). An extension of C by A is an exact sequence

ξ : 0 −→ Ai−→ X −→ C −→ 0

Note that C ∼= X/i(A) and A ∼= i(A)/0. Two extensions ξ, ξ′ are equivalent if there exists a map ϕsuch that we have the commutative diagram

ξ : 0A X C 0

ξ′ : 0 A X ′ C 0

= ϕ =

Observe by the Five-Lemma (or the Snake Lemma), ϕ is an isomorphism. We denote the equivalenceclass of ξ by [ξ].

Remark 5.54. Any split extension is equivalent to the trivial one:

0 −→ Ai

↪−→ A⊕ C π2−→ C −→ 0

Definition 5.55 (e(C,A)). For the moment, we define e(C,A) as the set of equivalence classes [ξ].

Theorem 5.56. There is a bijection e(C,A)θ−→ Ext1R(C,A), Ext1R(C,A)

ψ−→ e(C,A) in which thesplit extension corresponds to 0 ∈ Ext1R(C,A).

Proof: Choose a projection P· → C. Recall to compute Ext1R(C,A) = H1(Hom(P·, A)) = H1(0→Hom(P0, A)

d∗1−→ Hom(P1, A)d∗2−→ Hom(P2, A) −→ · · · ). By the Comparison Theorem, 1C lifts to a

chain map

· · · P2 P1 P0 C 0

· · · 0 A X C 0

0

d2

α1 α0 =

Commutativity gives d∗2(α1) = α1d2 = 0. So α1 is a cycle. To define θ, given an extension

ξ : 0 −→ Ai−→ X −→ C −→ 0

let θ(ξ)def= α1 ∈ ker d∗2/ im d∗1. Note that this is well defined as given another lift of 1C , say α′ ' α, so

there is a s such that α1 − α′ = 0s1 + s0d1 = s0d1 ∈ im d∗1. So α1 = α′ ∈ Ext1R(C,A). Furthermore,if [ξ] = [ξ′], one gets the commutative diagram

· · · P2 P1 P0 C 0

· · · 0 A X C 0

· · · 0 A X ′ 0 0

α2 α1 α0 =

= ϕ =

81 C. McWhorter

ϕα1 is the vertical composition for the lift for ξ′ So θ([ξ′])def= 1Aα1 and θ([ξ])

def= α1, which are equal.

Finally, θ takes trivial split extensions to 0 as

· · · P2 P1 P0 C 0

· · · 0 A A⊕ CC 0α1=0 (0 ε)T

=

π2

so that θ([ξ]) = α1 = 0.

We now need to define ψ. Given an element β ∈ Ext1R(C,A), we have β : P1 → A ∈ Hom(P1, A)such that β is a cycle, i.e. d∗2(β) = 0 so that βd2 = 0. So we have

· · · P2 P1 P0 C 0

· · · 0 A ? C 0

d2

β

d1 ε

=

i p

Let X be the pushout

P1 P0

A X

β

d1

β0

i

For an abelian category, X ∈ (A ⊕ P0)/L, where L = {(β(x),−d1(x))}. Let p : A ⊕ P0 → C be

given by (a, p) 7→ ε(p). Since L ⊆ ker p, we get an induced map (A⊕P0)/L = Xp−→ C. Putting all

this in the diagram, it is routine to verify that it commutes.

· · · P1 P0 C 0

· · · 0 A X C 0

β β0 =

p

Moreover, the bottom row is exact. Define ψ(β) to be the bottom row of the extension. It is simpleenough to show that this is well defined (if β = β′ + d∗1(J) for some J , we get isomorphic pushouts,possibly modifying β0 to β0 + Ji).

It is simple to see that θψ = 1 (we used β to construct the pushout so we clearly get β backunder θ). To see ψθ = 1, show that X is the pushout of the given α1 and d1.

Corollary 5.57. Ext1R(C,A) = 0 if and only if every extension of C by A splits.

Now under this bijection, we see that these are the same as sets. But are they the same asgroups? We know that Ext has an abelian structure. Does e(C,A) have an abelian structure? Wewant to set a group structure on e(C,A) “agreeing” with θ.


5.58 Baer Sum (1934)

Lets define addition on e(C,A). If ξ : 0 −→ Ai−→ X

p−→ C −→ 0 and ξ′ : 0 −→ Ai′−→ X

p′−→C −→ 0 are extensions of C by A, let X ′′ be the pullback of p, p′

X ′′ X ′

X C

so X ′′ = {(x, x′) ∈ X × X ′ | p(x) = p′(x′) in C}. We hope for a kernel but A is too “big”. LetY = X ′′/{(−i(a), i′(a))}a∈A, i.e. we identify two copies of A in X ×X ′. Then

0 −→ A −→ Y −→ C −→ 0

with maps a 7→ (i(a), 0) = (0, i(a)) and (x, x′) 7→ p(x). Note that p(x) = p′(x) is exact (one shouldverify this). This extension is called the Baer Sum: ξ + ξ′.

Theorem 5.59. The Baer sum in e(C,A) agrees under θ with the usual + in Ext1R(C,A), i.e.θ(ξ + ξ′) = θ(ξ) + θ(ξ′). Hence, the sum is well defined on equivalence classes [ξ] and under theBaer sum e(C,A) is a group.

Proof: This is far too long to present here.

5.60 Yoneda Extension

Similarly,

Theorem 5.61. We have an isomorphism (as sets and groups)

ExtnR(C,A) ∼= {Extensions ξ : 0 −→ A −→ Xn −→ · · · −→ X1 −→ C −→ 0}/ ∼

where ∼ is the equivalence relation ξ ∼ ξ′ if there are α’s such that

ξ : 0 A Xn · · · X1 C 0

ξ′ : 0 A X ′n · · · X ′1 C 0

= αn α1 =

commutes. The group structure is addition of extensions ξ + ξ′def= the extension

0 −→ A −→ Yn −→ Xn−1 ⊕X ′n−1 −→ · · · −→ X2 ⊕X ′2 −→ X ′′1 −→ C −→ 0

where Yn, X′′1 are the constructed pieces

X ′′1 = pullback X1, X′1 over C?

Yn = pushout X1, X′′1 over A?

83 C. McWhorter

6 The Derived Category

6.1 Derived Categories & Localization of Categories

To get better behavior (in the same way that the long exact sequence corrected non exactness),we want to replace modules M by their projective resolutions P·. Note P· is quasi-isomorphic toM . We naturally come to a category where quasi-isomorphic objects are somehow the same, i.e.quasi-isomorphisms are considered isomorphisms.

Let A be an abelian category. Let Ch(A) be the (abelian) category of cochain complexes.

Definition 6.2 (Derived Category). The derived category D(A) is a category equipped with a

functor Ch(A)i−→ D(A) with i(f) an isomorphism for all quasi-isomorphisms f satisfying the

universal mapping property: any functor Ch(A)F−→ C transforming quasi-isomorphisms factors

uniquely through D(A). That is, we have a unique G such that the following diagram commutes:

Ch(A) C

D(A)

F

i G

Note if such a category exists, it is unique up to isomorphism.

We show the existence of such a category by physically constructing D(A) by inverting allquasi-isomorphisms. Namely, we will “localize” Ch(A). One should read Gelfand-Manin chapters 3and 4 for more on this material generally.

In the proceeding lemma, the category B = Ch(A) and S = {quasi-iso in Ch(A) ⊆ morph Ch(A)}.We want to show D(A) = S−1B.

6.3 Localization of a Category

Let B be a category and let S be a collection of morphisms. Define S−1B by obj(S−1B)def= obj(B)

and morphisms by for each Xs−→ Y in S, adjoin a “morphism” X

s−1

←−− Y in S−1B. Note that s−1

is not a morphism in B but rather a symbol of sorts. The morphisms in S−1B are formal productsof morphisms in B and these formal inverses S−1 for s ∈ S such that the “codomain” of each is the“domain” of the next modulo an equivalence generated by

(i) consecutive morphisms in B can be replaced by compositions in B.

(ii) ss−1 /∈ S−1S can be replaced by 1.

(iii) 1s−1 = s−1 = s−11.

so that composition is just concatenation.

Proposition 6.4. (i) S−1B is a category

(ii) i : B → S−1B given on objects by obj : X 7→ X and on morphisms by f 7→ f is a functor.


(iii) The category satisfies the universal mapping property for all functors B F−→ C transformingmorphisms in S into isomorphism in C. That is, there is a unique functor G such that

B C

S−1B

F

iG

Proof: The proof of the first two parts is dull and routine verification. For the third, on objects letG(X) = F (X) of S−1B. Let G(s−11 f1s

−12 f2 · · · ) = F (s1)

−1F (f1)F (s2)−1F (f2) · · · . Note that i(s)

in S−1B is invertible, i(s) = s so ss−1 = 1 and s−1s = 1.

So we have D(A) = S−1(Ch(A)), where S = {quasi-iso}. We have a similar description of S−1Bwhen S is localizing.

Definition 6.5 (Localizing Morphisms). A collection S ⊆ morph of B is localizing if

(a) all 1X ∈ S and s, t ∈ S implies st ∈ S.

(b) for all f ∈ S with s ∈ S with the same codomain, there exist g, t with t ∈ S with a W such thatthe following square commutes

W Z

X Y

ε

g

s

f

(c) for g, t as given above, there exist s, f as above.

(d) For any f, g, there exist s ∈ S such that sf = sg if and only if there is a t ∈ S such that ft = gt.

Note the second condition above gives for all f, s, there is a g, t such that ft = sg so that inS−1B, s−1f = gt−1 so we can “rearrange” until all s’s (denominators) are on the right and all thef ’s (numerators) are on the left.

Proposition 6.6. Let S be a localizing collection of morphisms. Then S−1B can be definedequivalently as

85 C. McWhorter

(a) obj(S−1B) = obj(B)

(b) morphisms X → Y in S−1B are equivalence classes of “roofs”

with s ∈ S, written (s, f). This corresponds to fs−1 from the “old” description of S−1B. Thatis, (s, f) ∼ (t, g) if and only if there is a commutative diagram

with sr ∈ S (but not necessarily r ∈ S). Or that we have the diagram

(c) the identity morphism on X is the roof (1X , 1X).

(d) composition of roofs (s, f) and (t, g) is (st′, gf ′) from the diagram


where the square f ′ and t′ comes from the localizing property. [gt−1fs−1 = gf ′(t′)−1s−1]

(e) i : B → S−1B is given by on objects by X 7→ X and given on morphisms by f 7→ the roof (1, f).

Proof: We first show that ∼ is an equivalence relation. The reflective and symmetric properties aresimple. To see the transitive property, suppose (s, f) ∼ (t, g) and (t, g) ∼ (u, e), then

so we have the commutative diagram as given above with sr, tp ∈ S. By the localizing property,there are v, k with v ∈ S such that we have

Now t = pk = srv = thv so there is Z ′′′w−→W is S such that pkw = hvw so that we have the

diagram

commuting so (s, f) ∼ (u, e). So ∼ on roofs is an equivalence relation. Similarly, one can see thatcomposition on roofs is well defined. Furthermore, the category with objects being the objects of B

87 C. McWhorter

and morphisms given by roofs is a category. We show that this category R satisfies the universalmapping property for S−1B so that R = S−1B.

Given any functor B F−→ C such that F takes morphisms in S to isomorphism in C, we need toshow that there is a unique functor G such that the following diagram commutes

B C

R

F

iG

where i takes objects to objects via X 7→ X and on morphisms f 7→ (1, f). If G exists, then it is

unique as G(X)def= G(i(X)) = F (X) and

G(f) = G(i(f)) = G((s, t)) = G((1, s)−1(1, f)) = G((1, s))−1G(i(f)) = F (s)−1F (f)

Existence of such follows from the above definition. One only need show that it is indeed a functor.

6.7 Homotopy Category, K(A)

Note that for B = Ch(A) and S = {quasi-iso}, we have D(A)def= S−1B. Unfortunately, S is not a

localizing class. But it is if we go “mod homotopy”.

Definition 6.8 (Homotopy Category). The homotopy category K(A) is the category given byobj(K(A)) = obj(Ch(A)), that is cochain complexes, and morph(K(A)) is the cochain maps modulohomotopy, i.e. HomK(A)(X,Y ) = HomCh(A)(X,Y )/ ∼, where ∼ is the equivalence f ∼ g if and onlyif f ' g.

For a complex K ·, define K[x] to be K[n]idef= Kn+1 and dK[x] = (−1)ndk, called the shift, transla-

tion, or suspension. This gives a functor which is an equivalence of categories on Ch(A),K(A),D(A).

For any map K ·f−→ L·, we have C(f) = L·⊕K ·[1] with mixed differentials and Cyl(f) = L⊕K[1]⊗K

with differential from the “tot”


Lemma 6.9. For any cochian map, Kp−→ Y , there is a commutative diagram with exact rows

0 L C(f) K[1] 0

0 Cyl(f) C(f) ? 0

α

i p

where β(l, k, k′) = l + f(K) and other maps α, i, p, f , π are simple inclusions/projections. Further-more, α, β are homotopy equivalent: αβ = 1C and βα ' 1C(y) and the pictorial is functional inf .

Lemma 6.10. Let f, g : K → L with f ' g, i.e. f = g in K(A).

Proof: Note that Cyl(1K) = Tot(K ← K → K). It has two inclusions from K: α(k) = (k, 0, 0) withβα = 1, αβ ' 1 and α′(k) = (0, 0, k) with βα′ = 1, α′β ' 1.

Let t = αβ, t′ = α′β. Note that t, t′ ∈ S because they are homotopic to 1. Then tt′ = α′βαβ = t′.Then we have

i(t′t) = i(t′) = i(t′)i(t) = i(t′)

as they are isomorphic in S so isomorphic in D(A). Then we have i(t) = 1 in D(A). So

i(α′) = i(t)i(α′) = i(tα′) = i(αβα′) = i(α)

If f ' g via the homotopy {sn : Kn+1 → Ln}, we construct the map γ = (f, s, g) : Cyl(1k) → L.Check that γ is a cochain map so that it commutes with differentials (uses the fact that s is ahomotopy). Check also that γα = f and γα′ = g. So

i(f) = i(γα) = i(γ)i(α) = i(γ)i(α′) = i(g)

So we are getting a mapping cylinder to some object so that the map of the top and bottom arethe same somehow.

89 C. McWhorter

Theorem 6.11. (i) The functor

i : Ch(A) D(A) = S−1 Ch(A)

K(A)

π ∃

factors through K(A).

(ii) D(A) = S−1K(A) with S = {equiv classes of quasi-iso}. Note that if two morphisms are inthe same equivalence class and one is a quasi-iso then the other is.

(iii) S is a localizing class in K(A). So you can describe D(A) as “roofs of maps modulo homotopy”.

Proof: The first part is obvious from the proceeding lemma. For the second part, this is easy fromthe universal mapping property of localization R→ S−1R. For the third part, we need show thatS = {quasi-iso} is a localizing class in K(A). Note that 1X ∈ S and that if f, g are quasi-iso thenfg is a quasi-iso. Then given f, s with s a quasi-iso, we need to find g, t with t quasi-iso so that

Form the cone of s. Compose i with f to form the cone of if :

C(if)[−1] M C(s) C(if)

K L C(s)

t

f

if

=

s i

where t is the usual projection to M and g is the usual projection to K. The square does notcommute in Ch(A) as f(m) 6= s(k). But it is commutative up to homotopy. ft ' sg via thehomotopy σ(m, l, k) = ln−1. It is routine to check that this works. Hence, the diagram commutes inK(A). It is also easy to check that this is a quism and that the quism for the long exact sequence ofC(s) is exact so that t is a queasy. Similarly, g, t given as above shows there is a f, s as above. Forthe final part, suppose

K L Mfg s


with s ∈ K(A) with sf = sg in K(A), i.e. sf ' sg. Let σ : K → L[−1] be a homotopy

C(s)[−1] K C(h)[−1]

C(s)[−1] L M

= f−g

h t

p s

Define h by h(k) = (σ(k), (f − g)(k)). The square commutes. Take the cone of h then (f − g)t =pht = 0 in K(A) since ht ' 0 and s is a quism so that there is a long exact sequence giving C(s) isexact so that by the long exact sequence so that t is a quism. So ft ' gt so that they are equal inK(A) so that t ∈ S.

Corollary 6.12. For any Kf−→ L in Ch(A), f = 0 in D(A) if and only if there is a quism

Ls−→M such that sf ' 0 in Ch(A) (that is, f = 0 in K(A)).

Proof: Exercise. (By the theorem, going via K(A), D(A) = S−1K(A) and S is a localizing collectionin K(A) so the maps are roofs).

6.13 Distinguished Triangles

As D(A) is an additive category (see the book) but is almost never abelian, we can not talk aboutshort exact sequence and homology. We need to find a replacement for short exact sequences. Thereplacement will be something with similar behavior: distinguished triangles.

Definition 6.14 ((Distinguished) Triangles). A triangle in a category of complex is a diagram

K · −→ L· −→M ·w−→ K ·[1]

notice how this can be arranged in a triangle (ignoring shift). In fact, we also have

· · ·K u−→ Lv−→M

w−→ K[1]u[1]−→ L[1] −→ · · ·

This is not exact as we cannot even talk about exactness. A morphism of triangles is a commutativediagram

K L M K[1]

K ′ L′ M ′ K ′[1]

f g h f [1]

It is an isomorphism of triangles if f, g, h are isomorphisms in that category. A distinguished triangleis one isomorphic to

Kf−→ Cyl(f)

π−→ C(f)p−→ K[1]

91 C. McWhorter

for some Kf−→ L. That is, we have an isomorphism of triangles

K Cyl(f) C(f) K[1]

K L C(f) K[1]

=

f π

β =

p

=

f i=πα p

with β a homotopy equivalence so that the diagram commutes up to homotopy. iβ = παβ ' π1 = π?

Let us now see the relation to the short exact sequences of Ch(A).

Proposition 6.15. An exact sequence

0 −→ Kf−→ L

g−→M −→ 0

in Ch(A) is an isomorphism to a short exact sequence that can be extended to a distinguished trianglein K(A) or D(A). Conversely, every distinguished triangle is isomorphic to one that is an extensionof a short exact sequence in Ch(A).

Proof (D(A)): The following diagram commutes

0 K L M 0

0 K Cyl(f) C(f) 0

=

f g

β

where β is the usual map β(l, k, k′) = l + f(K), γ(l, k) = g(l) and has exact rows. Also, β is ahomotopy equivalence (in K(A), we need to show that γ is a homotopy equivalence, in D(A) needto show that γ is a quism). By the long exact sequence on homology and the Five Lemma, h(γ)is an isomorphism for all n so that γ is a quism. So then it is an isomorphism in D(A) and thebottom row now extends to a distinguished triangle

K −→ Cyl(f) −→ C(f) −→ K[1]

Conversely, given a distinguished triangle clearly

0 −→ K −→ Cyl(f) −→ C(f) −→ 0

is a short exact sequence in Ch(A).

Proposition 6.16 (Long Exact Sequences). If K −→ Lv−→M

w−→ K[1] is a distinguished trianglein D(A), then the induced sequence

· · · −→ H i(M [−1]) −→ H i(K)H(w)−→ H i(L)

H(v)−→ H i(M)H(w)−→ H i(K[1]) −→ · · ·

is exact.


Proof: The distinguished triangle is isomorphic to the triangle

Ku−→ L

i−→ C(u)p−→ K[1]

so the long exact sequence is really

· · · −→ H i(K)H(u)−→ H i(L)

H(l)−→ H i(C(u))H(p)−→ H i(K[1]) = H i+1(K) −→ · · ·

which is exactly the long exact sequence you get from the cone short exact sequence

0 −→ L −→ C(u) −→ K[1] −→ 0

so we get H(u) = δ from the long exact sequence.

6.17 Rotating Triangles

Can we rotate any triangle but does this preserve distinguished triangles?

Proposition 6.18. Distinguished triangles are preserved under rotation.

Proof: It is enough to show for a rotation in one direction and then the rest follows by induction.Without loss of generality, assume we rotate the triangle counter-clockwise. Up to isomorphism, adistinguished triangle of the form

Kf−→ L

i−→ C(f)p−→ K[1]

under rotation is

C(f)[−1] −→ Kf−→ L −→ C(f)

is L the cone of the first map?

C(f)[−1] K L C(f)

C(f)[−1] K Cyl(f) = C(p[−1]) C(f)

= =

f

=

p[−1] π

6.19 Ext & D(A)

We have an obvious functor A i−→ D(A) given by X maps to the complex · · · → 0→ X → 0→ · · · ,where X is in position (degree or index) 0. This is called the “stalk” of X and is denoted X[0]and takes morphisms to morphisms. But in D(A), i(X) = X[0] is isomorphic to any complexquasi-isomorphic to it. For example, X is quasi-iso to a projective resolution of it. We call anycomplex K with Hi(K) = 0 for all i 6= 0 a H0-complex.

93 C. McWhorter

Proposition 6.20. The functor i : A → D(A) gives an equivalence of categories of A with the fullsubcategory of D(A) consisting of all H0-complexes.

Proof: By definition, i is injective on objects. To see that i is onto modulo isomorphism, if K is anH0-complex, there are quasi-isomorphisms as follows:

· · · K−2 K−1 K0 K1 K2 · · ·

· · · K−2 K−1 ker d0 0 0 · · ·

d0

1 1

where we can extend ker d0 → H0(K) = ker d0/ im d−1. This is called the “soft truncation of K”.Why? We have cut off the sequence gently so that the homology does not change. So this isisomorphic to K in D(A), i(H0(K)) the stalk complex. To see i is fully faithful, oe only need showthat HomA(X,Y ) ∼= HomD(A)(i(X), i(Y )).

Definition 6.21 (Ext). For an object X ∈ A, let X[−i] denote the complex with X in position iand 0 elsewhere. For X,Y ∈ A, define

ExtiA(X,Y )def= HomD(A)(X[0], Y [i])

Remark 6.22. (i) Ext0A(X,Y )def= HomD(A)(i(X) = X[0], i(Y ) = Y [0]) = HomA(X,Y ) so that

Ext0R−mod(X,Y ) = HomR−mod(X,Y ) = HomR(X,Y ).

(ii) Since the shift [k] gives an equivalence of categories on Ch(A),K(A),D(A) (as [−k] is an

inverse), we get that ExtiA(X,Y )def= HomD(A)(X[0], Y [i]) ∼= HomD(A)(X[k], Y [i+ k]) for any

k ∈ Z. We can use this to define a composition Exti(X,Y )× Extj(Y, Z)→ Exti+j(X,Z) sothat ⊕Exti(X,X) is a graded ring. This is often used in Representation Theory.

(iii) ExtiA(X,Y ) = 0 for i < 0.To see this, if i < 0 and ϕ ∈ HomD(A)(X[0], Y [i]), then ϕ = s−1f aroof.

Let L be a soft truncation of K · at degree −i− 1 (note that K is quism to X[0] via s). It iseasy to check


commutes up to homotopy and s, r are quisms. So in D(A), we have (s, f) ∼ (sr, 0) to Y [i]for i < 0.

(iv) We have a relation to the Yoneda Extension: for i ≥ 0, we look at the two definitions ofExti(X,Y ). The “classic” definition gives any exact sequence

0 −→ Y −→ K−i+1 −→ · · · −→ K0 ε−→ X → 0

Call this sequence K. We have a roof

i.e. an element of ExtiA(X,Y ) = HomD(A)(X[0], Y [i]). It turns out to be all such elements of

ExtiA(X,Y ).

(v) Let I = the full subcategory of A of injective modules. Consider K+(I), complexes boundedbelow, of injective objects with morphisms being composition. Then the obvious functorK+(I)→ D+(A) is an equivalence of categories. This is proven on 179-183 of GM. We have asimilar statement for the category of projectives K−(P )→ D−(A). In the proof, one also getsif X ∈ Ch−(P ) or Y ∈ Ch−(I), then HomK(A)(X,Y ) ∼= HomD(A)(X,Y ) as sets. The reveresdirection here is exciting. It was first used to define Ext if the first component projectiveor second injective we are able to compute it. This gives a computation of Ext for objectsX,Y ∈ A. So we have a new definition

ExtiA(X,A)def= HomD(A)(X[0], Y [i])

= HomD(A)(P·, Y [i])

= HomK(A)(P·, Y [i])

= HomCh(A)(P·, Y [i])/homotopy

(vi) We have derived functors of F , RF,LF . See III.6 of GM. These are similar to Ext,Tor and

the class LnF and RnF but don’t take homology. RHom(X,Y )def= Hom(P·, Y ) as a complex

in Ch(A) or K(A) or D(A) which are nicer to work with then H i.

95 C. McWhorter

7 Triangulated Categories

7.1 Distinguished Triangles

We now formalize the property that we saw for K(A) and D(A). Let τ be an additive category. Fixan additive automorphism T : τ → τ , called the translate, translation, shift, or suspension of τ . Wewrite X[1] = T (X) and X[n] = Tn(X) for n ∈ Z. A triangle is a triple of morphisms (u, v, w) or

Xu−→ Y

v−→ Zw−→ X−→ [1] and a morphism of triangles is a commutative diagram

X Y Z X[1]

X ′ Y ′ Z ′ X ′[1]

f g h f [1]

Definition 7.2 (Triangulated Category). A triangulated category is an additive category τ with

• translation

• a collection of “distinguished” triangles

such that

(i) TR1a: X1−→ X −→ 0 −→ X[1] is a distinguished triangle.

(ii) TR1b: Any triangle which is isomorphic to a distinguished triangle is distinguished.

(iii) TR1c: Any morphism Xu−→ Y can be completed to a distinguished triangle (u, v, w), not

necessarily uniquely. [However, the axioms for a triangulated category force this triangle to beunique.]

(iv) TR2: If (u, v, w) is a distinguished triangle then its rotates (v, w,−u[1]) and (−w[−1], u, v)are distinguished triangles.

(v) TR3 (Completion of Morphisms): Given distinguished triangles given below and morphismsf, g

X Y Z X[1]

X ′ Y ′ Z ′ X ′[1]

f g h f [1]

such that the squares commute, there is a map h : Z → Z ′ completing the diagram to a map oftriangles. Note that h is not necessarily unique.

(vi) TR4 (Octahedral Axiom): Given morphisms Au−→ B, B

v−→ C, and distinguished triangles

Au−→ B

j−→ C ′β−→ A[1], B

v−→ Cx−→ A′

i−→ B[1], and Avu−→ C

y−→ B′γ−→ A[1], there


is a distinguished triangle (f, g, j[1] ◦ i): C ′f−→ B′

g−→ A′j[1]◦i−→ C ′[1] such that the following

diagram commutes

C ′ B A′

A C

B′∃f

β

j

v

i

j[1]i

u

vu

x

yγ∃g

Remark 7.3. Note that TR4 is often called the Octahedral Axiom as the commutative diagram isoften given as

B′

C ′ A′

A C

B′

g

γ

β

f

j[1]i

iu

vu

x

y

jv

The following proposition will show that the extension in TR1c is unique up to isomorphismand that in TR4, the octahedral maps f, g are unique up to isomorphism of triangles.

Proposition 7.4. A distinguished triangle is determined up to isomorphism by any one of its maps.

Proof: This follows almost immediately from TR2 and TR3.

Proposition 7.5. In TR3, if f, g are isomorphisms in τ , then h is an isomorphism in τ .

Proof: This follows similarly to the previous proof but makes use of the 5 Lemma for Triangles.

7.6 Cohomological Functors

Definition 7.7 (Cohomological Functor). An additive functor H : τ → A from a triangulatedtriangle τ to an abelian category A is called a cohomological functor if for any distinguished triangle

Xu−→ Y

v−→ Zw−→ X[1] the sequence F (X)

H(u)−→ F (Y )H(v)−→ F (Z)

H(w)−→ F (X[1]) is exact.

97 C. McWhorter

Example 7.8. There are two main examples of cohomological functors:

(i) The 0th homology

H0 :K(A)→ AH0 :D(A)→ A

where X0 7→ H0(X0). Note that H0(W [i]) = H i(W ) for any complex W in K(A) and D(A).

(ii) In any triangulated category τ , Homτ (U,−) is a cohomological functor.

Lemma 7.9. Let τ be a triangulated category. Fix an object U on τ then Homτ (U,−) is acohomological functor from the triangulated category τ to an abelian category. Similarly, Homτ (−, U)is a functor from τ to an abelian category.

Proof: Consider Homτ (U, 0) – the other direction is similar. It is enough to show that for anydistinguished triangle X

u−→ Yv−→ Z

w−→ X[1] that the sequence

Hom(U,X)u∗−→ Hom(U, Y )

v∗−→ Hom(U,Z)

is exact at Hom(U, Y ) as the exactness for the rest of the sequence is obtained by translation,rotation, or shift by TR2. It is a simple exercise to see that vu = 0 for any distinguished triangle.Applying Hom(U,−), we obtain v∗u∗ = 0∗ = 0 so imu∗ ⊂ ker v∗. Conversely, let f ∈ ker v∗ so that

v∗(f) = vf = 0. By TR1, we know that U1−→ U −→ 0 −→ U [1] is a distinguished triangle. There

is then a commutative diagram

U U 0 U [1]

X Y Z X[1]

1 0

f

u v w

Rotate, lift, rotate, to get a map of distinguished triangles, that is to obtain a map Uh−→ X such

that uh = f . But uh = u∗(h) so that f ∈ imu∗. But then the sequence is exact.We now prove a previous proposition.

Proposition 7.10. In TR3, if f, g are isomorphisms in τ , then h is an isomorphism in τ .

Proof: Given the diagram in TR3

X Y Z X[1]

X ′ Y ′ Z ′ X ′[1]

f g h f [1]

with f, g isomorphisms, apply Homτ (Z ′,−) (a cohomological functor) to get a commutative diagramwith exact rows (by the previous lemma)

Homτ (Z ′, X) Homτ (Z ′, Y ) Homτ (Z ′, Z) Homτ (Z ′, X[1])

Homτ (Z ′, X ′) Homτ (Z ′, Y ′) Homτ (Z ′, Z ′) Homτ (Z ′, X ′[1])

f g h f [1]


As f, g are isomorphisms and the fact that translation is an automorphism, we know that f [1] andg[1] are isomorphisms. So f∗, g∗, f [1]∗, and g[1]∗ are isomorphisms. Then by the 5 Lemma, we knowthat h∗ is an isomorphism.

7.11 The Cone in Triangulated Categories

If τ is a triangulated category then we know that any morphism Xu−→ Y extends – uniquely up

to isomorphism – to a distinguished triangle Xu−→ Y

v−→ Zw−→ X[1]. Then Z with maps v, w is

called the cone of u, written C(u). Note that in D(A) and K(A), C(u) is actually the mapping coneof u.

Theorem 7.12. Let A be an abelian category. The category K(A) with translation being shift [1] ofcomplexes and distinguished triangles being those isomorphic (up to homotopy) to a triangle of theform X

u−→ Y −→ C(u) −→ X[1] is a triangulated category. Likewise, K+(A),K−(A), and Kb(A)are also triangulated categories.

Proof: We show this for K(A) only. We need verify the axioms.

1 The last parts of TR1 follow from the definitions and from the fact that Xu−→ Y −→ C(u) −→

X[1] is distinguished. The first part of TR1 follows from the fact that

X X 0 X[1]

X X C(1) X[1]

1

1 1 1

1

One need check that the diagram commutes that that it is an isomorphism of triangles. The onlyreal work in this step is in checking that 0→ C(1) is homotopic to the identity of C(1) = X[1]⊕X.The homotopy is

1C(1) = hd+ dh, where, h =

(0 0

1X 0

)

2 Let Xu−→ Y

v−→ Zw−→ X[1] be a distinguished triangle. We need show that Y

v−→ Zw−→

X[1]−u[1]−→ Y [1] is a distinguished triangle (the converse can be proved by repeated translations).

To see that the second triangle is distinguished, we give an isomorphism between it and the

distinguished triangle Yv−→ Z

s−→ C(v)t−→ Y [1]. As the triangle X

u−→ Yv−→ Z

w−→ X[1]is distinguished, we can assume that Z = C(v) = X[1] ⊕ Y , or else compose with the naturalisomorphisms. Note that

dC(v) =

−dY 0 1Y0 −dX u[1]0 0 dY

Define θ : X[1]→ C(v) by θi(xi+1) = (−ui+1xi+1, xi+1, 0) for xi+1 ∈ X[1]i = Xi+1. We have the

99 C. McWhorter

diagram

Y Z X[1] Y [1]

Y Z C(v) Y [1]

v

1 1

w −u[1]

0 1

s t

We want to show that this diagram is a morphism of triangles in K(A). It is easy to check thisusing the explicit formulas for s, θw and that hi : Zi → C(v)i−1 given by hi(xi+1, yi) = (yi, 0, 0)is the necessary homotopy. One need also verify that this is an isomorphism of triangles but thiswas done earlier in a previous proof.

3 To see TR3, given distinguished triangles Xu−→ Y −→ C(u) −→ X[1] and X ′

u′−→ Y ′ −→C(u′) −→ X ′[1] and maps f : X → X ′ and g : Y → Y ′, there is a map h : C(u)→ C(u′) given by

hdef= g ⊕ f [1]. It is simple to check that h is indeed a chain map.

4 To see TR4, consider maps Au−→ B and B

v−→ C. We may assume that the distinguishedtriangles extending u, v, and vu are the standard ones with the cone in them. Otherwise, composethe result with the appropriate isomorphisms. Then we have

C(vu) C C(v)

A B

C(u)

g

(j[1]◦i)[1]

u

vu v

f

We want to show that there exists f, g so the outside “loop” is a distinguished triangle. Define

f(b, a)def= (v(b), a) and g(c, a)

def= (c, u(a)). One need check that f, g are chain maps (maps of

complexes): fd = df and gd = dg for differential d and that the two new regions in the diagramcommute: β = γf and x = gy using the previous notation.

Now to see that C(u)f−→ C(vu)

g−→ C(v)j[1]i−→ C(u)[1] is a distinguished triangle, note that

C(v)n = Cn ⊕Bn−1 and C(f) = (Cn ⊕An+1)⊕ (Bn−1 ⊕An and simply check that these “work”.Now if ϕ is a homotopy equivalence, we have ϕ : C(v) → C(u(f)) and ψ : C(u(f)) → C(v)where ϕ(c, b) = (c, 0, b, 0) and ψ(c, a, b, d) = (c, b). Check that ϕψ ' 1 and ψϕ ' 1C(v) and that

C(u)f−→ C(vu)

g−→ C(v)j[1]i−→ C(u)[1] is the same as C(u)

f−→ C(vu)ϕg−→ C(f)

(j[1]i)ψ−→ C(u)[1]by showing the appropriate diagram commutes. To see that this last triangle is distinguished,one need check that ϕg is the normal inclusion and (j[1]i)ψ is the usual projection (or homotopicto them).


Theorem 7.13. Let τ be a triangulated category, e.g. K(A), and S a localizing class of morphisms.Suppose S is compatible with the triangulation. That is,

(a) s ∈ S if and only if T (s) = s[1] ∈ S

(b) In TR3, if f, g ∈ S then h can be chosen to be in S as well

Then S−1τ with data

(i) translation as before (for τ) as obj τ = obj S−1τ

(ii) distinguished triangles being those isomorphic to image of a distinguished triangle from τ underthe functor τ → S−1τ given by x 7→ x and f 7→ (1, f)

Then S−1τ is a triangulated category.

Proof: The proof of this uses very careful tracking of roofs, see [GM] pp 251 - 256.

Corollary 7.14. The category D(A),D+(A),D−(A), and Db(A) are triangulated.

Proof: Using the theorem, one only need check (a) and (b). But (a) is obvious and to see (b),let S be the set of quasi isomorphisms. One cannot use the Triangulated Five Lemma as we do notknow that D(A) is triangulated. Instead, use the long exact sequence on homology from the twodistinguished triangles.

101 C. McWhorter

8 Spectral Sequences

8.1 Introduction

Spectral sequences were developed by Leray for use in Algebraic Topology. During World War II,he was imprisoned by the Nazis. Not wanting them to know he was an applied mathematician,he stated he was a pure mathematician who worked in Algebraic Topology. During this time inthe war, he developed spectral sequences. These were also independently developed by Lyndon ingroup cohomology. In 1945, Kozul made the procedure algebraic while in 1952 Massey created exactcouples.

We want to understand the homology of complexes: Hn(Tot(C··)). An example of this would bethe calculation of Tor from homologies of individual columns/rows and the maps between them.More generally, Hn(C·), calculating the homology of a complex from parts (meaning subcomplexesand quotients) of it via factors in a filtration of C· and maps between them. Spectral sequencesprovide an efficient machinery (more like bookkeeping) to paste these pieces together.

For some motivation, take a double complex with two columns.

Then Tot(E) is the mapping cone C(f). From the short exact sequence

0 −→ Y −→ C(f) −→ X[−1] −→ 0

we get a long exact sequence

· · · −→ Hn+1(X[−1])δn−1−→ Hn(Y )

β−→ Hn(C(f))α−→ Hn(X[−1])

δn−→−→ · · ·

where δn+1 = Hn(f), δn = Hn−1(f), Hn(X) = Hn+1(X[−1]), Hn(C(f)) = Hn(Tot(E)), andHn−1(X) = Hn(X[−1]). So we get an exact sequence

0 −→ coker(Hn(f)) −→ Hn(Tot(E))α−→ ker δ −→ 0

where ker δ = ker(Hn−1(f)) = im previous map and cokerHn(Y )/ kerβ = Hn(V )/ imHn(f). Thatis, we get a filtration

Hn(Tot(E)) ⊇ coker(Hn(f)) ⊇ 0


where factoring yields ker(Hn−1(f)) ∼= Hn(Tot(E))/ coker(Hn(f)) and coker(Hn(f)) ∼= coker(Hn(f))/0.Notice the similarity between this and composition series. Another way to track the same data is tolet E0 = E, called the E0 or initial page, be given as above. In each position, take homology alongthe vertical direction to get the E1 page

......

0 H2(Y ) H2(X) 0

0 H1(Y ) H1(X) 0

0 H0(Y ) H0(X) 0

0 0

where the vertical maps are now 0, so they are no longer useful. However, the horizontal maps stillcarry useful information. So we take horizontal homology to get the E2 page

......

0 coker(H2(f)) ker(H2(f)) 0



0 0

where now the horizontal maps are useful. However, we have maps along the diagonal (that is,along the lines y = n− x for n ∈ Z ∪ {0}. But these turn out to be not useful. Taking homologyalong these diagonal lines changes nothing so we have

E2 = E3 = E4 = · · · = E∞

103 C. McWhorter

Notice how we took the various maps RecallHn(Tot(E)) has filtration with factors coker(Hn(f)), kerHn−1(f))

which are in E2 above (if we took Tot with p + q = n, considering (p, q) = (x, y)). This is aninteresting connection.

As another example, take E = E0 the bicomplex with differentials dV the vertical maps and dH

the horizontal maps. To find Hn(Tot(E)) = Zn/Bn, we need to find the cycles. That is, we want

to find ordered tuples x + y = n, i + j = n, p + q = n, ⊕p+q=nEp,q with dV (Xp) = dH(Xp+1) (sothat they cancel out) and dH(X0) = 0 and dV (Xn) = 0. But how do we do this? We start withxn ∈ ker dV so that xn ∈ H0(column). Let zn = dH(xn). Now we have Zn = dV (some xn−1) if andonly if zn ∈ im dV if and only if zn = 0 in H0(column n−1) if and only if xn ∈ ker(H0(column n)→H0(column n−1). If so, we can assign to the xn the element xn−1 and look at the Zn−1 = dH(Xn−1),where the maps go between the Xn two left and one up. Note throughout we have dropped ± onelements to make things work for simplification. That is, we take “xn ∈ HH(HV (column))”. SoE1 is the bicomplex with vertical homologies in E0 and E2 is the bicomplex with the horizontalhomology of E1. Continuing this process, one can see any element xn surviving in spot (p, q) = (n, 0)on page En can be represented by an xn that can be extended to give a cycle (x0, · · · , xn) in theoriginal Hn(Tot(E)) with arrows (r, 0) 7→ (r, r − 1).

8.2 Homology Spectral Sequence

Definition 8.3 ((Homology) Spectral Sequence). A (homology, so the arrows go down and left,that is not a cochain complex) spectral sequence in an abelian category A is

(i) a family of pages: Er = {Erp,q} of objects for all p, q ∈ Z, r ≥ r0, i.e. the start page.


(ii) in each page, Er has maps dr : Erp,q → Erp−r,q+(r−1) with (dr)2 = 0.

(iii) the objects in page Er+1 satisfy Er+1p,q∼= Hp,q(E

r, dr). That is, the homology of the previouspage in each spot.

It is important to note that though we can draw E0 in a grid with d0, this does not have tobe a bicomplex. Furthermore, note that Er+1

p,q is a sub quotient, i.e. a quotient of a submodule ofErp,q. So one might ask if the Er+1 page inherits any properties from the Er page. This also showsthat the E’s are only getting “smaller”. In fact, at any spot (p, q) we have (via the CorrespondenceTheorem)

0 = Br0p,q ⊆ · · · ⊆ Br

p,q ⊆ Br+1p,q ⊆ · · · ⊆ Zr+1

p,q ⊆ Zrp,q ⊆ · · · ⊆ Zr0p,q = Erp,q

where the sequence of B’s are images of arrows lifted to E0p,q with Br

p,q/Zrp,q∼= Erp,q in E0.

Definition 8.4. A spectral sequence is bounded if for each n, the “p+ q = n” diagonal has finitelymany nonzero entries.

Note that if a spectral sequence is bounded, the differentials dr coming in and out of each spotwill eventually be zero maps. Why?

Erp+r,q−(r−1) −→ Erp,q −→ Ep−r,q+(r−1)

with sums n + 1, n, and n − 1 respectively, will be 0 for large n. So Erp,q = Er+1p,q = Er+2

p,q = · · · .We call this minimal r such that this occurs the stable value E∞p,q. If the spectral sequence is notbounded, we have to do other things:

E∞p,q =

⋂r Z

rp,q⋃

r Brp,q

Definition 8.5. We say a spectral sequences converges to objects {Hn} if for each n, the objects{E∞p,q | p+ q = n} give factors in some filtration of Hn, written Er0p,q → Hp+q or Er0p,q → Hn.

But this is useless unless {Hn} is relevant to something about the original set-up used to makethe spectral sequence, e.g. E0 or something about the E0 page.

8.6 Cohomology Spectral Sequence

Now we define the dual version:

Definition 8.7 ((Cohomology) Spectral Sequence). A (cohomology) spectral sequence is

(i) family of pages Er = {Ep,qr } for all p, q ∈ Z and r ≥ r0

(ii) on each page, maps dr : Ep,qr → Ep+q,q−(r−1)r with (dr)

2 = 0.

(iii) Ep,qr+1 = Hp+q(Ep,qr , dr) (that is the homology of the previous page at the spot p, q).

Note that if you reindex Ep,q as Ep,−q then this is the old definition of a homology spectralsequence. This shows that proving theorems for the homology spectral sequences shows them by adual argument for the cohomology spectral sequence.

105 C. McWhorter

Definition 8.8. We say that {Ep,qr } is bounded if in each diagonal p + q = n, there are finitelymany nonzero objects. If so, then eventually Ep,qr = Ep,qr+1 = · · · . We call this the stable value Ep,q∞ .The spectral sequence converges to {Hn} if for each n, objects in the diagonal p + q = n are thefactors in some filtration of Hn. We write Ep,qr → Hp+q or Hn.

Example 8.9. Any {Ep,q} spectral sequence converges to Ep,qr → ⊕p+q=nE∞p,qdef= Hn. That is,

Hn ⊇ ∗ ⊇ ∗ ⊇ · · · ⊇

with factors Es,n−s∞ , Ep,n−p∞ , · · · , Elast∞ . We have a reversed conclusion for a cohomology spectral

sequence.

8.10 Filtrations and Bounded Spectral Sequences

We have some “inclusion” of spectral sequences:

General Spectral Sequence ⊇ Filtration Spectral Sequence ⊇ Spectral Sequence of a Double Complex

The Loray-Serre spectral sequence of a fibration is a type of filtration of a spectral sequence: if F isa fiber (π−1(b) for some b ∈ B) and B is the base

F −→ Eπ−→ B

We will see how from a complex C with a filtration by subcomplexes to construct a spectral sequencefrom the homology of the factors converges to the homology of C.

Definition 8.11 (Filtration). A filtration of a chain complex C· is a chain of subcomplexes

· · · ⊆ Fp−1C ⊆ FpC ⊆ · · · ⊆ C

We will assume that the filtration is bounded (any position in the complex eventually stabilizesfor all n, i.e. there are only finitely many (FpC)n 6= 0) or bounded below or exhaustive:⋃

p

FpC = C

Now we look at the construction of the filtration.Let E0

p,q = (FpC/Fp−1C)p+q = FpCp+q/Fp+1Cp+q. That is, the pth column is the pth factorshifted down by p-steps. The differential d on C induces vertical maps (differential of factor complexd0 : E0

p+q → E0p,q−1). We know that E1

p,q = Hp+q(FpC/Fp−1C). For the page r (dropping the p+ q

subscripts), let πp : FpC → FpC/Fp−1C = E0p and Arp = {x ∈ FpC | d(x) ∈ Fp−rC}, which are the

“cycles modulo Fp−rC”. Then Zrp = π(Arp) ⊇ Brp = π(d(Ar−1p+r−1)). We have Zrp , B

rp ⊆ FpC/Fp+1C.

Note that we have

0 = B0p ⊆ B1

p ⊆ · · · ⊆ Brp ⊆ · · · ⊆ Zrp ⊆ · · · ⊆ Z1

p ⊆ Z0p = E0

p

and let Erp = Zr/Brp∼= Arp/

(d(Ar−1p+r−1) +Ar−1p−1

)(the congruence follows from the Second Isomor-

phism Theorem). Define dr to be the map induced by the original differential of C. One need check(dr)2 = 0 and Er+1 ∼= H(Er, dr). We have the following difficult (though not deep) Theorem:


Theorem 8.12 (Convergence Theorem). If the filtration on C is bounded or bounded below andexhaustive then the spectral sequence constructed above converges (this is trivial as it is bounded) tothe homology of C: E1

p,q = Hp+q(Fp/F−1)→ Hp+q(C).

We have the special case of the spectral sequence of a double complex. Then Tot(C··) has (at

least) two canonical filtrations. The first is Note E0p,q

def= Fp/Fp−1 = column p of C·· with d0 the

vertical differential dV . We have E1p,q = HV

q (Cp,∗) and d1 the differential induced by the original

horizontal differential d of Tot(C··) which is the horizontal differential induced by dH of C·· on thesub quotients (since dV (x) = 0 for all x ∈ HV (C··). So E2

p,q = HHp H

Vq (C). By the Convergence

Theorem, we have E2p,q = HHHV

q (C) → Hn=p+q(Tot(C··). The canonical construction follows

107 C. McWhorter

similarly by exchanging the roles of p, q.

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