MATH 613: HOMOLOGICAL ALGEBRA

LECTURES BY PROF. HARM DERKSEN; NOTES BY ALEKSANDER HORAWA

These are notes from the course Math 613: Homological Algebra taught by Prof. Harm

Derksen in Winter 2017 at the University of Michigan. They were L_{A}TEX’d by Aleksander

Horawa. This version is from July 1, 2017. Please check if a new version is available at

my website https://sites.google.com/site/aleksanderhorawa/. If you find a mistake,

please let me know at ahorawa@umich.edu.

specific reference. Citations are made where other resources were used.

Contents

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1

Page 2

1. Review of category theory

We begin with a short review of the necessary category theory.

(1) a class of objects, ObjC, and

(2) for all A, B ∈ ObjC, a set Hom_{C}(A, B) of morphisms from A to B,

(3) for any A,B,C ∈ ObjC, a composition map

Hom_{C}(A, B) × Hom_{C}(B,C) → Hom_{C}(A, C),

(f,g) ↦ g ◦ f = gf,

(4) for any A ∈ ObjC, a morphism id_{A }∈ Hom_{C}(A, A),

such that

(a) for any A, B ∈ ObjC and all f ∈ Hom_{C}(A, B)

id_{B}f = f = f id_{A},

(b) for any A, B, C, D ∈ ObjC and any f : A → B, g: B → C, h: C → D, the

composition is associative:

(hg)f = h(gf).

Note that ObjC may not be a set: for example, the category of sets cannot have the set of

all objects (Russel paradox). If ObjC is a set, then C is small.

• C = Sets: objects are sets, morphisms are functions,

• C = Groups: objects are groups, morphisms are group homomorphisms,

• C = Ring: rings and ring homomorphisms,

• C = Top: topological spaces and continuous maps,

• for a ring R, R-mod: left R-modules with R-module homomorphisms, and mod-R:

right R-modules with R-module homomorphisms,

• C = (A,≤), a poset: ObjC = A and

Hom_{A}(x, y) =

{1} if x ≤ y,

{∅ otherwise

• C = Ab: abelian groups and group homomorphisms.

g: B → A such that

gf = id_{A}, fg = id_{B}.

Inverses are unique: if g is another inverse, then

g = id_{A}g = (g f)g = g (fg) = g id_{B }= g .

inverse.

Page 3

Obj(C_{op}) = Obj(C)

Hom_{C}op (A, B) = {f_{op }| f ∈ Hom_{C}(B,A)}

and if in C

A

B

C

f

g

fg

then in C_{op}

A

B

C

fop

gop

(fg)op=gopfop

such that fe_{1 }= fe_{2}, we have e_{1 }= e_{2}.

such that g_{1}f = g_{2}f, we have g_{1 }= g_{2}.

The notions of monic and epi are dual: f is monic in C if and only if f_{op }is monic in C_{op}.

Let C be the category of metric (or at least Hausdorff) topological spaces. Then the inclusion

f : Q → R is not surjective but it is epi in C. Indeed, suppose g_{1},g_{2 }: R → X and g_{1}f = g_{2}f.

For any x ∈ R, there exists a sequence {x_{n}} ⊆ Q with lim_{n→∞} x_{n }= x. Then

g_{1}(x) = lim

n→∞

g_{1}(x_{n}) = lim

n→∞

g_{1}(f(x_{n})) = lim

n→∞

g_{2}(f(x_{n})) = lim

n→∞

g_{2}(x_{n}) = g_{2}(x)

since both g_{1 }and g_{2 }are continuous.

Similarly, in Rings, f : Z → Q is epi but not surjective.

morphism I → A.

If I,I are initial objects, there is a unique morphism f : I → I and a unique morphism

g: I → I. We then get morphisms fg: I → I and gf : I → I, but id_{I }: I → I and

id_{I }: I → I are also such morphisms and hence by uniqueness

fg = id_{I },

gf = id_{I}.

This shows initial objects are unique up to unique isomorphism.

morphism A → T.

Page 4

This is the dual notion to initial object:

I ∈ Obj(C) is initial if and only if I ∈ Obj(C_{op}) is terminal.

categories:

initial | terminal | |

Sets | ∅ | {0} |

Groups | {1} | {1} |

Ab | {0} | {0} = 0 |

Rings with 1 | Z | 0 |

If C has 0, A, B ∈ Obj(C), we have maps

A

0

B

0

so there is a unique morphism A → B that factors through 0 ∈ Obj(C), the zero morphism.

Two subobjects f : A → B, f : A → B are isomorphic if there is an isomorphism g: A → A

such that f = f g :

A

B

A

f

g

f

A

B

f(A)

f

f

and A → B and f(A) → B are isomorphic.

Page 5

gf = 0 and for every f : A → B with gf = 0, there is a unique morphism h: A → A such

that f = fh, i.e. the following diagram

A

B

C

A

f

0

g

f

h

0

commutes. The dual notion is the cokernel.

The above is an example of a universal property. We can restate it as follows. We define

a category G whose objects are pairs (A, f) with f : A → B and gf = 0 and a morphism

(A, f) → (A ,f ) in G is a morphism h: A → A in C such that f h = f. Then

(A, f) is the kernel of g if and only if (A, f) is a terminal object in G.

A kernel is a subobject: indeed, if e_{1},e_{2 }: A → A satisfy f = fe_{1 }= fe_{2}, then e_{1 }= e_{2 }by

uniqueness in the universal property

A

A

B

C

A

e1

f

0

f

0

g

e2

f

0

epimorphism but its cokernel is {1} → S_{n }(exercise).

morphisms π_{A }: A × B → A and π_{B }: A × B → B with universal property: if C ∈ Obj(C) and

f_{A }: C → A, f_{B }: C → B are morphisms, then there exists a unique morphism h: C → A×B

such that f_{A }= π_{A}h, f_{B }= π_{B}h, i.e. the following diagram

A

A × B

B

C

πA

πB

fA

fB

h

commutes.

Similarly, if A_{i}, i ∈ I are objects, their product is an object ∏_{i∈I} A_{i }together with morphisms

π_{i }: ∏_{i∈I }A_{i }→ A_{i }with analogous universal property.

Page 6

A ∐ B with the dual universal property, and similarly one can define the coproduct of any

family of objects, ∐_{i∈I }A_{i}.

π_{B}(a, b) = b, and the coproduct A ∐ B is the disjoint union with i_{A }: A → A ∐ B, i_{B }: B →

A ∐ B, the inclusion maps.

In Ab, the product and coproduct are the same for finite families of objects. However, for

infinite families we have

∏

i∈I

A_{i }= {(a_{1},a_{2 }...) | a_{i }∈ A_{i}},

∐

i∈I

A_{i }= {(a_{1},a_{2},...) | a_{i }∈ A_{i }and a_{i }= 0 for all but finitely many i}.

In Groups, G × H is the standard product and G ∐ H is G ∗ H, the free group product of

G and H. For example,

Z ∗ Z = 〈a, b〉.

In the category of rings with 1, Rings_{1}, A × B is the standard product and A ∐ B = A ⊗_{Z }B

with

i_{A }: A → A ⊗_{Z }B,

a ↦ a ⊗ 1.

Note that Z/2 ⊗_{Z }Z/3 = 0, so the inclusion map

i_{Z/2 }: Z/2 → Z/2 ⊗_{Z }Z/3

is not monic.

(1) assigns to A ∈ ObjC an object FA ∈ ObjC,

(2) assigns to f ∈ Hom_{C}(A, B) a morphism F ∈ Hom_{D}(FA,FB)

such that

(a) F(id_{A}) = id_{FA},

(b) F(gf) = F(g)F(f) if f : A → B, g: B → C in C.

Hom_{C}(A,−): C → Sets

such that for B ∈ ObjC

Hom_{C}(A,−)(B) = Hom_{C}(A, B),

and for f : B → C

Hom_{C}(A, f): Hom_{C}(A, B) → Hom_{C}(A, C),

g ↦ fg,

which can be represented by the following diagram:

Page 7

A

B

C

g

f

fg

Then Hom_{C}(A,id_{B}) = id_{Hom}C_{(A,B) }and

Hom_{C}(A, hg) = Hom(A, h) ◦ Hom(A, g),

which can be represented by the following diagram:

A

B

C

D

f

gf

h(gf)=(hg)f

g

h

hg

Hom_{R}(M,−): R-mod → Ab,

and if M is a right R-module then we have a functor

M ⊗_{R }−: R-mod → Ab.

F : Hom_{C}(A, B) → Hom_{D}(FA,FB)

is injective (resp. surjective).

What we defined above is actually a covariant functor.

(1) assigns to A ∈ ObjC an object FA ∈ ObjC,

(2) assigns to f ∈ Hom_{C}(A, B) a morphism F ∈ Hom_{D}(FB,FA)

such that

(a) F(id_{A}) = id_{FA},

(b) F(gf) = F(f)F(g) if f : A → B, g: B → C in C.

For a contravariant functor, a commuting triangle maps to a commuting triangle with arrows

reversed:

Page 8

A

B

C

f

g

gf

FA

FB

FC

Ff

Fg

F(gf)

A ∈ Obj(C), D(f) = f_{op }for f a morphism in C.

Hom_{C}(A,−) is covariant and Hom_{C}(−,A) is contravariant.

is a rule that assigns to A ∈ Obj(C) a morphism

η(A): F(A) → G(A)

such that for every morphism f : A → B the following diagram

FA

FB

GB

GA

Ff

η(A)

η(B)

Gf

commutes.

FA = {a ∈ A | there exists n ≥ 1 such that na = 0}.

Then η(A): FA ↪ A is a natural transformation between F and the identity functor on Ab.

ϵ: Hom_{C}(B,−) → Hom_{C}(A,−)

is a natural transformation given by

ϵ(C): Hom_{C}(B,C) ∋ f ↦ fe ∈ Hom_{C}(A, C),

since the following diagram

Hom(B,C)

Hom(B,D)

Hom(A,D)

Hom(A,C)

Hom_{C}(B,f)

ϵ(C)

ϵ(D)

Hom_{C}(A,f)

Page 9

commutes.

A functor F : C→D would be an isomorphism if there exists G : D→C such that FG = id_{D},

GF = id_{C}. However, such maps seldom exist so we weaken the notion slightly.

G(A) is an isomorphism for all A ∈ Obj(C).

G : D→C such that GF is naturally isomorphic to id_{C }and FG is naturally isomorphic to

id_{D}.

the full subcategory with Obj(C) = {R_{n }| n ≥ 0}. Then C is small while V is not small, so

C and V are not isomorphic. We will show that they are nonetheless equivalent. Define

F = id_{|C }: C→V,

the inclusion functor,

G : V→C,

G(V ) = R_{dim V }.

We now choose isomorphism

η(V ): R_{dim V }→ V

for every V (we use the meta axiom of choice here). Moreover, there is only one way to

define Gf for a linear map f : V → W to make η a natural transformation, i.e. making the

square

R

dim V

R

dim W

V

W

Gf

η(V )

η(W)

f

commute. We then have that

FG : V→V

and we note that η: FG → id_{V }is a natural isomorphism

η(V ): R_{dim V }= FG(V ) → V.

There is also a natural isomorphism η_{|C }: GF → id_{C}.

framework where we can develop homological algebra, generalizing categories such as Ab

and more generally R-mod.

abelian group, and

g(f_{1 }+ f_{2}) = gf_{1 }+ gf_{2},

f,f_{1},f_{2 }: A → B,

(g_{1 }+ g_{2})f = g_{1}f + g_{2}f,

g,g_{1},g_{2 }: B → C.

This makes Hom_{C}(A, A) a ring with 1 = id_{A}.

Page 10

object.

In additive categories, finite coproducts exist and are the same as product: if A, B ∈ ObjC,

then we have an object A ⊕ B = A × B = A ∐ B such that the following diagram

A ⊕ B

A

B

πA

πB

iA

iB

commutes, so (A ⊕ B,π_{A},π_{B}) is a product, (A ⊕ B,i_{A},i_{B}) is a coproduct, and

π_{A}i_{A }= id_{A}, π_{B}i_{B }= id_{B},

π_{A}i_{B }=0= π_{B}i_{A},

i_{A}π_{A }+ i_{B}π_{B }= id_{A⊕B}.

(1) every morphism has kernel and cokernel,

(2) every monic morphism is kernel of its cokernel,

(3) every epimorphism is cokernel of its kernel.

Let C be abelian from now on.

Lemma 1.37. A morphism f : A → B is monic if and only if kerf = 0 (i.e. 0 → A is a

kernel of A). Dually, f : A → B is epi if and only if B → 0 is a cokernel of f.

Proof. Suppose g: K → A is a kernel. If f is monic, then fg =0= f0, so g = 0. Hence g

factors through 0 → A, so 0 → A is a kernel of f.

Suppose conversely that 0 → A is a kernel of f : A → B. If fg_{1 }= fg_{2}, then

0 = fg_{1 }− fg_{2 }= f(g_{1 }− g_{2}),

so g_{1 }− g_{2 }factors through 0 → A, so g_{1 }− g_{2 }= 0 and g_{1 }= g_{2}.

□

A_{1}

A_{2}

B

f1

f2

then a pull-back is an object P together with maps g_{1 }: P → A_{1}, g_{2 }: P → A_{2 }such that

f_{1}g_{1 }= f_{2}g_{2 }and (P, g_{1},g_{2}) is universal with this property, i.e. if u_{1 }: C → A_{1 }and u_{2 }: C → A_{2}

satisfy fu_{1 }= fu_{2 }then there exists a unique h: C → P such that g_{1}h = u_{1 }and g_{2}h = u_{2},

i.e. the following diagram

Page 11

C

P

A_{1}

A_{2}

B

u1

u2

g1

g2

f1

f2

commutes. The dual notion is called push-out.

An abelian category has pull-backs and push-outs. Explicitly, for a map

A_{1 }⊕ A_{2}

B

(f1,−f2)

we have that

P

A_{1 }⊕ A_{2}

g1

g2

is the kernel of A_{1 }⊕A_{2 }→ B. The pushout is a cokernel of an analogous map A_{1 }⊕A_{2 }→ B.

Lemma 1.39. Suppose

P

A_{1}

A_{2}

B

g1

g2

f1

f2

is a pull-back. Then

(1) if g_{1 }is monic, then f_{2 }is monic,

(2) if f_{1 }is epi, then g_{2 }is epi.

Proof. We first prove (1). Indeed, suppose g_{1 }is monic and take u: C → A_{2 }with f_{2}u = 0.

Then there exists unique h: C → P such that g_{1}h = 0 and g_{2}h = u, i.e. the following diagram

C

P

A_{1}

A_{2}

B

h

0

u

g1

g2

f1

f2

commutes. But g_{1 }is monic, so this implies h = 0, and hence u = g_{2}h = 0. Hence 0 → A_{2 }is

a kernel of f_{2}, so f_{2 }is monic.

Page 12

For (2), suppose f_{1 }is epi. We recall that P → A_{1 }⊕ A_{2 }is kernel of A_{1 }⊕ A_{2 }→ B, and

A_{1 }⊕ A_{2 }→ B is epi, because A_{2 }→ B is epi. Hence A_{1 }⊕ A_{2 }→ B is cokernel of

P → A_{1 }⊕ A_{2}.

Hence the diagram

P

A_{1}

A_{2}

B

g1

g1

f1

f2

is a pushout. By the dual of (a), f_{1 }: A_{1 }→ B is epi implies that g_{2 }: P → A_{2 }is epi.

□

Lemma 1.40. If g: B → C is a morphism, then there exists a factorization g = vu where

u is epi and v is monic

B

I

C

u

g

v

Proof. Let f : A ↪ B be the kernel of g and u: B ↠ I be the cokernel of f. Then, because

gf = 0 and u is the cokernel of f, there exists v such that the following diagram

A

B

I

C

f

u

g

v

commutes. We have to show v is monic. Let w: K → I be the kernel of v and let P with

x: P → K, y: P → B be the pullback. Then, since f : A → B is the kernel of g and

gy = (vu)y = v(uy) = v(wx)=(vw)x = 0x = 0,

we get a unique map z: P → A such that the following diagram

A

B

I

C

P

K

f

u

g

v

x

y

z

w

commutes. Since P is the pullback and u: B → I is epi, x: P → K is epi. Now,

wx = uy = u(fz)=(uf)z = 0z = 0,

and since x is epi, we have that w = 0. This shows the kernel of I → C is 0 → I and hence v

is monic.

□

The image of g is the kernel of the cokernel or equivalently the cokernel of the kernel.

Page 13

Remark 1.41. For a ring R, R_{op}, the opposite ring is R_{op }= R as a set with multiplication

∗ in R_{op }defined by a ∗ b = b · a. If M is a left R-module, them M is a right R_{op}-module: if

m ∈ M, a ∈ R_{op }= R, then m ∗ a = a · m. This indeed gives a right module, for example:

(m ∗ a) ∗ b = b · (a · m)=(b · a) · m = m ∗ (b · a) = m ∗ (a ∗ b).

The category R-mod is isomorphic to mod-R_{op }and the category R_{op}-mod is isomorphic to

mod-R. Moreover, if M is a left R-module, we write _{R}M, if M is a right R-module, we write

M_{R}, and if M is a R-S-bimodule, we write _{R}M_{S}.

We work for now in the category mod-R.

A_{0}

A_{1}

A_{2}

···

A_{n+1}

f0

f1

f2

fn

is exact if imf_{i−1 }= kerf_{i }for i = 1,2,...,n.

Then

• 0

A

B

f

is exact if and only if f is injective,

• B

C

0

g

is exact if and only if g is surjective,

• 0

A

B

0

f

is exact if and only if f is an isomorphism,

• 0

A

B

C

0

f

g

is exact (we call it a short exact sequence) if

and only if f is injective, g is surjective, and C

≅ B/A,

• 0

kerf

A

B

cokerf

0

f

is exact for any f.

Theorem 1.43 (Snake Lemma). Suppose the diagram

0

0

0

kerp

kerq

kerr

A

B

C

0

0

A

B

C

cokerp

cokerq

cokerr

0

0

0

p

q

r

has exact rows, exact columns, and commuting squares. Then we have an exact sequence

kerp → kerq → kerr → cokerp → cokerq → cokerr,

Page 14

i.e. the red sequence in the following diagram

0

0

0

kerp

kerq

kerr

A

B

C

0

0

A

B

C

cokerp

cokerq

cokerr

0

0

0

p

q

r

Moreover, if we add zeros at the end of the two middle exact sequences, then we can add

zeros at the end of the “snake”, i.e. the red sequence in the following diagram exists and is

exact

0

0

0

0

kerp

kerq

kerr

0

A

B

C

0

0

A

B

C

0

cokerp

cokerq

cokerr

0

0

0

0

p

q

r

Proof. The proof is diagram chasing, and we omit it here.

□

Theorem 1.44 (Five Lemma). If the diagram

A_{1}

A_{2}

A_{3}

A_{4}

A_{5}

B_{1}

B_{2}

B_{3}

B_{4}

B_{5}

f1

f2

f3

f4

f5

Page 15

has exact rows and commuting squares, and f_{1},f_{2},f_{4},f_{5 }are isomorphisms, then f_{3 }is also

an isomorphism.

Proof. This proof is diagram chasing again, and we omit it here.

□

We can generalize both of these lemma to an abelian category C. We present the main ideas

Suppose B ∈ ObjC. Consider pairs (A, f) where f : A → B is a morphism. Than (A, f) ∼

(A ,f ) if there exists C ∈ ObjC and epimorphisms g: C ↠ A and g : C ↠ A such that

fg = f g :

C

A

A

B

g

g

f

f

Then ∼ is an equivalence relation. For example, transitivity is proved as follows: if (A, f) ∼

(A ,f ) and (A ,f ) ∼ (A ,f ), there exist C and C and C ↠ A, C ↠ A , C ↠ A , C ↠ A

and hence taking the pullback we get the following commutative diagram

P

C

C

A

A

A

B

f

f

f

and hence (A, f) ∼ (A ,f ).

Then we say a ∈ B if a is a congruence class of (A, f) for some f : A → B, i.e. a = [(A, f)].

Suppose g: B → C. With this definition, if a = [(A, f)] then g(a) = [(A, gf)]:

A

B

C

f

gf

g

Then 0 = [(0,f)] for the unique map 0 → B.

Lemma 1.45. A morphism g: B → C is monic if and only if for all a ∈ B, g(a)=0 implies

that a = 0.

Page 16

Lemma 1.46. A morphism g: B → C is epi if and only if for all c ∈ C, there exists b ∈ B

such that g(b) = c.

Lemma 1.47. A morphism f : A → B is equal to 0 if and only if f(a)=0 for all a ∈ A.

Lemma 1.48. If f : A → B is a morphism and a, a ∈ A such that f(a) = f(a ) then there

exists b ∈ A such that f(b)=0 and for every g: A → C with g(a)=0 we have g(b) = −g(a )

and for every g: A → C with g(a )=0 we have g(b) = g(a).

Proof. If a = [(D, h)] and a = [(D ,h )], then take b = [D ⊕ D ,(h,−h )].

□

These lemmas suffice to proceed with the diagram chasing arguments, so they show that the

2. Algebraic topology

In this chapter, we review the motivating examples of homology and cohomology from

2.1. Singular Homology.

∆_{n }=

{

(x_{0},x_{1},...,x_{n}) ∈ R_{n+1 }| x_{i }≥ 0,

n

∑

i=0

x_{i }= 1

}

.

For a topological space X, a singular n-simplex is a continuous function σ: ∆_{n }→ X.

We let

S_{n}(X) = free Z-module with basis of all singular n-simplices on X

and define

f_{i }= f_{n}

i : ∆n−1 → ∆n

by

f_{i}(x_{0},...,x_{n−1})=(x_{0},x_{1},...,x_{i−1},0,x_{i+1},...,x_{n−1}).

Then we define a map d_{n }: S_{n}(X) → S_{n−1}(X) by

d_{n}(σ) =

n

∑

i=0

(−1)_{i}σ ◦ f_{i}.

Lemma 2.2. For any n, d_{n−1 }◦ d_{n }= 0, so d_{2 }= 0.

Page 17

Proof. If j<i, then f_{i }◦ f_{j }= f_{j }◦ f_{i−1}, so

d_{n−1}(d_{n}(σ)) =

n−1

∑

j=0

n

∑

i=0

(−1)_{i+j}σ ◦ f_{n}

i ◦ fn−1

j

=

∑

0≤i≤j≤n−1

(−1)_{i+j}σ ◦ f_{i }◦ f_{j }+

∑

0≤j<i≤n−1

(−1)_{i+j}σ ◦ f_{i }◦ f_{j}

=

∑

0≤i≤j≤n−1

(−1)_{i+j}σ ◦ f_{i }◦ f_{j }+

∑

0≤j<i≤n−1

(−1)_{i+j}σ ◦ f_{j }◦ f_{i+1}

=

∑

0≤i≤j≤n−1

(−1)_{i+j}σ ◦ f_{i }◦ f_{j }−

∑

0≤j≤i≤n−1

(−1)_{i+j}σ ◦ f_{j }◦ f_{i}

setting i → i + 1

= 0

showing d_{2 }= 0.

□

We hence get a chain complex S_{•}(X):

···

S_{n+1}(X)

S_{n}(X)

S_{n−1}(X)

S_{n−2}(X)

···

dn+1

dn

dn−1

with d_{n−1}d_{n }= 0 for all n. (By convention, S_{n}(X) = 0 for n < 0.)

Z_{n}(X) = kerd_{n},

the module of cycles,

B_{n}(X) = imd_{n+1}, the module of boundaries.

Note that B_{n}(X) ⊆ Z_{n}(X) ⊆ S_{n}(X) by the lemma above.

a

b

∂(

) = b − a

>

while the boundary of a circle treated as a singular 1-simplex is a − a = 0

a∂

= 0

>

This justifies the names cycle and boundary.

H_{n}(X) = H_{sing,n}(X) = Z_{n}(X)/B_{n}(X) = kerd_{n}/imd_{n+1}.

One can show that H_{n}(X) is a topological invariant: if X is homeomorphic to Y then

H_{n}(X)

≅ H_{n}(Y ).

≅ Z_{d }where d is the number of path-connected

components of X.

Page 18

The homology of a contractible space is trivial, so for example

H_{n}(∗) = H_{n}(R_{n}).

However, we can distinguish between R_{n }and R_{m }using homology, because

H_{j}(R_{n }\ {∗}) =

{

Z for j = n − 1,

0 for 0 <j<n − 1.

2.2. Relative homology. Suppose Y ⊆ X is a subspace. Then S_{n}(Y ) ⊆ S_{n}(X) and we

can define

S_{n}(X, Y ) = S_{n}(X)/S_{n}(Y )

and we get the sequences

0

S_{n}(Y )

S_{n}(X)

S_{n}(X, Y )

0

0

S_{n−1}(Y )

S_{n−1}(X)

S_{n−1}(X, Y )

0

dn

dn

dn

We then define

Z_{n}(X, Y ) = ker(d_{n }: S_{n}(X, Y ) → S_{n−1}(X, Y )),

B_{n}(X, Y ) = im(d_{n+1 }: S_{n+1}(X, Y ) → S_{n}(X, Y )),

H_{n}(X, Y ) = Z_{n}(X, Y )/B_{n}(X, Y ).

2.3. Homology with coefficients. If M is a Z-module, we can define homology with

coefficients in M by setting

S_{n}(X;M) = S_{n}(X) ⊗_{Z }M,

Z_{n}(X;M) = ker(d_{n }: S_{n}(X;M) → S_{n−1}(X;M)),

B_{n}(X;M) = im(d_{n+1 }: S_{n+1}(X;M) → S_{n}(X;M)),

H_{n}(X;M) = Z_{n}(X;M)/B_{n}(X;M).

Note that − ⊗_{Z }M and H_{n}(−) do not commute. Hence taking homology with coefficients is

a non-trivial procedure.

We can write concisely

H_{n}(X) = H_{n}(S_{•}(X)),

H_{n}(X, Y ) = H_{n}(S_{•}(X, Y )),

H_{n}(X;M) = H_{n}(S_{•}(X;M)),

H_{n}(X, Y ;M) = H_{n}(S_{•}(X, Y ;M)).

Page 19

2.4. Simplicial homology.

is a set of finite nonempty subsets of V such that

(1) if v ∈ V then {v} ∈ S,

(2) if ∅ = τ ⊆ σ ∈ S then τ ∈ S.

If σ ∈ S then dimσ = |σ| − 1.

|K| =

∐

σ∈S

∆_{σ}

/

∼

where ∆_{σ }= ∆_{dim σ }= ∆_{|σ|−1 }with vertices labeled with 〈v〉, v ∈ σ, and if τ ⊆ σ then we have

a linear map f : ∆_{τ }→ ∆_{σ }given by f(〈v〉) = 〈v〉, then

∆_{τ}

x ∼ f(x) ∈ ∆_{σ}.

triangle with vertices 1,2,3 which is homeomorphic to a circle

1

2

3

We set

C_{n}(K) = Z-module with basis of all n-simplices.

To define d_{n }: C_{n}(K) → C_{n−1}(K), we choose a total ordering on V and write

C_{n}(K) 〈v_{0}v_{1 }...v_{n}〉 if v_{0 }< v_{1 }<...<v_{n}.

Then we define

d_{n}(〈v_{0}v_{1 }...v_{n}〉) =

n

∑

i=0

(−1)_{n}〈v_{0 }... ˆv_{i }...v_{n}〉.

Once again, we set

Z_{n}(K) = kerd_{n}

B_{n}(K) = imd_{n+1}

H_{simp,n}(K) = Z_{n}(K)/B_{n}(K).

Proposition 2.10. For any n, we have that H_{simp,n}(K)

≅ H_{sing,n}(|K|).

1

2

3

Page 20

so that |K| = S_{1}. We have that

C_{0}(K) = Z〈1〉 + Z〈2〉 + Z〈3〉 ∼= Z_{3}

C_{1}(K) = Z〈12〉 + Z〈13〉 + Z〈23〉 ∼= Z_{3}

d_{1 }:

〈12〉 ↦ 〈2〉−〈1〉

〈13〉 ↦ 〈3〉−〈1〉

〈23〉 ↦ 〈3〉−〈2〉

so

Z_{1}(K) = Z(〈13〉−〈23〉−〈12〉)

B_{0}(K) = {〈a〈1〉 + b〈2〉 + c〈3〉 | a + b + c = 0} ∼= Z_{2}

B_{1}(K)=0

Z_{0}(K) = Z_{3}

and hence

H_{1}(K)

≅ Z,

H_{0}(K)

≅ Z.

on opposite edges, so P_{2}(R) = |K| where K is the following simplicial complex

1

2

3

4

5

6

1

2

3

4

5

6

Then we have that

C_{2 }= Z_{18}, C_{1 }= Z_{27}, C_{0 }= Z_{10}

and the sequence

0 → Z_{18 }→ Z_{27 }→ Z_{10 }→ 0

has kerd_{2 }= 0 so H_{2}(K) = 0.

However, the map

d_{2 }: C_{2}(K;Z/2) → C_{1}(K,Z/2)

has

d_{2}

( sum of all 2-simplices

with right orientation

)

= 2(〈12〉 + 〈23〉 + 〈34〉 + 〈45〉 + 〈56〉−〈16〉)=0

so

H_{2}(K;Z/2) = Z/2.

Hence indeed homology with coefficients is a non-trivial construction.

Page 21

2.5. Functoriality. For singular homology, if f : X → Y is continuous , we get an induced

map

S_{n}(f) = f_{∗ }: S_{n}(X) → S_{n}(Y ): σ ↦ f ◦ σ.

Moreover, the square

S_{n}(X)

S_{n}(Y )

S_{n−1}(X)

S_{n−1}(Y )

f∗

d

d

f∗

commutes, and hence

f_{∗}(B_{n}(X)) ⊆ B_{n}(Y ),

f_{∗}(Z_{n}(X)) ⊆ Z_{n}(Y ),

so f_{∗ }induces

H_{n}(f) = f_{∗ }: H_{n}(X) = Z_{n}(X)/B_{n}(X) → H_{n}(Y ).

Altogether, we have a functor

H_{n }: Top → Ab.

2.6. Cohomology. Let M be a Z-module and let

S_{n}(X;M) = Hom_{Z}(S_{n}(X),M).

We have the following commutative triangle

S_{n}(X)

M

S_{n+1}(X)

dn+1

which gives a map

δ_{n }: S_{n}(X;M) → S_{n+1}(X;M).

Concisely, we apply the functor Hom_{Z}(−,M) to both S_{n}(X) and d_{n }and write

δ_{n }= Hom(d_{n+1},M).

We then get the following cochain complex

···

S_{n−1}(X;M)

S_{n}(X;M)

S_{n+1}(X;M)

···

δn−1

δn

δn+1

We then define

Z_{n}(X;M) = kerδ_{n},

the cocycles,

B_{n}(X;M) = imδ_{n−1},

the coboundaries,

H_{n}(X;M) = Z_{n}(X;M)/B_{n}(X;M), the cohomology group.

We admit the convention

S_{n}(X) = S_{n}(X;Z), H_{n}(X) = H_{n}(X;Z).

Page 22

H_{0 }= Z,

H_{0 }= Z,

H_{1 }= Z/2, H_{1 }= 0,

H_{2 }= 0,

H_{2 }= Z/2.

Suppose now X is a smooth manifold and let Ω_{n}(X) be the set of n-forms on X, so an

element of Ω_{n}(X) is gdf_{1 }∧ df_{2 }∧ ... ∧ df_{n}. We have maps

d: Ω_{n}(X) → Ω_{n+1}(X)

with d_{2 }= 0, which give a cochain complex

···

Ω_{n−1}(X)

Ω_{n}(X)

Ω_{n+1}(X)

···

dn−1

dn

called the de Rham complex. Its cohomology is called the de Rham cohomology

H_{•}

DR(X) = H•(Ω•(X)).

Then

Ω_{0}(X) = {smooth functions X → R}.

But then

ω =

dx

y

= −

dy

x

but ω = df for any f which shows H_{1}

DR(S1) = 0.

3. Homological algebra

In this chapter, we are in the category of right R-modules, mod-R.

C_{• }:

···

C_{n+1}

C_{n}

C_{n−1}

···

dn+1

dn

with d_{n}d_{n+1 }= 0 for all n.

We say C_{• }is bounded from below (above) if there exists a such that for any n<a (n>a),

C_{n }= 0. Moreover, C_{• }is bounded if it is bounded both from below and from above.

bounded from below. For a finite simplicial complex, the chain complex is moreover bounded.

B_{n}(C_{•}) = imd_{n+1}, Z_{n}(C_{•}) = kerd_{n}

and the nth homology group of C_{• }to be

H_{n}(C_{•}) = Z_{n}(C_{•})/B_{n}(C_{•}).

u_{n }: C_{n }→ D_{n }for all n ∈ Z such that u_{n−1}d_{n }= d_{n}u_{n}, so the following diagram

Page 23

···

C_{n+1}

C_{n}

C_{n−1}

···

···

D_{n+1}

D_{n}

D_{n−1}

···

dn+1

un+1

dn

un

un−1

dn+1

dn

commutes. We sometimes write more concisely that ud = du.

For a chain complex map u_{n}(B_{n}(C_{•})) ⊆ B_{n}(D_{•}) and u_{n}(Z_{n}(C_{•})) → Z_{n}(D_{•}) and hence u_{n}

induces a map

u_{∗,n }: H_{n}(C_{•}) → H_{n}(D_{•}).

R-modules, but we could apply this construction to more general categories) with chain complex

maps as morphisms. This is an abelian category.

We moreover have a homology functor

H_{n}(−): Ch(mod-R) → mod-R.

f_{∗,n }: S_{n}(X) → S_{n}(Y ): σ ↦ f ◦ σ

form a chain complex map

f_{∗ }: S_{•}(X) → S_{•}(Y ).

Then f_{∗ }induces

f_{∗,n }: H_{n}(X) → H_{n}(Y )

and in fact we have a functor

H_{n}(−): Top → Ab.

Theorem 3.7. If we have a short exact sequence of chain complexes

0 → C_{• }→ D_{• }→ E_{• }→ 0

(or, equivalently, 0 → C_{n }→ D_{n }→ E_{n }→ 0 is exact for any n). Then we have a long exact

sequence

···

H_{n+1}(E_{•})

H_{n}(C_{•})

H_{n}(D_{•})

H_{n}(E_{•})

H_{n−1}(C_{•})

···

∂

∂

Proof. We apply Snake Lemma 1.43 to get the red maps below

Page 24

0

0

0

0

Z_{n}(C)

Z_{n}(D)

Z_{n}(E)

0

C_{n}

D_{n}

E_{n}

0

0

C_{n−1}

D_{n−1}

E_{n−1}

0

C_{n−1}/B_{n−1}(C)

D_{n−1}/B_{n−1}(D)

E_{n−1}/B_{n−1}(D)

0

0

0

0

d

d

d

Then using the exactness of the red sequence above, we get the following commutative

diagram, and apply Snake Lemma 1.43 again to get the red maps

0

0

0

H_{n}(C)

H_{n}(D)

D_{n}(E)

C_{n}/B_{n}(C)

D_{n}/B_{n}(D)

E_{n}/B_{n}(E)

0

0

Z_{n−1}(C)

Z_{n−1}(D)

Z_{n−1}(E)

H_{n−1}(C)

H_{n−1}(D)

H_{n−1}(D)

0

0

0

d

d

d

This completes the proof.

□

···

C_{n−1}

C_{n}

C_{n+1}

C_{n+2}

···

dn−1

dn

dn+1

where d_{n+1}d_{n }= 0 for all n.

If C_{• }is a chain complex, setting C_{n }= C_{−n }with d_{n }= d_{−n }makes C_{•}, d_{• }a cochain complex.

chain complex C[p], d[p] with

C[p]_{n }= C_{p+n}, d[p]_{n }= (−1)_{p}d_{p+n}.

Page 25

Analogously, for a cochain complex C_{•}, we define the shift of C_{• }by p to be the cochain

complex

C[p]_{n }= C_{n−p}, d[p]_{n }= (−1)_{p}d_{n−p}.

This notation is used for example in the following context: instead of saying that for any n,

the sequence

0 → Z_{n}(C_{•}) → C_{n }→ B_{n−1}(C_{•}) → 0

is exact, we can say that the sequence of chain complexes

0 → Z_{•}(C_{•}) → C_{• }→ B[−1]_{•}(C_{•}) → 0

is exact.

We get the following analogous statement to Theorem 3.7 for cohomology.

Theorem 3.10. If 0 → C_{• }→ D_{• }→ E_{• }→ 0 is exact, then we have a long exact sequence

···→ H_{n−1}(E_{•}) → H_{n}(C_{•}) → H_{n}(D_{•}) → H_{n}(E_{•}) → H_{n+1}(C_{•}) →···

sequence

0 → S_{•}(Y ) → S_{•}(X) → S_{•}(X, Y ) → 0

and by Theorem 3.7, we get the long exact sequence of a pair

···→ H_{n+1}(X, Y ) → H_{n}(Y ) → H_{n}(X) → H_{n}(X, Y ) →···

In some cases, this allows us to calculate homology groups of topological spaces.

U, V ⊆ X open, we have a short exact sequence

0

Ω_{n}(X)

Ω_{n}(U) ⊕ Ω_{n}(V )

Ω_{n}(U ∩ V )

0

ω

(ω_{|U },ω_{|V })

(ω_{1},ω_{2})

(ω_{1})_{|U∩V }− (ω_{2})_{|U∩V}

Applying Theorem 3.10 to this short exact sequence, we get the Mayer-Vietoris sequence for

de Rham cohomology

···→ H_{n}

DR(X) → Hn

DR(U) ⊕ Hn

DR(V ) → Hn

DR(U ∩ V ) → Hn−1

DR (X) →···

3.1. Homotopy. We will define a homotopy of maps of chain complexes, using the

motivation of homotopy from topology.

Let X, Y be topological spaces and f,g: X → Y be continuous maps.

that h(0,x) = f(x) and h(1,x) = g(x). We then say f and g are homotopic and being

homotopic is an equivalence relation.

Page 26

If σ ∈ X is a 0-simplex, then

h(t, σ): [0,1]

︸︷︷︸

∼=∆1

→ Y,

so h(t, σ) is a 1-simplex in Y . This gives a group homomorphism

s_{0 }: S_{0}(X) → S_{1}(X).

If σ is a 1-simplex in X, then h(t, σ): ∆_{1 }× [0,1] → Y , so if ∆_{1 }= [a, b] whence d∆_{1 }= b − a

we get a map

Y

h(1, σ) = g∗σ

h(t, b)

h(0, σ) = f∗σ

h(t, a)

σ1

σ2

Then there exist two 2-simplices σ_{1 }+ σ_{2 }such that

d(σ_{1 }+σ_{2}) = g_{∗}σ−h(t, b)−f_{∗}σ+h(t, a)=(g_{∗ }−f_{∗})(σ)+s_{0}(a)−s_{0}(b)=(g_{∗ }−f_{∗})(σ)−s_{0}(dσ).

Define s_{1 }: S_{1}(X) → S_{2}(Y ) by setting

s_{1}(σ) = σ_{1 }+ σ_{2 }∈ S_{2}(Y ).

Then ds_{1}(σ)=(g_{∗ }− f_{∗})(σ) − s_{0}(dσ), so

ds_{1 }+ s_{0}d = g_{∗ }− f_{∗}.

In general, there exists s_{n }: S_{n}(X) → S_{n+1}(Y ) such that

sd + sd = g_{∗ }− f_{∗}.

continuous f : X → Y and g: Y → X such that gf is homotopic to id_{X }and fg is homotopic to

id_{Y }.

f : {0} → R_{n},

g: R_{n }→ {0},

we have gf = id_{{0} }and

fg: R_{n }→ R_{n}

is constant equal to 0 and a homotopy from fg to id_{R}

n is

h(t, x) = tx.

maps. A homotopy from f to g is a collection of functions s_{n }: C_{n }→ D_{n+1 }for n ∈ Z such

that

d_{n+1}s_{n }+ s_{n}d_{n }= g_{n }− f_{n}

(or, concisely, ds + sd = g − f),

which can be represented by the diagram

Page 27

···

C_{n+1}

C_{n}

C_{n−1}

···

···

D_{n+1}

D_{n}

D_{n−1}

···

dn+1

gn+1

dn

fn

gn

sn

fn−1

sn−1

dn+1

dn

Recall that f,g induce f_{∗,n},g_{∗,n }: H_{n}(C_{•}) → H_{n}(D_{•}). Explicitly, if a ∈ H_{n}(C_{•}), say a =

x + B_{n}(C) for x ∈ Z_{n}(C), then

f_{∗}(a) = f(x) + B_{n}(D) ∈ H_{n}(D),

g_{∗}(a) = g(x) + B_{n}(D) ∈ H_{n}(D).

If f,g are homotopic, then have that

g_{∗}(a) − f_{∗}(a) = g_{n}(x) − f_{n}(x) + B_{n}(D) = d_{n+1}s_{n}(x) + s_{n−1}d_{n}(X) ∈ B_{n}(D),

so g_{∗}(a) = f_{∗}(a). Hence if f,g are homotpic, then f_{∗ }= g_{∗}.

g: D_{• }→ C_{• }such that fg and gf are homotopic to the appropriate identity.

By the above, if C_{•}, D_{• }are homotopy equivalent, then for any n, f_{∗,n }: H_{n}(C_{•}) → H_{n}(D_{•}) is

an isomorphism because

g_{∗}f_{∗ }= (gf)_{∗ }= id_{∗ }= id,

f_{∗}g_{∗ }= (fg)_{∗ }= id_{∗ }= id.

H_{n}(X) = H_{n}({∗}) =

{

Z if n = 0,

0 otherwise.

For example,

H_{n}(R_{n}) =

{

Z if n = 0,

0 otherwise.

Moreover, R_{m }\ {0} is homotopy equivalent to S_{m−1}, the (m − 1)-dimensional sphere.

3.2. Split exact sequences.

Proposition 3.20. Suppose we have a short exact sequence

0

A

B

C

0

f

g

Then the following are equivalent

(1) there exists f : B → A with f f = id_{A}

(2) there exists g : C → B with gg = id_{C}

(3) there exists a submodule C ⊆ B with B

≅ A ⊕ C

≅ f(A) ⊕ C .

Proof. We only show (1) implies (3). The rest are similar and left as exercises. Take C =

kerf . For b ∈ B, we have

b = ff (b)

︸ ︷︷ ︸

∈f(A)

+(b − ff (b))

︸

︷︷

︸

∈C

,

Page 28

so f (b − ff (b)) = f (b) − f ff (b) = 0.

□

hold.

0

Z/p

Z/p

2

Z/p

0

1+(p)

p + (p_{2})

1+(p_{2})

1+(p)

that d_{n}s_{n−1}d_{n }= d_{n }for all n (i.e. dsd = d).

Suppose C_{• }is split. Let b ∈ B_{n−1}. Then b = d(a) for some a ∈ C_{n }and ds(b) = dsd(a) =

d(a) = b, so ds = id_{B}n−1 .

Then

0 → B_{n }→ Z_{n }→ H_{n }→ 0

is also split: if b ∈ B_{n }then b = d(a) for a ∈ C_{n+1 }and ds(b) = dsd(a) = d(a) = b, so

ds_{|B}n = id_{B}n . Hence

C_{n}

≅ Z_{n }⊕ B_{n−1}

≅ B_{n }⊕ H_{n }⊕ B_{n−1}

and we have the following diagram

···

C_{n+1}

C_{n}

C_{n−1}

···

B_{n}

⊕

B_{n+1}

⊕

H_{n+1}

B_{n−1}

⊕

B_{n}

⊕

H_{n}

B_{n−2}

⊕

B_{n−1}

⊕

H_{n−1}

∼=

∼=

∼=

∼=

0

∼=

0

The lowest level has maximal homology in the sense that all boundary maps are zero.

3.3. Mapping cone. A cone C(X) over X is

C(X) = [0,1] × X/ ∼,

where (0,x) ∼ (0,y) for all x, y ∈ X.

{0}×[0,1]

∼

= [0,1]

{1}×[0,1]

∼

= {∗}

Page 29

For X = S_{1}, C(X) is an actual cone, justifying the name

{0}×S1

∼

= S_{1}

{1}×S1

∼

= {∗}

C(f) = C(X) ∐ Y/ ∼

where C(x) (1,x) ∼ f(x) ∈ Y .

circle attached to [0,1] via f:

Let f : S_{1 }→ S_{1 }be the map z ↦ z_{2}. Then C(f) = P_{2}(R), because it is a hemisphere with

the antipodal identification on the boundary circle.

We generalize the topological notion of a cone to a purely algebraic one.

by setting

cone(f)_{n }= B_{n−1 }⊕ C_{n}

d(b, c)=(−d(b),d(c) − f(b))

or in matrix form

d_{cone }=

( −d_{B}

0

−f d_{C}

)

.

This is indeed a chain complex:

d_{2}

cone =

( −d_{B}

0

−f d_{C}

)( −d_{B}

0

−f d_{C}

)

=

(

d_{2}

B

0

fd_{B }− d_{C}f d_{2}

C

)

= 0.

We have an exact sequence

Page 30

0

C

cone(f)

B[−1]

0

c

(0,c)

(b, c)

−b

0

idC

(−idB 0)

···

H_{n}(B)

H_{n}(C)

H_{n}(cone(f))

H_{n−1}(B)

H_{n−1}(C)

···

∂

∂

Lemma 3.28. The boundary map ∂ above is f_{∗}.

be a cycle. This lifts to (−b,0) ∈ cone(f)_{n+1}. Then

d(−b,0) = (0,f(b)) ∈ cone(f)_{n}.

This lifts to f(b) ∈ C_{n}, and actually f(b) ∈ Z_{n}. Hence

∂[b]=[f(b)] = f_{∗}[b],

completing the proof.

□

f_{∗ }: H_{n}(B_{•}) → H_{n}(C_{•})

is an isomorphism for all n.

Corollary 3.30. A chain map f : B_{• }→ C_{• }is a quasi-isomorphism if and only if cone(f) is

exact (i.e. H_{n}(cone(f)) = 0 for all n).

Suppose

0

B_{•}

C_{•}

D_{•}

0

f

is a short exact sequence. Then we have two long exact sequences and in fact they are

isomorphic by the Five Lemma 1.44 applied to the diagram:

···

H_{n}(B)

H_{n}(C)

H_{n}(D)

H_{n−1}(B)

H_{n−1}(C)

···

···

H_{n}(B)

H_{n}(C)

H_{n}(cone(f))

H_{n−1}(B)

H_{n−1}(C)

···

=

f∗

=

f∗

=

=

f∗

∼=

f∗

Page 31

4. Homological δ-functors

a group homomorphism.

0 → A → B → C → 0

in C we have that

0 → FA → FB → FC.

One can similarly define right-exact and analogous notions for contravariant functors.

Lemma 4.3. If F is left exact and 0 → A → B → C is exact, then

0 → FA → FB → FC

is exact.

Proof. Let D be the image of g: B → C. Then the following diagram has an exact row and

column

0

0

A

B

D

0

C

C/D

0

and hence so does the following diagram

0

0

FA

FB

FD

FC

FC/D

Page 32

We only have to show exactness at FB. If a is in the kernel of F→FC then since FD → FC

is a injective, a is in the kernel of FB → FD and hence a is in the image of FA → FB. □

T_{n }: C→D, n ≥ 0,

together with a morphism δ_{n }: T_{n}(C) → T_{n−1}(A) for every short exact sequence

0

A

B

C

0

f

g

such that

(1) there exists a long exact sequence

···

T_{n+1}(C)

T_{n}(A)

T_{n}(B)

T_{n}(C)

T_{n−1}(A)

···

δn+1

Tn(f)

Tn(g)

δn

where we set T_{n}(A) = 0 for n < 0, so T_{0 }is right exact,

(2) for every morphism of exact sequences

0

A

B

C

0

0

A

B

C

0

fA

fB

fC

we have a commuting square

T_{n}(C )

T_{n−1}(A)

T_{n}(C)

T_{n−1}(A)

δn

Tn(fC )

Tn−1(fA)

δn

complexes, then the homology functor

H_{• }: Ch_{≥0}(C) → C

is a homological δ-functor.

T_{0 }: R-mod → Ab, T_{0}(M) = M/rM,

T_{1}(M)=[r]M = {a ∈ M | ra = 0},

T_{n}(M) = 0 for all n > 1.

Then T = {T_{n}} is a homological δ-functor. Applying Snake Lemma 1.43 to the following

diagram

Page 33

0

0

0

0

A

B

C

0

A

B

C

0

0

A

B

C

0

A/rA

B/rB

C/rC

0

0

0

0

r·

r·

r·

we obtain the long exact sequence

0 → T_{1}(A) → T_{1}(B) → T_{1}(C) → T_{0}(A) → T_{0}(B) → T_{0}(C) → 0,

showing property (1) holds. Property (2) is trivial.

transformations ϵ_{n }: S_{n }→ T_{n }for all n ≥ 0 such that if

0 −→ A −→ B −→ C −→ 0

is exact, then

S_{n}(C)

S_{n−1}(A)

T_{n}(C)

T_{n−1}(A)

δn

ϵn(C)

ϵn−1(A)

δn

commutes.

natural transformation ϵ_{0 }: S_{0 }→ T_{0}, there exists a unique natural transformation ϵ_{n }: S_{n }→

T_{n}, n ≥ 1, such that ϵ: S → T is a morphism of δ-functors.

5. Projectives and left derived functors

Let C be an abelian category.

morphism γ: P → C, there exists β: P → B such that g ◦ β = j

Page 34

P

B

C

0

β

γ

g

(Equivalently, if B → C → 0 is exact, then Hom_{C}(P, B) → Hom_{C}(P, C) → 0 is exact.)

The functor Hom_{C}(M,−) is always left exact. For projective modules, this functor is

moreover exact.

Lemma 5.2. An element P is projective if and only if Hom_{C}(P,−) is right-exact (and hence

exact).

Proof. The ‘if’ implication is clear. For the ‘only if’ implication, suppose 0 → A → B →

C → 0 is exact. Then

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

is exact by left-exactness together with the projective property.

□

Lemma 5.4. If 0 → A → B → P → 0 is exact and P is projective, then this short exact

sequence splits.

Proof. For the map id: P → P, there is a unique map P → B as above, showing that the

sequence splits.

□

Lemma 5.5. A direct summand of a projective object is projective.

Proof. Suppose P = P_{1 }⊕ P_{2 }and P is projective. Then since

Hom_{C}(P,−) = Hom_{C}(P_{1},−) ⊕ Hom_{C}(P_{2},−)

is exact, both Hom_{C}(P_{1},−) and Hom_{C}(P_{2},−) are both exact. Indeed, if one of them was

not, then the counterexample would also work for the direct product.

□

if it is projective.

Theorem 5.7 (Quillen–Suslin). Projective modules over F[x_{1},...,x_{n}] where F is a field are

free.

are projective R-modules.

Lemma 5.9. In R-mod, P is projective if and only if it is a direct summand of a free module.

Proof. The ‘if’ implication is clear from Lemma 5.5. For the converse, if P is a projective

module, let F(P) be the free module with generators [a] for a ∈ P. Define

f : F(P) ↠ P

by f([a]) = a. The sequence

0 → kerf → f(P) → P → 0

Page 35

is exact, so it splits by Lemma 5.4. Hence f(P)

≅ kerf ⊕ P.

□

≅ M_{1 }⊕ M_{2 }implies M_{1 }= 0 or

M_{2 }= 0.

Moreover, M is simple if for a submodule M_{1 }⊆ M we have M_{1 }= 0 or M_{1 }= M.

Theorem 5.11 (Krull-Schmidt). Let F be a field and R be a finite-dimensional F-algebra.

If M is a finite-dimensional R-module, then M

≅ M_{1 }⊕ M_{2 }⊕···⊕ M_{r}, where M_{1}, ..., M_{r}

are indecomposable and if M

≅ M_{1 }⊕ M_{2 }⊕···⊕ M_{s }with M_{1},...,M_{s }indecomposable, then

r = s and, after reordering, M_{i}

≅ M_{i}.

R = P_{1 }⊕ P_{2 }⊕···⊕ P_{r}

where P_{1},P_{2},...,P_{r }are projective indecomposables. If P is any projective finite-dimensional

R-modules, then for some M

P ⊕ M

≅ R

d ∼

= R

d ∼

= P

d

1 ⊕ Pd

2 ⊕···⊕ Pd

r ,

so P_{d}1

1

⊕ P_{d}2

2

⊕···⊕ P_{d}r

r .

We will write R-fdmod for the category of finite-dimensional R-modules.

R = P ⊕ P ⊕···⊕ P

︸

︷︷

︸

n

where

P = F_{n }are the ith columns.

The only indecomposables are actually P.

R = P_{1 }⊕ P_{2 }⊕···⊕ P_{n}

where

P_{i }= ith columns =

∗

∗

...

∗

0

...

0

Then the short exact sequence

0 → P_{1 }→ P_{2 }→ S_{2 }→ 0

is non-split. Hence S_{2 }is not projective.

exists an epimorphism f : P → M where P is projective.

Page 36

Indeed, the exact sequence

0 → Z/2 → Z/2n → Z/n → 0

is non-split, so Z/n is not projective. But every other non-zero finite abelian group has direct

summand Z/n.

then we have a map F(M) ↠ M as described above.

If I_{1},I_{2 }are ideals in a commutative ring with 1, then we have an exact sequence

0 → I_{1 }∩ I_{2 }→ I_{1 }⊕ I_{2 }→ I_{1 }+ I_{2 }→ 0.

R =

C[x, y, z]

(xy − z_{2 }− 1)

and I_{1 }= (z − 1,x), I_{2 }= (z + 1,x). These are ideals which are not principal but they are

indecomposable. Then using the equation xy = z_{2 }− 1=(z − 1)(z + 1) we obtain

I_{1 }∩ I_{2 }= (x).

Since the short exact sequence

0 → (x) → I_{1 }⊕ I_{2 }→ R → 0

splits (R is projective), we have that

I_{1 }⊕ I_{2 }= R ⊕ (x)

≅ R

2

and so I_{1}, I_{2 }are projective. Hence we obtained two decompositions of R_{2 }into

indecomposables.

R =

C[x, y]

(y_{2 }− x_{3 }+ x)

.

Then the maximal ideals of R are projective.

morphism ϵ: P_{0 }→ M such that

···→ P_{3 }→ P_{2 }→ P_{1 }→ P_{0 }→ M → 0

is exact. If all P are projective, this is a projective resolution; if all P are free, this is a free

resolution, etc.

In this case,

H_{n}(P_{•}) =

{ M if n = 0,

0

if n = 0.

Proposition 5.21. If C has enough projectives, then every object has a projective resolution.

Proof. Take any M ∈ Obj(C). There is an epi P_{0 }↠ M from a projective P_{0 }and, taking the

kernel K_{0 }→ P_{0}, we have a projective P_{1 }with an epi P_{1 }→ K_{0}. We take its kernel K_{1 }→ P_{1}

and again get a projective P_{2 }↠ K_{1}. This way, we get that the diagram

Page 37

0

0

K_{1}

···

P_{2}

P_{1}

P_{0}

M

0

K_{0}

0

0

commutes. Continuing, this gives a projective resolution of M.

□

Let C be an abelian category with enough projectives, D be an abelian category, and F : C →

D be a right exact functor. We present the idea behind derived functors first. Given M ∈

Obj(C), choose projective resolution of M:

···→ P_{3 }→ P_{2 }→ P_{1 }→ P_{0 }→ M → 0.

Apply F to P to obtain

···→FP_{3 }→ FP_{2 }→ FP_{1 }→ FP_{0 }→ 0.

Define

L_{i}(FM) = H_{i}(F(P_{•})).

Note that L_{0}F(M) = H_{0}(F(P_{•})) = F(M), since

F(P_{1}) → F(P_{0}) → F(M) → 0

is exact.

Questions: Is L_{i}F(M) well-defined? Is L_{i}F a functor?

Theorem 5.22 (Comparison Theorem). Suppose P_{• }→ M is a projective resolution and

Q_{• }→ N is any resolution, and f : M → N be a morphism. Then there exists a chain map

f : P_{• }→ Q_{• }such that

P_{•}

M

Q_{•}

N

f

f

commutes. Moreover, f is unique up to homotopy.

Proof. We construct the family f_{0},f_{1},..., making the diagram

Page 38

···

P_{2}

P_{1}

P_{0}

M

0

···

Q_{2}

Q_{1}

Q_{0}

N

0

d2

γ2

f2

d1

γ1

f1

ϵ

f0

γ0

f

d2

d1

η

commute, step by step as follows. The map γ_{0 }= f ϵ: P_{0 }→ N is surjective, so it lifts to

f_{0 }: P_{0 }→ Q_{0}. Next,

γ_{1 }= f_{0}d_{1 }: P_{1 }↠ imd_{1 }= kerη

lifts to f_{1 }: P_{1 }→ Q_{1}, since P_{1 }is projective. In the next step,

γ_{2 }= f_{1}d: P_{2 }↠ imd_{2 }= kerd_{1}

lifts to f_{2 }: P_{2 }→ Q_{2}, since P_{2 }is projective. We continue this way to construct a chain map

f : P_{• }→ Q_{• }such that the approporiate diagram commutes.

For uniqueness, suppose g: P_{• }→ Q_{• }is another chain map such that

P_{•}

M

Q_{•}

N

g

f

commutes. We can replace the pair f, g by the pair h = f − g, 0: if we construct a suitable

collection of maps s_{i }for h and 0, then ds + sd = h − 0 = f − g, as required. We let

s_{0 }= 0: M → Q_{0}. The map h_{0 }factors through kerη, since ηh_{0 }= 0, and since the map

Q_{1 }→ kerη is surjective and P_{0 }is projective, there is a unique map s_{1 }: P_{0 }→ Q_{1 }making the

diagram

···

P_{1}

P_{0}

M

0

···

Q_{1}

kerη

Q_{0}

N

0

h1

s1

ϵ

h0

0

0

η

commute. This shows d_{1}s_{1 }+ s_{0}ϵ = d_{1}s_{1 }= h_{0}. Now, d_{1}(s_{1}d_{1 }− h_{1}) = h_{0}d_{1 }− d_{1}h_{1 }= 0, so

s_{1}d_{1 }− h_{1 }factors through kerd_{1}, and as Q_{2 }→ kerd_{1 }is surjective and P_{1 }is projective, we

get a unique map s_{2 }: P_{1 }→ Q_{2 }making the triangle in the diagram

···

P_{2}

P_{1}

P_{0}

M

0

···

Q_{2}

kerd_{1}

Q_{1}

Q_{0}

N

0

h2

d1

s2

h1

s1

ϵ

h0

0

0

d1

η

commute. Continuing this way, we construct the family s such that ds + sd = f − g, as

required.

□

For every object M ∈ ObjC, we can choose a projective resolution. Then L_{i}F is a functor,

called the left derived functor of F.

Page 39

As a consequence of the Comparison Theorem 5.22, if f : M → N, we get a map

···

P_{1}

P_{0}

M

···

Q_{1}

Q_{0}

N

f1

f0

f

which is unique up to homotopy, so we get a well-defined map

L_{i}F(f )=(f_{∗})_{i }: H_{i}(F(P_{•}))

︸ ︷︷ ︸

LiFM

→ H_{i}(F(Q_{•}))

︸

︷︷

︸

LiFN

.

Moreover, if P_{• }and P_{• }are two resolutions of M, then id: M → M gives rise to unique maps

(up to homotopy)

···

P_{1}

P_{0}

M

···

P_{1}

P_{0}

M

f1

f0

idM

g1

g0

idM

so we get maps

f_{∗ }: H_{•}(FP_{•}) → H_{∗}(FP_{•})

g_{∗ }: H_{•}(FP_{•}) → H_{∗}(FP_{•})

such that (gf)_{∗ }= g_{∗}f_{∗ }= id

(fg)_{∗ }= f_{∗}g_{∗ }= id

(by uniqueness by to homotopy). Hence the functor is well-defined: for two choices of

projective resolutions of objects, the construction yields isomorphic functors.

M = R/m = F. Then a projective resolution of M is

0

R

R_{2}

R

M

0

−y

x

(

x

y

)

and we apply F = M ⊗_{R }− to this resolution to get

0

F

F_{2}

F

0

0

0

so

H_{n}(F(P_{•})) =

F,

if n =0or2,

F_{2}, if n = 1,

0,

otherwise.

Theorem 5.24 (Horseshoe Lemma). Suppose 0 → A → A → A → 0 is exact and P_{•}

ϵ

→ A ,

P_{•}

ϵ

→ A are projective resolutions. Then there exists a unique projective resolution P_{•}

ϵ

→ A

and an exact sequence

0 → P_{• }→ P_{• }→ P_{• }→ 0

such that

Page 40

0

0

P_{•}

A

P_{•}

A

P_{•}

A

0

0

ϵ

ϵ

ϵ

commutes.

Proof. We recursively construct a projective resolution P_{n }= P_{0 }⊕P_{0 }, using projectivity and

from ϵ ), we get a map ϵ: P_{0 }⊕ P_{0 }→ A (using the fact that P_{0 }⊕ P_{0 }is both a product and

a coproduct) such that the following diagram

0

0

0

kerϵ

P_{0}

A

0

P_{0 }⊕ P_{0}

A

0

kerϵ

P_{0}

A

0

0

0

ϵ

ϵ

ϵ

commutes. Note that P_{0 }⊕ P_{0 }is projective as a direct sum of projectives and as both ϵ

and ϵ are surjective, so is ϵ. Hence we get the following diagram and we apply the Snake

Lemma 1.43 (the cokernels here are all 0) to see that the sequence of kernels is exact

Page 41

0

0

0

0

kerϵ

P_{0}

A

0

0

kerϵ

P_{0 }⊕ P_{0}

A

0

0

kerϵ

P_{0}

A

0

0

0

0

ϵ

ϵ

ϵ

We then apply the same procedure to the diagram with kernels to construct d_{1 }: P_{1 }⊕ P_{1 }→

kerϵ, where the product is projective and the map is epi onto the kernel

0

0

0

0

P_{1}

kerϵ

P_{0}

A

0

P_{1 }⊕ P_{1}

kerϵ

P_{0 }⊕ P_{0}

A

0

P_{1}

kerϵ

P_{0}

A

0

0

0

0

0

d1

ϵ

d1

ϵ

d1

ϵ

Continuing this way, this constructs the sequence P_{n }= P_{n }⊕ P_{n }of projectives with the

desired properties.

□

Theorem 5.25. The derived functor L_{i}F is a universal homological δ-functor.

Proof. For an exact sequence

0

A

A

A

0,

the Horseshoe Lemma 5.24 gives an exact sequence

0

P_{•}

P_{•}

P_{•}

0,

and we apply F to get

0

FP_{•}

FP_{•}

FP_{•}

0.

Page 42

On every level n, we have a splitting

0

P_{n}

P_{n }= P_{n }⊕ P_{n}

P_{n}

0

which shows that there is a map FP_{n }→ FP_{n}. When we take homology, we get the long

exact sequence

···

H_{1}(F(P ))

H_{1}(F(P))

H_{1}(F(P ))

H_{0}(F(P ))

H_{0}(F(P))

H_{0}(F(P ))

0,

δ

or, in other words,

···

L_{1}FA

L_{1}FA

L_{0}FA

L_{0}FA

L_{0}FA

0.

δ

Hence this δ is an approporiate map and we just have to check naturality and universality.

□

6. Injectives and right derived functors

Let C abelian category. For an object M, Hom_{C}(−,M) is a left-exact contravariant functor.

Lemma 6.1. The following are equivalent

(1) for every monic f : A → B and every morphism g: A → I, there exists h: B → I

such that hf = g, i.e. the following diagram

0

A

B

I

f

g

h

commutes,

(2) Hom_{C}(−,I) is exact,

(3) I is projective in C_{op}.

Proposition 6.3 (Baer’s criterion). In R-mod, I is injective if and only if for every left

idea J ⊆ R and every R-module homomorphism g: J → I, there exists g: R → I such that

hf = g, i.e. the following diagram

0

J

R

I

f

g

g

commutes

Page 43

Proof. The ‘only if’ implication follows directly from the definition. We will use Zorn’s lemma

to prove the ‘if’ implication. Suppose we have

0

A

B

I

f

g

and we want to construct a map B → I. Consider the set

S = {(C, h) | f(A) ⊆ C ⊆ B submodule, and hf = g}

and set (C, h) ≤ (C ,h ) if and only if C ⊆ C and h

|C

= h. Note that the set is non-empty,

because we can choose C = f(A). Suppose {(C_{x},h_{x}) | x ∈ X} is a chain, so for any x, y ∈ X

(C_{x},h_{x}) ≤ (C_{y},h_{y}) or (C_{x},h_{x}) ≥ (C_{y},h_{y}).

Define C = ⋃

x∈X

C_{x }and h: C → I by setting h(a) = h_{x}(a) if a ∈ C_{x }for some x ∈ X. This is

well-defined since this is a chain, and h_{|C}x = h_{x }for any x ∈ X. Hence (C_{x},h_{x}) ≤ (C, h) for

any x ∈ X, showing that (C, h) is an upper bound.

By Zorn’s lemma, S has a maximal element, call it (C, h). Let b ∈ B. We have an exact

sequence

0

J

R ⊕ C

Rb + C

0

a

(a,−ab)

(a, c)

(ab + c)

where J = {a ∈ R | ab ∈ C}. We let g: J → I, g(a) = h(ab) and hence there exists a g such

that the diagram

0

J

R

I

f

g

g

commutes. We then have the diagram

0

J

R ⊕ C

ˆ

C = Rb + C

0

I

(g,h)

h

Now, (C, h) ≤ ( ˆC, h), so (C, h)=(ˆC, h). Hence b ∈ C. This shows that B ⊆ C, and hence

B = C, so we get the diagram

Page 44

0

A

B = C

I

f

g

h

completing the proof.

□

Lemma 6.4. If R is a PID, then I is injective if and only if Iis divisible, i.e. for any a ∈ I,

r ∈ R \ {0}, there exists b ∈ I such that rb = a.

Q/Z =

⊕

p prime

Z_{p}∞ ,

where Z_{p}∞ = Z

[

1

p

]

/Z.

exists a monic M → I where I is injective. So C has enough injectives if and only if C_{op }has

enough projectives.

m+ Z. This

does not exactly prove that Ab has enough injectives: it only shows it for finitely generated

abelian groups. The general statement takes some more work.

If M is a left R-modules, N is a Z-module, then

Hom_{Ab}(M,N)

Hom_{R}(M,Hom_{Ab}(R, N))

f

[(m, r) ↦ f(rm)]

∼=

If N is an injective Z-module, then Hom_{Ab}(−,N) is exact, and hence

Hom_{R}(−,Hom_{Ab}(R, N))

is exact, showing that Hom_{Ab}(R, N) is an injective R-module.

Notation. We denote by

A_{B }= the set of functions B → A,

C_{D }= the category of functors D→C.

By the above, the object I_{0 }= Hom_{Ab}(R,Q/Z) is an injective R-module. The map

Φ: M → I

HomR(M,I0)

0

,

Φ(m)(f) = f(m) ∈ I_{0}.

is injective (as a set map). Hence the category R-mod has enough injectives.

0

M

I_{0}

I_{1}

···

Page 45

and an injective resolution of M is a coresolution such that I_{n }are injective objects.

Proposition 6.9. If C has enough projectives, then any object has an injective resolution.

Proof. This is the dual to Proposition 5.21.

□

0

Z

Q

Q/Z

0

and an injective resolution of Z/n is

0

Z/n

Q/Z

Q/Z

0

If F is a left exact functor and C has enough injectives, then we define R_{n}F, right derived

functors, where for an injective resolution

0

M

I_{0}

I_{1}

···

we set

R_{n}F(M) = H_{n}(F(I_{•})),

R_{0}F(M) = F(M).

All the results about the left derived functors from Chapter 5 hold dually for right derived

functors.

7. Limits

Let A be an abelian category and I be another category (often a poset).

morphisms π_{i }: L → F(i) for all i ∈ ObjI such that

L

F(i)

F(j)

πi

πj

F(f)

commutes for all f : i → j, and (L, π) is universal with this property, i.e. if (L ,π ) satisfy

L

F(i)

F(j)

πi

πj

F(f)

for all f_{i }: i → j, then there exists a unique h: L → L such that π_{i}h = π_{i }for all i

Page 46

L

L

F(i)

F(j)

πi

πj

h

πi

πj

F(f)

We write (L, π) = lim

i∈I

F(i).

i∈I

F(i) is a product in A.

i_{j }: F(j) → L such that the dual universal property holds.

with i ≤ k, j ≤ k (i.e. i → k, j → k). Then

colim_{i∈I }F(i) = colimF(i) = lim

←−

F(i)

is called the direct limit.

Dually, a poset I is cofiltered if for all i, j ∈ I there exists k such that k ≤ i, k ≤ j and

lim

i∈I

F(i) = lim

−→

F(i)

is called the inverse limit.

0 → 1 → 2 → 3 →···

and then for F(i) = Z/2_{i }we get

Z/Z → Z/2 → Z/4 → Z/8 →···

with maps 1 ↦ 2 everywhere, and hence

lim

←−

F(i) = lim

←−Z/2

n = Z[1/2]/Z.

Direct limits “feel like unions”.

Conversely, for

···→ 3 → 2 → 1 → 0

and F(i) = Z/2_{i}, we get

···Z/8 → Z/4 → Z/2 → Z/1 = {0}

with the maps 1 ↦ 1 everywhere, and hence (in Rings)

lim

−→Z/2

n = Z2,

the 2-adic numbers. Inverse limits “feel like some sort of completion” and oftentimes one can

also define a topology on them.

i∈I

A_{i }exists for every set I. The dual

notion is called cocomplete.

Page 47

Lemma 7.7. Suppose A is complete, I is a small category, and F : I→A. Then limF(i)

exists. Similarly, if A is cocomplete, colimits exist.

Proof. For a morphism f : j → k,

∏

i∈Obj I

F(i)

F(k)

πk−F(f)πj

so that

∏

i∈Obj I

F(i)

F(j)

F(k)

πj

πk

F(f)

commutes if the above morphism is 0. Hence we can let K be the kernel to get the diagram

0

K

∏

i∈Obj I

F(i)

∏

f : j→k

F(k)

F(i)

F(k)

πi

Then K = limF(i) satisfies the universal property.

□

8. Sheaves and sheaf cohomology

In this section, we present a brief review the theory of sheaves together with an application

We begin with a motivating example.

F(U) = set of continuous functions U → R.

If V ⊆ U, then we have a restriction map ϱ_{UV }: F(U) → F(V ) which maps f ∈ F(U) to

f_{|V }∈ F(V ).

We define a category Top(X) whose objects are open sets U ⊆ X and the partial ordering

⊆ gives morphisms: for V ⊆ U, i_{V U }: V → U is the inclusion. Define F(i_{V U }) = ϱ_{UV }. This

makes F into a contravariant functor Top(X) → Ab (or even R-mod).

with F(∅) = 0. We have a category Presheaves(X) which is a subcategory of Ab_{Top(X)}op

.

A morphism η: F→G is a collection of morphisms

η(U): F(U) → G(U) for U ∈ Top(X)

such that

Page 48

F(U)

G(U)

F(V )

G(V )

η(U)

η(V )

i ∈ I, of U we have

(1) if f ∈ F(U) and f_{|U}i = 0 for all i, then f = 0,

(2) if f_{i }∈ F(U_{i}) for all i and for all i, j we have (f_{i})_{|U}i_{∩U}j = (f_{j})_{U}i_{∩U}j then there exists

f ∈ F(U) with f_{|U}i = f_{i }for all i.

Note that condition (1) can be restated by requiring uniqueness in condition (2).

If I is totally ordered, then (1) and (2) are equivalent to

0

F(U)

⊕

i

F(U_{i})

⊕

i<j

F(U_{i }∩ U_{j})

f

(f_{U}i ,i ∈ I)

(g_{i},i ∈ I)

(g_{i }− g_{j},i<j)

being exact.

F_{x }= lim

−→

{F(U) | U

x}

or, equivalently,

F_{x }= {(U, f) | f ∈ F(U), x ∈ U}/ ∼

where (U, f) ∼ (V,g) if and only if there exists W with x ∈ W ⊆ U ∩ V and f_{|W }= g_{|W }.

F_{+}(U) = set of all function U → ∐

x∈U

F_{x }with f(x) ∈ F_{x }such that

for every y ∈ U there exists V ⊆ U and y ∈ V and g ∈ F(V )

such that g maps to f(x) ∈ F_{x }for all x ∈ V

.

The universal property of F_{+}: F_{+ }is a sheaf and if G is a sheaf and ϕ: F→G is a morphism

of pre-sheaves, then there exists a unique morphism ϕ_{+ }: F_{+ }→ G such that

F

F_{+}

G

commutes. Then F_{+ }is called the sheafification of F.

Lemma 8.6. Sheafification is exact.

Page 49

D = P_{1}(C) or D is finite.

Define a sheaf O_{X }on P_{1}(C) by setting

O_{X}(U) = {f ∈ C(t) | f regular on U},

a subring of the function field C(t).

Write the affine line as A_{1 }= P_{1 }\ {∞} ∼= C. Then

O_{X}(P_{1}) = C, O_{X}(A_{1}) = C[t].

Moveover,

O_{X}(P_{1 }\ {a_{1},...,a_{n}}) = C

[ 1

t − a_{1}

,...,

1

t − a_{n}

]

for a_{1},...,a_{n }∈ A_{1}.

Define a subsheaf I of O_{X }by setting

I(U) = {f ∈ O_{X}(U) | f_{|U∩{0,∞} }= 0}.

There is an exact sequence

0

I

O_{X}

O_{X}/I

0

but the last of these, O_{X}/I, is not a sheaf. Indeed:

(O_{X}/I)(A_{1})

≅ C[t]/(t),

(O_{X}/I)(P_{1 }\ {0}) = C

[1

t

]/(1

t

)

.

Then 0 + (t) and 1 + (_{1}

t) agree on A1 \ {0}, since

(O_{X}/I)(A_{1 }\ {0})=0,

but since (O_{X}/I)(P_{1}) = C, there is no element there mapping to f_{1 }and f_{2}.

Instead, sheafify to get an exact sequence of sheaves

0

I

O_{X}

(O_{X}/I)_{+}

0

where (O_{X}/I)_{+}(P_{1}) = C_{2}.

R_{i}Γ(X,F) = H_{i}(X,F)

is called the sheaf cohomology.

0

Γ(X, I)

︸ ︷︷ ︸

=0

Γ(X,O_{X})

︸ ︷︷ ︸

=C

Γ(X,(O_{X}/I)_{+})

︸

︷︷

︸

=C2

H_{1}(X, I)

︸ ︷︷ ︸

=0

···

Page 50

9. Adjoint functors

R: B → A are adjoint if there exists a natural isomorphism

T : Hom_{B}(L(A),B)

≅ Hom_{A}(A, R(B))

of groups.

Proposition 9.2 (Yoneda Lemma). Let A be an abelian category. A sequence

A

B

C

α

β

is exact if for all M,

Hom_{A}(M,A)

Hom_{A}(M,B)

Hom_{A}(M,C)

α∗

β∗

is exact.

Proposition 9.3. If L and R are adjoint, then L is right-exact and R is left-exact.

Proof. Suppose

0

B_{1}

B_{2}

B_{3}

0

is a short exact sequence in B. We apply

Hom_{B}(L(A),−)

≅ Hom_{A}(A, R(−))

to get

0

Hom_{B}(L(A),B_{1})

Hom_{B}(L(A),B_{2})

Hom_{B}(L(A),B_{3})

0

Hom_{A}(A, R(B_{1}))

Hom_{A}(A, R(B_{2}))

Hom_{A}(A, R(B_{3}))

∼=

∼=

∼=

for all A ∈ ObjA. By Yoneda lemma, we get that

0 → R(B_{1}) → R(B_{2}) → R(B_{3})

is exact, and so R is left-exact. Similarly, L is right-exact.

□

(1) A right R-module,

(2) B R-S bimodule,

(3) C right S-module.

Then A ⊗_{R }B is a right S-module with (a ⊗ b)s = (a ⊗ bs) and Hom_{S}(B,C) is a right

R-module with (fr)(b) = f(rb). There is a natural isomorphism

T : Hom_{S}(A ⊗_{R }B,C)

∼=

→ Hom_{R}(A,Hom_{S}(B,C)).

Page 51

Indeed, if f : A ⊗_{R }B → C, then for we get

a ↦ [f(a ⊗ −): B → C]

Conversely, if A → Hom_{S}(B,C) is any map, then we get a bilinear map A × B to C which

factors through a map A ⊗_{R }B → C.

We have that

− ⊗_{R }B: mod-R → mod-S

Hom_{S}(B,−): mod-S → mod-R

are an adjoint pair, so − ⊗_{R }B is right exact and Hom_{S}(B,−) is left exact.

Then we set

L_{i}(− ⊗_{R }B) = Tor_{i}(−,B),

and we will show that

L_{i}(A ⊗_{R }−) = Tor_{i}(A,−),

because

L_{i}(− ⊗_{R }B)

≅ L_{i}(− ⊗ B)(A).

d_{h }: C_{p,q }→ C_{p−1,q}

and vertical maps

d_{v }: C_{p,q }→ C_{p,q−1}

such that d_{h }◦ d_{h }= 0, d_{v }◦ d_{v }= 0, and d_{v}d_{h }+ d_{h}d_{v }= 0

...

...

...

···

C_{p−1,q+1}

C_{p,q+1}

C_{p+1,q+1}

···

···

C_{p−1,q}

C_{p,q}

C_{p+1,q}

···

···

C_{p−1,q−1}

C_{p,q−1}

C_{p+1,q−1}

···

...

...

...

The map d_{v }: C_{•q }→ C_{•q−1 }is almost a chain map: if we set f_{pq }= (−1)_{p}d_{v}

pq : Cpq → Cpq−1,

then f_{•q }: C_{•q }→ C_{•q−1 }is a chain map. Hence f_{•q }is in Ch(C), and f_{•• }is in Ch(Ch(C)).

Tot

⊕

(C_{••})_{n }=

⊕

p+q=n

C_{p,q}

with d = d_{v }+ d_{h}, whence

d_{2 }= (d_{v }+ d_{h})_{2 }= (d_{v})_{2 }+ (d_{v}d_{h }+ d_{h}d_{v})+(d_{h})_{2 }=0+0+0=0.

Page 52

If C is complete,

Tot_{π}(C_{••})_{n }=

∏

p+q=n

C_{p,q}

with d = d_{v }+ d_{h}.

Proposition 9.7 (Acyclic Assembly Lemma). Suppose C_{•• }is a double complex in mod-R.

Suppose C is an upper half plane complex (i.e. C_{p,q }= 0 if q < 0) and columns are exact.

Then Tot_{π}(C) is acyclic (exact).

Proof. We claim that H_{0}(Tot_{π}(C_{••})) = 0. By symmetry, it is enough to restrict our attention

to the p = 0 portion of the diagram

...

...

...

···

C_{−1,2}

C_{0,2}

C_{1,2}

···

···

C_{−1,1}

C_{0,1}

C_{1,1}

···

···

C_{−1,0}

C_{0,0}

C_{1,0}

···

0

0

0

Suppose

(c_{0},c_{1},c_{2},...) ∈ C_{00 }× C_{−1,1 }× C_{−2,2 }×···

is a cycle. Then

d(c_{0},c_{1},c_{2},...)=(d_{h}c_{0 }+ d_{v}c_{1},d_{h}c_{1 }+ c_{v}c_{2},...) ∈ C_{−1,0 }× C_{−2,1 }×··· .

We want to find

(b_{0},b_{1},...) ∈ C_{10 }× C_{01 }× C_{−12 }×···

with

d(b_{0},b_{1},...)=(d_{h}b_{0 }+ d_{v}b_{1},d_{h}b_{1 }+ d_{v}b_{2},...)=(c_{0},c_{1},...).

Pick b_{0 }= 0. Then d_{v}c_{0 }= 0 so c_{0 }= d_{v}b_{1 }for some b_{1}. Then

0 = d_{h}c_{0 }+ d_{v}c_{1 }= d_{h}d_{v}b_{1 }+ d_{v}c_{1 }= −d_{v}d_{h}b_{1 }+ d_{v}c_{1 }= d_{v}(c_{1 }− d_{h}b_{1}).

Hence there exists b_{2 }∈ C_{−1,2 }such that d_{v}b_{2 }= c_{1 }−d_{h}b_{1}. Then c_{1 }= d_{v}b_{2 }+d_{h}b_{1}, and proceed

by induction to construct b_{3}, b_{4}, .... This completes the proof.

□

Corollary 9.8. If C_{•• }is a double complex in the right half plane with exact rows, then

Tot_{π}(C_{••}) is exact.

Corollary 9.9. If C_{•• }is a double complex in the right half plane with exact columns, then

Tot

⊕

(C_{•,•}) is exact.

Page 53

Proof. Define new complex τ_{n}C_{•• }by

(τ_{n}C)_{p,q }=

C_{pq}

if q > n,

kerd_{v }: C_{pn }→ C_{pn−1}

if q = n,

0

if q < n.

Then we get the diagram

...

...

...

0

C_{0,n+2}

C_{1,n+2}

C_{1,n+2}

···

0

C_{0,n+1}

C_{1,n+1}

C_{1,n+1}

···

0

kerd_{v}

kerd_{v}

kerd_{v}

···

0

0

0

Hence Tot_{π}(τ_{n}C) = Tot

⊕

(τ_{n}C) is exact. This shows that Tot

⊕

(C) is exact.

□

Theorem 9.10. We have that

L_{n}(A ⊗_{R }−)(B)

≅ L_{n}(− ⊗_{R }B)(A)

and we call them Tor_{R}

n (A, B).

actually a homework exercise, but we include it here for completeness.)

Theorem 9.11. Suppose C is an abelian category and

T : C×···×C→D

is an additive functor in p variables, some of the covariant and some contravariant. Assume

moreover that T is right-balanced:

(1) when any covariant variable is replaced by an injective module, T becomes exact in

the other variables,

(2) when any contravariant variable is replaced by a projective module, T becomes exact

in the other variables.

Then for any i, j there is a natural isomorphism

R_{∗}T(A_{1},...,

ˆ

A_{i},...,A_{p})(A_{i})

≅ R

∗T(A1,...,

ˆ

A_{j},...,A_{p})(A_{j}).

Proof. Note that if we fix modules A_{1},...,

ˆ

A_{i},...,

ˆ

A_{j},...,A_{p}, then the functor

T(A_{1},...,

ˆ

A_{i},...,

ˆ

A_{j},...,A_{p}): C×C→D

Page 54

is right-balanced, and it is enough to show the assertion for this functor in 2 variables. Hence

suppose that

T : C×C→D

R_{∗}T(A,−)(B)

≅ R

∗T(−,B)(A).

If the first variable is covariant, choose an injective resolution ϵ: A → P_{• }and if it is

contravariant, choose a projective resolution ϵ: P_{• }→ A. Similarly, if the second variable is

covariant, choose an injective resolution η: A → Q_{• }and if it is contravariant, choose a

projective resolution η: Q_{• }→ A. We then get the double complex:

...

...

...

...

0

T(A, Q_{2})

T(P_{0},Q_{2})

T(P_{1},Q_{2})

T(P_{2},Q_{2})

···

0

T(A, Q_{1})

T(P_{0},Q_{1})

T(P_{1},Q_{1})

T(P_{2},Q_{1})

···

0

T(A, Q_{0})

T(P_{0},Q_{0})

T(P_{1},Q_{0})

T(P_{2},Q_{0})

···

0

T(P_{0},B)

T(P_{1},B)

T(P_{2},B)

0

0

0

T(ϵ,1)

T(ϵ,1)

T(ϵ,1)

T(1,η)

T(1,η)

T(1,η)

We will show that the maps

T(ϵ,1): Tot(T(P_{•},Q_{•})) → Tot(T(A, Q_{•})) = T(A, Q_{•})

T(1,η): Tot(T(P_{•},Q_{•})) → Tot(T(P_{•},B)) = T(P_{•},B)

are quasi-isomorphisms, and hence they induce natural isomorphisms

H_{∗}(Tot(T(P_{•},Q_{•})))

≅ R

∗(T(A,−))(B),

H_{∗}(Tot(T(P_{•},Q_{•})))

≅ R

∗(T(−,B))(A),

which gives the result.

We only show T(ϵ,1) is a quasi-isomorphism using the Acyclic Assembly Lemma 9.7; the

fact that T(1,η) is a quasi-isomorphism is symmetric. Let C_{•• }be the double complex

Page 55

...

...

...

...

0

T(A, Q_{2})

T(P_{0},Q_{2})

T(P_{1},Q_{2})

T(P_{2},Q_{2})

···

0

T(A, Q_{1})

T(P_{0},Q_{1})

T(P_{1},Q_{1})

T(P_{2},Q_{1})

···

0

T(A, Q_{0})

T(P_{0},Q_{0})

T(P_{1},Q_{0})

T(P_{2},Q_{0})

···

0

0

0

0

T(ϵ,1)

T(ϵ,1)

T(ϵ,1)

and note that Tot(C_{•,•})[1] is the mapping cone of ϵ⊗1: Tot(T(P_{•},Q_{•})) → T(A, Q_{•}). Hence,

to show that ϵ ⊗ 1 is a quasi-isomorphism, it is enough to show that the mapping cone

Tot(C_{•,•})[1] is acyclic. Finally, since Q_{i }are injective if the second variable is covariant and

projecitve if the second variable is contravariant, the right-balanced condition shows that

acyclic.

This completes the proof.

□

10. Tor and Ext

We restrict our attention to A = Ab.

0

Z

Z

Z/n

0

n·

For an abelian group B, apply − ⊗_{Z }B to P_{• }to get

0

B

B

0

n·

Hence

Tor_{0}(Z/n, B) = Z/n ⊗ B = H_{0}(P_{• }⊗ B) = B/nB,

Tor_{1}(Z/n, B) = B[n] = {b ∈ B : nb = 0},

Tor_{k}(Z/n, B) = 0 for k ≥ 2,

Tor_{0}(Z,B) = B,

Tor_{k}(Z,B) = 0 for k ≥ 1,

In general,

Tor_{i}(Z/n,Z/m) = Z/(n, m) for i = 0,1.

To calculate Ext, we apply Hom_{Ab}(−,B) to P_{• }to get

0

B

B

0

n·

Page 56

and then

Ext_{0}(Z/n, B) = H_{0}(P_{•}) = B[n]

Ext_{1}(Z/n, B) = H_{1}(P_{•}) = B/nB

and in particular

Ext_{i}(Z/n,Z/m) = Z/(n, m) for i = 0,1.

If A = lim

−→

A_{α }= colimA_{α}, then

Tor_{i}(A, B) = Tor_{i}(lim

−→

A_{α},B) = lim

−→

Tor_{i}(A_{α},B).

Proposition 10.2. Suppose A and B are abelian group. Then

(1) Tor_{1}(A, B) is a torsion group,

(2) Tor_{n}(A, B)=0 for n ≥ 2.

Proof. If A is finitely generated, then

A

≅ Z

r ⊕ Z/n1 ⊕···⊕ Z/nk

and the proposition is clear.

Otherwise, A = lim

−→

A_{α }for {A_{α}} finitely generated subgroups. The limit of torsion groups is

torsion, so

Tor_{1}(A, B) = lim

−→

Tor_{1}(A_{α},B)

is torsion.

□

Tor_{1}(Q/Z,B) = lim

−→

Tor_{1}

(

Z

[1

n

]

/

Z

︸ ︷︷ ︸

Z/n

,B

)

= lim

−→

B[n],

the torsion subgroup of B.

If A is torsion free then A = lim

−→Z

m, so

Tor_{1}(A, B) = lim

−→

Tor_{1}(Z_{m},B)=0.

Hence A is torsion-free if and only if Tor_{1}(A,−) = 0 if and only if A ⊗_{Z }− is exact, or by

definition, A is a flat Z-module.

flat if A ⊗_{R }− is exact.

In general, a projective module is flat but the converse is not true. For example, Q is a flat

Z-module but it is not projective.

Suppose R is a ring with 1 and

S ⊆ Z(R)

︸ ︷︷ ︸

center

⊆ R.

Suppose 1 ∈ S and S is closed under multiplication. The localization of R with respect to

S, S_{−1}R, is

S_{−1}R = {(r, s) | r ∈ R, s ∈ S}/ ∼

Page 57

where

(r_{1},s_{1}) ∼ (r_{2},s_{2}) if and only if there exists s_{3 }∈ S such that (r_{1}s_{2 }− r_{2}s_{1})s_{3 }= 0.

We think of (r, s) as _{r}

s

. If [(r, s)] is an equivalent class, then

[(r_{1},s_{1})] + [(r_{2},s_{2})] = [(r_{1}s_{2 }+ r_{2}s_{1},s_{1}s_{2})],

[(r_{1},s_{1})] · [(r_{2},s_{2})] = [(r_{1}r_{2},s_{1}s_{2})].

We have a map ϕ: R → S_{−1}R, setting ϕ(r) = [(r,1)]. We then set

S_{−1}M := S_{−1}R ⊗_{R }M, the localization of M at S.

Theorem 10.5. The localization S_{−1}R is a flat R-module, i.e. S_{−1}R⊗_{R}− is an exact functor

from R-mod to S_{−1}R-mod.

Proof. Define category I with

Obj(I) = S,

for s_{1},s_{2 }∈ S, we set Hom_{I}(s_{1},s_{2}) = {s ∈ S | ss_{1 }= s_{2}}.

This is a filtered category (see definition below)

s_{1}

s_{1}s_{2}

s_{2}

s2

s1

s_{1}

s_{2}

s_{1}s_{2}

t1

t2

s1

We then have a functor F : I → R-mod given by F(s) = R for s ∈ Obj(I) = S and if

s_{1}

s

→ s_{2 }is a morphism then

F(s_{1}) = R

F(s_{2}) = R

F(s)=s·

We claim that colim_{s∈I }F(s) exists and in fact colim_{s∈I }F(s) = S_{−1}R. Indeed, we define

ϕ_{s }: F(s) = R → S_{−1}R

by ϕ_{s}(r) = [(r, s)]. Then one can easily check that this gives a map from the colimit to

S_{−1}(R) which is an isomorphism by the universal property of localization.

Therefore, for n ≥ 1

Tor_{n}(S_{−1}R, B) = colim Tor_{n}(F(s),B) = colim Tor_{n}(R, B)

︸

︷︷

︸

=0

= 0,

and hence S_{−1}R is flat.

□

(1) for any A, B, there exists C with morphisms α: A → C and β: B → C,

(2) if A

B

α

β

then there exists γ: B → C such that γα = γβ.

Page 58

Exercise. The following conditions are equivalent

(1) A is a flat right R-module,

(2) A ⊗_{R }− is an exact functor,

(3) Tor_{1}(A, B) = 0 for all left R-modules B,

(4) Tor_{n}(A, B) = 0 for all left R-modules B and n ≥ 1.

(The first equivalence is the definition.)

dual of B.

If B = 0, let C be a maximal subgroup. Then B/C

≅ Z/p for p prime, and hence

Hom_{Ab}(B/C,Q/Z) = 0,

and thus B_{∗ }= Hom(B,Q/Z) = 0.

Lemma 10.8. A morphism f : B → C is injective if and only if f_{∗ }: C_{∗ }→ B_{∗ }is surjective,

where f_{∗ }= Hom_{Ab}(f,Q/Z).

Proof. Suppose A → B is a kernel of f, so 0 → A → B → C is exact, and hence

C_{∗ }→ B_{∗ }→ A_{∗ }→ 0

is exact, since Q/Z is injective. Hence f is injective if and only if A = 0 if and only if A_{∗ }= 0

if and only if f_{∗ }is surjective.

□

Proposition 10.9. The following are equivalent:

(1) B is a flat left R-module,

(2) B_{∗ }is an injective right R-module,

(3) I ⊗_{R }B

∼=

→ IB for every right ideal I ⊆ R,

(4) Tor_{1}(R/I,B)=0 for every right ideal I ⊆ R.

Proof. We first check that (3) is equivalent to (4). Apply − ⊗_{R }B to the exact sequence

0

I

R

R/I

0

to get an exact sequence

Tor_{1}(R, B)

︸

︷︷

︸

=0

Tor_{1}(R/I,B)

I ⊗_{R }B

R ⊗_{R }B

︸ ︷︷ ︸

∼=B

R/I ⊗_{R }B

︸ ︷︷ ︸

B/IB

0

ϕ

and hence

0

Tor_{1}(R/I,B)

I ⊗_{R }B

IB

︸︷︷︸

ker ϕ

0

Page 59

is exact. This shows (3) is equivalent to (4).

We now show (1) is equivalent to (2). We have

Hom_{R}(A, B_{∗}) = Hom_{R}(A,Hom_{Z}(B,Q/Z)))

≅ Hom(A ⊗_{R }B,Q/Z)=(A ⊗_{R }B)_{∗}.

If A ⊆ A is a submodule, then the following diagram

Hom_{R}(A, B_{∗})

Hom(A ,B_{∗})

(A ⊗_{R }B)_{∗}

(A ⊗_{R }B)_{∗}

∼=

∼=

commutes. Now,

B_{∗ }is injective if and only if Hom_{R}(A, B_{∗}) → Hom_{R}(A ,B_{∗}) is surjective for all A ⊆ A

if and only if (A ⊗_{R }B)_{∗ }→ (A ⊗_{R }B)_{∗ }is surjective for all A ⊆ A

if and only if A ⊗_{R }B → A ⊗_{R }B is injective for all A ⊆ A

if and only if − ⊗_{R }B is exact

if and only if B is flat.

We finally show (2) is equivalent to (3). Note that B_{∗ }is injective if and only if

B∗

︷

︸︸

︷

Hom_{R}(R, B_{∗}) →

(I⊗B)∗

︷

︸︸

︷

Hom_{R}(I,B_{∗})

is surjective for all right ideals I ⊆ R. But this is equivalent to I ⊗_{R }B → B is injective for

all I, which holds if and only if I ⊗_{R }B

≅ IB.

□

R_{m }→ R_{n }→ M → 0.

Note that projectivity is not equivalent to flatness. Indeed, Q is a flat Z-module (it is a

localization of Z), but it is not projective.

We will show that for finitely presented modules, flat modules are projective.

For M,A left R-modules, define

σ: A_{∗ }⊗_{R }M → Hom_{R}(M,A)_{∗}

σ(f ⊗ m)(h) = f(h(m))

for f ∈ A_{∗}, m ∈ M, h ∈ Hom_{R}(M,A).

Proposition 10.11. If M is finitely presented, then σ is an isomorphism for all A.

Proof. Clear if M = R_{n}. Now, if

R_{m}

R_{n}

M

0

then we have the following commutative diagram

Page 60

A_{∗ }⊗_{R }R_{m}

A_{∗ }⊗_{R }R_{n}

A_{∗ }⊗_{R }M

0

0

Hom_{R}(R_{m},A)_{∗}

Hom_{R}(R_{n},A)_{∗}

Hom_{R}(M,A)_{∗}

0

0

∼=

∼=

σ

∼=

∼=

which has exact rows, because Hom(−,A)_{∗ }and A ⊗_{R }− are right-exact covariant functors.

By Five Lemma 1.44, we obtain that σ is an isomorphism.

□

Theorem 10.12. A finitely presented flat R-module is projective.

Proof. Suppose M is finitely presented, flat, and f : B → C is surjective, so f_{∗ }: C_{∗ }→ B_{∗ }is

injective. Hence the square

C_{∗ }⊗_{R }M

B_{∗ }⊗ M

Hom(M,C)_{∗}

Hom(M,B)_{∗}

f∗⊗1

∼=

∼=

commutes. Since M is flat, f_{∗ }⊗ 1 is injective, and hence

Hom_{R}(M,B) → Hom_{R}(M,C)

is surjective. This shows that Hom_{R}(M,−) is exact, so M is projective.

□

Lemma 10.13 (Flat resolution lemma). If ··· → F_{2 }→ F_{1 }→ F_{0 }→ A → 0 is a flat

resolution, then

Tor_{R}

n (A, B) = Hn(F• ⊗ B)

for a right R-module A and a left R-module B.

If F_{n }are in fact projective, this is how we defined Tor_{R}

n (A, B), and this lemma shows that

we can compute Tor_{R}

n (A, B) by considering only a flat resolution.

Proof. For n = 0, we have an exact sequence

F_{1 }⊗_{R }B

F_{0 }⊗_{R }B

A ⊗_{R }B

0

and hence H_{0}(F_{• }⊗_{R }B) = A ⊗_{R }B = Tor_{0}(A, B).

We have an exact sequence

0

K

F_{0}

A

0.

We then get a long exact sequence

Page 61

=0

︷

︸︸

︷

Tor_{2}(F_{0},B)

Tor_{2}(A, B)

Tor_{1}(K, B)

=0

︷

︸︸

︷

Tor_{1}(F_{0},B)

Tor_{1}(A, B)

K ⊗ B

F_{0 }⊗ B

A ⊗ B

0

and hence Tor_{n}(A, B)

≅ Tor_{n−1}(K, B). We have the exact sequence

F_{2}

F_{1}

K

0

which gives the exact sequence

F_{2 }⊗ B

F_{1 }⊗ B

K ⊗ B

0

and we get that

Tor_{1}(A, B) = ker(K ⊗ B → F_{0 }⊗ B) = ker

( F_{1 }⊗ B

d_{2}(F_{2 }⊗ B)

→ F_{0 }⊗ B

)

= H_{1}(F_{• }⊗ B).

By induction on n, we finally obtain:

Tor_{n}(A, B) = Tor_{n−1}(K, B) = H_{n−1}(F_{•}[1] ⊗ B) = H_{n}(F_{• }⊗ B),

as required.

□

Proposition 10.14. Suppose R → T is a ring homomorphism and T is flat as a left

R-module. Then for all right R-modules A and left T-modules C, we have that

Tor_{R}

n (A, C)

≅ Tor

T

n (A ⊗R T,C).

Proof. Let P_{• }→ A be a projective resolution so that

Tor_{R}

n (A, C) = Hn(P• ⊗R C).

Note that P_{n }⊗_{R }T is a projective T-module: P_{n }⊕Q = F for some free R-module F, whence

F ⊗ T = (P_{n }⊕ Q) ⊗ T = (P_{n }⊗ T) ⊕ (Q ⊗ T),

so P_{n }⊗ T is a direct summand of the free T-module F ⊗ T. Hence

P_{• }⊗_{R }T → A ⊗_{R }T

is a projective resolution (since − ⊗_{R }T is exact). Hence

Tor_{T}

n (A ⊗R T,C) = Hn(P• ⊗R T ⊗T C) = Hn(P• ⊗R C) = TorR

n (A, C).

This completes the proof.

□

Corollary 10.15. Let R be commutative, T be a flat R-algebra with ϕ: R → T and ϕ(R) ⊆

Z(T). Then

T ⊗_{R }Tor_{R}

n (A, B)

≅ Tor

T

n (A ⊗R T,T ⊗R B).

Page 62

Proof. We have that:

Tor_{T}

n (A ⊗R T,T ⊗R B) = TorR

n (A, T ⊗R B) = Hn(P• ⊗R T ⊗R B) = Hn(T ⊗R P• ⊗R B)

= T ⊗_{R }H_{n}(P_{• }⊗_{R }B) = T ⊗_{R }Tor_{R}

n (A, B),

as required.

□

for S = R \ p. If M is an R-module, M_{p }= R_{p }⊗_{R }M, then

Tor_{R}

n (A, B)p = Rp ⊗ TorR

n (A, B) = TorRp

n (A ⊗R Rp,B ⊗R Rp) = TorRp

n (Ap,Bp).

In general,

S_{−1 }Tor_{R}

n (A, B)

≅ Tor

S−1R

n

(S_{−1}A, S_{−1}B).

0 → B → I_{0 }→ I_{0}/B → 0

is exact, but since I_{0 }is divisible, I_{1 }= I_{0}/B is also divisible, so it is injective. Hence

0 → B → I_{0 }→ I_{1 }→ 0

is an injective resolution. This shows that

Ext_{n}(A, B) = 0 for n ≥ 2.

For B = Z, we get

0 → Z → Q → Q/Z → 0,

and hence Ext

∗

(A,Z) is the homology of

0 → Hom(A,Q) → Hom(A,Q/Z) → 0.

If A is torsion, then

Hom(A,Q)=0.

In that case,

Ext_{1}(A,Z) = Hom(A,Q/Z) = A_{∗},

the Pontryagin dual we defined before.

Proposition 10.18. Let A be a finitely generated R-module over a commutative Noetherian

ring R, B be an R-module, and S ⊆ R be a multiplicative system. Then

S_{−1 }Ext_{n}

R(A, B) = Extn

S−1R(S−1A, S−1B).

In particular, for a prime p, we have

Ext_{n}

R(A, B)p = Extn

Rp

(A_{p},B_{p}).

Recall that, to check if an R-module M is 0, it is enough to check that for any prime ideal p,

M_{p }= 0. So in this case, to check that Ext_{n}

R(A, B) = 0, it is enough to check that

Ext_{n}

Rp

(A_{p},B_{p})=0

for any prime p.

We know Ext

•

(M,N) as a measure of failure to Extend maps. It is a derived functor of Hom

in multiple ways. On the one hand, we see the same Ext in lots of different places, but on

the other hand this also means objects of Ext are “slippery”.

Page 63

We will compare Ext to something more concrete by asking the following question: When

does a short exact sequence (of R-modules) split?

Given a short exact sequence, it splits if there is a section

E :

0

A

B

C

0

C

1C

In the long exact sequence, we get

Hom_{R}(C, B)

Hom_{R}(C, C)

Ext_{1}

R(C, A)

?

1_{C}

1_{C}

0

δ

?

Answer: The short exact sequence E splits if and only if δ(1_{C})=0 ∈ Ext_{1}

R(C, A).

R(C, A).

Remark 10.20. We can also compute the obstruction of E as follows. Take 1_{C }and lift it

to a projective resolution of C, and a map to E

···

P_{2}

P_{1}

P_{0}

C

0

A

B

C

0

d2

d1

α

β

1C

Such a lift is unique up to chain homotopy.

We claim that the map α ∈ Hom_{R}(P_{1},A) defines the same class as θ(E) = δ(1_{C}). We have

the following diagram:

δ(1_{C}) = u ∈ Hom(P_{1},A)

Hom(P_{1},B)

v ∈ Hom(P_{0},B)

Hom(P_{0},C))

Hom(C, C) 1_{C}

We choose 1_{C }∈ Hom(C, C), map it to Hom(P_{0},C), lift it to v ∈ Hom(P_{0},B), map it to

Hom(P_{1},B), and lift it to u ∈ Hom(P_{1},A).

The maps u, v,1_{C }give a commutative diagram as above, with α replaced with u and β

replaced with v. The chain map is chain homotopic to the original map, and hence u and v

give the same class in Ext_{1}

R(C, A).

Page 64

E :

0

A

B

C

0

(an extension of C by A).

An isomorphism of extensions is a commutative diagram

E :

0

A

B

C

0

E :

0

A

B

C

0

=

=

By Five Lemma 1.44, the map B → B is an isomorphism.

0

A

A ⊕ C

C

0

(1,0)

0

Z

?

Z/p

0

i

j

• ? = Z ⊕ Z/p, the trivial extension.

• ? = Z, i = ·p, j = ·k for any k ∈ (Z/p)_{×}.

If two of these are isomorphic extensions,

0

Z

Z

Z/p

0

0

Z

Z

Z/p

0

=

p

·k

=

·k

then k = k . So there can be nonisomorphic extensions with isomorphic middles.

We will write ext_{1}(C, A) for the set of isomorphism classes of extensions of C by A.

If there are enough projectives in A, the obstruction map

θ: ext_{1}(C, A) → Ext_{1}(C, A)

is well-defined. (For an isomorphism E

≅ E , we get an isomorphism of long exact sequenes.)

In fact, more is true.

Theorem 10.24. The map θ: ext_{1}(C, A) → Ext_{1}(C, A) is a bijection if there are enough

projectives in A.

Proof. Let us construct an inverse (we work with R-modules, but the same construction

works in general).

Given η ∈ Ext_{1}(C, A), choose a projective resolution P_{• }→ C, and a representative φ ∈

Hom_{R}(P_{1},A):

Page 65

P_{2}

P_{1}

P_{0}

C

0

A

d1

φ

d0

Take the pushout and extend it to an isomorphism of its cokernel:

P_{2}

P_{1}

P_{0}

C

0

A

B

C

0

d1

φ

d0

=

Since φ represents a cocycle, it factors through S = P_{1}/im(P_{2}). Hence we have another

pushout square (we give it the same name by abuse of notation):

0

S

P_{0}

C

0

0

A

B

C

0

d1

φ

d0

=

where the map A → B is injective, because S → P_{0 }is injective.

We use this short exact sequence in the setting above:

P_{2}

P_{1}

P_{0}

C

0

E :

0

A

B

C

0

d1

φ

d0

=

By the Remark 10.20 earlier, θ(E)=[φ] ∈ Ext_{1}

R(C, A).

To conclude that this inverse construction is well-defined, we need to show that the same

Ext_{1}-classes of maps give the same extension.

This follows from the fact that the construction of our extension from φ came as a pushout

of

0

S

P_{0}

C

0

0

A

B

C

0

d1

φ

d0

=

since the maps S → A are independent of coboundary.

□

Remark 10.25.

• We can generalize this construction to higher Ext’s (bijective to isomorphism classes

of longer exact sequences).

Page 66

• There is a way to add extensions that is compatible with θ.

• We can multiply Ext_{i}(C, A) ⊗ Ext_{j}(D, C) → Ext_{i+j}(D, A) that comes from splicing

exact sequences.

• We have a notion of Ext_{i }in any abelian category.

11. Universal coefficients theorem

Recall that a right R-module is flat if and only if Tor_{1}(A, M) = 0 for all M if and only if

Tor_{n}(A, M) = 0 for all M and all n ≥ 1. Moreover, if

0

A

B

C

0

is exact, then

(1) if A, C are flat then B is flat,

(2) if B,C are flat then A is flat.

To see this, we just look at the long exact sequence for Tor.

Set up: let R be a ring, P_{• }a complex of flat R-modules, M an R-module.

Theorem 11.1 (Künneth). Assume B_{n}(P_{•}) is R-flat for all n (for example, if R = Z or

any PID or a field). There is a natural short exact sequence

0

H_{n}(P_{•}) ⊗_{R }M

H_{n}(P_{• }⊗_{R }M)

Tor_{R}

1 (Hn−1(P•),M)

0

(1) If R is a field, Tor_{R}

1 (−,−) = 0, we get the obvious isomorphism

H_{n}(P_{•}) ⊗_{R }M

≅ H_{n}(P_{• }⊗_{R }M),

because − ⊗_{R }M is exact when R is a field.

(2) Let R = Z, P_{• }= Z

2

→ Z, M = Z/2. Then

H_{i}(P_{•}) =

{

Z/2 if i = 0

0

otherwise

But

P_{• }⊗_{R }M = Z/2

2

→ Z/2

and hence

H_{i}(P_{• }⊗_{R }M) =

{

Z/2 if i = 0,1

0

otherwise

To see this via Künneth Theorem 11.1, we note that

0

H_{1}(P_{• }⊗ M)

Tor_{Z}

1 (Z/2,Z/2)

︸

︷︷

︸

=Z/2

∼=

(3) (Non-example). Let R = Z/4, M = Z/2, and P_{• }= Z/4

2

→ Z/4. If Künneth was

true, we would get

Page 67

0

H_{1}(P_{•}) ⊗_{R }M

︸

︷︷

︸

=Z/2⊗Z/4Z/2=Z/2

H_{1}(P_{• }⊗_{R }M)

︸

︷︷

︸

H1(Z/2

2

→Z/2)=Z/2

Tor_{R}

1 (H0(P•),M)

︸

︷︷

︸

TorZ/4

1

(Z/2,Z/2)=Z/2

0

This is impossible for cardinality reasons.

Above, to find Tor_{Z/4}

1

(Z/2,Z/2), we take the resolution

K_{• }:

···

Z/4

Z/4

Z/4

Z/2

0

·2

·2

and note that

Tor_{Z/4}

1

(Z/2,Z/2) = H_{1}(K_{• }⊗_{Z/4 }Z/2) = Z/2.

Z_{•}(P) and B_{•}(P) have trivial boundary maps.

We claim that Z_{n}(P_{•}) is a flat R-module. We have a short exact sequence

0

Z_{n}(P_{•})

P_{n}

B_{n−1}(P_{•})

0

d

As each B_{n−1}(P_{•}) is flat, this also shows that Z_{n}(P_{•}) is flat. The long exact sequence in

homology gives

···

H_{n}(B_{•}(P) ⊗ M)

H_{n}(Z_{•}(P) ⊗ M)

H_{n}(P_{• }⊗ M)

H_{n−1}(B_{•}(P) ⊗ M)

H_{n−1}(Z_{•}(P) ⊗ M)

···

αn

αn

and hence we have the short exact sequence

(∗)

0

coker(α_{n})

H_{n}(P_{•}⊗_{R})

ker(α_{n−1})

0

As Z_{•}(P) has a trivial differential, the same is true for Z_{•}(P) ⊗ M and B_{•}(P) ⊗ M. This

shows that

H_{n}(Z_{•}(P) ⊗ M) = Z_{n}(P) ⊗ M,

H_{n}(B_{•}(P) ⊗ M) = B_{n}(P) ⊗ M.

But since B_{n}(P) and Z_{n}(P) are flat, and we have the short exact sequence

0

B_{n}(P)

Z_{n}(P)

H_{n}(P)

0,

this is a flat resolution of H_{n}(P). Therefore

H_{•}(B_{n}(P) ⊗ M

αn

→ Z_{n}(P) ⊗ M) = Tor_{R}

• (Hn(P),M),

since Tor can be calculated using flat resolutions 10.13. This shows that

coker(α_{n}) = Tor_{R}

0 (Hn(P),M) = Hn(P) ⊗ M,

ker(α_{n}) = Tor_{R}

1 (Hn(P),M).

Substituting this into the short exact sequence (∗) completes the proof.

□

Page 68

Corollary 11.3. Assume R = Z (or B_{n}(P_{•}) is free). Then for all M, the Künneth sequence

splits (non-canonically), i.e. we have isomorphisms

(H_{n}(P_{•}) ⊗ M) ⊕ Tor_{R}

1 (Hn−1(P),M)

≅ H_{n}(P_{• }⊗ M).

Proof. We know that each d(P_{n}) is free. Hence the short exact sequence

0

Z_{n}

P_{n}

d(P_{n})

0

splits (non-canonically!), and so P_{n}

≅ Z_{n }⊕ d(P_{n}). Taking − ⊗ M of both sides, we obtain

Z_{n }⊗ M ⊆ ker(d_{n }⊗ 1) ⊆ P_{n }⊗ M

≅ Z_{n }⊗ M ⊕ d(P_{n}) ⊗ M.

We hence get that ker(d_{n }⊗ 1)

≅ Z_{n }⊗ M ⊕ C, for some complement C. Taking the quotient

by im(d_{n+1 }⊗ 1) = im(d_{n+1}) ⊗ M, we get

H_{n}(P_{• }⊗ M)

≅ H_{n}(P_{•}) ⊗ M ⊕ C.

Hence the Künneth exact sequence

0

H_{n}(P_{•}) ⊗_{Z }M

H_{n}(P_{• }⊗_{Z }M)

Tor_{Z}

1 (Hn−1(P•),M)

0

splits, with C

≅ Tor_{Z}

1 (Hn−1(P),M).

□

M is some abelian group,

H_{n}(X;M) = H_{n}(S_{•}(X) ⊕ M) = H_{n}(S_{•}(X)) ⊗ M ⊕ Tor_{Z}

1 (Hn−1(S•(x)),M),

and hence

H_{n}(X;M) = H_{n}(X) ⊗ M ⊕ Tor_{Z}

1 (Hn−1(X),M).

So, to calculate the homology groups with coefficients in M, it is enough to calculate them

with coefficients in Z (but the splitting is not functorial, so this does not tell us anything

about the maps between the homology groups).

For example, let X = P_{2}(R) and M = Z/2. We then have:

H_{0}(X) = Z, H_{1}(X) = Z/2, H_{2}(X)=0,

and hence

H_{2}(X;Z/2) = H_{2}(X) ⊗_{Z }Z/2

︸

︷︷

︸

=0

⊕Tor_{1}(H_{1}(X),Z/2) = Tor_{1}(Z/2,Z/2) = Z/2.

There is an analog of the corollary for cohomology.

Theorem 11.5. Suppose P_{• }is a chain complex of left R-modules such that d(P_{n}) is projective

for all n. Then we have

H_{n}(Hom_{R}(P, M))

≅ Hom_{R}(H_{n}(P_{•}),M) ⊕ Ext_{1}

R(Hn−1(P•),M).

Page 69

groups of X, we obtain

H_{0}(X) = Z,

H_{1}(X) = H_{1}(X;Z) = Hom_{Z}(Z/2,Z)

︸

︷︷

︸

=0

⊕Ext_{1}(Z,Z)

︸ ︷︷ ︸

=0

,

H_{2}(X) = Hom(0,Z) ⊕ Ext_{1}(Z/2,Z) = Z/2.

Let P_{• }be a complex of right R-modules and Q_{• }be a complex of left R-modules. We have

the double complex

...

...

...

···

P_{0 }⊗ Q_{2}

P_{1 }⊗ Q_{2}

P_{2 }⊗ Q_{2}

···

···

P_{0 }⊗ Q_{1}

P_{1 }⊗ Q_{1}

P_{2 }⊗ Q_{1}

···

···

P_{0 }⊗ Q_{0}

P_{1 }⊗ Q_{0}

P_{2 }⊗ Q_{0}

···

...

...

...

and the total complex is

(P ⊗_{R }Q)_{n }=

⊕

p+q=n

P_{p }⊗ Q_{q}.

There is an analog of Künneth Theorem for the total complex.

Theorem 11.7. If P_{n}, d(P_{n}) are flat for all n, then

0

⊕

p+q=n

H_{p}(P_{•}) ⊗ H_{q}(Q_{•})

H_{n}(P_{• }⊗ Q_{•})

⊕

p+q=n−1

Tor_{1}(H_{p}(P_{•}),H_{q}(Q_{•}))

0

is exact.

For topological spaces X, Y , (after some work) this gives the result

H_{n}(X × Y )

≅

⊕

p+q=n

H_{p}(X) ⊗ H_{q}(Y ) ⊕

⊕

p+q=n−1

Tor_{Z}

1 (Hp(X),Hq(Y )).

12. Quivers

• Q_{0 }is a finite set of vertices,

• Q_{1 }is a finite set of arrows,

Page 70

• h: Q_{1 }→ Q_{0 }is the head map; h(a) ∈ Q_{0 }is the head of arrow a ∈ Q_{1},

• t: Q_{1 }→ Q_{0 }is the tail map; t(a) ∈ Q_{0 }is the tail of arrow a ∈ Q_{1}.

b

a

c

d

1

2

represents a quiver with Q_{0 }= {1,2}, Q_{1 }= {a, b, c, d} and

t(a) = t(b) = h(c) = t(d) = h(d)=1,

h(a) = h(b) = t(c)=2.

p = a_{d}a_{d−1 }...a_{2}a_{1}

where t(a_{i+1}) = h(a_{i}) for i = 1,2,...,d − 1. Then h(p) = h(a_{d}) is the head of path p, and

t(p) = t(a_{1}) is the tail of path p. Also, for every x ∈ Q_{0}, we have a path e_{x }of length 0 with

h(e_{x}) = t(e_{x}) = x.

h(p) = h(b) = 2.

While the order in which the arrows in a path are written may seem strange at first, note

that it is the same as composition of functions. This will be useful later on, when we discuss

representations of quivers — the arrows will be represented by certain functions and paths

indeed become compositions of them.

p = a_{d}a_{d−1 }...a_{1}, q = b_{e}b_{e−1 }...b_{1},

then

pq = a_{d}a_{d−1 }...a_{1}b_{e}b_{e−1 }...b_{1}

is the composition. If t(p) = x, then pe_{x }= p, and if h(p) = y, then e_{y}p = p.

We can associate a category P_{Q }to a quiver Q:

• objects are elements of Q_{0},

• Hom_{P}Q (y, x) = {paths p from x to y}, i.e. h(p) = y, t(p) = x,

• id_{x }= e_{x},

• the composition map Hom_{P}Q (z,y) × Hom_{P}Q (y, x) → Hom_{P}Q (z,x) is given by path

composition, as defined above: (p, q) ↦ pq.

Throughout the rest of this section, we will make the following distinction: for a field K, we

will write K-mod for the category of finite-dimensional K-vector spaces, and K-Mod for

the category of all K-vector spaces.

Page 71

Rep_{K}(Q)=(K-mod)_{P}Q .

Explicitly, the objects in Rep_{K}(Q) are determined by a set of finite-dimensional vector spaces

V (x) for each x ∈ Q_{0}, and K-linear maps

V (a): V (ta) → V (ha)

for each a ∈ Q_{1}. Moreover,

V (e_{x}) = id_{V (x)},

V (a_{d}a_{d−1 }...a_{2}a_{1}) = V (a_{d})V (a_{d−1})...V (a_{2})V (a_{1}).

If V , W are representations, a morphism ϕ: V → W is a collection of linear maps

ϕ(x): V (x) → W(x)

such that

V (t(a))

V (h(a))

W(t(a))

W(h(a))

ϕ(t(a))

V (a)

ϕ(h(a))

W(a)

commutes for all arrows a.

a

b

1

2

3

The paths are e_{1},e_{2},e_{2}, a, b, ba. The representations of Q are triples of finite-dimensional

K-vector spaces V (1), V (2), V (3) together with maps

V (a): V (1) → V (2),

V (b): V (2) → V (3).

• K-vector space with a basis consisting of all paths in Q,

• if p, q are paths, we define

p · q =

{ pq (the composition) if t(p) = h(q),

0

otherwise.

Then KQ is an associative K-algebra with 1 = ∑

x∈Q0

e_{x}.

1

a

The paths are e_{1}, a, a_{2},a_{2},..., and hence KQ = K[a], the polynomial ring in a.

Page 72

a

a

Then KQ = K〈a, b〉 is the free associative algebra generated by a and b (non-commutative).

...

1

2

3

n − 1

n

Then KQ is isomorphic to the algebra of lower-triangular n × n matrices.

Theorem 12.12. The categories KQ-mod (finite-dimensional left KQ-modules) and Rep_{K}(Q)

are equivalent.

Sketch of the proof. If M is a finite-dimensional KQ-module, then

M =

∑

x∈Q0

e_{x}M =

⊕

x∈Q0

e_{x}M,

and we can define V (x) = e_{x}M. Then

e_{x}e_{y }=

{ e_{x}

x = y,

0 otherwise,

and for a ∈ Q_{1}, with t(a) = x, h(a) = y, the multiplication by a map restricts to

V (a): e_{x}M

︸︷︷︸

V (x)

→ e_{y}M

︸︷︷︸

V (y)

.

Then we can define F : KQ-mod → Rep_{K}(Q) by F(M) = V . Conversely, if V is a

representation of Q, let

M =

⊕

x∈Q0

V (x),

and an arrow a ∈ Q_{1 }acts on M by

M

M

V (x)

V (y)

V (a)

This defines a map G : Rep_{K}(Q) → KQ-mod by letting G(V ) = M. Checking the axioms

and that F ◦ G, G ◦ F are naturally isomorphic to the identities, the result follows.

□

Note that dim_{K }KQ < ∞ if and only if there are finitely many paths if and only if there are

no oriented cycles.

Page 73

If M is a finite-dimensional KQ-module, then M = ⊕

x∈Q0

M(x), where M(x) = e_{x}M and the

map

a·: M → M

restricts to

M(a): M(t(a)) → M(h(a)).

Then

KQ =

⊕

x∈Q0

KQe_{x }=

⊕

x∈Q0

P_{x}

as left KQ-modules. Then P_{x }= KQe_{x }is a projective KQ-module and P_{x }has a basis

consisting of all paths starting at x. Note also that P_{x}(y) = e_{y}P_{x }= e_{y}KQe_{x }is spanned by

all paths from x to y.

1

a

the category Rep_{K}(Q) is naturally isomorphic to K[a]-mod.

1

2

3

we have

P_{1 }:

Ke_{1}

Ka

Kba

K

K

K

1

1

P_{2 }:

0

K

K

P_{3 }:

0

0

K

Then note that KQ will be

∗ ∗ ∗

0 ∗ ∗

0 0 ∗

with

P_{1 }=

∗

∗

∗

,P_{2 }=

∗

∗

0

,P_{3 }=

∗

0

0

.

The map a: x → y corresponds to P_{z}(a): P_{z}(x) → P_{z}(y) which maps a path p from z to x

to the path ap.

Moreover,

Hom_{KQ}(P_{x},M) → M(x)

(ϕ: P_{x }→ M) ↦ ϕ(e_{x}) ∈ M(X)

Page 74

is an isomorphism, and Hom_{KQ}(P_{x},−) is an exact functor.

Consider m ⊆ KQ, the (two-sided) ideal generated by all arrow. Then m is spanned by all

paths of length ≥ 1, and, in general, m_{d }is the ideal spanned by all paths of length ≥ d.

An ideal J ⊆ KQ is admissible if m_{d }⊆ J ⊆ m_{2 }for some d. Then A = KQ/J is a

finite-dimensional K-algebra.

equivalent categories.

Theorem 12.16. A finite-dimensional K-algebra is Morita equivalent to KQ/J where J is

an admissible ideal for some quiver Q.

Denote e_{x }+ J by e_{x }and m + J/J by m. Then

A =

⊕

x∈Q0

Ae_{x }=

⊕

x∈Q0

P_{x}

where P_{x }= Ae_{x }is projective. Again,

Hom_{A}(P_{x},M) = M(x) = e_{x}M

and P_{x }is indecomposable (in fact, the only indecomposable ones).

We then see that A-mod (the category of finite-dimensional left A-modules) has enough

projectives.

Note that A_{op}-mod = mod-A is equivalent to A-mod via the map

M ↦ D(M) = M_{∗ }= Hom_{K}(M,K),

f ↦ D(f) = f_{∗}.

Then I_{x }= (e_{x}A)_{∗}, x ∈ Q_{0 }are the indecomposable injectives.

If Q is a quiver, the simple representations are P_{x}/mP_{x }= S_{x }with

S_{x}(y) =

{ K if y = x,

0 if y = x,

and mS_{x }= 0.

We then have an exact sequence

0

⊕

a, t(a)=x

P_{h(a)}

P_{x}

S_{x}

0

P_{h(a)}

p

pa

which gives a projective resolution of the simple module S_{x}. In general, if M is any module,

a projective resolution is

Page 75

0

⊕

a∈Q1

P(h(a)) ⊗_{K }M(t(a))

⊕

x∈Q0

P_{x }⊗_{K }M(x)

M

0

p ⊗ w

pa ⊗ w − p ⊗ aw

p ⊗ v

pv

Then

Ext_{n}

KQ(M,N)=0if n ≥ 2.

1

2

3

again and let A = KQ/(ba). Then

P_{1 }:

K

K

0,

P_{2 }:

0

K

K,

P_{3 }:

0

0

K.

We then have that

0

P_{3}

P_{2}

P_{1}

S_{1}

0

and applying Hom_{A}(−,S_{3}) to P_{•}, we get

0

Hom_{A}(P_{1},S_{3})

︸

︷︷

︸

=0

Hom_{A}(P_{2},S_{3})

︸

︷︷

︸

=0

Hom_{A}(P_{3},S_{3})

︸

︷︷

︸

∼=K

0.

Hence

Ext_{2}

A(S1,S3) = K.

1

a

and consider J = (a_{2}) ⊆ KQ = K[a]. Then

P_{1 }= A = KQ/J = K[a]/(a_{2}).

In this case,

···

P_{1}

P_{1}

P_{1}

S_{1}

0

·a

·a

and Ext_{n}(S_{1},S_{1}) = K for any n ≥ 0.

If we look at A-mod for A = KQ/J, how can we recover the quiver Q?

Page 76

• The simple representations are S_{x}, x ∈ Q_{0}.

• Ext_{1}

A(Sx,Sy) = Kl where l is the number of arrows x → y.

13. Homological dimension

(1) The projective (resp. flat) dimension , pd(A) = n (resp. fd(A)) is the smallest n such

that there is a resolution

0

P_{n}

···

P_{2}

P_{1}

P_{0}

A

0

such that P_{0},...,P_{n }are projective (resp. flat).

(2) The injective dimension, id(A) is the smallest n such that there is an injective

resolution

0

A

E_{0}

E_{1}

···

E_{n}

0.

Lemma 13.2. The following are equivalent

(1) pd(A) ≤ d,

(2) Ext_{n}

R(A, B)=0 for n>d and all right R-modules B,

(3) Ext_{d+1}

R

(A, B)=0 for all B,

(4) if

0

A_{d}

P_{d−1}

···

P_{2}

P_{1}

P_{0}

A

0

is a resolution of A with P_{0},...,P_{d−1 }projective, then A_{d }is projecitve.

Proof. We note that trivially, (4) implies (1) implies (2) implies (3). We show (3) implies (4).

Suppose (3) is true. Let A_{0 }= A and define P_{k }projective and A_{k+1 }recursively so that

0

A_{k+1}

P_{k}

A_{k}

0

is exact. Then the long exact sequence for Ext gives

Ext_{l}(P_{k},B)

︸

︷︷

︸

=0

Ext_{l}(A_{k+1},B)

Ext_{l+1}(A_{k},B)

Ext_{l+1}(P_{k},B)

︸

︷︷

︸

=0

.

Then Ext_{1}(A_{d},B) = ··· = Ext_{d+1}(A_{0},B) = 0 for all B, and hence Ext_{1}(A_{d},B) = 0, so A_{d }is

projective.

□

Dually, we get the following statement.

Lemma 13.3. The following are equivalent:

(1) id(B) ≤ d,

(2) Ext_{n}

R(A, B)=0 for n>d and all A,

(3) Ext_{d+1}

R

(A, B)=0 for all A,

(4) if

Page 77

0

A

E_{0}

E_{1}

···

E_{d−1}

A_{d}

0

is exact and E_{0},...,E_{d−1 }are injective then A_{d }is injective.

Note that

sup{id(B) | B right R-module} = sup{d | Ext_{d}(A, B) = 0 for some right R-modules A, B}

= sup{pd(A) | A right R-module}.

If R is left and right Noetherian, then lgldim(R) = rgldim(R).

Recall that for a path algebra KQ, there is a 2-step resolution of any M:

0 → P_{1 }→ P_{0 }→ M → 0,

and hence the global dimension KQ is at most 1. Moreover, KQ is semisimple if the global

dimension is 0.

We immediately get the following corollary to Baer’s criterion for injectivity 6.3.

Corollary 13.5. We have that

rgldim(R) = sup{pd(R/I) | I right R-ideal}.

We also have a similar construction for Tor (and A ⊗_{R }B). The following numbers are the

same:

sup{fd(A) | A right R-module}

= sup{d | Tor_{R}

d (A, B) = 0 for some A ∈ mod-R, B ∈ R-mod}

= sup{fd(B) | B left R-module}

= sup{fd(R/J) | J right ideal}

= sup{fd(R/J) | J left ideal}

Proposition 13.7. Assume that R is right Noetherian. Then:

(1) for every finitely-generated right R-module A, pd(A) = fd(A),

(2) tordim(R) = rgldim(R).

Proof. We first prove (1). Note that any finitely generated projective module is flat, so

fd(A) ≤ pd(A). If fd(A) = d < ∞, take a resolution of A

0

A_{d}

P_{d−1}

···

P_{1}

P_{0}

A

0

with P_{0},...,P_{d−1 }finitely generated and free. Then A_{d }is finitely generated (by a lemma

finitely presented and flat, which shows that it is projective. Hence pd(A) ≤ d.

Then (2) immediately follows:

rgldim(R) = sup{pd(R/J) | J right ideal}

= sup{fd(R/J) | J right ideal}

= tordim(R),

Page 78

completing the proof.

□

Global dimension 0.

summand of R.

Theorem 13.9 (Wedderburn’s Theorem). If R is semi-simple, then

R

≅

r

∏

i=1

Mat_{n}i_{,n}i (D_{i}),

for division rings D_{i}.

Theorem 13.10. The following are equivalent:

(1) R is semi-simple,

(2) R has right (left) global dimension 0,

(3) every R-module is projective,

(4) every R-module is injective,

(5) all exact sequences split.

Proof. The proof is clear.

□

KQ-module A, we have a projective resolution

0

P_{1}

P_{0}

A

0

and hence lgldim(KQ) = rgldim(KQ) ≤ 1.

If R = KQ/J (a finite-dimensional K-algebra) with m_{d }⊆ J ⊆ m_{2 }(so J is admissible). If

J = 0, gldimKQ ≤ 1 and in fact

gldimKQ =

{ 0 if Q has no arrows,

1 if Q has arrows.

If J = 0, then in fact gldimKQ/J ≥ 2.

13.1. Von Neumann regular rings.

such that aba = a.

For a: X → k, define

b(x) =

{ _{1}

a(x)

if a(x) = 0,

0

if a(x)=0,

whence aba = a_{2}b = a.

and we can choose a splitting B: imA → k_{n}. Extend B to k_{n }→ k_{n }to get ABA = A.

Page 79

Suppose R is von Neumann regular. If a ∈ R, then there exists b ∈ R such that aba = a.

Then e = ab is an idempotent, e_{2 }= ababa = ab = e. We then have that

aR ⊇ abR = eR ⊇ abaR = aR,

so aR = eR.

Lemma 13.15. A finitely generated right (or left) ideal is generated by one idempotent.

Proof. Suppose R is commutative. If e, f are both idempotent,

(e + f − ef)=(e, f),

since e(e+f−ef) = e_{2}+ef−e_{2}f = e+ef−ef = e and similarly for f. The non-commutative

case is similar.

□

R = {f | f almost constant},

so for f ∈ R, there exists c such that f(x) = c for all but finitely many x. Then R is von

Neumann regular. However,

I = {f ∈ R | f(x) = 0 for x ∈ R \ Z}

is not finitely generated; indeed, we need elements with f(a) = 0 for arbitrarily large a ∈ Z,

but for finitely many f ∈ I will have f(a) = 0 for a large enough.

If I is a finitely generated right ideal then for an idempotent e

I = eR

and hence

R = eR ⊕ (1 − e)R

≅ I ⊕ R/I,

so R/I is projective, and hence flat.

If I is not finitely generated, then

I = lim

−→

I_{α }for I_{α }finitely generated ideal

and

R/I = lim

−→

R/I_{α},

For all left R-modules M, we have

Tor_{1}(R/I,M) = lim

−→

Tor_{1}(R/I_{α},M)=0

and hence R/I is flat. Therefore, fd(M) = 0, and hence

tordim(R)=0.

Page 80

13.2. Global dimension of polynomial rings. We will show that for a field k,

(1)

gldim(k[x_{1},...,x_{n}]) = n.

Hilbert showed that if M is finitely generated, then it has a free resolution of length n

0

F_{n}

···

F_{2}

F_{1}

F_{0}

M

0

showing that the global dimension is at most n.

Writing R = k[x_{1},...,x_{n}], the R-module k has the Koszul resolution

0

R(

n

n)

···

R(

n

2)

R_{n}

R

k

0

of length n. Taking Hom(−,k) of this sequence of this resolution, we get

0

k(

n

n)

···

k(

n

2)

k(

n

1)

k(

n

0)

0

0

0

0

0

0

showing that

Tor_{R}

j (k, k) = k(

n

j ).

Proposition 13.17. If f : R → S is a ring homomorphism and M is an S-module, then

pd_{R}(M) ≤ pd_{S}(M) + pd_{R}(S).

Proof. Let pd_{S}(M) = n, pd_{R}(S) = d and choose a projective S-resolution of M,

0

Q_{n}

···

Q_{1}

Q_{0}

M

0

and let M_{0 }= M and

0 −→ M_{i+1 }−→ Q_{i }−→ M_{i }−→ 0.

Choose projective R-resolutions of M_{i}’s. Then the Horseshoe Lemma 5.24 gives projective

resolutions P_{•j }→ Q_{j}. We then have a double complex (by adjusting the signs of the maps

appropriately)

Page 81

...

...

...

0

Q_{2}

P_{02}

P_{12}

···

0

Q_{1}

P_{01}

P_{11}

···

0

Q_{0}

P_{01}

P_{10}

···

M

0

with exact rows. The total complex gives a map Tot(P_{••}) → M but this projective resolution

could be large, even infinite. However, note that pd_{R}(Q_{1}) ≤ pd_{R}(S) = d because Q_{i }is a

direct summand of a free S-module. We replace P_{d,i }by P_{d,i}/imP_{d+1,i }to get

Tot(P_{••}) → M,

a projective R-resolution of M. Then we obtain

pd_{R}(M) ≤ n + d = pd_{S}(M) + pd_{R}(S),

as required.

□

Lemma 13.18. Suppose

0

A

B

C

0

is an exact sequence of R-modules. Then

pd_{R}(B) ≤ max{pd_{R}(A),pd_{R}(C)}

and if the inequality is strict, then pd_{R}(C) = pd_{R}(A)+1.

Proof. By the long exact sequence for Ext, we get

···

Ext_{i}(C, M)

Ext_{i}(B,M)

Ext_{i}(A, M)

Ext_{i+1}(C, M)

Ext_{i+1}(B,M)

Ext_{i+1}(A, M)

···

If i = pd_{R}(B), then for some R-module M, we obtain Ext_{i}(B,M) = 0, and hence one of

the neighboring terms in the long exact sequence above are non-zero, so Ext_{i}(C, M) = 0 or

Ext_{i}(A, M) = 0, showing that pd_{R}(C) ≥ i or pd_{R}(A) ≥ i.

Page 82

If the inequality is strict, then for any i > pd_{R}(B), we get Ext_{i}(B,M) = Ext_{i+1}(B,M) = 0,

so Ext_{i}(A, M)

≅ Ext

i+1(C, M) by the long exact sequence above, which shows that pdR(C) =

pd_{R}(A) + 1.

□

Let R be a ring, x ∈ R be central, A a left R-module.

all y ∈ A.

Suppose x is a nonzero divisor on R. We have a short exact sequence

0

R

R

R/x

0.

x·

We apply − ⊗_{R }A to get

Tor_{1}(R, A)

︸ ︷︷ ︸

=0

Tor_{1}(R/x, A)

︸

︷︷

︸

{y∈A | xy=0}

A

A

A/xA

0

x·

and x is a nonzero divisor on A if and only if Tor_{1}(R/x, A) = 0.

Let (R,m) be a commutative Noetherian local ring with m its unique maximal ideal.

x_{1},...,x_{n }∈ m such that x_{i }is a nonzero divisor on A/(x_{1},...,x_{i−1})A.

The depth of A, depth(A) is the largest n such that there exists a regular sequence of length

n on A.

Theorem 13.21 (Auslander–Buchsbaum). If R is a commutative Noetherian local ring and

A is a finitely generated R-module with pd(A) < ∞, then

depth(R) = depth(A) + pd(A).

a chain of prime ideals

p_{0 }⊂ p_{1 }⊂···⊂ p_{n }⊂ R.

If k = R/m, a field, we get

dim_{k}(m/m_{2}) ≥ dimR.

We also have that

depth(R) ≤ dimR.

There are various relationships between regular rings and Cohen–Maccaulay rings, even

though they are not equivalent.

happens when we go from an R-module A to the R/x-module A/xA.

Page 83

Theorem 13.25 (First Change of Rings Theorem). Let R be a ring, x ∈ R be central,

nonzero divisor, A is an R/x-module with pd_{R/x}(A) finite. Then pd_{R}(A)=1+pd_{R/x}(A).

Sketch of proof. If pd_{R/x}(A) = 0, A is a projective R/x-module, and then A is not a

projective R-module, because xA = 0. Then

1 ≤ pd_{R}(A) ≤ pd_{R}(A/x)=1,

since 0 → R → R → R/x → 0 is a projective resolution of R/x.

The general argument now goes by induction of pd_{R/x}(A). Assume pd_{R/x}(A) ≥ 1. Take P

projective R/x-module with an exact sequence

0

M

P

A

0.

Since 1 + pd_{R/x}(M) = pd_{R/x}(A), we can apply the inductive hypothesis to get

pd_{R}(M)=1+pd_{R/x}(M).

By Lemma 13.18, we get that

pd_{R}(P) ≤ max{pd(M),pd_{R}(A)}

and either equality holds or pd_{R}(A) = pd_{R}(M) + 1. In the first case, we get a contradiction.

In the second case,

pd_{R}(A) = pd_{R}(M)+1=pd_{R/x}(A)+1,

as required.

□

Theorem 13.26 (Second Change of Rings Theorem). Let x ∈ R be a central nonzero divisor

on R and on A. Then

pd_{R}(A) ≥ pd_{R/x}(A/xA).

Corollary 13.27. If A is an R-module and we write A[x] = R[x] ⊗_{R }A, we get that

pd_{R[x]}(A[x]) = pd_{R}(A).

Proof. The ≥ inequality follows from Second Change of Rings Theorem 13.26. The ≤ is

immediate, since a projective resolution P_{• }→ A of A gives a projective resolution

P_{•}[x] → A[x]

of A[x].

□

Theorem 13.28. We have that gldimR[x] = gldimR + 1.

Proof. If M is an R[x]-module, then M is an R-module, and we will write ˜

M for M as an

R-module. We have the following exact sequence

0

R[x] ⊗_{R }˜

M

R[x] ⊗_{R }˜

M

R[x] ⊗_{R[x] }M

︸

︷︷

︸

=M

0

p ⊗ v

px ⊗ v − p ⊗ xv

Page 84

By a similar result to Lemma 13.18: if we have a short exact sequence 0 → A → B → C → 0,

we get

pd(C) ≤ max{pd(B),pd(A)+1}.

Hence

pd_{R[x]}(M) ≤ pd_{R[x]}(R[x] ⊗_{R }˜

M)+1 ≤ pd_{R}(˜

M)+1=pd_{R}(M)+1 ≤ gldimR + 1.

Taking the supremum over all M, we get one inequality. We skip the proof of the other

inequality.

□

Corollary 13.29. For a field k, we have that

gldimk[x_{1},...,x_{n}] = n.

Let R be a ring and R_{∗ }be the group of unites of R. We define the Jacobson radical as

J(R) = {r ∈ R | for any s ∈ R, 1 − rs ∈ R_{∗}}

and one can prove that

J(R) = {r ∈ R | for any s ∈ R, 1 − sr ∈ R_{∗}}

J(R) =

⋂

m

m

where the intersection can be over left maximal ideals m or over right maximal ideals m.

J(R) = m.

If (R,m) is a local commutative ring, clearly J(R) = m.

Proposition 13.31 (Nakayama Lemma). Let m be the Jacobson radical of R. If B is a

finitely generated left R-module and mB = B then B = 0.

Proof. Suppose B = 0, {b_{1},...,b_{n}} minimal set of generators over B. Then b_{n }∈ B = mB,

and we can write

b_{n }=

n

∑

i=1

r_{i}b_{i}

for r_{i }∈ m.

Then

(1 − r_{n})b =

n−1

∑

i=1

r_{i}b_{i }∈ Rb_{1 }+ ··· + Rb_{n−1}

but r_{n }∈ m so 1 − r_{n }∈ R_{∗}, and so

b_{n }∈ Rb_{1 }+ ··· + Rb_{n−1},

and hence b_{1},...,b_{n−1 }generate B. This contradicts minimality of the set of generators. □

In what follows, assume (R,m) is a local Noetherian, commutative ring and k = R/m.

Corollary 13.32. Let B be a left finitely generated R-module. Elements b_{1},...,b_{n }∈ B

generate B if and only if the images b_{1},...,b_{n }in B/mB span B/mB as a k-vector space.

Page 85

Note that the finitely-generated assumption is necessary: Q_{2 }is a Z_{2}-module with

mQ_{2 }= (2)Q_{2 }= Q_{2},

so Q_{2}/mQ_{2 }= 1 which is spanned by the only element as a k-vector space, but Q_{2 }over Z_{2 }is

not finitely generated.

Proof. For the ‘only if’ direction, if A = Rb_{1 }+ ··· + Rb_{n }and B = A + mB. Then m · B/A =

B/A, so B/A = 0 and hence A = B.

□

Corollary 13.33. Elements b_{1},...,b_{n }∈ B are a minimal set of generators if and only if

the images of b_{1},...,b_{n }in B/mB form a basis of B/mB as a k-vector space.

Proposition 13.34. If P is a finitely generated projective R-module, then P is free.

Proof. Let n = dim_{k}(P/mP). By lifting the generators of P/mP as a k-vector space, we get

a minimal set of generators for P, giving a short exact sequence

0

K

R_{n}

P

0

where K is the kernel of the map R_{n }→ P. Since P is projective, this sequence splits, so

R_{n }∼

= P ⊕ K.

Taking − ⊗_{R }R/m, we get

k_{n }∼

= P/mP ⊕ K/mK

≅ k

n ⊕ K/mK

= P.

□

If A is a finitely generated R-module. Let A_{0 }= A and, recursively, having defined A_{i}, let

β_{i }= dim_{k }A_{i}/mA_{i }< ∞

and define A_{i+1 }as the following kernel

0

A_{i+1}

R_{β}i

A_{i}

0.

This gives a free resolution of A:

···

R_{β}2

R_{β}1

R_{β}0

A

0.

Apply − ⊗_{R }R/m to

···

R_{β}2

R_{β}1

R_{β}0

0

to get

···

k_{β}2

k_{β}1

k_{β}0

0

0

0

(checking that the maps are indeed 0 is an exercise). This shows that

Tor_{R}

i (A, R/m) = kβi .

Page 86

Lemma 13.36. If depth(R)=0 and A is a finitely generated R-module, then pd(A)=0 or

pd(A) = ∞.

Proof. Suppose 0 < pd(A) = n < ∞. Take a resolution

0

B

F_{n−2}

···

F_{2}

F_{1}

F_{0}

A

0

with F_{0},...,F_{n−2 }free. Then

pd(B) = n − (n − 1) = 1.

Let t = dim_{k }B/mB and consider

0

P

R_{t}

B

0.

Then P is projective (since pd(P) = pd(B)−1 = 0), so it finitely generated, and hence free.

s ∈ m,

{r ∈ R | rs = 0} = m.

Now, P ⊆ mR_{t}, and hence sP ⊆ smR_{t }= 0, but P is free, so P = 0. Since P = 0, this shows

that pd(B) = 0, a contradiction.

□

The Auslander–Buchsbaum Theorem 13.21 follows from similar arguments to this lemma

If depth(R) = 0, then pd(A) = 0, so A is projective, and hence free, so depth(A) =

depth(R) = 0.

Theorem 13.37. A ring R is regular if and only if gldimR < ∞, and in that case

gldimR = dimR = pd_{R}(R/m).

Note that R = k[x]/(x_{2}) has infinite global dimension and it is not regular, and in this case

the Krull dimension is 0.

Theorem 13.38. A regular local ring is Cohen-Macaulay.

Proof. In general, depth(R) ≤ dimR. If R is regular, let

m = (x_{1},...,x_{n}),

where n = dim_{R }m/m_{2 }and x_{1},...,x_{n }is a regular sequence, so

depth(R) ≥ n = dimR,

and hence depth(R) = dim(R).

□

Page 87

13.3. Koszul resolution. Let R be regular and pd_{R}(R/m) = n. The minimal free

R-resolution of R/m is the Koszul resolution.

Let x ∈ m be a nonzero divisor and consider

K(x) :

0

R

R

0

x·

a resolution of R/(x). Suppose x_{1}, x_{2 }is a regular sequence, and consider K(x_{1}) ⊗_{R }K(x_{2}):

R

R

R

R

x2

x1

−x2

x1

and the total complex gives a resolution of R/(x_{1},x_{2}):

0

R

R ⊕ R

R

0

(−x2

x1

)

(x1,x2)

In general, if x = (x_{1},...,x_{n}) is a regular sequence, we let K(x) = K(x_{1},...,x_{n}) be the

total complex of K(x_{1}) ⊗_{R }K(x_{2}) ⊗_{R }···⊗_{R }K(x_{n}). Explicitly, we can write it as

0

R(

n

n)

···

R(

n

2)

R(

n

1)

R(

n

0)

0

∂

∂

∂

∂

where we identify

R(

n

1)= ⊕Re_{i},

R(

n

2)= ⊕

i<j

R(e_{i }∧ e_{j}),

...

R(

n

n)= R(e_{1 }∧···∧ e_{n}),

and ∂ = ∑x_{i}

∂

∂ei

, so

e_{i}1 ∧···∧ e_{i}k ↦

∑

j

(−1)_{j−1}x_{j}e_{i}1 ∧···∧ ̂e_{i}j ∧···∧ e_{i}k .

Theorem 13.39. The resolution K(x) is a free resolution of R/(x_{1},...,x_{n}) and

H_{q}(K(x)) =

{ 0

if q > 0,

R/(x_{1},...,x_{n}) if q = 0.

We first prove the theorem in the n = 1 case.

Proposition 13.40. If x ∈ m is a nonzero divisor, C_{• }is a chain complex of R-module, then

we have an exact sequence

0

H_{0}(K(x) ⊗ H_{q}(C_{•}))

H_{q}(K(x) ⊗_{R }C_{•})

H_{1}(K(x) ⊗_{R }H_{q−1}(C_{•}))

0.

Proof. We have an exact sequence of complexes

Page 88

0

R

···

0

0

R

0

···

K(x)

···

0

R

R

0

···

R[−1]

···

0

R

0

0

···

0

x

Taking − ⊗_{R }C_{•}, we get

0

C_{•}

K(x) ⊗_{R }C_{•}

C_{•}[−1]

0

and the long exact sequence of this complex gives the desired result.

□

Proposition 13.40 with

C_{• }= K(x_{1},...,x_{n−1})

x = x_{n}

to get the exact sequence

0

H_{0}(K(x_{n}) ⊗ H_{q}(K(x_{1},...,x_{n−1})))

H_{q}(K(x))

H_{1}(K(x_{n}) ⊗_{R }H_{q−1}(K(x_{1},...,x_{n−1})))

0.

For q ≥ 2, both the kernel and the cokernel in this exact sequence are 0, so H_{q}(K(x)) = 0.

For q = 1, the kernel is 0 and the cokernel is

H_{1}(K(x_{n}) ⊗ R/(x_{1},...,x_{n−1})) = ker(R/(x_{1},...,x_{n−1})

·xn

−→ R/(x_{1},...,x_{n−1})) = 0,

and hence H_{1}(K(x)) = 0.

For q_{0}, we have

0

R/(x_{n}) ⊗_{R }R/(x_{1},...,x_{n−1})

︸

︷︷

︸

∼=R/(x1,...,xn)

H_{0}(K(x))

0

∼=

which completes the proof.

□

Page 89

Tensoring the resolution

0

R(

n

n)

···

R(

n

2)

R(

n

1)

R(

n

0)

0

∂

∂

∂

∂

with − ⊗_{R }R/m, we get

0

k(

n

n)

···

k(

n

2)

k(

n

1)

k(

n

0)

0

0

0

0

0

and hence

Tor_{i}(R/(x_{1},...,x_{n}),k) = k(

n

k).

If m = (x_{1},...,x_{n}), this gives

Tor_{i}(k, k) = k(

n

i ).

The resolution

0

R(

n

n)

···

R(

n

2)

R(

n

1)

m

0

∂

∂

∂

is called the Koszul resolution of m.

R-module. We say x_{1},...,x_{n }∈ m is a maximal A-sequence if

A/(x_{1},...,x_{n})A

has no nonzero divisors in m.

Proposition 13.42. All maximal A-sequences have same length, and this length is equal to

depth(A).

□

14. Local cohomology

Let R be a commutative ring, I ⊆ R be an ideal. We define a functor

F_{I }: R−mod → R−mod

by

F_{I}(A) = {a ∈ A | there exists d such that I_{d}a = 0}.

and if f : A → B is an R-module homomorphism, the square

F_{I}(A)

F_{I}(B)

A

B

FI (f)

f

commutes. Then F_{I }is left exact and

H_{•}

I (A) = R•FI(A)

is called the local cohomology. If I_{d }⊆ J and J_{e }⊆ I, then

F_{I }= F_{J}

Page 90

so

H_{•}

I = H•

J .

Moreover, we have that

F_{I}(A) = H_{0}

I (A) = {a ∈ A | there exists d such that Ida = 0} = lim

−→

Hom_{R}(R/I_{d},A).

In general,

H_{n}

I (A) = lim

−→

Ext_{n}

R(R/Id,A).

15. Spectral sequences

In this chapter, we work in the category of R-modules.

• objects R_{r}

pq for p, q ∈ Z, r ≥ a,

• differentials d_{r}

pq : Er

pq → Er

p−r,q+r−1 satisfying dr ◦ dr = 0,

• R_{r+1}

pq

=

ker dr

pq

im(dr

p+r,q−r+1_{)}

.

On a diagram, we can represent d_{0}, d_{1}, d_{2 }as follows

d_{0}

•

E_{0}

p,q+1

•

•

E_{0}

pq

E_{0}

p+1

•

•

•

d_{1}

•

E_{1}

p,q+1

•

•

E_{1}

pq

E_{1}

p+1

•

•

•

d_{2}

•

E_{2}

p,q+1

•

•

E_{2}

pq

E_{2}

p+1

•

•

•

Page 91

and similarly for d_{r }for r ≥ 3. We can think of these pictures as “pages” or “sheets”: for each

r, there is a “page” with arrows d_{r }as follows

•

•

•

•

•

•

d0

d1

d2

d4

We let E_{r }= ⊕

p,q∈Z

E_{r}

pq. Then Er+1 is a subquotient of Er.

Suppose that for r = a, we have B_{a }= 0, Z_{a }= E_{a}.

Let Z_{r+1 }⊇ B_{r }such that kerd_{r }= _{Z}r+1

Br

and B_{r+1 }⊇ B_{r }such that imd_{r }= _{B}r+1

Br

, and then

B_{r+1 }⊆ Z_{r+1}. We then have

B_{a }⊆ B_{a+1 }⊆ B_{a+2 }⊆···⊆ B_{∞ }⊆ Z_{∞ }⊆···⊆ Z_{a−a }⊆ Z_{a−1 }⊆ Z_{a }= R_{a}

where

B_{∞ }=

⋃

i

B_{i}, Z_{∞ }=

⋂

i

Z_{i}, E_{∞ }= Z_{∞}/B_{∞}.

that E_{a}

p,n−p = 0.

filtration

0 = F_{s}H_{n }···⊆ F_{p}H_{n }⊆ F_{p+1}H_{n }⊆···F_{t}H_{n }= H_{n}

such that

E_{∞}

pq = FpHp+q/Fp−1Hp+q.

In that case, we write E_{1}

pq ⇒ Hp+q.

Note that E_{1}

pq converges to the (p+q)th homology group, which can be represented as follows:

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

H_{n}

p + q = n

p

q

Page 92

15.1. Homology spectral sequences. We construct a spectral sequence from a filtration

of a chain complex. Let C_{• }be a chain complex with a filtration

···⊆ F_{p}C_{• }⊆ F_{p+1}C_{• }⊆···

and assume

⋃

p

F_{p}C_{• }= C_{•}.

We construct a spectral sequence with

E_{0}

pq =

F_{p}C_{p+q}

F_{p−1}C_{p+q}

and

E_{1}

pq = Hp+q(E0

p•).

Note that E_{0}

pq in E0

p• has degree p + q, and, to compute the homology, we note that the

boundary maps to compute the homology are induced by the boundary maps in C_{•}:

E_{0}

pq =

F_{p}C_{p+q}

F_{p−1}C_{p+q}

→

F_{p}C_{p+q−1}

F_{p−1}C_{p+q−1}

= E_{0}

pq−1.

In general, E_{r+1}

pq

will be the homology of E_{r}

pq.

In what follows, we drop the “q” from the notation and simply write

E_{0}

p =

F_{p}C

F_{p−1}C

and so on. We let

η_{p }: F_{p}C →

F_{p}C

F_{p−q}C

= E_{0}

p

be the projection and set

A_{r}

p = {c ∈ FpC | d(c) ∈ Fp−rC},

Z_{r}

p = ηp(At

p) =

A_{r}

p + Fp−1C

F_{p−1}C

∈ E_{0}

p,

B_{r+1}

p−r = ηp−r(d(Ar

p)) =

d(A_{r}

p) + Fp−r−1C

F_{p−r−1}C

∈ E_{0}

p−r,

B_{r}

p = ηp(d(Ar−1

p+r−1)) =

d(A_{r−1}

p+r−1) + Fp−1C

F_{p−1}(C)

⊆ Z_{r}

p.

This will simplify calculations, for example:

Z_{r}

p =

A_{r}

p + Fp−1C

F_{p−1}C

≅

A_{r}

p

A_{r}

p ∩ Fp−1C

=

A_{r}

p

A_{r−1}

p−1

.

In what follows, we will use that, for B ⊆ A, we have A ∩ (B + C) = B + (A ∩ C). We set

E_{r}

p =

Z_{r}

p

B_{r}

p

=

A_{r}

p + Fp−1C

d(A_{r−1}

p+r−1) + Fp−1C

=

A_{r}

p

A_{r}

p ∩ (d(Ar−1

p+r−1) + Fp−1C)

=

A_{r}

p

d(A_{r−1}

p+r−1) + Ar−1

p−1

and since

E_{r}

p =

A_{r}

p

dA_{r−1}

p+r−1 + Ar−1

p−1

,

Page 93

E_{r}

p−r =

A_{r}

p−r

dA_{r−1}

p−1 + Ar−1

p−r−1

,

the boundary map d induces

d_{r}

p : Er

p → Er

p−r.

We claim that

E_{r+1}

pq

=

kerd_{r}

pq

imd_{r}

p+rq−r+1

.

For that sake, we will compute the kernel of the map d_{r}

p. If

a + dA_{r−1}

p+r−1 + Ar−1

p−1 ∈ kerdr

p,

then without loss of generality, assume that

d(a) ∈ A_{r−1}

p−r−1

(otherwise, we could choose a different representative a that would satisfy this). Then

a ∈ A_{r+1}

p

. This shows that

kerd_{r}

p =

A_{r+1}

p

+ A_{r−1}

p

d(A_{r−1}

p+r−1

≅

A_{r+1}

p

+ F_{p−1}C

dA_{r−1}

p+r−1 + Fp−1C

≅

Z_{r+1}

p

B_{r}

p

.

Then

d_{r}

p : Er

p =

Z_{r}

p

B_{r}

p

↠

Z_{r}

p

Z_{r+1}

p

≅

B_{r+1}

p−r

B_{r}

p−r

↪

Z_{r}

p−r

B_{r}

p−r

= E_{r}

p−r

and so

imd_{r}

p =

B_{r+1}

p−r

B_{r}

p−r

and shifting the index

imd_{r}

p+r =

B_{r+1}

p

B_{r}

p

⊆

Z_{r+1}

p

B_{r}

p

= kerd_{r}

p.

Hence

kerd_{r}

p

imd_{r}

p+r

=

Z_{r+1}

p

B_{r+1}

p

= E_{r+1}

p

.

This shows that E_{r+1}

p

is the homology of E_{r}

p.

One can also show that

Zr

p

Zr+1

p

≅

Br+1

p−r

Br

p−r

with a similar calculation, but we omit it here.

Assume that a filtration is bounded, i.e. for every n there exist s, t such that

0 = F_{s}C_{n }⊆ F_{s+1}C_{n }⊆···⊆ F_{t}C_{n }= C_{n}.

Then E_{0}

pq is bounded and for any n, there are only finitely many p such that E0

p,n−p = 0.

Theorem 15.4 (Spectral Convergence). There exists a filtration on H_{n}(C_{•}),

···⊆ F_{p}H_{n}(C_{•}) ⊆ F_{p+1}H_{n}(C_{•}) ⊆···

such that

F_{p}H_{p+q}(C_{•})/F_{p−1}H_{p+q}(C_{•})

≅ E

∞

pq .

Concisely, we write

E_{1}

pq = Hp+q(FpC•/Fp−1C•) −→ Hp+q(C•).

Page 94

Here, E_{1}

pq is homology of the associated graded, E∞

pq is the associated graded of homology.

Proof. Recall that we had

0 ⊆ B_{0}

p ⊆ B1

p ⊆···⊆ B∞

p ⊆ Z∞

p ⊆···⊆ Z1

p ⊆ Z0

p = FpC,

where B_{∞}

p = ⋃

i

B_{i}

p, Z∞

p = ⋂Zi

p, and

E_{r}

p = Er

p• = Zr

p/Br

p,

E_{∞}

p = Z∞

p /B∞

p .

Suppose p + q = n. Then

A_{r}

pq = {c ∈ FpCn | d(c) ∈ Fp−rCn}.

For r ≥ r_{0}(n, p) (when F_{p−r}C_{n }= 0, i.e. p − r ≤ s), so

A_{r}

pq = kerd ∩ FpCn = A∞

pq.

Then

Z_{r}

pq = ηp(Ar

pq) = ηp(A∞

pq) = Z∞

pq =

kerd ∩ F_{p}C_{n }+ F_{p−1}C_{n}

F_{p−1}C_{n}

.

Note that in general, f

(

⋂

i

A_{i}

)

= ⋂

i

f(A_{i}), but because of boundedness from below these

are finite intersections, so equality does hold.

Moreover,

B_{r+1}

pq

= η_{p}(A_{r}

p+r,q−r) =

d(A_{r}

p+r,q−r) + Fp−1Cn

F_{p−1}C_{n}

,

B_{∞}

pq =

d(⋃A_{r}

p+r,q−r + Fp−1(Cn))

F_{p−1}C_{n}

.

Define:

F_{p}H_{n}(C) =

kerd_{n }∩ f_{p}C_{n}

imd_{n+1 }∩ f_{p}C_{n}

=

A_{∞}

pq

d

(

⋃

r

A_{r}

pr,p−r+1

).

Then we have that

F_{p}H_{n}(C))

F_{p−1}H_{n}(C)

=

A_{∞}

pq

d(⋃

r

A_{r}

p+q,q−r+1) + A∞

p−1,q−1

.

Applying η_{p }to the right hand side of the above, we get

η_{p}(A_{∞}

pq)

η_{p}(d(⋃

r

A_{r}

p+q,q−r+1) + A∞

p−1,q−1

) ∼=

η_{p}(A_{∞}

pq)

η_{p}

(⋃

r

A_{r}

p+q,q−r+1

) =

Z_{∞}

pq

B_{∞}

pq

.

We claim that η_{p }actually gives an isomorphism above. Indeed, consider

η_{p }: A_{∞}

pq →

η_{p}(A_{∞}

pq

η_{p}(d(⋃

r

A_{r}

p+q,q−r+1

)).

Suppose a ∈ A_{∞}

pq, so

a + f_{p−1}C ∈ d

(⋃

A_{r}

p+r,q−r+1

)

+ F_{p−1}C

Page 95

and we can write a = b + c for

b ∈ d

(⋃

A_{r}

p+r,q−r+1

)

c ∈ F_{p−1}C ∩ A_{∞}

pq = A∞

p+q−1

which completes the proof.

□

0

A_{•}

B_{•}

C_{•}

0

is an exact sequence of complexes. We will recover the long exact sequence in homology

using the Convergence Theorem 15.4. Consider the following filtration on B:

0 = F_{−1}B ⊆ F_{0}B

︸︷︷︸

=A

⊆ F_{1}B = B.

Then

E_{0}

0q =

F_{0}B_{q}

F_{−1}B_{q}

= A_{q}, E_{0}

1q =

F_{1}B_{q+1}

F_{0}B_{q}

=

B_{q+1}

A_{q+1}

= C_{q+1},

so E_{0}

pq can be represented as

...

...

...

...

0

A_{2}

C_{3}

0

0

A_{1}

C_{2}

0

0

A_{0}

C_{1}

0

0

0

C_{0}

0

0

0

0

0

d0

d0

d0

d0

d0

Hence E_{1}

0q = Hq(A), E1

1q = Hq+1(C) and in general E1

pq can be represented as

...

...

...

...

0

H_{1}(A)

H_{2}(C)

0

0

H_{0}(A)

H_{1}(C)

0

0

0

H_{0}(C)

0

d1

d1

Page 96

Moreover, by definition of E_{2}

1q, we get the following exact sequence

0

E_{2}

1q

H_{q+1}(C)

H_{q}(A)

E_{2}

0q

0.

Finally, looking at the diagram for E_{0}

pq, we see that the maps dr : Er

pq → Er

p−r,q+r−1 for r ≥ 2

are all 0, and hence

E_{2}

pq = E∞

pq .

F_{0}H_{q}(B) =

F_{0}H_{q}(B)

F_{−1}H_{q}(B)

= E_{∞}

0p,

H_{q+1}(B)

F_{0}H_{q+1}(B)

=

F_{1}H_{q+1}(B)

F_{0}H_{q+1}(B)

= E_{∞}

1q ,

since F_{1}H_{q+1}(B) = H_{q+1}(B) and F_{−1}H_{q}(B) = 0. We then have

0

F_{0}H_{1}(B)

︸ ︷︷ ︸

=E∞

0

H_{q}(B)

H_{q}(B)

F_{0}H_{q}(B)

︸ ︷︷ ︸

=H∞

q,q−1

0.

We then obtain

···

H_{q}(A)

H_{q}(B)

H_{q}(C)

H_{q−1}(A)

···

E_{∞}

0q

E_{∞}

1,q−1

which recovers the long exact sequence of homology.

15.2. Cohomology spectral sequences. One can dualize all the results in the previous

section to cohomology.

The objects are E_{pq}

r , r ≥ a, and the maps are

d_{pq}

r : Epq

r

→ E_{p+r,q−r+1}

r

and

E

pq

r+1 =

kerd_{pq}

r

imd

p−r,q+r−1

r

.

Similarly to the Convergence Theorem 15.4, one can prove that if the spectral sequence is

bounded, then

E_{pq}

r

−→ H_{p+q},

i.e. there exists filtration F_{t}H_{n }⊆ F_{t−1}H_{n }⊆··· with quotients E_{pq}

∞.

Page 97

15.3. Spectral sequences in topology. We show an example of a spectral sequence in

topology and an application of that spectral sequence.

Y if given a homotopy G: Y × [0,1] → B and a mapping g: Y × {0} → E with p ◦ g(y,0) =

G(y,0), then there exists a homotopy ˜G: Y × [0,1] → E with ˜G(y,0) = g and p ◦ ˜G = G,

i.e. the two triangles

Y × {0}

E

Y × [0,1]

B

g

p

G

˜G

commute.

Serre fibration if p has HLP for all n-cells.

Proposition 15.8. Suppose p is a fibration and if B is path-connected then all fibers p_{−1}(b),

b ∈ B are homotopy equivalent.

In particular, p is surjective. Moreover, H_{•}(p_{−1}(b)) does not depend on b.

Theorem 15.9 (Leray Spectral Sequence). Suppose π: E → B is a fibration with F =

π_{−1}(b) for some b ∈ B, with B simply connected. Let M be an abelian group. There is a

spectral sequence

E_{2}

pq = Hp(B;Hq(F;M) −→ Hp+q(E;M).

Corollary 15.10. Suppose π: E → S_{n }is a fibration and F is a fiber, with n ≥ 2. Then

there exists a long exact sequence

···

H_{q}(F)

H_{q}(E)

H_{q−n}(F)

H_{q−1}(F)

H_{q−1}(E)

···

Proof. We have

E_{2}

pq = Hp(Sn;Hq(F)) =

{ H_{q}(F) if p = 0,n

0

otherwise

which gives the following diagram of E_{2}

pq

Page 98

...

...

...

...

...

H_{2}(F)

0

···

0

H_{2}(F)

0

···

H_{1}(F)

0

···

0

H_{1}(F)

0

···

H_{0}(F)

0

···

0

H_{0}(F)

0

···

p = 0

p = 1

···

p = n − 1

p = n

p = n + 1

and if n > 2 then d_{2 }= 0 and E_{3}

pq = E2

pq, and in general

E_{2}

pq = E3

pq = ··· = En

pq.

For E_{n}

pq, the map dn is non-trivial:

H_{n−1}(F)

0

...

...

...

...

...

H_{2}(F)

0

···

0

H_{2}(F)

0

···

H_{1}(F)

0

···

0

H_{1}(F)

0

···

H_{0}(F)

0

···

0

H_{0}(F)

0

···

p = 0

p = 1

···

p = n − 1

p = n

p = n + 1

dn

This gives the exact sequence

(∗)

0

E_{n+1}

np

H_{q}(F)

H_{q+n−1}(F)

E_{n+1}

0,q+n−1

0

dn

Moreover, for r>n, we have that d_{r }= 0 again. Therefore,

E_{∞}

pq = En+1

pq

.

This gives the diagram

Page 99

E_{∞}

0n

0

···

0

E_{∞}

nn

...

...

···

...

E_{∞}

01

0

···

0

E_{∞}

n1

E_{∞}

00

0

···

0

E_{∞}

n0

which gives a short exact sequence

(∗∗)

0

E_{∞}

0q

H_{q}(E)

E_{∞}

n,q−n

0

We then splice the exact sequences (∗) and (∗∗) to get the long exact sequence required. □

15.4. Spectral sequences of double complexes and their applications. Suppose C_{••}

is a double complex, with nonzero terms only in the first quadrant. Let Tot(C) be the total

complex. We define a bounded filtration on Tot(C)

···⊆ F_{p }Tot(C) ⊆ F_{p+1 }Tot(C) ⊆···

where

F_{k }Tot(C)_{n }=

k

⊕

p=0

C_{p,n−p}.

Let d_{h }be the horizontal maps and d_{v }be the vertical maps in the double complex, with

d = d_{h }+ d_{v }and d_{h}d_{v }+ d_{v}d_{h }= 0. We set

E_{0}

pq =

F_{p}(Tot(C)_{p+q}

F_{p−1}(Tot(C))_{p+q}

= C_{pq}

and d_{0 }= d_{v}. Then

E_{1}

pq = Hq(Cp•) −→ Hp+q(Tot(C••))

by Theorem 15.4. We have

d_{1 }: E_{1}

pq = Hv

q (Cp) → E1

p−1,q = Hv

q (Cp−q,•)

is induced by d_{h}, and so we write d_{1 }= d_{h}. Then also

E_{2}

pq = Hh

p Hv

p (C••) −→ Hp+q(Tot(C••)).

We could also define D_{pq }= C_{qp }(the transposition of the double complex) and apply the

above construction to that. This gives the two spectral sequences

IIE2

qp = Hv

q Hh

p (C••) −→ Hp+q(Tot(C••))

IE2

pq = Hh

p Hv

q (C••) −→ Hp+q(Tot(C••))

Page 100

Let A be a right R-module and B be a left R-module. Then

F = A ⊗_{R }−

is right exact

G = − ⊗_{R }B

is right exact

We already proved (Theorem 9.10) that

L_{p}F(B)

≅ L_{p}G(A) = Tor_{p}(A, B)

but the proof can be reinterpreted using spectral sequences of double complexes.

and P ⊗ Q is the double complex. Applying the above result to this double complex, we

obtain

IIE2

qp = Hv

q Hh

p (P ⊗ Q) −→ Hp+q(Tot(P ⊗ Q)),

IE2

pq = Hh

p Hv

q (P ⊗ Q) −→ Hp+q(Tot(P ⊗ Q)).

We have that

H_{v}

q (P• ⊗ Q•) = P• ⊗ Hq(Q•) =

{ P_{• }⊗ B if q = 0,

0

if q = 0.

Then

IE2

pq =

{ H_{h}

p (P• ⊗ B) = LpG(A) if q = 0

0

if q = 0

Hence the diagram for _{I}E_{2}

pq is

...

...

...

0

0

0

···

0

0

0

···

L_{0}G(A)

L_{1}G(A)

L_{2}G(A)

···

from which it is clear that d_{n }= 0 for n ≥ 2. Hence

E_{2}

pq = E3

pq = ··· = E∞

pq .

This shows that

H_{n}(Tot(P ⊗ Q)) = E_{∞}

n0 = LnG(A).

Similarly, we obtain that

IIE2

pq = Hn(Tot(P ⊗ Q)) = LnF(B),

which proves the theorem.

□

Page 101

One can also prove Künneth’s Formula 11.1 using spectral sequences. Suppose P is a complex

of flat R-modules, bounded from below. Let M be an R-module. Then there is a convergent

spectral sequence

E_{2}

pq = Torp(Hq(P),M) −→ Hp+q(P ⊗ M),

called the Künneth spectral sequence.

Alternative proof of Kunneth’s Formula 11.1. Let P be a complex of flat R-modules, bounded

from below. Assume that B_{n }= d(P_{n+1}) is flat for all n. We show that there is an exact

sequence

0

H_{q}(P) ⊗ M

H_{q}(P ⊗ M)

Tor_{R}

1 (Hq−1(P),M)

0

Let Z_{n }= ker(d: P_{n }→ P_{n−1}). We showed before that Z_{n }is also flat, and we have a short

exact sequence

0

B_{n}

Z_{n}

H_{n}(P)

0

which gives a flat resolution of H_{n}(P), showing that the tor dimension of H_{n}(P) is at most 1.

Then

E_{2}

pq = Torp(Hq(P),M)=0,

for p ≥ 2 and p < 0.

The diagram for E_{2}

pq is

...

...

...

...

0

H_{2}(P) ⊗ M

Tor_{1}(H_{2}(P),M)

0

0

H_{1}(P) ⊗ M

Tor_{1}(H_{1}(P),M)

0

0

H_{0}(P) ⊗ M

Tor_{1}(H_{0}(P),M)

0

which shows that d_{2 }= d_{3 }= ··· = 0, and hence E_{∞}

pq = E2

pq. Thus Hq(P ⊗ M) has a filtration

with quotients E_{2}

0q and E2

1,q−1. This gives the short exact sequence

0

H_{q}(P) ⊗ M

H_{q}(P ⊗ M)

Tor_{R}

1 (Hq−1(P),M)

0

as required.

□

Another application is the base change for Tor.

Page 102

Theorem 15.11 (Base change for Tor). Let f : R → S be a ring homomorphism. If A is a

right R-module, B is a left S-module, then

E_{2}

pq = TorS

p (TorR

q (A, S),B) −→ TorR

p+q(A, B).

Proof. Let P_{• }→ A be a projective R-resolution and Q_{• }→ B be a projective S-resolution.

Consider the double complex P ⊗_{R }Q. Then

IE2

pq = Hv

p Hh

q (P ⊗R Q) = Hv

p (P ⊗R Hh

q (Q)) =

{ H_{p}(P ⊗_{R }B) = Tor_{R}

p (A, B) if q = 0,

0

if q = 0.

Note that P ⊗_{R }− commutes with homology, because P is a projective and hence flat

R-module.

Then, by the usual argument, d_{2 }= d_{3 }= ··· = 0 and so _{I}E_{2}

pq =I E∞

pq , and hence

H_{p}(Tot(P ⊗ Q)) = E_{∞}

p0 = Hp(P ⊗R B) = TorR

p (A, B).

Moreover,

IIE2

pq = Hv

p Hh

q (P ⊗R Q) = Hv

p Hh

q ((P ⊗R S) ⊗S Q) = Hv

p (Hh

q (P ⊗R S) ⊗S Q).

Here, Q is a projective and hence flat S-module, but not necessarily a flat R-module, so we

have to tensor with S first. Hence

IIE2

pq = Hv

p (Hh

q (P ⊗R S) ⊗S Q) = Hv

p (TorR

q (A, S) ⊗S Q) = TorS

p (TorR

q (A, S),B),

completing the proof.

□

15.5. Hyperhomology and hyperderived functors. Let A be an abelian category with

enough projectives and A_{• }be a chain complex in A.

double complex P_{•• }of projectives together with augmentation ϵ: P_{•0 }→ A_{•}

...

...

...

...

···

P_{−11}

P_{01}

P_{11}

P_{21}

···

···

P_{−10}

P_{00}

P_{10}

P_{20}

···

···

A_{−1}

A_{0}

A_{1}

A_{2}

···

such that

(1) P_{p• }→ A_{p }is a resolution and if we define

B_{p}(P_{•q},d_{h}) = im(d_{h }: P_{p+1,q }→ P_{pq})

Z_{p}(P_{•q},d_{h}) = ker(d_{h }: P_{pq }→ P_{p−1,q})

Page 103

H_{p}(P_{•q},d_{h}) = Z_{p}(P_{•q},d_{h})/B_{p}(P_{•q},d_{h})

then

B_{p}(P_{•q},d_{h}) → B_{p}(A_{•})

H_{p}(P_{•q},d_{h}) → H_{p}(A_{•})

are projective resolutions (which implies that Z_{p}(P_{•q},d_{h}) → Z_{p}(A_{•}) is a projective

resolution),

(2) if A_{p }= 0 then P_{p• }is the zero complex.

Lemma 15.13. Every chain complex has a CE-resolution.

A ∈ A has a projective resolution 5.21.

□

hyperderived functor L_{p}F : Ch(A) → B of F as follows: if A_{• }is a chain complex in A, then

L_{p}F(A_{•}) = H_{p}(Tot(F(P_{•})))

where P_{•• }is a CE-resolution of A_{•}. Dually, we can define the right hyperderived functor R_{p}G

for a left exact functor G.

There are a lot of details which we will leave out: the fact that this functor is well-defined,

what this functor does to morphisms and so on. These are analogous to these properties for

left derived functors presented in Chapter 5.

Suppose for simplicity that A is bounded from below. We consider the two spectral sequences

IE2

pq = Hh

p Hv

q (F(P)) = Hh

p (LqF(A•)),

since P_{p• }→ A_{p }is a projecitve resolution, so

H_{v}

q (F(Pp•)) = Lq(FAp)

by definition. For the other spectral sequence, we note that

H_{q}(F(P_{••})) = FH_{v}

q (P••)

because the exact sequences

0

Z_{h}

pq

P_{pq}

B_{h}

p−1q

0

0

B_{pq}

Z_{pq}

H_{h}

pq

0

split. We then have

IIE2

pq = Hv

p Hh

q (F(P••)) = Hv

p (FHh

q (P••)) = (LpF)(Hh

q (A•))

because

H_{h}

q (P••) → Hq(A•)

is a projective resolution.

Page 104

Both of these spectral sequences converge to

H_{p+q}(Tot(F(P_{••}))) = L_{p+q}F(A_{•}),

i.e.

IE2

pq = Hh

p (LqF(A•)) −→ Lp+qF(A•),

IIE2

pq = (LpF)(Hh

q (A•)) −→ Lp+qF(A•).

Dually, if F : A→B is a left exact functor, A_{• }is a cochain complex, bounded from below,

then

IE

pq

2 = Hp(RqF(A•)) −→ Rp+qF(A•),

IIE

pq

2 = RpF(Hq(A•)) −→ Rp+qF(A•).

Theorem 15.15 (Grothendieck spectral sequence). Suppose A, B, C be abelian categories

where A, B have enough injectives, and

G : A→B

F : B→C

be left exact functors where G sends injectives to F-acyclic objects (R_{p}F(A)=0 for p > 0):

I, injective

GI,F-acyclic

A

B

C

R_{p}F(G(I)) = 0, p> 0

G

FG

F

We then have that

E

pq

2 = (RpF)(RqG)(A) −→ Rp+q(FG)(A).

The idea is that we can compute the derived functors using acyclic objects, instead of

projective resolutions. For example, we showed that to compute Tor, it is enough to consider

flat resolutions, and, indeed, flat objects are acyclic with respect to tensor products.

Proof. Suppose A → I_{• }is an injective resolution. Then G(I_{•}) is a cocomplex, and we can

apply the above construction to it. We obtain

IE

pq

2 = Hp((RqF)(G(I•))) −→ (Rp+qF)(G(I•)).

Now, G(I_{•}) is F-acyclic by assumption, and hence

R_{q}F(G(I_{p})) =

{ FG(I_{p}) if q = 0,

0

if q = 0.

This shows that

IE

pq

2 =

{ R_{p}(FG)(A) if q = 0,

0

if q = 0.

Page 105

Hence

(R_{p+q}F)(G(I_{•})) = R_{p+q}(FG)(A).

Using the second spectral sequence, we immediately get that

IIE

pq

2 = RpF(Hq(G(I•))) = RpF(RqG(A)).

Altogether, this shows that

R_{p}F(R_{q}G(A)) −→ R_{p+q}(FG)(A),

as required.

□

Then the functor

f_{∗ }: Sheaves(X) → Sheaves(Y ),

(f_{∗}F)(U) = F(f_{−1}(U)),

for a sheaf F on X, sends injectives to injectives. Then

Γ(X;F) = F(X)

gives a functor

Γ(X;−): Sheaves(X) → Ab

and

R_{p}Γ(X,F) = H_{p}(X;F)

the sheaf cohomology.

Then

(R_{p}Γ)(R_{q}f_{∗})(F) = H_{p}(Y ;R_{q}f_{∗}F),

and since

Γf_{∗}F = f_{∗}F(Y ) = F(f_{−1}(Y )) = F(X) = Γ(X),

we get that

R_{p+q}(Γf_{∗})(F) = H_{p+q}(X;F).

Then the Grothendieck spectral sequence 15.15 gives

H_{p}(Y ;R_{q}f_{∗}F) −→ H_{p+q}(X;F).

In particular, if R_{q}f_{∗}F = 0 for q > 0, then

H_{p}(Y ;f_{∗}F) = H_{p}(X;F).

Page 106

16. Triangulated categories

Let A be an abelian category and Ch(A) be the category of cochain complexes on A. We

recall a few definitions.

Suppose f : A_{• }→ B_{•}. The cone of f is given by

A_{n+2}

A_{n+1}

⊕

⊕

B_{n+1}

B_{n}

cone(f)_{n+1}

cone(f)_{n}

−dA

−f

dB

d

and the boundary map d: cone(f)_{n }→ cone(f)_{n+1 }is given by the matrix

d =

(−d_{A}

0

−f d_{B}

)

.

We then have an exact sequence

0

B_{•}

cone(f)_{•}

A[−1]_{•}

0.

δ

Similarly, we define the cylinder of f:

A_{n+1}

A_{n}

⊕

⊕

A_{n+2}

A_{n+1}

⊕

⊕

B_{n+1}

B_{n}

cyl(f)_{n+1}

cyl(f)_{n}

dA

−dA

idA

−f

dB

d

and the boundary map d: cyl(f)_{n }→ cyl(f)_{n+1 }is given by the matrix

d =

d_{A}

id_{A}

0

0 −d_{A}

0

0

−f d_{B}

.

We have exact sequences

0

A_{•}

cyl(f)_{•}

cone(f)_{•}

0

0

B_{•}

cyl(f)_{•}

cone(−id_{A})_{•}

0

α

β

Page 107

with

α =

0

0

id_{B}

,

β = (id_{A }0 id_{B})

and αβ ∼ id_{B}, βα = id_{B}, so α and β are homotopy equivalences.

We construct a quotient category K = K(A) of Ch = Ch(A) by

Obj(K) = Obj(Ch)

Hom_{K}(A_{•},B_{•}) = Hom_{Ch}(A_{•},B_{•})/ ∼

where ∼ is the chain homotopy equivalence. It is easy to check that composition in K is

well-defined.

This makes K into an additive category with an additive functor

Ch → K.

The cohomology functor H_{n }: Ch(A) → A factors through K because homotopy equivalent

maps induce the same maps on cohomology, and hence the triangle

Ch

A

K

Hn

Hn

commutes.

The category K is universal with this property. Suppose F : Ch → B is a functor such that if

f : A_{• }→ B_{• }is a chain homotopy equivalence, then F(f) is an isomorphism, then F factors

through K.

To show this, we first note that we have maps

B_{•}

cyl(id_{B})

α

α

β

where

α =

0

0

id

,

α =

id

0

0

.

We then have that

id = F(id) = F(βα) = F(β)F(α),

so F(α) and F(β) are inverses, and similarly F(α ) and F(β). Hence:

F(α ) = F(α)F(β)F(α ) = F(α).

Suppose f,g: B → C and f ∼ g, so

f − g = ds + sd.

Page 108

Then γ = (fsg): cyl(B) → C is a chain map, and

γα = g, γα = f,

so

F(g) = F(γα) = F(γ)F(α),

F(f) = F(γα ) = F(γ)F(α ).

If i: A_{• }→ B_{• }is a map, then we have a triangle

cone(u)

A_{•}

B_{•}

δ

u

v

We will call this a strict exact triangle.

C_{•}

A_{•}

B_{•}

w

u

v

is called an exact triangle if there exists ˜u: ˜

A_{• }→ ˜

B_{• }and an isomorphism in K

f : A → ˜A, g: B → ˜B, h: C → cone(˜u)

such that the diagram

A

B

C

A[−1]

˜A

˜B

cone(˜u)

˜A[−1]

u

f

v

g

w

h

f[−1]

˜u

˜v

commutes.

0

A_{•}

A_{•}

idA

is an exact triangle, because the diagram

A

A

0

A[−1]

A

A

cone(id_{A})

A[−1]

=

idA

=

=

idA

Page 109

commutes and letting C = cone(id_{A})

s =

(0 −id

0

0

)

, d_{C }=

(−d_{A}

0

id_{A}

d_{A}

)

we get

sd_{C }+ d_{C}s =

(id_{A}

0

0 id_{A}

)

= id_{C},

so id_{C }∼ 0.

C

A

B

w

u

v

is exact. We show that

A[−1]

B

C

−u[−1]

v

w

is exact. Assume without loss of generality that C = cone(u) and

v =

( 0

id_{B}

)

, w = δ = (id_{A }0).

Letting D = cone(v), we get that

A

B

C

A[−1]

B[−1]

B

C

D

B[−1]

u

v

=

δ

=

h

−u[−1]

=

v

π

where

π = (0 id_{A }0), h =

−u

id_{A}

0

,

and the map C → D is given by the matrix

0

0

id_{A}

0

0 id_{B}

.

We have that

id_{D }− hπ =

id u 0

0 0 0

0 0 id

= sd_{D }+ d_{D}s

Page 110

for the map

s =

0 0 −id

0 0

0

0 0

0

.

Similarly, we obtain that

B

C[−1]

A

v

−w[1]

u

is an exact triangle.

automorphism T : K→K and distinguished triples (u, v, w), called exact triangles, where

u: A → B, v: B → C, w: C → TA

for some A,B,C, such that the following axioms are satisfied:

(TR 1) Every u: A → B can be embedded in a triangle (u, v, w)

C

A

B

w

u

v

and

0

A

A

idA

is an exact triangle, and if (u, v, w) is isomorphic to (u ,v ,w ) and (u, v, w) is an

exact triangle, then (u ,v ,w ) is an exact triangle.

(TR 2) If (u, v, w) is an exact triangle, then

(v, w,−Tu) and (−T_{−1}w, u, v)

are exact triangles.

(TR 3) If

C

A

B

w

u

v

C

A

B

w

u

v

are exact triangles and f : A → A , g: B → B with gu = u f are morphisms, then

there exists h: C → C such that

(f,g,h): (u, v, w) → (u ,v ,w )

is a morphism of exact triangles, i.e. the following diagram

Page 111

A

B

C

TA

A

B

C

TA

f

g

h

Tf

commutes.

(TR 4) Suppose A, B, C, A ,B ,C are objects in K and

• (u, j, ∂) is an exact triangle on (A,B,C ),

• (v, x, i) is an exact triangle on (B,C,A ),

• (vu, y, δ) is an exact triangle on (A, C, B ),

then there exists an exact triangle (f,g,(Tj)i) on (C ,B ,A ) such that

∂ = δf, x = gy, ig = (Tu)δ.

We can represent this as the diagram

B

C

A

C

B

A

δ

g

∂

f

vu

u

x

y

j

v

(Tj)i

i

in which all the triangles commute. (Note that, as described above, only some of

these triangles are exact. We distinguish in blue the arrows that go to T applied to

the objects.)

Theorem 16.5. For an abelian category A, the quotient K(A) is a triangulated category

with the automorphism T(A) = A[−1].

Proof. Axioms (TR1), (TR2), (TR3) have already been verified in the discussion above. We

only have to show that (TR4) holds.

Without loss of generality,

C = cone(u), B = cone(vu), A = cone(v)

and we can represent the maps as follows

j =

( 0

id_{B}

)

: B → C , ∂ = (id_{A}

0) : C → A,

y =

( 0

id_{C}

)

: C → B , δ = (id_{A}

0) : B → A,

Page 112

x =

( 0

id_{C}

)

: C → A , i = (id_{B}

0) : A → B.

Taking

f =

(id_{A}

0

0

v

)

, g =

(u

0

0 id_{C}

)

,

one can easily verify that

δf = (id_{A}

0) = ∂, gy = x,

yv =

(0

v

)

= fj, ig = (Tu)δ.

We have to prove that (f,g,(Tj)i). We have

C

B

A

C [−1]

C

B

D

C [−1]

=

idA

0

0

v

=

u

0

0

idC

h=

0

0

idB

0

0

0

0

idC

0

0

idB

0

=

idA

0

0

v

0

0

0

0

idA

0

0

idC

idA

0

0

0

0

idB

0

0

where D = cone(f). We construct a chain map

π =

(0 id_{b}

u

0

0 0 0 id_{C}

)

: cone(f) → A ,

and claim that this gives an inverse map in the quotient category. We have that

d_{A }=

(−id_{b}

0

−v d_{C}

)

, d_{D }=

d_{A}

0

0

0

u

−d_{B}

0

0

−id_{A}

0

−d_{A}

0

0

−v −vu d_{C}

and d_{D}h = hd_{A }, d_{A }π = πd_{D}, πh = id_{A},

hπ =

0 0 0 0

0 id_{B}

u

0

0 0 0 0

0 0 0 id_{C}

.

Page 113

Setting

s =

0 0 −id_{B}

0

0 0

0

0

0 0

0

0

0 0

0

0

we get that

sd_{D }+ d_{D}s =

id_{A}

0 0 0

0 0 −u 0

0 0 id_{A}

0

0 0 0 0

= id_{D }− hπ.

This shows that id_{D }∼ hπ, and hence h, π are homotopy equivalences, i.e. isomorphisms

in K. Finally, πp = g, p = hπp = hg (in K).

□

Similarly, K_{b }= Ch_{b }/ ∼, bounded chain complexes, K_{+}, positive chain complexes, K_{− }

negative chain complexes, are all triangulated categories.

and A is an abelian category, then H is called a cohomological functor if for every exact

triangle ∆

C

A

B

w

u

v

we have a long exact sequence

···

H(T_{i−1}C)

H(T_{i}A)

H(T_{i}B)

H(T_{i}C)

H(T_{i+1}A)

···

H(Tiu)

H(Tiv)

H(Tiw)

We will then write H_{i}(A) = H(T_{i}A) and H_{i}(u) = H(T_{i}u).

K(A)

A

A_{•}

H_{0}(A_{•})

is a cohomological functor. Indeed, for an exact triangle

cone(u)

A

B

u

we have a long exact sequence

···

H_{i}(A_{•})

H_{i}(B_{•})

H_{i}(cone(u))

H_{i+1}(A_{•}) = H_{i}(A[−1])

···

Page 114

Proposition 16.8. If K is a triangulated category, X ∈ ObjK, then Hom_{K}(X,−) is a

cohomological functor K→A.

Proof. Suppose

C

A

B

w

u

v

is an exact triangle. By (TR 1), there is an extra triangle

0

A

A

idA

and by (TR 3), there exists h such that the squares in

A

A

0

TA

A

B

C

TA

id

id

0

u

h

0

id

u

v

w

commute, so that vu = h0 = 0. By (TR 2), we also get that wv = (Tu)w = 0.

We have a chain complex

···

A

B

C

TA

TB

TC

···

and this gives a chain complex

···

Hom_{K}(X, A)

Hom_{K}(X, B)

Hom_{K}(X, C)

Hom_{K}(X, T A)

Hom_{K}(X, T B)

Hom_{K}(X, T C)

···

Suppose b: X → B with vb = 0. We have

X

X

0

TX

TX

A

B

C

TA

TB

id

T−1h

b

−id

h

TB

u

v

w

−Tu

Page 115

then by (TR 3) we get −(Tu)h = −Tb, so u(T−1h) = b. This shows exactness at

Hom_{K}(X, B). We can first rotate the triangles, and then shift using T to show exactness

everywhere.

□

Z[0]

···

0

Z

0

0

···

···

A_{−1}

A_{0}

A_{1}

A_{2}

···

d0

and so

Hom_{Ch}(Z[0],A_{•}) = ker(d_{0})

Hom_{K}(Z[0],A_{•}) = ker(d_{0})/ ∼ = ker(d_{0})/im(d_{−1}) = H_{0}(A).

This shows that the cohomology functor is representable.

17. Derived categories

Suppose C is a category and S is a collection of morphisms.

functor q: C → S_{−1}C such that

(1) q(s) is an isomorphism in S_{−1}C for all s ∈ S,

(2) (S_{−1}C,q) is universal with property (1), i.e. for every category D and functor

F : C→D

such that F(s) is an isomorphism for all s ∈ S, there exists a unique functor

˜F: S_{−1}C→D such that the diagram

C

S_{−1}C

D

q

F

˜F

commutes.

If C is a small category, then S_{−1}C exists. Indeed, let S_{−1}C be the free category on ObjC

generated by all morphisms in C and all ˜s, s ∈ S, modulo relations from C and s˜s = id,

˜ss = id for all s ∈ S. Morphisms in S

−1C are of the form ˜s4s3s2

˜s_{1 }and so on.

K = S_{−1 }Ch(A).

category of A is

D(A) = Q_{−1 }Ch(A).

Page 116

This definition is very abstract so we will try to go via the quotient category K to get a

better understanding.

Let R be the collection of quasi-isomorphisms in K(A), then R_{−1}K(A) = D(A):

Ch(A)

S_{−1 }Ch(A) = K(A)

R_{−1}K(A)

Q_{−1 }Ch(A)

where we get the dotted arrows from the universal properties, and they are unique so they

are inverses.

(1) S is closed under composition,

(2) id_{X }∈ S so any x ∈ ObjC,

(3) Ore condition: if t: Z → Y is in S and g: X → Y (in C), then there exists a

commuting square

W

Z

X

Y

f

s

t

g

for s ∈ S, f ∈ Mor(C) and also the dual statement holds,

(4) if f,g: X → Y in C, then: there exists s ∈ S such that sf = sg if and only if there

exists t ∈ S such that ft = gt.

The idea behind this definition is to represent morphisms in S_{−1}C in the form fs_{−1 }= f˜s

with s ∈ S, f ∈ C:

A

B

C

in S−1C

s

f

The Ore condition shows that we can write s_{−1}

1 f2 where s1 ∈ S as f3s−1

3

where s_{3 }∈ S, and

we can write the composition in the same form again

(f_{1}s_{−1}

1 )(f2s−1

2 ) = f1f3s−1

3 s−1

1

= (f_{1}f_{3})(s_{1}s_{3})_{−1}.

Define a category D with Obj(D) = Obj(C) and Hom_{D}(A, B) as the set of all diagrams

A

C

B

s

f

with s ∈ S and f ∈ C, modulo ≡ where

[A

s1

←− C_{1}

f1

−→ B] ≡ [A

s2

←− C_{2}

f2

−→ B]

if there exist t_{1},t_{2 }∈ S with f_{1}t_{1 }= f_{2}t_{2 }and s_{1}t_{2 }= s_{2}t_{2}:

Page 117

C_{1}

A

D

B

C_{2}

s1

f1

t1

t2

s2

f2

The composition is defined by considering the following diagram

D

C_{2}

A_{3}

C_{1}

A_{2}

A_{1}

˜f1

˜s2

s2

f2

f1

s1

h2∈D

h1∈D

and letting

A_{1}

D

A_{3}

˜s2s1

f2 ˜f1

to be the composition. One can check that ≡ is an equivalence relation and composition is

well-defined.

We have a functor F : C→D, sending f : A → B in C to

F(f)=[A

idA

←− A

f

−→ B] ∈ Hom_{D}(A, B).

We claim that if s ∈ S, then F(s) is an isomorphism in D. Indeed,

F(s)=[A

idA

←− A

s

−→ B]

has inverse

[B

s

←− A

idA

−→ A]

because the composition

[B

s

←− A

s

−→ B]

is equivalent to

[B

idB

←− B

idB

−→ B]

via the diagram

A

B

A

B

B

s

a

idA

s

idB

idB

Page 118

Finally, one can show that (D,F) has the universal property of localization. This gives a

very concrete description of a localization with respect to a multiplicative system.

For an abelian category A, we can hence describe the derived category as follows:

(1) K(A) = U_{−1 }Ch(A) where U is a collection of homotopy equivalences (so we replace

morphisms by equivalence classes, making the set of morphisms smaller),

(2) D(A) = S_{−1}K(A) where S is a collection of quasi-isomorphisms (this S is actually a

multiplicative system, and hence morphisms in D(A) can be described as fractions

of morphisms in K(A)).

We still have to show that the collection of quasi-isomorphisms is a mulitplicative system.

We do this in more generality.

Proposition 17.5. Suppose K is a triangulated category, A is an abelian category, and

H : K→A is a cohomological functor. Let S be the collection of all s such that H_{n}(s) is an

isomorphism for all n. Then S is a multiplicative system.

Proof. We check the axioms:

(1), (2) By functoriality of H, S is closed under composition and id_{x }∈ S.

(3) Ore property. Given s ∈ S, f ∈ K, we want to find t ∈ S, g ∈ K such that

W

Z

X

Y

t

g

f

s

commutes. Embed s in an exact triangle

C

Z

Y

δ

s

u

and then embed uf in an exact triangle and rotate it to get an exact triangle

C

W

X

v

t

uf

Together, by (TR 3), there exists g: W → Z such that the following diagram

W

X

C

TW

TX

Z

Y

C

TZ

TY

t

g

uf

f

=

v

Tg

Tf

s

u

commutes. Since H_{n}(s) is an isomorphism for all n, we get

H_{n}(Z)

H_{n}(Y )

H_{n}(C)

H_{n+1}(Z)

H_{n+1}(Y )

∼=

∼=

so H_{n}(C) = 0, and hence H_{n}(t) is an isomorphism for all n. The dual property holds

by considering the dual category (the dual category of a triangulated category is also

triangulated).

Page 119

(4) If f,g: X → Y , we show that sf = sg for some s ∈ S if and only if ft = gt for some

t ∈ S.

We show the ‘only if’ implication; the other implication is symmetric. Suppose

s: Y → Y satisfies sf = sg and s ∈ S. Let h = f − g and embed s in an exact

triangle

Y

Z

Y

δ

u

s

From the long exact sequence, as above, we get H_{•}(Z) = 0. Since Hom_{K}(X,−) is a

cohomological functor, we have the exact sequence

Hom_{K}(X, Z)

Hom_{K}(X, Y )

Hom_{K}(X, Y )

v

uv = h

sh = 0

and, since sh, by exactness, there exists v: X → Z such that uv = h. Now, v lies in

an exact triangle

Z

X

X

w

t

v

But vt = 0, so 0 = uvt = ht = ft − gt. Hence ft = gt. Finally, since H_{•}(Z) = 0,

H_{n}(t) is an isomorphism for all n, and hence t ∈ S.

This shows that S is a multiplicative system.

□

By the above discussion, this shows that morphisms in S_{−1}K are of the form fs_{−1 }for s ∈ S,

f ∈ K. A morphism between X and Y is

X

X

Y

s

f

and is sometimes referred to as a roof.

Proposition 17.6. The derived category D(A) = S_{−1}K(A) is a triangulated category with

T(fs_{−1}) = T(f)T(s)_{−1}.

Proof. First note that T is well-defined: if f_{1}s_{−1}

1

= f_{2}s_{−1}

2 , then T(f1)T(s1)−1 = T(f2)T(s2)−1.

An exact triangle in D(A) is a triangle that is isomorphic to an exact triangle in K(A). We

need to check the 4 axioms, but we will only check some of them, the rest can be found

(TR 1) Suppose f = us_{−1 }is a morphism X → Y in D(A):

Page 120

Z

X

Y

s

u

We have that u lies in a exact triangle in K(A)

U

Z

Y

w

u

v

and we have the diagram

Z

Y

U

Z[−1]

X

Y

U

X[−1]

u

s−1

f

v

=

=

s[−1]

s−1[−1]

Then

U

X

Y

s[−1]w

f

v

is an exact triangle.

The (TR 2) axioms is clear. The axioms (TR 3) and (TR 4) require a proof, but we omit it

here.

□

Similarly, we can define D_{b}(A) from bounded chain complexes and D_{+}(A) from positive

chain complexes.

Proposition 17.7. Suppose I_{• }is a cochain complex of injectives, bounded from below, Z_{•}

is a cochain complex. If t: I_{• }→ Z_{• }is a quasi-isomorphism in K(A), then there exists

s: Z_{• }→ I_{• }with st = id in K(A).

Corollary 17.8. Suppose I_{• }is a cochain complex of injectives, bounded from below, in D(A).

Then Hom_{D(A)}(X, I) = Hom_{K(A)}(X, I).

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