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 LATEX’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.
The textbook for this course was [Wei94], and the notes largely follow this book without
specific reference. Citations are made where other resources were used.
Contents
1. Review of category theory
2
2. Algebraic topology
16
3. Homological algebra
22
4. Homological δ-functors
31
5. Projectives and left derived functors
33
6. Injectives and right derived functors
42
7. Limits
45
8. Sheaves and sheaf cohomology
47
9. Adjoint functors
50
10. Tor and Ext
55
11. Universal coefficients theorem
66
12. Quivers
69
13. Homological dimension
76
14. Local cohomology
89
15. Spectral sequences
90
16. Triangulated categories
106
17. Derived categories
115
References
120
1
Page 2
1. Review of category theory
We begin with a short review of the necessary category theory.
Definition 1.1. A category C is
(1) a class of objects, ObjC, and
(2) for all A, B ∈ ObjC, a set HomC(A, B) of morphisms from A to B,
(3) for any A,B,C ∈ ObjC, a composition map
HomC(A, B) × HomC(B,C) → HomC(A, C),
(f,g) ↦ g ◦ f = gf,
(4) for any A ∈ ObjC, a morphism idA ∈ HomC(A, A),
such that
(a) for any A, B ∈ ObjC and all f ∈ HomC(A, B)
idBf = f = f idA,
(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.
Examples 1.2.
• 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
HomA(x, y) =
{1} if x ≤ y,
{∅ otherwise
• C = Ab: abelian groups and group homomorphisms.
Definition 1.3. Fix a category C. If f : A → B is a morphism, an inverse of f is a morphism
g: B → A such that
gf = idA, fg = idB.
Inverses are unique: if g is another inverse, then
g = idAg = (g f)g = g (fg) = g idB = g .
Definition 1.4. If f has an inverse, we call it an isomorphism, and we write f−1 for that
inverse.
Page 3
Definition 1.5. Suppose C is a category. The opposite category, Cop, is defined by
Obj(Cop) = Obj(C)
HomCop (A, B) = {fop | f ∈ HomC(B,A)}
and if in C
A
B
C
f
g
fg
then in Cop
A
B
C
fop
gop
(fg)op=gopfop
Definition 1.6. A morphism f : B → C is monic if for any A ∈ ObjC and any e1,e2 : A → B
such that fe1 = fe2, we have e1 = e2.
Example 1.7. In Groups, Sets, Top, a morphism is monic if and only if it is injective.
Definition 1.8. A morphism f : A → B is epi if for any C ∈ ObjC and all g1,g2 : B → C
such that g1f = g2f, we have g1 = g2.
The notions of monic and epi are dual: f is monic in C if and only if fop is monic in Cop.
Example 1.9. In Sets, an epimorphism is a surjective map.
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 g1,g2 : R → X and g1f = g2f.
For any x ∈ R, there exists a sequence {xn} ⊆ Q with limn→∞ xn = x. Then
g1(x) = lim
n→∞
g1(xn) = lim
n→∞
g1(f(xn)) = lim
n→∞
g2(f(xn)) = lim
n→∞
g2(xn) = g2(x)
since both g1 and g2 are continuous.
Similarly, in Rings, f : Z → Q is epi but not surjective.
Definition 1.10. An object I ∈ Obj(C) is initial if for every A ∈ Obj(C) there is a unique
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 idI : I → I and
idI : I → I are also such morphisms and hence by uniqueness
fg = idI ,
gf = idI.
This shows initial objects are unique up to unique isomorphism.
Definition 1.11. An object T ∈ Obj(C) is terminal if for all A ∈ Obj(C), there is a unique
morphism A → T.
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This is the dual notion to initial object:
I ∈ Obj(C) is initial if and only if I ∈ Obj(Cop) is terminal.
Examples 1.12. We provide a few examples of initial and terminal objects in a few
categories:
initial terminal
Sets {0}
Groups {1} {1}
Ab {0} {0} = 0
Rings with 1 Z 0
Definition 1.13. A zero object is initial and terminal. We denote it by 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.
Definition 1.14. A monic morphism f : A → B is called a subobject of B.
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
Example 1.15. In Sets: if f : A → B injective, it is a subobject and we have that
A
B
f(A)
f
f
and A → B and f(A) → B are isomorphic.
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Definition 1.16. Suppose C has zero object. We say f : A → B is a kernel of g: B → C if
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 e1,e2 : A → A satisfy f = fe1 = fe2, then e1 = e2 by
uniqueness in the universal property
A
A
B
C
A
e1
f
0
f
0
g
e2
f
0
Example 1.17. In Groups, consider S2 → Sn sending (12) to (12). This map is not an
epimorphism but its cokernel is {1} → Sn (exercise).
Definition 1.18. If A, B ∈ Obj(C) then a product is an object A × B together with
morphisms πA : A × B → A and πB : A × B → B with universal property: if C ∈ Obj(C) and
fA : C → A, fB : C → B are morphisms, then there exists a unique morphism h: C → A×B
such that fA = πAh, fB = πBh, i.e. the following diagram
A
A × B
B
C
πA
πB
fA
fB
h
commutes.
Similarly, if Ai, i ∈ I are objects, their product is an object ∏i∈I Ai together with morphisms
πi : ∏i∈I Ai → Ai with analogous universal property.
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Definition 1.19. A coproduct is an object A∐B together with iA : A → A∐B and iB : B →
A ∐ B with the dual universal property, and similarly one can define the coproduct of any
family of objects, ∐i∈I Ai.
Examples 1.20. In Sets, the product A × B is the Cartesian product with πA(a, b) = a,
πB(a, b) = b, and the coproduct A ∐ B is the disjoint union with iA : A → A ∐ B, iB : 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
Ai = {(a1,a2 ...) | ai ∈ Ai},
i∈I
Ai = {(a1,a2,...) | ai ∈ Ai and ai = 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, Rings1, A × B is the standard product and A ∐ B = A ⊗Z B
with
iA : A → A ⊗Z B,
a ↦ a ⊗ 1.
Note that Z/2 ⊗Z Z/3 = 0, so the inclusion map
iZ/2 : Z/2 → Z/2 ⊗Z Z/3
is not monic.
Definition 1.21. A functor F : C→D from category C to D is a rule that
(1) assigns to A ∈ ObjC an object FA ∈ ObjC,
(2) assigns to f ∈ HomC(A, B) a morphism F ∈ HomD(FA,FB)
such that
(a) F(idA) = idFA,
(b) F(gf) = F(g)F(f) if f : A → B, g: B → C in C.
Example 1.22. If A ∈ ObjC, we have a functor
HomC(A,−): C → Sets
such that for B ∈ ObjC
HomC(A,−)(B) = HomC(A, B),
and for f : B → C
HomC(A, f): HomC(A, B) → HomC(A, C),
g ↦ fg,
which can be represented by the following diagram:
Page 7
A
B
C
g
f
fg
Then HomC(A,idB) = idHomC(A,B) and
HomC(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
Example 1.23. Suppose R is a ring. If M is a left R-module, then we have a functor
HomR(M,−): R-mod → Ab,
and if M is a right R-module then we have a functor
M ⊗R −: R-mod → Ab.
Definition 1.24. A functor F : C→D is faithful (resp. full) if for any A, B ∈ ObjC,
F : HomC(A, B) → HomD(FA,FB)
is injective (resp. surjective).
What we defined above is actually a covariant functor.
Definition 1.25. A contravariant functor F : C→D from category C to D is a rule that
(1) assigns to A ∈ ObjC an object FA ∈ ObjC,
(2) assigns to f ∈ HomC(A, B) a morphism F ∈ HomD(FB,FA)
such that
(a) F(idA) = idFA,
(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:
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A
B
C
f
g
gf
FA
FB
FC
Ff
Fg
F(gf)
Example 1.26. There is a contravariant functor D: C → Cop given by D(A) = A for
A ∈ Obj(C), D(f) = fop for f a morphism in C.
Example 1.27. If A ∈ ObjC then HomC(−,A): C → Sets is a contravariant functor. So
HomC(A,−) is covariant and HomC(−,A) is contravariant.
Definition 1.28. Suppose F,G : C→D are functors. A natural transformation η: F→G
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.
Example 1.29. Let F : Ab → Ab be given by
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.
Example 1.30. Fix a category C. Suppose e: A → B is a morphism. Then
ϵ: HomC(B,−) → HomC(A,−)
is a natural transformation given by
ϵ(C): HomC(B,C) ∋ f ↦ fe ∈ HomC(A, C),
since the following diagram
Hom(B,C)
Hom(B,D)
Hom(A,D)
Hom(A,C)
HomC(B,f)
ϵ(C)
ϵ(D)
HomC(A,f)
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commutes.
A functor F : C→D would be an isomorphism if there exists G : D→C such that FG = idD,
GF = idC. However, such maps seldom exist so we weaken the notion slightly.
Definition 1.31. A natural transformation η: F→G is a natural isomorphism if η(A): F(A) →
G(A) is an isomorphism for all A ∈ Obj(C).
Definition 1.32. A functor F : C→D is an equivalence of categories if there exists a functor
G : D→C such that GF is naturally isomorphic to idC and FG is naturally isomorphic to
idD.
Example 1.33. Let V be the category of finite-dimensional R-vector spaces and C⊆V be
the full subcategory with Obj(C) = {Rn | 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 ) = Rdim V .
We now choose isomorphism
η(V ): Rdim 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 → idV is a natural isomorphism
η(V ): Rdim V = FG(V ) → V.
There is also a natural isomorphism η|C : GF → idC.
1.1. Abelian Categories. The reference for this section is [Fre03]. We introduce a general
framework where we can develop homological algebra, generalizing categories such as Ab
and more generally R-mod.
Definition 1.34. A category C is an Ab-category if for all A, B ∈ ObjC, HomC(A, B) is an
abelian group, and
g(f1 + f2) = gf1 + gf2,
f,f1,f2 : A → B,
(g1 + g2)f = g1f + g2f,
g,g1,g2 : B → C.
This makes HomC(A, A) a ring with 1 = idA.
Page 10
Definition 1.35. An additive category is an Ab-category with finite products and a zero
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,πAB) is a product, (A ⊕ B,iA,iB) is a coproduct, and
πAiA = idA, πBiB = idB,
πAiB =0= πBiA,
iAπA + iBπB = idA⊕B.
Definition 1.36. An additive category is abelian if
(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 fg1 = fg2, then
0 = fg1 − fg2 = f(g1 − g2),
so g1 − g2 factors through 0 → A, so g1 − g2 = 0 and g1 = g2.
Definition 1.38. If we have maps
A1
A2
B
f1
f2
then a pull-back is an object P together with maps g1 : P → A1, g2 : P → A2 such that
f1g1 = f2g2 and (P, g1,g2) is universal with this property, i.e. if u1 : C → A1 and u2 : C → A2
satisfy fu1 = fu2 then there exists a unique h: C → P such that g1h = u1 and g2h = u2,
i.e. the following diagram
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C
P
A1
A2
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
A1 ⊕ A2
B
(f1,−f2)
we have that
P
A1 ⊕ A2
g1
g2
is the kernel of A1 ⊕A2 → B. The pushout is a cokernel of an analogous map A1 ⊕A2 → B.
Lemma 1.39. Suppose
P
A1
A2
B
g1
g2
f1
f2
is a pull-back. Then
(1) if g1 is monic, then f2 is monic,
(2) if f1 is epi, then g2 is epi.
Proof. We first prove (1). Indeed, suppose g1 is monic and take u: C → A2 with f2u = 0.
Then there exists unique h: C → P such that g1h = 0 and g2h = u, i.e. the following diagram
C
P
A1
A2
B
h
0
u
g1
g2
f1
f2
commutes. But g1 is monic, so this implies h = 0, and hence u = g2h = 0. Hence 0 → A2 is
a kernel of f2, so f2 is monic.
Page 12
For (2), suppose f1 is epi. We recall that P → A1 ⊕ A2 is kernel of A1 ⊕ A2 → B, and
A1 ⊕ A2 → B is epi, because A2 → B is epi. Hence A1 ⊕ A2 → B is cokernel of
P → A1 ⊕ A2.
Hence the diagram
P
A1
A2
B
g1
g1
f1
f2
is a pushout. By the dual of (a), f1 : A1 → B is epi implies that g2 : P → A2 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, Rop, the opposite ring is Rop = R as a set with multiplication
∗ in Rop defined by a ∗ b = b · a. If M is a left R-module, them M is a right Rop-module: if
m ∈ M, a ∈ Rop = 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-Rop and the category Rop-mod is isomorphic to
mod-R. Moreover, if M is a left R-module, we write RM, if M is a right R-module, we write
MR, and if M is a R-S-bimodule, we write RMS.
We work for now in the category mod-R.
Definition 1.42. A sequence
A0
A1
A2
···
An+1
f0
f1
f2
fn
is exact if imfi−1 = kerfi 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
A1
A2
A3
A4
A5
B1
B2
B3
B4
B5
f1
f2
f3
f4
f5
Page 15
has exact rows and commuting squares, and f1,f2,f4,f5 are isomorphisms, then f3 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
below. For more details, see [GM03, Chap. II.5] [Gelfand Manin, II.5 exercises].
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.
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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
Snake Lemma 1.43 and the Five Lemma 1.44 hold in an abelian category.
2. Algebraic topology
In this chapter, we review the motivating examples of homology and cohomology from
algebraic topology. For a more detailed introduction to the area, see [Hat02].
2.1. Singular Homology.
Definition 2.1. A geometric n-simplex is
n =
{
(x0,x1,...,xn) ∈ Rn+1 | xi ≥ 0,
n
i=0
xi = 1
}
.
For a topological space X, a singular n-simplex is a continuous function σ: ∆n → X.
We let
Sn(X) = free Z-module with basis of all singular n-simplices on X
and define
fi = fn
i : ∆n−1 → ∆n
by
fi(x0,...,xn−1)=(x0,x1,...,xi−1,0,xi+1,...,xn−1).
Then we define a map dn : Sn(X) → Sn−1(X) by
dn(σ) =
n
i=0
(−1)iσ ◦ fi.
Lemma 2.2. For any n, dn−1 ◦ dn = 0, so d2 = 0.
Page 17
Proof. If j<i, then fi ◦ fj = fj ◦ fi−1, so
dn−1(dn(σ)) =
n−1
j=0
n
i=0
(−1)i+jσ ◦ fn
i ◦ fn−1
j
=
0≤i≤j≤n−1
(−1)i+jσ ◦ fi ◦ fj +
0≤j<i≤n−1
(−1)i+jσ ◦ fi ◦ fj
=
0≤i≤j≤n−1
(−1)i+jσ ◦ fi ◦ fj +
0≤j<i≤n−1
(−1)i+jσ ◦ fj ◦ fi+1
=
0≤i≤j≤n−1
(−1)i+jσ ◦ fi ◦ fj
0≤j≤i≤n−1
(−1)i+jσ ◦ fj ◦ fi
setting i → i + 1
= 0
showing d2 = 0.
We hence get a chain complex S(X):
···
Sn+1(X)
Sn(X)
Sn−1(X)
Sn−2(X)
···
dn+1
dn
dn−1
with dn−1dn = 0 for all n. (By convention, Sn(X) = 0 for n < 0.)
Definition 2.3. We define
Zn(X) = kerdn,
the module of cycles,
Bn(X) = imdn+1, the module of boundaries.
Note that Bn(X) ⊆ Zn(X) ⊆ Sn(X) by the lemma above.
Example 2.4. The boundary of a segment from a to b is b − a
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.
Definition 2.5. The nth singular homology group is defined as
Hn(X) = Hsing,n(X) = Zn(X)/Bn(X) = kerdn/imdn+1.
One can show that Hn(X) is a topological invariant: if X is homeomorphic to Y then
Hn(X)
≅ Hn(Y ).
Examples 2.6. We have that H0(X)
≅ Zd where d is the number of path-connected
components of X.
Page 18
The homology of a contractible space is trivial, so for example
Hn(∗) = Hn(Rn).
However, we can distinguish between Rn and Rm using homology, because
Hj(Rn \ {∗}) =
{
Z for j = n − 1,
0 for 0 <j<n − 1.
2.2. Relative homology. Suppose Y ⊆ X is a subspace. Then Sn(Y ) ⊆ Sn(X) and we
can define
Sn(X, Y ) = Sn(X)/Sn(Y )
and we get the sequences
0
Sn(Y )
Sn(X)
Sn(X, Y )
0
0
Sn−1(Y )
Sn−1(X)
Sn−1(X, Y )
0
dn
dn
dn
We then define
Zn(X, Y ) = ker(dn : Sn(X, Y ) → Sn−1(X, Y )),
Bn(X, Y ) = im(dn+1 : Sn+1(X, Y ) → Sn(X, Y )),
Hn(X, Y ) = Zn(X, Y )/Bn(X, Y ).
2.3. Homology with coefficients. If M is a Z-module, we can define homology with
coefficients in M by setting
Sn(X;M) = Sn(X) ⊗Z M,
Zn(X;M) = ker(dn : Sn(X;M) → Sn−1(X;M)),
Bn(X;M) = im(dn+1 : Sn+1(X;M) → Sn(X;M)),
Hn(X;M) = Zn(X;M)/Bn(X;M).
Note that − ⊗Z M and Hn(−) do not commute. Hence taking homology with coefficients is
a non-trivial procedure.
We can write concisely
Hn(X) = Hn(S(X)),
Hn(X, Y ) = Hn(S(X, Y )),
Hn(X;M) = Hn(S(X;M)),
Hn(X, Y ;M) = Hn(S(X, Y ;M)).
Page 19
2.4. Simplicial homology.
Definition 2.7. An abstract simplicial complex is a pair K = (V,S) where V is a set and S
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.
Definition 2.8. A geometric realization |K| of K = (V,S) is
|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.
Example 2.9. If V = {1,2,3} and S = {{1,2},{2,3},{1,3},{1},{2},{3}}, then |K| is a
triangle with vertices 1,2,3 which is homeomorphic to a circle
1
2
3
We set
Cn(K) = Z-module with basis of all n-simplices.
To define dn : Cn(K) → Cn−1(K), we choose a total ordering on V and write
Cn(K) 〈v0v1 ...vn〉 if v0 < v1 <...<vn.
Then we define
dn(〈v0v1 ...vn〉) =
n
i=0
(−1)n〈v0 ... ˆvi ...vn〉.
Once again, we set
Zn(K) = kerdn
Bn(K) = imdn+1
Hsimp,n(K) = Zn(K)/Bn(K).
Proposition 2.10. For any n, we have that Hsimp,n(K)
≅ Hsing,n(|K|).
Example 2.11. We compute the homology of
1
2
3
Page 20
so that |K| = S1. We have that
C0(K) = Z〈1〉 + Z〈2〉 + Z〈3〉 ∼= Z3
C1(K) = Z〈12〉 + Z〈13〉 + Z〈23〉 ∼= Z3
d1 :
〈12〉 ↦ 〈2〉−〈1〉
〈13〉 ↦ 〈3〉−〈1〉
〈23〉 ↦ 〈3〉−〈2〉
so
Z1(K) = Z(〈13〉−〈23〉−〈12〉)
B0(K) = {〈a〈1〉 + b〈2〉 + c〈3〉 | a + b + c = 0} ∼= Z2
B1(K)=0
Z0(K) = Z3
and hence
H1(K)
≅ Z,
H0(K)
≅ Z.
Example 2.12. Recall that P2(R) can be thought of as a square with opposite identifications
on opposite edges, so P2(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
C2 = Z18, C1 = Z27, C0 = Z10
and the sequence
0 → Z18 → Z27 → Z10 → 0
has kerd2 = 0 so H2(K) = 0.
However, the map
d2 : C2(K;Z/2) → C1(K,Z/2)
has
d2
( sum of all 2-simplices
with right orientation
)
= 2(〈12〉 + 〈23〉 + 〈34〉 + 〈45〉 + 〈56〉−〈16〉)=0
so
H2(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
Sn(f) = f: Sn(X) → Sn(Y ): σ ↦ f ◦ σ.
Moreover, the square
Sn(X)
Sn(Y )
Sn−1(X)
Sn−1(Y )
f
d
d
f
commutes, and hence
f(Bn(X)) ⊆ Bn(Y ),
f(Zn(X)) ⊆ Zn(Y ),
so finduces
Hn(f) = f: Hn(X) = Zn(X)/Bn(X) → Hn(Y ).
Altogether, we have a functor
Hn : Top → Ab.
2.6. Cohomology. Let M be a Z-module and let
Sn(X;M) = HomZ(Sn(X),M).
We have the following commutative triangle
Sn(X)
M
Sn+1(X)
dn+1
which gives a map
δn : Sn(X;M) → Sn+1(X;M).
Concisely, we apply the functor HomZ(−,M) to both Sn(X) and dn and write
δn = Hom(dn+1,M).
We then get the following cochain complex
···
Sn−1(X;M)
Sn(X;M)
Sn+1(X;M)
···
δn−1
δn
δn+1
We then define
Zn(X;M) = kerδn,
the cocycles,
Bn(X;M) = imδn−1,
the coboundaries,
Hn(X;M) = Zn(X;M)/Bn(X;M), the cohomology group.
We admit the convention
Sn(X) = Sn(X;Z), Hn(X) = Hn(X;Z).
Page 22
Example 2.13. For X = P2(R) we have that
H0 = Z,
H0 = Z,
H1 = Z/2, H1 = 0,
H2 = 0,
H2 = Z/2.
Suppose now X is a smooth manifold and let Ωn(X) be the set of n-forms on X, so an
element of Ωn(X) is gdf1 ∧ df2 ∧ ... ∧ dfn. We have maps
d: Ωn(X) → Ωn+1(X)
with d2 = 0, which give a cochain complex
···
n−1(X)
n(X)
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
0(X) = {smooth functions X → R}.
Example 2.14. For X = S1, we have that 2xdx+2ydy = d(x2+y2) = d1=0so xdx = −ydy.
But then
ω =
dx
y
= −
dy
x
but ω = df for any f which shows H1
DR(S1) = 0.
3. Homological algebra
In this chapter, we are in the category of right R-modules, mod-R.
Definition 3.1. A chain complex is a diagram
C:
···
Cn+1
Cn
Cn−1
···
dn+1
dn
with dndn+1 = 0 for all n.
We say Cis bounded from below (above) if there exists a such that for any n<a (n>a),
Cn = 0. Moreover, Cis bounded if it is bounded both from below and from above.
Example 3.2. In the topological examples we have seen for homology, all sequences were
bounded from below. For a finite simplicial complex, the chain complex is moreover bounded.
Definition 3.3. We define
Bn(C) = imdn+1, Zn(C) = kerdn
and the nth homology group of Cto be
Hn(C) = Zn(C)/Bn(C).
Definition 3.4. A chain complex map C→ Dis a collection of module homomorphisms
un : Cn → Dn for all n ∈ Z such that un−1dn = dnun, so the following diagram
Page 23
···
Cn+1
Cn
Cn−1
···
···
Dn+1
Dn
Dn−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 un(Bn(C)) ⊆ Bn(D) and un(Zn(C)) → Zn(D) and hence un
induces a map
u∗,n : Hn(C) → Hn(D).
Definition 3.5. We define Ch(mod-R) to be the category of chain complexes (of right
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
Hn(−): Ch(mod-R) → mod-R.
Example 3.6. If X, Y are topological spaces and f : X → Y is continuous, the maps
f∗,n : Sn(X) → Sn(Y ): σ ↦ f ◦ σ
form a chain complex map
f: S(X) → S(Y ).
Then finduces
f∗,n : Hn(X) → Hn(Y )
and in fact we have a functor
Hn(−): Top → Ab.
Theorem 3.7. If we have a short exact sequence of chain complexes
0 → C→ D→ E→ 0
(or, equivalently, 0 → Cn → Dn → En → 0 is exact for any n). Then we have a long exact
sequence
···
Hn+1(E)
Hn(C)
Hn(D)
Hn(E)
Hn−1(C)
···
Proof. We apply Snake Lemma 1.43 to get the red maps below
Page 24
0
0
0
0
Zn(C)
Zn(D)
Zn(E)
0
Cn
Dn
En
0
0
Cn−1
Dn−1
En−1
0
Cn−1/Bn−1(C)
Dn−1/Bn−1(D)
En−1/Bn−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
Hn(C)
Hn(D)
Dn(E)
Cn/Bn(C)
Dn/Bn(D)
En/Bn(E)
0
0
Zn−1(C)
Zn−1(D)
Zn−1(E)
Hn−1(C)
Hn−1(D)
Hn−1(D)
0
0
0
d
d
d
This completes the proof.
Definition 3.8. A cochain complex is a diagram
···
Cn−1
Cn
Cn+1
Cn+2
···
dn−1
dn
dn+1
where dn+1dn = 0 for all n.
If Cis a chain complex, setting Cn = C−n with dn = d−n makes C, da cochain complex.
Definition 3.9. For a chain complex Cand p ∈ Z, we define the shift of Cby p to be the
chain complex C[p], d[p] with
C[p]n = Cp+n, d[p]n = (−1)pdp+n.
Page 25
Analogously, for a cochain complex C, we define the shift of Cby p to be the cochain
complex
C[p]n = Cn−p, d[p]n = (−1)pdn−p.
This notation is used for example in the following context: instead of saying that for any n,
the sequence
0 → Zn(C) → Cn → Bn−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
···→ Hn−1(E) → Hn(C) → Hn(D) → Hn(E) → Hn+1(C) →···
Example 3.11. If X is a topological space and Y ⊆ X is a subset, we have a short exact
sequence
0 → S(Y ) → S(X) → S(X, Y ) → 0
and by Theorem 3.7, we get the long exact sequence of a pair
···→ Hn+1(X, Y ) → Hn(Y ) → Hn(X) → Hn(X, Y ) →···
In some cases, this allows us to calculate homology groups of topological spaces.
Example 3.12. Let X be a manifold and Ωn(X) be the n-forms on X. If X = U ∪ V for
U, V ⊆ X open, we have a short exact sequence
0
n(X)
n(U) ⊕ Ωn(V )
n(U ∩ V )
0
ω
|U |V )
12)
1)|U∩V − (ω2)|U∩V
Applying Theorem 3.10 to this short exact sequence, we get the Mayer-Vietoris sequence for
de Rham cohomology
···→ Hn
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.
Definition 3.13. A homotopy from f to g is a continuous function h: [0,1] × X → Y such
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
s0 : S0(X) → S1(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) = gσ
h(t, b)
h(0, σ) = fσ
h(t, a)
σ1
σ2
Then there exist two 2-simplices σ1 + σ2 such that
d(σ1 2) = gσ−h(t, b)−fσ+h(t, a)=(g−f)(σ)+s0(a)−s0(b)=(g−f)(σ)−s0(dσ).
Define s1 : S1(X) → S2(Y ) by setting
s1(σ) = σ1 + σ2 ∈ S2(Y ).
Then ds1(σ)=(g− f)(σ) − s0(dσ), so
ds1 + s0d = g− f.
In general, there exists sn : Sn(X) → Sn+1(Y ) such that
sd + sd = g− f.
Definition 3.14. Topological spaces X and Y are homotopy equivalent if there exist
continuous f : X → Y and g: Y → X such that gf is homotopic to idX and fg is homotopic to
idY .
Definition 3.15. We say X is contractible if X is homotopy equivalent to a point.
Example 3.16. The Euclidean space Rn is contractible: for
f : {0} → Rn,
g: Rn → {0},
we have gf = id{0} and
fg: Rn → Rn
is constant equal to 0 and a homotopy from fg to idR
n is
h(t, x) = tx.
Definition 3.17. Suppose C, Dare chain complexes and f,g: C→ Dare chain complex
maps. A homotopy from f to g is a collection of functions sn : Cn → Dn+1 for n ∈ Z such
that
dn+1sn + sndn = gn − fn
(or, concisely, ds + sd = g − f),
which can be represented by the diagram
Page 27
···
Cn+1
Cn
Cn−1
···
···
Dn+1
Dn
Dn−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 : Hn(C) → Hn(D). Explicitly, if a ∈ Hn(C), say a =
x + Bn(C) for x ∈ Zn(C), then
f(a) = f(x) + Bn(D) ∈ Hn(D),
g(a) = g(x) + Bn(D) ∈ Hn(D).
If f,g are homotopic, then have that
g(a) − f(a) = gn(x) − fn(x) + Bn(D) = dn+1sn(x) + sn−1dn(X) ∈ Bn(D),
so g(a) = f(a). Hence if f,g are homotpic, then f= g.
Definition 3.18. We say C, Dare homotopy equivalent if there exist f : C→ Dand
g: D→ Csuch that fg and gf are homotopic to the appropriate identity.
By the above, if C, Dare homotopy equivalent, then for any n, f∗,n : Hn(C) → Hn(D) is
an isomorphism because
gf= (gf)= id= id,
fg= (fg)= id= id.
Example 3.19. If X is contractible, then
Hn(X) = Hn({∗}) =
{
Z if n = 0,
0 otherwise.
For example,
Hn(Rn) =
{
Z if n = 0,
0 otherwise.
Moreover, Rm \ {0} is homotopy equivalent to Sm−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 = idA
(2) there exists g : C → B with gg = idC
(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.
Definition 3.21. A short exact sequence is split if any of the above equivalent conditions
hold.
Example 3.22. The following short exact sequence is not split
0
Z/p
Z/p
2
Z/p
0
1+(p)
p + (p2)
1+(p2)
1+(p)
Definition 3.23. A chain complex Cis split if there exists sn : Cn → Cn+1 for any n such
that dnsn−1dn = dn for all n (i.e. dsd = d).
Suppose Cis split. Let b ∈ Bn−1. Then b = d(a) for some a ∈ Cn and ds(b) = dsd(a) =
d(a) = b, so ds = idBn−1 .
Then
0 → Bn → Zn → Hn → 0
is also split: if b ∈ Bn then b = d(a) for a ∈ Cn+1 and ds(b) = dsd(a) = d(a) = b, so
ds|Bn = idBn . Hence
Cn
≅ Zn ⊕ Bn−1
≅ Bn ⊕ Hn ⊕ Bn−1
and we have the following diagram
···
Cn+1
Cn
Cn−1
···
Bn
Bn+1
Hn+1
Bn−1
Bn
Hn
Bn−2
Bn−1
Hn−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.
Example 3.24. For X = [0,1], C(X) is a 1-simplex
{0}×[0,1]
= [0,1]
{1}×[0,1]
= {∗}
Page 29
For X = S1, C(X) is an actual cone, justifying the name
{0}×S1
= S1
{1}×S1
= {∗}
Definition 3.25. Suppose f : X → Y is a continuous map. The mapping cone is
C(f) = C(X) ∐ Y/ ∼
where C(x) (1,x) ∼ f(x) ∈ Y .
Examples 3.26. Let f : [0,2π] → S1, f(t) = (cost,sint). Then C(f) is a 1-simplex with a
circle attached to [0,1] via f:
Let f : S1 → S1 be the map z ↦ z2. Then C(f) = P2(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.
Definition 3.27. Let f : B→ Cbe a chain map. We define a new chain complex cone(f)
by setting
cone(f)n = Bn−1 ⊕ Cn
d(b, c)=(−d(b),d(c) − f(b))
or in matrix form
dcone =
( −dB
0
−f dC
)
.
This is indeed a chain complex:
d2
cone =
( −dB
0
−f dC
)( −dB
0
−f dC
)
=
(
d2
B
0
fdB − dCf d2
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)
Since Hn(B[−1]) = Hn−1(B), we have the long exact sequence (Theorem 3.7)
···
Hn(B)
Hn(C)
Hn(cone(f))
Hn−1(B)
Hn−1(C)
···
Lemma 3.28. The boundary map ∂ above is f.
Proof. To prove this statement, we trace through the proof of Theorem 3.7. Let b ∈ B[−1]n+1
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) ∈ Cn, and actually f(b) ∈ Zn. Hence
∂[b]=[f(b)] = f[b],
completing the proof.
Definition 3.29. A chain map f : C→ Dis a quasi-isomorphism if
f: Hn(B) → Hn(C)
is an isomorphism for all n.
Corollary 3.30. A chain map f : B→ Cis a quasi-isomorphism if and only if cone(f) is
exact (i.e. Hn(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:
···
Hn(B)
Hn(C)
Hn(D)
Hn−1(B)
Hn−1(C)
···
···
Hn(B)
Hn(C)
Hn(cone(f))
Hn−1(B)
Hn−1(C)
···
=
f
=
f
=
=
f
∼=
f
Page 31
4. Homological δ-functors
Definition 4.1. A function F : C→D is additive if F : HomC(A, B) → HomD(FA,FB) is
a group homomorphism.
Definition 4.2. An additive functor F : C→D is left-exact if for every short exact sequence
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. □
Definition 4.4. A homological δ-functor C→D is a collection of additive functors
Tn : C→D, n ≥ 0,
together with a morphism δn : Tn(C) → Tn−1(A) for every short exact sequence
0
A
B
C
0
f
g
such that
(1) there exists a long exact sequence
···
Tn+1(C)
Tn(A)
Tn(B)
Tn(C)
Tn−1(A)
···
δn+1
Tn(f)
Tn(g)
δn
where we set Tn(A) = 0 for n < 0, so T0 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
Tn(C )
Tn−1(A)
Tn(C)
Tn−1(A)
δn
Tn(fC )
Tn−1(fA)
δn
Example 4.5. If C is an abelian category and Ch≥0(C) is the category of non-negative chain
complexes, then the homology functor
H: Ch≥0(C) → C
is a homological δ-functor.
Example 4.6. Let R be a ring and r ∈ R. Set
T0 : R-mod → Ab, T0(M) = M/rM,
T1(M)=[r]M = {a ∈ M | ra = 0},
Tn(M) = 0 for all n > 1.
Then T = {Tn} 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
we obtain the long exact sequence
0 → T1(A) → T1(B) → T1(C) → T0(A) → T0(B) → T0(C) → 0,
showing property (1) holds. Property (2) is trivial.
Definition 4.7. A morphism of homological δ-functors from S to T is a collection of natural
transformations ϵn : Sn → Tn for all n ≥ 0 such that if
0 −→ A −→ B −→ C −→ 0
is exact, then
Sn(C)
Sn−1(A)
Tn(C)
Tn−1(A)
δn
ϵn(C)
ϵn−1(A)
δn
commutes.
Definition 4.8. A homological δ-functor T is universal if for every δ-functor S and every
natural transformation ϵ0 : S0 → T0, there exists a unique natural transformation ϵn : Sn
Tn, n ≥ 1, such that ϵ: S → T is a morphism of δ-functors.
Example 4.9. The homology functor H: Ch≥0(C) → C is universal.
5. Projectives and left derived functors
Let C be an abelian category.
Definition 5.1. An element P is projective if for every epimorphism g: B → C and every
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 HomC(P, B) → HomC(P, C) → 0 is exact.)
The functor HomC(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 HomC(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.
Example 5.3. In R-mod, free modules are projective.
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 = P1 ⊕ P2 and P is projective. Then since
HomC(P,−) = HomC(P1,−) ⊕ HomC(P2,−)
is exact, both HomC(P1,−) and HomC(P2,−) are both exact. Indeed, if one of them was
not, then the counterexample would also work for the direct product.
Example 5.6. For R equal to Z or a field or a division ring, an R-module is free if and only
if it is projective.
Theorem 5.7 (Quillen–Suslin). Projective modules over F[x1,...,xn] where F is a field are
free.
Examples 5.8. Suppose R1,R2 are nonzero rings with 1. If R = R1 × R2 then R1 and R2
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.
Definition 5.10. An R-module M is indecomposable if M
≅ M1 ⊕ M2 implies M1 = 0 or
M2 = 0.
Moreover, M is simple if for a submodule M1 ⊆ M we have M1 = 0 or M1 = 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
≅ M1 ⊕ M2 ⊕···⊕ Mr, where M1, ..., Mr
are indecomposable and if M
≅ M1 ⊕ M2 ⊕···⊕ Ms with M1,...,Ms indecomposable, then
r = s and, after reordering, Mi
≅ Mi.
Example 5.12. Taking M = R above, we obtain
R = P1 ⊕ P2 ⊕···⊕ Pr
where P1,P2,...,Pr 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 Pd1
1
⊕ Pd2
2
⊕···⊕ Pdr
r .
We will write R-fdmod for the category of finite-dimensional R-modules.
Example 5.13. Let R = Mn(F) be the ring of n × n matrices over F. Then
R = P ⊕ P ⊕···⊕ P
︷︷
n
where
P = Fn are the ith columns.
The only indecomposables are actually P.
Example 5.14. Let R ⊆ Mn(F) be the subset of upper-triangular matrices. Then
R = P1 ⊕ P2 ⊕···⊕ Pn
where
Pi = ith columns =


...
0
...
0


Then the short exact sequence
0 → P1 → P2 → S2 → 0
is non-split. Hence S2 is not projective.
Definition 5.15. An abelian category C has enough projectives if every M ∈ Obj(C) there
exists an epimorphism f : P → M where P is projective.
Page 36
Example 5.16. In the category of finite abelian groups, there are no projectives except 0.
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.
Example 5.17. If R is a ring with 1, then R-mod has enough projectives. If M is a module,
then we have a map F(M) ↠ M as described above.
If I1,I2 are ideals in a commutative ring with 1, then we have an exact sequence
0 → I1 ∩ I2 → I1 ⊕ I2 → I1 + I2 → 0.
Example 5.18. Let
R =
C[x, y, z]
(xy − z2 − 1)
and I1 = (z − 1,x), I2 = (z + 1,x). These are ideals which are not principal but they are
indecomposable. Then using the equation xy = z2 − 1=(z − 1)(z + 1) we obtain
I1 ∩ I2 = (x).
Since the short exact sequence
0 → (x) → I1 ⊕ I2 → R → 0
splits (R is projective), we have that
I1 ⊕ I2 = R ⊕ (x)
≅ R
2
and so I1, I2 are projective. Hence we obtained two decompositions of R2 into
indecomposables.
Example 5.19. Take the equation y2 = x3 − x, so
R =
C[x, y]
(y2 − x3 + x)
.
Then the maximal ideals of R are projective.
Definition 5.20. A resolution of M ∈ Obj(C) is a nonnegative complex Ptogether with a
morphism ϵ: P0 → M such that
···→ P3 → P2 → P1 → P0 → 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,
Hn(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 P0 ↠ M from a projective P0 and, taking the
kernel K0 → P0, we have a projective P1 with an epi P1 → K0. We take its kernel K1 → P1
and again get a projective P2 ↠ K1. This way, we get that the diagram
Page 37
0
0
K1
···
P2
P1
P0
M
0
K0
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:
···→ P3 → P2 → P1 → P0 → M → 0.
Apply F to P to obtain
···→FP3 → FP2 → FP1 → FP0 → 0.
Define
Li(FM) = Hi(F(P)).
Note that L0F(M) = H0(F(P)) = F(M), since
F(P1) → F(P0) → F(M) → 0
is exact.
Questions: Is LiF(M) well-defined? Is LiF 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→ Qsuch that
P
M
Q
N
f
f
commutes. Moreover, f is unique up to homotopy.
Proof. We construct the family f0,f1,..., making the diagram
Page 38
···
P2
P1
P0
M
0
···
Q2
Q1
Q0
N
0
d2
γ2
f2
d1
γ1
f1
ϵ
f0
γ0
f
d2
d1
η
commute, step by step as follows. The map γ0 = f ϵ: P0 → N is surjective, so it lifts to
f0 : P0 → Q0. Next,
γ1 = f0d1 : P1 ↠ imd1 = kerη
lifts to f1 : P1 → Q1, since P1 is projective. In the next step,
γ2 = f1d: P2 ↠ imd2 = kerd1
lifts to f2 : P2 → Q2, since P2 is projective. We continue this way to construct a chain map
f : P→ Qsuch that the approporiate diagram commutes.
For uniqueness, suppose g: P→ Qis 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 si for h and 0, then ds + sd = h − 0 = f − g, as required. We let
s0 = 0: M → Q0. The map h0 factors through kerη, since ηh0 = 0, and since the map
Q1 → kerη is surjective and P0 is projective, there is a unique map s1 : P0 → Q1 making the
diagram
···
P1
P0
M
0
···
Q1
kerη
Q0
N
0
h1
s1
ϵ
h0
0
0
η
commute. This shows d1s1 + s0ϵ = d1s1 = h0. Now, d1(s1d1 − h1) = h0d1 − d1h1 = 0, so
s1d1 − h1 factors through kerd1, and as Q2 → kerd1 is surjective and P1 is projective, we
get a unique map s2 : P1 → Q2 making the triangle in the diagram
···
P2
P1
P0
M
0
···
Q2
kerd1
Q1
Q0
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 LiF 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
···
P1
P0
M
···
Q1
Q0
N
f1
f0
f
which is unique up to homotopy, so we get a well-defined map
LiF(f )=(f)i : Hi(F(P))
︸ ︷︷ ︸
LiFM
→ Hi(F(Q))
︷︷
LiFN
.
Moreover, if Pand Pare two resolutions of M, then id: M → M gives rise to unique maps
(up to homotopy)
···
P1
P0
M
···
P1
P0
M
f1
f0
idM
g1
g0
idM
so we get maps
f: H(FP) → H(FP)
g: H(FP) → H(FP)
such that (gf)= gf= id
(fg)= fg= 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.
Example 5.23. In R-mod, for R = F[x, y] where F is a field, let m = (x, y) ⊆ R with
M = R/m = F. Then a projective resolution of M is
0
R
R2
R
M
0
−y
x
(
x
y
)
and we apply F = M ⊗R − to this resolution to get
0
F
F2
F
0
0
0
so
Hn(F(P)) =
F,
if n =0or2,
F2, 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 Pn = P0 ⊕P0 , using projectivity and
Snake Lemma 1.43. We note that since we have map P0 → A (composition), P0 → A (lift
from ϵ ), we get a map ϵ: P0 ⊕ P0 → A (using the fact that P0 ⊕ P0 is both a product and
a coproduct) such that the following diagram
0
0
0
kerϵ
P0
A
0
P0 ⊕ P0
A
0
kerϵ
P0
A
0
0
0
ϵ
ϵ
ϵ
commutes. Note that P0 ⊕ P0 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
P0
A
0
0
kerϵ
P0 ⊕ P0
A
0
0
kerϵ
P0
A
0
0
0
0
ϵ
ϵ
ϵ
We then apply the same procedure to the diagram with kernels to construct d1 : P1 ⊕ P1
kerϵ, where the product is projective and the map is epi onto the kernel
0
0
0
0
P1
kerϵ
P0
A
0
P1 ⊕ P1
kerϵ
P0 ⊕ P0
A
0
P1
kerϵ
P0
A
0
0
0
0
0
d1
ϵ
d1
ϵ
d1
ϵ
Continuing this way, this constructs the sequence Pn = Pn ⊕ Pn of projectives with the
desired properties.
Theorem 5.25. The derived functor LiF 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
Pn
Pn = Pn ⊕ Pn
Pn
0
which shows that there is a map FPn → FPn. When we take homology, we get the long
exact sequence
···
H1(F(P ))
H1(F(P))
H1(F(P ))
H0(F(P ))
H0(F(P))
H0(F(P ))
0,
δ
or, in other words,
···
L1FA
L1FA
L0FA
L0FA
L0FA
0.
δ
Hence this δ is an approporiate map and we just have to check naturality and universality.
We omit this here, but the full proof can be found in [Wei94].
6. Injectives and right derived functors
Let C abelian category. For an object M, HomC(−,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) HomC(−,I) is exact,
(3) I is projective in Cop.
Definition 6.2. If any (and hence all) conditions in Lemma 6.1 hold, M is called injective.
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 {(Cx,hx) | x ∈ X} is a chain, so for any x, y ∈ X
(Cx,hx) ≤ (Cy,hy) or (Cx,hx) ≥ (Cy,hy).
Define C = ⋃
x∈X
Cx and h: C → I by setting h(a) = hx(a) if a ∈ Cx for some x ∈ X. This is
well-defined since this is a chain, and h|Cx = hx for any x ∈ X. Hence (Cx,hx) ≤ (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.
Example 6.5. In Ab = Z-mod, the objects Q and Q/Z are injective, and in fact
Q/Z =
p prime
Zp,
where Zp= Z
[
1
p
]
/Z.
Definition 6.6. An abelian category C has enough injectives if for every M ∈ ObjC there
exists a monic M → I where I is injective. So C has enough injectives if and only if Cop has
enough projectives.
Example 6.7. In Ab, we can embed Z → Q and Z/m → Q/Z via 1 + (m) ↦ 1
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
HomAb(M,N)
HomR(M,HomAb(R, N))
f
[(m, r) ↦ f(rm)]
∼=
If N is an injective Z-module, then HomAb(−,N) is exact, and hence
HomR(−,HomAb(R, N))
is exact, showing that HomAb(R, N) is an injective R-module.
Notation. We denote by
AB = the set of functions B → A,
CD = the category of functors D→C.
By the above, the object I0 = HomAb(R,Q/Z) is an injective R-module. The map
Φ: M → I
HomR(M,I0)
0
,
Φ(m)(f) = f(m) ∈ I0.
is injective (as a set map). Hence the category R-mod has enough injectives.
Definition 6.8. A coresolution of M ∈ ObjC is
0
M
I0
I1
···
Page 45
and an injective resolution of M is a coresolution such that In 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.
Example 6.10. In Ab, an injective resolution of Z is
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 RnF, right derived
functors, where for an injective resolution
0
M
I0
I1
···
we set
RnF(M) = Hn(F(I)),
R0F(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).
Definition 7.1. A limit is a functor F : I → A is an object L ∈ ObjA together with
morphisms πi : 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 fi : i → j, then there exists a unique h: L → L such that πih = πi 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).
Example 7.2. If I is a set with no morphisms, then L = ∏
i∈I
F(i) is a product in A.
Definition 7.3. Dually, the colimit, colimi∈I F(i), is an object L together with morphisms
ij : F(j) → L such that the dual universal property holds.
Definition 7.4. A poset I = (I,≤) is directed (filtered) if for all i, j ∈ I, there exists k ∈ I
with i ≤ k, j ≤ k (i.e. i → k, j → k). Then
colimi∈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.
Example 7.5. We have I = N is a poset with
0 → 1 → 2 → 3 →···
and then for F(i) = Z/2i 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/2i, 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.
Definition 7.6. An abelian category A is complete if ∏
i∈I
Ai 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
of the above theory, sheaf cohomology. For more details, see [Har77, Chap. 2].
We begin with a motivating example.
Example 8.1. Let X be a topological space and, for U ⊆ X open, let
F(U) = set of continuous functions U → R.
If V ⊆ U, then we have a restriction map ϱUV : 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, iV U : V → U is the inclusion. Define F(iV U ) = ϱUV . This
makes F into a contravariant functor Top(X) → Ab (or even R-mod).
Definition 8.2. A pre-sheaf of abelian groups is a contravariant functor F : Top(X) → Ab
with F(∅) = 0. We have a category Presheaves(X) which is a subcategory of AbTop(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 )
Definition 8.3. A presheaf F is a sheaf if for every U ⊆ X open and open covering Ui,
i ∈ I, of U we have
(1) if f ∈ F(U) and f|Ui = 0 for all i, then f = 0,
(2) if fi ∈ F(Ui) for all i and for all i, j we have (fi)|Ui∩Uj = (fj)Ui∩Uj then there exists
f ∈ F(U) with f|Ui = fi 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(Ui)
i<j
F(Ui ∩ Uj)
f
(fUi ,i ∈ I)
(gi,i ∈ I)
(gi − gj,i<j)
being exact.
Definition 8.4. Let F be a pre-sheaf. The stalk of F at x is the direct limit
Fx = lim
−→
{F(U) | U
x}
or, equivalently,
Fx = {(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 .
Definition 8.5. If F is a pre-sheaf, then we define F+ by
F+(U) = set of all function U → ∐
x∈U
Fx with f(x) ∈ Fx such that
for every y ∈ U there exists V ⊆ U and y ∈ V and g ∈ F(V )
such that g maps to f(x) ∈ Fx 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
Example 8.7. If X = P1(C), the Zariski topology on X is: D ⊆ X closed if and only if
D = P1(C) or D is finite.
Define a sheaf OX on P1(C) by setting
OX(U) = {f ∈ C(t) | f regular on U},
a subring of the function field C(t).
Write the affine line as A1 = P1 \ {∞} ∼= C. Then
OX(P1) = C, OX(A1) = C[t].
Moveover,
OX(P1 \ {a1,...,an}) = C
[ 1
t − a1
,...,
1
t − an
]
for a1,...,an ∈ A1.
Define a subsheaf I of OX by setting
I(U) = {f ∈ OX(U) | f|U∩{0,∞} = 0}.
There is an exact sequence
0
I
OX
OX/I
0
but the last of these, OX/I, is not a sheaf. Indeed:
(OX/I)(A1)
≅ C[t]/(t),
(OX/I)(P1 \ {0}) = C
[1
t
]/(1
t
)
.
Then 0 + (t) and 1 + (1
t) agree on A1 \ {0}, since
(OX/I)(A1 \ {0})=0,
but since (OX/I)(P1) = C, there is no element there mapping to f1 and f2.
Instead, sheafify to get an exact sequence of sheaves
0
I
OX
(OX/I)+
0
where (OX/I)+(P1) = C2.
Definition 8.8. The global sections functor Γ(X,−): F↦ F(X) is left exact, and
RiΓ(X,F) = Hi(X,F)
is called the sheaf cohomology.
Example 8.9. For the example above, we have the long exact sequence
0
Γ(X, I)
︸ ︷︷ ︸
=0
Γ(X,OX)
︸ ︷︷ ︸
=C
Γ(X,(OX/I)+)
︷︷
=C2
H1(X, I)
︸ ︷︷ ︸
=0
···
Page 50
9. Adjoint functors
Definition 9.1. Let A, B be abelian categories. Then additive functors L: A → B and
R: B → A are adjoint if there exists a natural isomorphism
T : HomB(L(A),B)
≅ HomA(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,
HomA(M,A)
HomA(M,B)
HomA(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
B1
B2
B3
0
is a short exact sequence in B. We apply
HomB(L(A),−)
≅ HomA(A, R(−))
to get
0
HomB(L(A),B1)
HomB(L(A),B2)
HomB(L(A),B3)
0
HomA(A, R(B1))
HomA(A, R(B2))
HomA(A, R(B3))
∼=
∼=
∼=
for all A ∈ ObjA. By Yoneda lemma, we get that
0 → R(B1) → R(B2) → R(B3)
is exact, and so R is left-exact. Similarly, L is right-exact.
Example 9.4. Let
(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 HomS(B,C) is a right
R-module with (fr)(b) = f(rb). There is a natural isomorphism
T : HomS(A ⊗R B,C)
∼=
→ HomR(A,HomS(B,C)).
Page 51
Indeed, if f : A ⊗R B → C, then for we get
a ↦ [f(a ⊗ −): B → C]
Conversely, if A → HomS(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
HomS(B,−): mod-S → mod-R
are an adjoint pair, so − ⊗R B is right exact and HomS(B,−) is left exact.
Then we set
Li(− ⊗R B) = Tori(−,B),
and we will show that
Li(A ⊗R −) = Tori(A,−),
because
Li(− ⊗R B)
≅ Li(− ⊗ B)(A).
Definition 9.5. A double complex is a set {Cp,q}p,q∈Z of objects with horizontal maps
dh : Cp,q → Cp−1,q
and vertical maps
dv : Cp,q → Cp,q−1
such that dh ◦ dh = 0, dv ◦ dv = 0, and dvdh + dhdv = 0
...
...
...
···
Cp−1,q+1
Cp,q+1
Cp+1,q+1
···
···
Cp−1,q
Cp,q
Cp+1,q
···
···
Cp−1,q−1
Cp,q−1
Cp+1,q−1
···
...
...
...
The map dv : C•q → C•q−1 is almost a chain map: if we set fpq = (−1)pdv
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)).
Definition 9.6. Assume C is cocomplete. We set define the total complex
Tot
(C••)n =
p+q=n
Cp,q
with d = dv + dh, whence
d2 = (dv + dh)2 = (dv)2 + (dvdh + dhdv)+(dh)2 =0+0+0=0.
Page 52
If C is complete,
Totπ(C••)n =
p+q=n
Cp,q
with d = dv + dh.
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. Cp,q = 0 if q < 0) and columns are exact.
Then Totπ(C) is acyclic (exact).
Proof. We claim that H0(Totπ(C••)) = 0. By symmetry, it is enough to restrict our attention
to the p = 0 portion of the diagram
...
...
...
···
C−1,2
C0,2
C1,2
···
···
C−1,1
C0,1
C1,1
···
···
C−1,0
C0,0
C1,0
···
0
0
0
Suppose
(c0,c1,c2,...) ∈ C00 × C−1,1 × C−2,2 ×···
is a cycle. Then
d(c0,c1,c2,...)=(dhc0 + dvc1,dhc1 + cvc2,...) ∈ C−1,0 × C−2,1 ×··· .
We want to find
(b0,b1,...) ∈ C10 × C01 × C−12 ×···
with
d(b0,b1,...)=(dhb0 + dvb1,dhb1 + dvb2,...)=(c0,c1,...).
Pick b0 = 0. Then dvc0 = 0 so c0 = dvb1 for some b1. Then
0 = dhc0 + dvc1 = dhdvb1 + dvc1 = −dvdhb1 + dvc1 = dv(c1 − dhb1).
Hence there exists b2 ∈ C−1,2 such that dvb2 = c1 −dhb1. Then c1 = dvb2 +dhb1, and proceed
by induction to construct b3, b4, .... This completes the proof.
Corollary 9.8. If C•• is a double complex in the right half plane with exact rows, then
Totπ(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 τnC•• by
nC)p,q =
Cpq
if q > n,
kerdv : Cpn → Cpn−1
if q = n,
0
if q < n.
Then we get the diagram
...
...
...
0
C0,n+2
C1,n+2
C1,n+2
···
0
C0,n+1
C1,n+1
C1,n+1
···
0
kerdv
kerdv
kerdv
···
0
0
0
Hence TotπnC) = Tot
nC) is exact. This shows that Tot
(C) is exact.
Theorem 9.10. We have that
Ln(A ⊗R −)(B)
≅ Ln(− ⊗R B)(A)
and we call them TorR
n (A, B).
We will actually prove a more general statement, following [Wei94, Exer. 2.7.4]. (This was
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
RT(A1,...,
ˆ
Ai,...,Ap)(Ai)
≅ R
T(A1,...,
ˆ
Aj,...,Ap)(Aj).
Proof. Note that if we fix modules A1,...,
ˆ
Ai,...,
ˆ
Aj,...,Ap, then the functor
T(A1,...,
ˆ
Ai,...,
ˆ
Aj,...,Ap): 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
is right-balanced. We mimic the proof of [Wei94, Theorem 2.7.2] to show
RT(A,−)(B)
≅ R
T(−,B)(A).
If the first variable is covariant, choose an injective resolution ϵ: A → Pand if it is
contravariant, choose a projective resolution ϵ: P→ A. Similarly, if the second variable is
covariant, choose an injective resolution η: A → Qand if it is contravariant, choose a
projective resolution η: Q→ A. We then get the double complex:
...
...
...
...
0
T(A, Q2)
T(P0,Q2)
T(P1,Q2)
T(P2,Q2)
···
0
T(A, Q1)
T(P0,Q1)
T(P1,Q1)
T(P2,Q1)
···
0
T(A, Q0)
T(P0,Q0)
T(P1,Q0)
T(P2,Q0)
···
0
T(P0,B)
T(P1,B)
T(P2,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, Q2)
T(P0,Q2)
T(P1,Q2)
T(P2,Q2)
···
0
T(A, Q1)
T(P0,Q1)
T(P1,Q1)
T(P2,Q1)
···
0
T(A, Q0)
T(P0,Q0)
T(P1,Q0)
T(P2,Q0)
···
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 Qi are injective if the second variable is covariant and
projecitve if the second variable is contravariant, the right-balanced condition shows that
C•• has exact rows. Then by Acyclic Assemply Lemma 9.7, we obtain that Tot(C•,•)[1] is
acyclic.
This completes the proof.
10. Tor and Ext
We restrict our attention to A = Ab.
Example 10.1. Consider the projective resolution
0
Z
Z
Z/n
0
For an abelian group B, apply − ⊗Z B to Pto get
0
B
B
0
Hence
Tor0(Z/n, B) = Z/n ⊗ B = H0(P⊗ B) = B/nB,
Tor1(Z/n, B) = B[n] = {b ∈ B : nb = 0},
Tork(Z/n, B) = 0 for k ≥ 2,
Tor0(Z,B) = B,
Tork(Z,B) = 0 for k ≥ 1,
In general,
Tori(Z/n,Z/m) = Z/(n, m) for i = 0,1.
To calculate Ext, we apply HomAb(−,B) to Pto get
0
B
B
0
Page 56
and then
Ext0(Z/n, B) = H0(P) = B[n]
Ext1(Z/n, B) = H1(P) = B/nB
and in particular
Exti(Z/n,Z/m) = Z/(n, m) for i = 0,1.
If A = lim
−→
Aα = colimAα, then
Tori(A, B) = Tori(lim
−→
Aα,B) = lim
−→
Tori(Aα,B).
Proposition 10.2. Suppose A and B are abelian group. Then
(1) Tor1(A, B) is a torsion group,
(2) Torn(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
−→
Aα for {Aα} finitely generated subgroups. The limit of torsion groups is
torsion, so
Tor1(A, B) = lim
−→
Tor1(Aα,B)
is torsion.
Example 10.3. We have that
Tor1(Q/Z,B) = lim
−→
Tor1
(
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
Tor1(A, B) = lim
−→
Tor1(Zm,B)=0.
Hence A is torsion-free if and only if Tor1(A,−) = 0 if and only if A ⊗Z − is exact, or by
definition, A is a flat Z-module.
Definition 10.4. A left R-module B is flat if − ⊗R B is exact, and a right R-module A is
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−1R, is
S−1R = {(r, s) | r ∈ R, s ∈ S}/ ∼
Page 57
where
(r1,s1) ∼ (r2,s2) if and only if there exists s3 ∈ S such that (r1s2 − r2s1)s3 = 0.
We think of (r, s) as r
s
. If [(r, s)] is an equivalent class, then
[(r1,s1)] + [(r2,s2)] = [(r1s2 + r2s1,s1s2)],
[(r1,s1)] · [(r2,s2)] = [(r1r2,s1s2)].
We have a map ϕ: R → S−1R, setting ϕ(r) = [(r,1)]. We then set
S−1M := S−1R ⊗R M, the localization of M at S.
Theorem 10.5. The localization S−1R is a flat R-module, i.e. S−1R⊗R− is an exact functor
from R-mod to S−1R-mod.
Proof. Define category I with
Obj(I) = S,
for s1,s2 ∈ S, we set HomI(s1,s2) = {s ∈ S | ss1 = s2}.
This is a filtered category (see definition below)
s1
s1s2
s2
s2
s1
s1
s2
s1s2
t1
t2
s1
We then have a functor F : I → R-mod given by F(s) = R for s ∈ Obj(I) = S and if
s1
s
→ s2 is a morphism then
F(s1) = R
F(s2) = R
F(s)=s·
We claim that colims∈I F(s) exists and in fact colims∈I F(s) = S−1R. Indeed, we define
ϕs : F(s) = R → S−1R
by ϕs(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
Torn(S−1R, B) = colim Torn(F(s),B) = colim Torn(R, B)
︷︷
=0
= 0,
and hence S−1R is flat.
Definition 10.6. A category C is filtered if
(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) Tor1(A, B) = 0 for all left R-modules B,
(4) Torn(A, B) = 0 for all left R-modules B and n ≥ 1.
(The first equivalence is the definition.)
Definition 10.7. If B is a left R-module, then B= HomAb(B,Q/Z) is the right Pontryagin
dual of B.
If B = 0, let C be a maximal subgroup. Then B/C
≅ Z/p for p prime, and hence
HomAb(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→ Bis surjective,
where f= HomAb(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 fis surjective.
Proposition 10.9. The following are equivalent:
(1) B is a flat left R-module,
(2) Bis an injective right R-module,
(3) I ⊗R B
∼=
→ IB for every right ideal I ⊆ R,
(4) Tor1(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
Tor1(R, B)
︷︷
=0
Tor1(R/I,B)
I ⊗R B
R ⊗R B
︸ ︷︷ ︸
∼=B
R/I ⊗R B
︸ ︷︷ ︸
B/IB
0
ϕ
and hence
0
Tor1(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
HomR(A, B) = HomR(A,HomZ(B,Q/Z)))
≅ Hom(A ⊗R B,Q/Z)=(A ⊗R B).
If A ⊆ A is a submodule, then the following diagram
HomR(A, B)
Hom(A ,B)
(A ⊗R B)
(A ⊗R B)
∼=
∼=
commutes. Now,
Bis injective if and only if HomR(A, B) → HomR(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 Bis injective if and only if
B
︸︸
HomR(R, B) →
(I⊗B)
︸︸
HomR(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.
Definition 10.10. A module M is finitely presented if there exists an exact sequence
Rm → Rn → 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: AR M → HomR(M,A)
σ(f ⊗ m)(h) = f(h(m))
for f ∈ A, m ∈ M, h ∈ HomR(M,A).
Proposition 10.11. If M is finitely presented, then σ is an isomorphism for all A.
Proof. Clear if M = Rn. Now, if
Rm
Rn
M
0
then we have the following commutative diagram
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AR Rm
AR Rn
AR M
0
0
HomR(Rm,A)
HomR(Rn,A)
HomR(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→ Bis
injective. Hence the square
CR M
B⊗ M
Hom(M,C)
Hom(M,B)
f⊗1
∼=
∼=
commutes. Since M is flat, f⊗ 1 is injective, and hence
HomR(M,B) → HomR(M,C)
is surjective. This shows that HomR(M,−) is exact, so M is projective.
Lemma 10.13 (Flat resolution lemma). If ··· → F2 → F1 → F0 → A → 0 is a flat
resolution, then
TorR
n (A, B) = Hn(F⊗ B)
for a right R-module A and a left R-module B.
If Fn are in fact projective, this is how we defined TorR
n (A, B), and this lemma shows that
we can compute TorR
n (A, B) by considering only a flat resolution.
Proof. For n = 0, we have an exact sequence
F1 R B
F0 R B
A ⊗R B
0
and hence H0(FR B) = A ⊗R B = Tor0(A, B).
We have an exact sequence
0
K
F0
A
0.
We then get a long exact sequence
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=0
︸︸
Tor2(F0,B)
Tor2(A, B)
Tor1(K, B)
=0
︸︸
Tor1(F0,B)
Tor1(A, B)
K ⊗ B
F0 ⊗ B
A ⊗ B
0
and hence Torn(A, B)
≅ Torn−1(K, B). We have the exact sequence
F2
F1
K
0
which gives the exact sequence
F2 ⊗ B
F1 ⊗ B
K ⊗ B
0
and we get that
Tor1(A, B) = ker(K ⊗ B → F0 ⊗ B) = ker
( F1 ⊗ B
d2(F2 ⊗ B)
→ F0 ⊗ B
)
= H1(F⊗ B).
By induction on n, we finally obtain:
Torn(A, B) = Torn−1(K, B) = Hn−1(F[1] ⊗ B) = Hn(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
TorR
n (A, C)
≅ Tor
T
n (A ⊗R T,C).
Proof. Let P→ A be a projective resolution so that
TorR
n (A, C) = Hn(PR C).
Note that Pn R T is a projective T-module: Pn ⊕Q = F for some free R-module F, whence
F ⊗ T = (Pn ⊕ Q) ⊗ T = (Pn ⊗ T) ⊕ (Q ⊗ T),
so Pn ⊗ T is a direct summand of the free T-module F ⊗ T. Hence
PR T → A ⊗R T
is a projective resolution (since − ⊗R T is exact). Hence
TorT
n (A ⊗R T,C) = Hn(PR T ⊗T C) = Hn(PR 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 TorR
n (A, B)
≅ Tor
T
n (A ⊗R T,T ⊗R B).
Page 62
Proof. We have that:
TorT
n (A ⊗R T,T ⊗R B) = TorR
n (A, T ⊗R B) = Hn(PR T ⊗R B) = Hn(T ⊗R PR B)
= T ⊗R Hn(PR B) = T ⊗R TorR
n (A, B),
as required.
Example 10.16. Suppose p ⊆ R is a prime ideal, R is a commutative ring and Rp = S−1R
for S = R \ p. If M is an R-module, Mp = Rp R M, then
TorR
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 TorR
n (A, B)
≅ Tor
S−1R
n
(S−1A, S−1B).
Example 10.17. Suppose A, B are abelian groups. Then there exists I0 injective such that
0 → B → I0 → I0/B → 0
is exact, but since I0 is divisible, I1 = I0/B is also divisible, so it is injective. Hence
0 → B → I0 → I1 → 0
is an injective resolution. This shows that
Extn(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,
Ext1(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 Extn
R(A, B) = Extn
S−1R(S−1A, S−1B).
In particular, for a prime p, we have
Extn
R(A, B)p = Extn
Rp
(Ap,Bp).
Recall that, to check if an R-module M is 0, it is enough to check that for any prime ideal p,
Mp = 0. So in this case, to check that Extn
R(A, B) = 0, it is enough to check that
Extn
Rp
(Ap,Bp)=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”.
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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
HomR(C, B)
HomR(C, C)
Ext1
R(C, A)
?
1C
1C
0
δ
?
Answer: The short exact sequence E splits if and only if δ(1C)=0 ∈ Ext1
R(C, A).
Definition 10.19. The obstruction of E as above is θ(E) := δ(1C) ∈ Ext1
R(C, A).
Remark 10.20. We can also compute the obstruction of E as follows. Take 1C and lift it
to a projective resolution of C, and a map to E
···
P2
P1
P0
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 ∈ HomR(P1,A) defines the same class as θ(E) = δ(1C). We have
the following diagram:
δ(1C) = u ∈ Hom(P1,A)
Hom(P1,B)
v ∈ Hom(P0,B)
Hom(P0,C))
Hom(C, C) 1C
We choose 1C ∈ Hom(C, C), map it to Hom(P0,C), lift it to v ∈ Hom(P0,B), map it to
Hom(P1,B), and lift it to u ∈ Hom(P1,A).
The maps u, v,1C 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 Ext1
R(C, A).
Definition 10.21. An extension in an abelian category A is a short exact sequence
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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.
Definition 10.22. The trivial extension of C by A is
0
A
A ⊕ C
C
0
(1,0)
Example 10.23. What are the extensions of Z/p by Z (in Z-mod)?
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 ext1(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: ext1(C, A) → Ext1(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: ext1(C, A) → Ext1(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 ∈ Ext1(C, A), choose a projective resolution P→ C, and a representative φ ∈
HomR(P1,A):
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P2
P1
P0
C
0
A
d1
φ
d0
Take the pushout and extend it to an isomorphism of its cokernel:
P2
P1
P0
C
0
A
B
C
0
d1
φ
d0
=
Since φ represents a cocycle, it factors through S = P1/im(P2). Hence we have another
pushout square (we give it the same name by abuse of notation):
0
S
P0
C
0
0
A
B
C
0
d1
φ
d0
=
where the map A → B is injective, because S → P0 is injective.
We use this short exact sequence in the setting above:
P2
P1
P0
C
0
E :
0
A
B
C
0
d1
φ
d0
=
By the Remark 10.20 earlier, θ(E)=[φ] ∈ Ext1
R(C, A).
To conclude that this inverse construction is well-defined, we need to show that the same
Ext1-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
P0
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).
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• There is a way to add extensions that is compatible with θ.
• We can multiply Exti(C, A) ⊗ Extj(D, C) → Exti+j(D, A) that comes from splicing
exact sequences.
• We have a notion of Exti in any abelian category.
11. Universal coefficients theorem
Recall that a right R-module is flat if and only if Tor1(A, M) = 0 for all M if and only if
Torn(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, Pa complex of flat R-modules, M an R-module.
Theorem 11.1 (Künneth). Assume Bn(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
Hn(P) ⊗R M
Hn(PR M)
TorR
1 (Hn−1(P),M)
0
Examples 11.2.
(1) If R is a field, TorR
1 (−,−) = 0, we get the obvious isomorphism
Hn(P) ⊗R M
≅ Hn(PR M),
because − ⊗R M is exact when R is a field.
(2) Let R = Z, P= Z
2
→ Z, M = Z/2. Then
Hi(P) =
{
Z/2 if i = 0
0
otherwise
But
PR M = Z/2
2
→ Z/2
and hence
Hi(PR M) =
{
Z/2 if i = 0,1
0
otherwise
To see this via Künneth Theorem 11.1, we note that
0
H1(P⊗ M)
TorZ
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
H1(P) ⊗R M
︷︷
=Z/2⊗Z/4Z/2=Z/2
H1(PR M)
︷︷
H1(Z/2
2
→Z/2)=Z/2
TorR
1 (H0(P),M)
︷︷
TorZ/4
1
(Z/2,Z/2)=Z/2
0
This is impossible for cardinality reasons.
Above, to find TorZ/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
TorZ/4
1
(Z/2,Z/2) = H1(KZ/4 Z/2) = Z/2.
Proof of Künneth Theorem 11.1. Note that B(P) ⊆ Z(P) ⊆ Pare complexes where
Z(P) and B(P) have trivial boundary maps.
We claim that Zn(P) is a flat R-module. We have a short exact sequence
0
Zn(P)
Pn
Bn−1(P)
0
d
As each Bn−1(P) is flat, this also shows that Zn(P) is flat. The long exact sequence in
homology gives
···
Hn(B(P) ⊗ M)
Hn(Z(P) ⊗ M)
Hn(P⊗ M)
Hn−1(B(P) ⊗ M)
Hn−1(Z(P) ⊗ M)
···
αn
αn
and hence we have the short exact sequence
(∗)
0
coker(αn)
Hn(PR)
ker(αn−1)
0
As Z(P) has a trivial differential, the same is true for Z(P) ⊗ M and B(P) ⊗ M. This
shows that
Hn(Z(P) ⊗ M) = Zn(P) ⊗ M,
Hn(B(P) ⊗ M) = Bn(P) ⊗ M.
But since Bn(P) and Zn(P) are flat, and we have the short exact sequence
0
Bn(P)
Zn(P)
Hn(P)
0,
this is a flat resolution of Hn(P). Therefore
H(Bn(P) ⊗ M
αn
→ Zn(P) ⊗ M) = TorR
(Hn(P),M),
since Tor can be calculated using flat resolutions 10.13. This shows that
coker(αn) = TorR
0 (Hn(P),M) = Hn(P) ⊗ M,
ker(αn) = TorR
1 (Hn(P),M).
Substituting this into the short exact sequence (∗) completes the proof.
Page 68
Corollary 11.3. Assume R = Z (or Bn(P) is free). Then for all M, the Künneth sequence
splits (non-canonically), i.e. we have isomorphisms
(Hn(P) ⊗ M) ⊕ TorR
1 (Hn−1(P),M)
≅ Hn(P⊗ M).
Proof. We know that each d(Pn) is free. Hence the short exact sequence
0
Zn
Pn
d(Pn)
0
splits (non-canonically!), and so Pn
≅ Zn ⊕ d(Pn). Taking − ⊗ M of both sides, we obtain
Zn ⊗ M ⊆ ker(dn ⊗ 1) ⊆ Pn ⊗ M
≅ Zn ⊗ M ⊕ d(Pn) ⊗ M.
We hence get that ker(dn ⊗ 1)
≅ Zn ⊗ M ⊕ C, for some complement C. Taking the quotient
by im(dn+1 ⊗ 1) = im(dn+1) ⊗ M, we get
Hn(P⊗ M)
≅ Hn(P) ⊗ M ⊕ C.
Hence the Künneth exact sequence
0
Hn(P) ⊗Z M
Hn(PZ M)
TorZ
1 (Hn−1(P),M)
0
splits, with C
≅ TorZ
1 (Hn−1(P),M).
Example 11.4. Let X be a topological space and S(X) be the singular chain complex. If
M is some abelian group,
Hn(X;M) = Hn(S(X) ⊕ M) = Hn(S(X)) ⊗ M ⊕ TorZ
1 (Hn−1(S(x)),M),
and hence
Hn(X;M) = Hn(X) ⊗ M ⊕ TorZ
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 = P2(R) and M = Z/2. We then have:
H0(X) = Z, H1(X) = Z/2, H2(X)=0,
and hence
H2(X;Z/2) = H2(X) ⊗Z Z/2
︷︷
=0
⊕Tor1(H1(X),Z/2) = Tor1(Z/2,Z/2) = Z/2.
There is an analog of the corollary for cohomology.
Theorem 11.5. Suppose Pis a chain complex of left R-modules such that d(Pn) is projective
for all n. Then we have
Hn(HomR(P, M))
≅ HomR(Hn(P),M) ⊕ Ext1
R(Hn−1(P),M).
Page 69
Example 11.6. Again, let X = P2(R). By the above result and knowing the homology
groups of X, we obtain
H0(X) = Z,
H1(X) = H1(X;Z) = HomZ(Z/2,Z)
︷︷
=0
⊕Ext1(Z,Z)
︸ ︷︷ ︸
=0
,
H2(X) = Hom(0,Z) ⊕ Ext1(Z/2,Z) = Z/2.
Let Pbe a complex of right R-modules and Qbe a complex of left R-modules. We have
the double complex
...
...
...
···
P0 ⊗ Q2
P1 ⊗ Q2
P2 ⊗ Q2
···
···
P0 ⊗ Q1
P1 ⊗ Q1
P2 ⊗ Q1
···
···
P0 ⊗ Q0
P1 ⊗ Q0
P2 ⊗ Q0
···
...
...
...
and the total complex is
(P ⊗R Q)n =
p+q=n
Pp ⊗ Qq.
There is an analog of Künneth Theorem for the total complex.
Theorem 11.7. If Pn, d(Pn) are flat for all n, then
0
p+q=n
Hp(P) ⊗ Hq(Q)
Hn(P⊗ Q)
p+q=n−1
Tor1(Hp(P),Hq(Q))
0
is exact.
For topological spaces X, Y , (after some work) this gives the result
Hn(X × Y )
p+q=n
Hp(X) ⊗ Hq(Y ) ⊕
p+q=n−1
TorZ
1 (Hp(X),Hq(Y )).
12. Quivers
Definition 12.1. A quiver is a finite direction graph, consisting of Q = (Q0,Q1, h, t), where
• Q0 is a finite set of vertices,
• Q1 is a finite set of arrows,
Page 70
• h: Q1 → Q0 is the head map; h(a) ∈ Q0 is the head of arrow a ∈ Q1,
• t: Q1 → Q0 is the tail map; t(a) ∈ Q0 is the tail of arrow a ∈ Q1.
Example 12.2. The picture
b
a
c
d
1
2
represents a quiver with Q0 = {1,2}, Q1 = {a, b, c, d} and
t(a) = t(b) = h(c) = t(d) = h(d)=1,
h(a) = h(b) = t(c)=2.
Definition 12.3. A path p of length d ≥ 1 is a sequence
p = adad−1 ...a2a1
where t(ai+1) = h(ai) for i = 1,2,...,d − 1. Then h(p) = h(ad) is the head of path p, and
t(p) = t(a1) is the tail of path p. Also, for every x ∈ Q0, we have a path ex of length 0 with
h(ex) = t(ex) = x.
Example 12.4. In the example above, p = bdca is a path with t(p) = t(a) = 1 and
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.
Definition 12.5. If p, q are paths and t(p) = h(q), say
p = adad−1 ...a1, q = bebe−1 ...b1,
then
pq = adad−1 ...a1bebe−1 ...b1
is the composition. If t(p) = x, then pex = p, and if h(p) = y, then eyp = p.
We can associate a category PQ to a quiver Q:
• objects are elements of Q0,
• HomPQ (y, x) = {paths p from x to y}, i.e. h(p) = y, t(p) = x,
• idx = ex,
• the composition map HomPQ (z,y) × HomPQ (y, x) → HomPQ (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
Definition 12.6. The category of representations of Q over a field K is
RepK(Q)=(K-mod)PQ .
Explicitly, the objects in RepK(Q) are determined by a set of finite-dimensional vector spaces
V (x) for each x ∈ Q0, and K-linear maps
V (a): V (ta) → V (ha)
for each a ∈ Q1. Moreover,
V (ex) = idV (x),
V (adad−1 ...a2a1) = V (ad)V (ad−1)...V (a2)V (a1).
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.
Example 12.7. Consider the quiver
a
b
1
2
3
The paths are e1,e2,e2, 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).
Definition 12.8. Let Q be a quiver and K be a field. The path algebra KQ is defined as
• 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
ex.
Example 12.9. Consider the quiver Q:
1
a
The paths are e1, a, a2,a2,..., and hence KQ = K[a], the polynomial ring in a.
Page 72
Example 12.10. Consider the quiver Q:
a
a
Then KQ = K〈a, b〉 is the free associative algebra generated by a and b (non-commutative).
Example 12.11. Consider the quiver
...
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 RepK(Q)
are equivalent.
Sketch of the proof. If M is a finite-dimensional KQ-module, then
M =
x∈Q0
exM =
x∈Q0
exM,
and we can define V (x) = exM. Then
exey =
{ ex
x = y,
0 otherwise,
and for a ∈ Q1, with t(a) = x, h(a) = y, the multiplication by a map restricts to
V (a): exM
︸︷︷︸
V (x)
→ eyM
︸︷︷︸
V (y)
.
Then we can define F : KQ-mod → RepK(Q) by F(M) = V . Conversely, if V is a
representation of Q, let
M =
x∈Q0
V (x),
and an arrow a ∈ Q1 acts on M by
M
M
V (x)
V (y)
V (a)
This defines a map G : RepK(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 dimK 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) = exM and the
map
a·: M → M
restricts to
M(a): M(t(a)) → M(h(a)).
Then
KQ =
x∈Q0
KQex =
x∈Q0
Px
as left KQ-modules. Then Px = KQex is a projective KQ-module and Px has a basis
consisting of all paths starting at x. Note also that Px(y) = eyPx = eyKQex is spanned by
all paths from x to y.
Example 12.13. For the quiver Q given by
1
a
the category RepK(Q) is naturally isomorphic to K[a]-mod.
Example 12.14. For the quiver
1
2
3
we have
P1 :
Ke1
Ka
Kba
K
K
K
1
1
P2 :
0
K
K
P3 :
0
0
K
Then note that KQ will be
∗ ∗ ∗
0 ∗ ∗
0 0 ∗
with
P1 =
,P2 =
0
,P3 =
0
0
.
The map a: x → y corresponds to Pz(a): Pz(x) → Pz(y) which maps a path p from z to x
to the path ap.
Moreover,
HomKQ(Px,M) → M(x)
(ϕ: Px → M) ↦ ϕ(ex) ∈ M(X)
Page 74
is an isomorphism, and HomKQ(Px,−) 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, md is the ideal spanned by all paths of length ≥ d.
An ideal J ⊆ KQ is admissible if md ⊆ J ⊆ m2 for some d. Then A = KQ/J is a
finite-dimensional K-algebra.
Definition 12.15. Two rings A, B are called Morita equivalent if A-Mod and B-Mod are
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 ex + J by ex and m + J/J by m. Then
A =
x∈Q0
Aex =
x∈Q0
Px
where Px = Aex is projective. Again,
HomA(Px,M) = M(x) = exM
and Px 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 Aop-mod = mod-A is equivalent to A-mod via the map
M ↦ D(M) = M= HomK(M,K),
f ↦ D(f) = f.
Then Ix = (exA), x ∈ Q0 are the indecomposable injectives.
If Q is a quiver, the simple representations are Px/mPx = Sx with
Sx(y) =
{ K if y = x,
0 if y = x,
and mSx = 0.
We then have an exact sequence
0
a, t(a)=x
Ph(a)
Px
Sx
0
Ph(a)
p
pa
which gives a projective resolution of the simple module Sx. 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
Px K M(x)
M
0
p ⊗ w
pa ⊗ w − p ⊗ aw
p ⊗ v
pv
Then
Extn
KQ(M,N)=0if n ≥ 2.
Example 12.17. Take the quiver
1
2
3
again and let A = KQ/(ba). Then
P1 :
K
K
0,
P2 :
0
K
K,
P3 :
0
0
K.
We then have that
0
P3
P2
P1
S1
0
and applying HomA(−,S3) to P, we get
0
HomA(P1,S3)
︷︷
=0
HomA(P2,S3)
︷︷
=0
HomA(P3,S3)
︷︷
∼=K
0.
Hence
Ext2
A(S1,S3) = K.
Example 12.18. Take the quiver Q
1
a
and consider J = (a2) ⊆ KQ = K[a]. Then
P1 = A = KQ/J = K[a]/(a2).
In this case,
···
P1
P1
P1
S1
0
·a
·a
and Extn(S1,S1) = 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 Sx, x ∈ Q0.
• Ext1
A(Sx,Sy) = Kl where l is the number of arrows x → y.
13. Homological dimension
Definition 13.1. Let A be a right R-module.
(1) The projective (resp. flat) dimension , pd(A) = n (resp. fd(A)) is the smallest n such
that there is a resolution
0
Pn
···
P2
P1
P0
A
0
such that P0,...,Pn are projective (resp. flat).
(2) The injective dimension, id(A) is the smallest n such that there is an injective
resolution
0
A
E0
E1
···
En
0.
Lemma 13.2. The following are equivalent
(1) pd(A) ≤ d,
(2) Extn
R(A, B)=0 for n>d and all right R-modules B,
(3) Extd+1
R
(A, B)=0 for all B,
(4) if
0
Ad
Pd−1
···
P2
P1
P0
A
0
is a resolution of A with P0,...,Pd−1 projective, then Ad is projecitve.
Proof. We note that trivially, (4) implies (1) implies (2) implies (3). We show (3) implies (4).
Suppose (3) is true. Let A0 = A and define Pk projective and Ak+1 recursively so that
0
Ak+1
Pk
Ak
0
is exact. Then the long exact sequence for Ext gives
Extl(Pk,B)
︷︷
=0
Extl(Ak+1,B)
Extl+1(Ak,B)
Extl+1(Pk,B)
︷︷
=0
.
Then Ext1(Ad,B) = ··· = Extd+1(A0,B) = 0 for all B, and hence Ext1(Ad,B) = 0, so Ad is
projective.
Dually, we get the following statement.
Lemma 13.3. The following are equivalent:
(1) id(B) ≤ d,
(2) Extn
R(A, B)=0 for n>d and all A,
(3) Extd+1
R
(A, B)=0 for all A,
(4) if
Page 77
0
A
E0
E1
···
Ed−1
Ad
0
is exact and E0,...,Ed−1 are injective then Ad is injective.
Note that
sup{id(B) | B right R-module} = sup{d | Extd(A, B) = 0 for some right R-modules A, B}
= sup{pd(A) | A right R-module}.
Definition 13.4. This is called the right global dimension of R, rgldim(R).
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 → P1 → P0 → 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 | TorR
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}
Definition 13.6. This number is the Tor-dimension of R, tordim(R).
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
Ad
Pd−1
···
P1
P0
A
0
with P0,...,Pd−1 finitely generated and free. Then Ad is finitely generated (by a lemma
analogous to Lemmas 13.2 and 13.3 but for flat dimension), and Ad is flat. Hence Ad is
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.
Definition 13.8. A ring R is semi-simple if every right (equivalently, left) ideal is a direct
summand of R.
Theorem 13.9 (Wedderburn’s Theorem). If R is semi-simple, then
R
r
i=1
Matni,ni (Di),
for division rings Di.
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.
Example 13.11. Let Q be a quiver and KQ be a path algebra. We have seen that for any
KQ-module A, we have a projective resolution
0
P1
P0
A
0
and hence lgldim(KQ) = rgldim(KQ) ≤ 1.
If R = KQ/J (a finite-dimensional K-algebra) with md ⊆ J ⊆ m2 (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.
Definition 13.12. A ring R is von Neumann regular if for any a ∈ R, there exists b ∈ R
such that aba = a.
Example 13.13. Let kX be the set of functions X → k where k is a field and X is a set.
For a: X → k, define
b(x) =
{ 1
a(x)
if a(x) = 0,
0
if a(x)=0,
whence aba = a2b = a.
Example 13.14. The ring Matn×n(k) for a field k. For a matrix A, we have a map kn → imA
and we can choose a splitting B: imA → kn. Extend B to kn → kn 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, e2 = 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) = e2+ef−e2f = e+ef−ef = e and similarly for f. The non-commutative
case is similar.
Example 13.16. Define R ⊆ RR be the ring
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
−→
Iα for Iα finitely generated ideal
and
R/I = lim
−→
R/Iα,
For all left R-modules M, we have
Tor1(R/I,M) = lim
−→
Tor1(R/Iα,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[x1,...,xn]) = n.
Hilbert showed that if M is finitely generated, then it has a free resolution of length n
0
Fn
···
F2
F1
F0
M
0
showing that the global dimension is at most n.
Writing R = k[x1,...,xn], the R-module k has the Koszul resolution
0
R(
n
n)
···
R(
n
2)
Rn
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
TorR
j (k, k) = k(
n
j ).
In particular, this will show equation (1).
Proposition 13.17. If f : R → S is a ring homomorphism and M is an S-module, then
pdR(M) ≤ pdS(M) + pdR(S).
Proof. Let pdS(M) = n, pdR(S) = d and choose a projective S-resolution of M,
0
Qn
···
Q1
Q0
M
0
and let M0 = M and
0 −→ Mi+1 −→ Qi −→ Mi −→ 0.
Choose projective R-resolutions of Mi’s. Then the Horseshoe Lemma 5.24 gives projective
resolutions P•j → Qj. We then have a double complex (by adjusting the signs of the maps
appropriately)
Page 81
...
...
...
0
Q2
P02
P12
···
0
Q1
P01
P11
···
0
Q0
P01
P10
···
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 pdR(Q1) ≤ pdR(S) = d because Qi is a
direct summand of a free S-module. We replace Pd,i by Pd,i/imPd+1,i to get
Tot(P••) → M,
a projective R-resolution of M. Then we obtain
pdR(M) ≤ n + d = pdS(M) + pdR(S),
as required.
Lemma 13.18. Suppose
0
A
B
C
0
is an exact sequence of R-modules. Then
pdR(B) ≤ max{pdR(A),pdR(C)}
and if the inequality is strict, then pdR(C) = pdR(A)+1.
Proof. By the long exact sequence for Ext, we get
···
Exti(C, M)
Exti(B,M)
Exti(A, M)
Exti+1(C, M)
Exti+1(B,M)
Exti+1(A, M)
···
If i = pdR(B), then for some R-module M, we obtain Exti(B,M) = 0, and hence one of
the neighboring terms in the long exact sequence above are non-zero, so Exti(C, M) = 0 or
Exti(A, M) = 0, showing that pdR(C) ≥ i or pdR(A) ≥ i.
Page 82
If the inequality is strict, then for any i > pdR(B), we get Exti(B,M) = Exti+1(B,M) = 0,
so Exti(A, M)
≅ Ext
i+1(C, M) by the long exact sequence above, which shows that pdR(C) =
pdR(A) + 1.
Let R be a ring, x ∈ R be central, A a left R-module.
Definition 13.19. An element x is a nonzero divisor on A if xy = 0 implies that y = 0 for
all y ∈ A.
Suppose x is a nonzero divisor on R. We have a short exact sequence
0
R
R
R/x
0.
We apply − ⊗R A to get
Tor1(R, A)
︸ ︷︷ ︸
=0
Tor1(R/x, A)
︷︷
{y∈A | xy=0}
A
A
A/xA
0
and x is a nonzero divisor on A if and only if Tor1(R/x, A) = 0.
Let (R,m) be a commutative Noetherian local ring with m its unique maximal ideal.
Definition 13.20. A regular sequence in a finitely-generated R-module A is a sequence
x1,...,xn ∈ m such that xi is a nonzero divisor on A/(x1,...,xi−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).
Definition 13.22. The Krull dimension, dimR, of R is the maximal n such that there exists
a chain of prime ideals
p0 ⊂ p1 ⊂···⊂ pn ⊂ R.
If k = R/m, a field, we get
dimk(m/m2) ≥ dimR.
Definition 13.23. The local ring R is regular if dimk(m/m2) = dimR.
We also have that
depth(R) ≤ dimR.
Definition 13.24. A local ring R is called Cohen–Maccaulay if depth(R) = dimR.
There are various relationships between regular rings and Cohen–Maccaulay rings, even
though they are not equivalent.
For a proof of Theorem 13.21, see [Eis95, Chap. 19]. The general idea is to understand what
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 pdR/x(A) finite. Then pdR(A)=1+pdR/x(A).
Sketch of proof. If pdR/x(A) = 0, A is a projective R/x-module, and then A is not a
projective R-module, because xA = 0. Then
1 ≤ pdR(A) ≤ pdR(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 pdR/x(A). Assume pdR/x(A) ≥ 1. Take P
projective R/x-module with an exact sequence
0
M
P
A
0.
Since 1 + pdR/x(M) = pdR/x(A), we can apply the inductive hypothesis to get
pdR(M)=1+pdR/x(M).
By Lemma 13.18, we get that
pdR(P) ≤ max{pd(M),pdR(A)}
and either equality holds or pdR(A) = pdR(M) + 1. In the first case, we get a contradiction.
In the second case,
pdR(A) = pdR(M)+1=pdR/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
pdR(A) ≥ pdR/x(A/xA).
Corollary 13.27. If A is an R-module and we write A[x] = R[x] ⊗R A, we get that
pdR[x](A[x]) = pdR(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
pdR[x](M) ≤ pdR[x](R[x] ⊗R ˜
M)+1 ≤ pdR
M)+1=pdR(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[x1,...,xn] = n.
Let R be a ring and Rbe 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.
Example 13.30. If R = KQ/I for an admissible ideal I so that md ⊆ I ⊆ m2, then
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, {b1,...,bn} minimal set of generators over B. Then bn ∈ B = mB,
and we can write
bn =
n
i=1
ribi
for ri ∈ m.
Then
(1 − rn)b =
n−1
i=1
ribi ∈ Rb1 + ··· + Rbn−1
but rn ∈ m so 1 − rn ∈ R, and so
bn ∈ Rb1 + ··· + Rbn−1,
and hence b1,...,bn−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 b1,...,bn ∈ B
generate B if and only if the images b1,...,bn in B/mB span B/mB as a k-vector space.
Page 85
Note that the finitely-generated assumption is necessary: Q2 is a Z2-module with
mQ2 = (2)Q2 = Q2,
so Q2/mQ2 = 1 which is spanned by the only element as a k-vector space, but Q2 over Z2 is
not finitely generated.
Proof. For the ‘only if’ direction, if A = Rb1 + ··· + Rbn and B = A + mB. Then m · B/A =
B/A, so B/A = 0 and hence A = B.
Corollary 13.33. Elements b1,...,bn ∈ B are a minimal set of generators if and only if
the images of b1,...,bn 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 = dimk(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
Rn
P
0
where K is the kernel of the map Rn → P. Since P is projective, this sequence splits, so
Rn
= P ⊕ K.
Taking − ⊗R R/m, we get
kn
= P/mP ⊕ K/mK
≅ k
n ⊕ K/mK
and hence K = mK, so by Nakayama Lemma 13.31, K = 0. This shows Rn
= P.
If A is a finitely generated R-module. Let A0 = A and, recursively, having defined Ai, let
βi = dimk Ai/mAi < ∞
and define Ai+1 as the following kernel
0
Ai+1
Rβi
Ai
0.
This gives a free resolution of A:
···
Rβ2
Rβ1
Rβ0
A
0.
Apply − ⊗R R/m to
···
Rβ2
Rβ1
Rβ0
0
to get
···
kβ2
kβ1
kβ0
0
0
0
(checking that the maps are indeed 0 is an exercise). This shows that
TorR
i (A, R/m) = kβi .
Definition 13.35. The numbers βi are called Betti numbers of A.
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
Fn−2
···
F2
F1
F0
A
0
with F0,...,Fn−2 free. Then
pd(B) = n − (n − 1) = 1.
Let t = dimk B/mB and consider
0
P
Rt
B
0.
Then P is projective (since pd(P) = pd(B)−1 = 0), so it finitely generated, and hence free.
Since depth(R) = 0, every element in m is a zero divisor. By [Eis95, Cor. 3.2], there exists
s ∈ m,
{r ∈ R | rs = 0} = m.
Now, P ⊆ mRt, and hence sP ⊆ smRt = 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
and Change of Rings Theorems 13.25 and 13.26.
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 = pdR(R/m).
Note that R = k[x]/(x2) 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 = (x1,...,xn),
where n = dimR m/m2 and x1,...,xn 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 pdR(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
a resolution of R/(x). Suppose x1, x2 is a regular sequence, and consider K(x1) ⊗R K(x2):
R
R
R
R
x2
x1
−x2
x1
and the total complex gives a resolution of R/(x1,x2):
0
R
R ⊕ R
R
0
(−x2
x1
)
(x1,x2)
In general, if x = (x1,...,xn) is a regular sequence, we let K(x) = K(x1,...,xn) be the
total complex of K(x1) ⊗R K(x2) ⊗R ···⊗R K(xn). 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)= ⊕Rei,
R(
n
2)= ⊕
i<j
R(ei ∧ ej),
...
R(
n
n)= R(e1 ∧···∧ en),
and ∂ = ∑xi
∂ei
, so
ei1 ∧···∧ eik
j
(−1)j−1xjei1 ∧···∧ ̂eij ∧···∧ eik .
Theorem 13.39. The resolution K(x) is a free resolution of R/(x1,...,xn) and
Hq(K(x)) =
{ 0
if q > 0,
R/(x1,...,xn) if q = 0.
We first prove the theorem in the n = 1 case.
Proposition 13.40. If x ∈ m is a nonzero divisor, Cis a chain complex of R-module, then
we have an exact sequence
0
H0(K(x) ⊗ Hq(C))
Hq(K(x) ⊗R C)
H1(K(x) ⊗R Hq−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.
Proof of Theorem 13.39. The case n = 1 is the Proposition 13.40. We apply the
Proposition 13.40 with
C= K(x1,...,xn−1)
x = xn
to get the exact sequence
0
H0(K(xn) ⊗ Hq(K(x1,...,xn−1)))
Hq(K(x))
H1(K(xn) ⊗R Hq−1(K(x1,...,xn−1)))
0.
For q ≥ 2, both the kernel and the cokernel in this exact sequence are 0, so Hq(K(x)) = 0.
For q = 1, the kernel is 0 and the cokernel is
H1(K(xn) ⊗ R/(x1,...,xn−1)) = ker(R/(x1,...,xn−1)
·xn
−→ R/(x1,...,xn−1)) = 0,
and hence H1(K(x)) = 0.
For q0, we have
0
R/(xn) ⊗R R/(x1,...,xn−1)
︷︷
∼=R/(x1,...,xn)
H0(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
Tori(R/(x1,...,xn),k) = k(
n
k).
If m = (x1,...,xn), this gives
Tori(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.
Definition 13.41. Suppose (R,m) is a local Noetherian ring and A is a finitely generated
R-module. We say x1,...,xn ∈ m is a maximal A-sequence if
A/(x1,...,xn)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).
Proof. The proof is in [Wei94] and we omit it here.
14. Local cohomology
Let R be a commutative ring, I ⊆ R be an ideal. We define a functor
FI : R−mod → R−mod
by
FI(A) = {a ∈ A | there exists d such that Ida = 0}.
and if f : A → B is an R-module homomorphism, the square
FI(A)
FI(B)
A
B
FI (f)
f
commutes. Then FI is left exact and
H
I (A) = RFI(A)
is called the local cohomology. If Id ⊆ J and Je ⊆ I, then
FI = FJ
Page 90
so
H
I = H
J .
Moreover, we have that
FI(A) = H0
I (A) = {a ∈ A | there exists d such that Ida = 0} = lim
−→
HomR(R/Id,A).
In general,
Hn
I (A) = lim
−→
Extn
R(R/Id,A).
15. Spectral sequences
This chapter largely follows [Wei94], but another reference for this topic is [McC01].
In this chapter, we work in the category of R-modules.
Definition 15.1. A spectral sequence consists of
• objects Rr
pq for p, q ∈ Z, r ≥ a,
• differentials dr
pq : Er
pq → Er
p−r,q+r−1 satisfying dr ◦ dr = 0,
• Rr+1
pq
=
ker dr
pq
im(dr
p+r,q−r+1)
.
On a diagram, we can represent d0, d1, d2 as follows
d0
E0
p,q+1
E0
pq
E0
p+1
d1
E1
p,q+1
E1
pq
E1
p+1
d2
E2
p,q+1
E2
pq
E2
p+1
Page 91
and similarly for dr for r ≥ 3. We can think of these pictures as “pages” or “sheets”: for each
r, there is a “page” with arrows dr as follows
d0
d1
d2
d4
We let Er = ⊕
p,q∈Z
Er
pq. Then Er+1 is a subquotient of Er.
Suppose that for r = a, we have Ba = 0, Za = Ea.
Let Zr+1 ⊇ Br such that kerdr = Zr+1
Br
and Br+1 ⊇ Br such that imdr = Br+1
Br
, and then
Br+1 ⊆ Zr+1. We then have
Ba ⊆ Ba+1 ⊆ Ba+2 ⊆···⊆ B⊆ Z⊆···⊆ Za−a ⊆ Za−1 ⊆ Za = Ra
where
B=
i
Bi, Z=
i
Zi, E= Z/B.
Definition 15.2. A spectral sequence is bounded if for any n, there are finitely many p such
that Ea
p,n−p = 0.
Definition 15.3. A bounded spectral sequence converges to Hif for every n there is a
filtration
0 = FsHn ···⊆ FpHn ⊆ Fp+1Hn ⊆···FtHn = Hn
such that
E
pq = FpHp+q/Fp−1Hp+q.
In that case, we write E1
pq ⇒ Hp+q.
Note that E1
pq converges to the (p+q)th homology group, which can be represented as follows:
Hn
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 Cbe a chain complex with a filtration
···⊆ FpC⊆ Fp+1C⊆···
and assume
p
FpC= C.
We construct a spectral sequence with
E0
pq =
FpCp+q
Fp−1Cp+q
and
E1
pq = Hp+q(E0
p•).
Note that E0
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:
E0
pq =
FpCp+q
Fp−1Cp+q
FpCp+q−1
Fp−1Cp+q−1
= E0
pq−1.
In general, Er+1
pq
will be the homology of Er
pq.
In what follows, we drop the “q” from the notation and simply write
E0
p =
FpC
Fp−1C
and so on. We let
ηp : FpC →
FpC
Fp−qC
= E0
p
be the projection and set
Ar
p = {c ∈ FpC | d(c) ∈ Fp−rC},
Zr
p = ηp(At
p) =
Ar
p + Fp−1C
Fp−1C
∈ E0
p,
Br+1
p−r = ηp−r(d(Ar
p)) =
d(Ar
p) + Fp−r−1C
Fp−r−1C
∈ E0
p−r,
Br
p = ηp(d(Ar−1
p+r−1)) =
d(Ar−1
p+r−1) + Fp−1C
Fp−1(C)
⊆ Zr
p.
This will simplify calculations, for example:
Zr
p =
Ar
p + Fp−1C
Fp−1C
Ar
p
Ar
p ∩ Fp−1C
=
Ar
p
Ar−1
p−1
.
In what follows, we will use that, for B ⊆ A, we have A ∩ (B + C) = B + (A ∩ C). We set
Er
p =
Zr
p
Br
p
=
Ar
p + Fp−1C
d(Ar−1
p+r−1) + Fp−1C
=
Ar
p
Ar
p ∩ (d(Ar−1
p+r−1) + Fp−1C)
=
Ar
p
d(Ar−1
p+r−1) + Ar−1
p−1
and since
Er
p =
Ar
p
dAr−1
p+r−1 + Ar−1
p−1
,
Page 93
Er
p−r =
Ar
p−r
dAr−1
p−1 + Ar−1
p−r−1
,
the boundary map d induces
dr
p : Er
p → Er
p−r.
We claim that
Er+1
pq
=
kerdr
pq
imdr
p+rq−r+1
.
For that sake, we will compute the kernel of the map dr
p. If
a + dAr−1
p+r−1 + Ar−1
p−1 ∈ kerdr
p,
then without loss of generality, assume that
d(a) ∈ Ar−1
p−r−1
(otherwise, we could choose a different representative a that would satisfy this). Then
a ∈ Ar+1
p
. This shows that
kerdr
p =
Ar+1
p
+ Ar−1
p
d(Ar−1
p+r−1
Ar+1
p
+ Fp−1C
dAr−1
p+r−1 + Fp−1C
Zr+1
p
Br
p
.
Then
dr
p : Er
p =
Zr
p
Br
p
Zr
p
Zr+1
p
Br+1
p−r
Br
p−r
Zr
p−r
Br
p−r
= Er
p−r
and so
imdr
p =
Br+1
p−r
Br
p−r
and shifting the index
imdr
p+r =
Br+1
p
Br
p
Zr+1
p
Br
p
= kerdr
p.
Hence
kerdr
p
imdr
p+r
=
Zr+1
p
Br+1
p
= Er+1
p
.
This shows that Er+1
p
is the homology of Er
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 = FsCn ⊆ Fs+1Cn ⊆···⊆ FtCn = Cn.
Then E0
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 Hn(C),
···⊆ FpHn(C) ⊆ Fp+1Hn(C) ⊆···
such that
FpHp+q(C)/Fp−1Hp+q(C)
≅ E
pq .
Concisely, we write
E1
pq = Hp+q(FpC/Fp−1C) −→ Hp+q(C).
Page 94
Here, E1
pq is homology of the associated graded, E
pq is the associated graded of homology.
Proof. Recall that we had
0 ⊆ B0
p ⊆ B1
p ⊆···⊆ B
p ⊆ Z
p ⊆···⊆ Z1
p ⊆ Z0
p = FpC,
where B
p = ⋃
i
Bi
p, Z
p = ⋂Zi
p, and
Er
p = Er
p• = Zr
p/Br
p,
E
p = Z
p /B
p .
Suppose p + q = n. Then
Ar
pq = {c ∈ FpCn | d(c) ∈ Fp−rCn}.
For r ≥ r0(n, p) (when Fp−rCn = 0, i.e. p − r ≤ s), so
Ar
pq = kerd ∩ FpCn = A
pq.
Then
Zr
pq = ηp(Ar
pq) = ηp(A
pq) = Z
pq =
kerd ∩ FpCn + Fp−1Cn
Fp−1Cn
.
Note that in general, f
(
i
Ai
)
= ⋂
i
f(Ai), but because of boundedness from below these
are finite intersections, so equality does hold.
Moreover,
Br+1
pq
= ηp(Ar
p+r,q−r) =
d(Ar
p+r,q−r) + Fp−1Cn
Fp−1Cn
,
B
pq =
d(⋃Ar
p+r,q−r + Fp−1(Cn))
Fp−1Cn
.
Define:
FpHn(C) =
kerdn ∩ fpCn
imdn+1 ∩ fpCn
=
A
pq
d
(
r
Ar
pr,p−r+1
).
Then we have that
FpHn(C))
Fp−1Hn(C)
=
A
pq
d(⋃
r
Ar
p+q,q−r+1) + A
p−1,q−1
.
Applying ηp to the right hand side of the above, we get
ηp(A
pq)
ηp(d(⋃
r
Ar
p+q,q−r+1) + A
p−1,q−1
) ∼=
ηp(A
pq)
ηp
(⋃
r
Ar
p+q,q−r+1
) =
Z
pq
B
pq
.
We claim that ηp actually gives an isomorphism above. Indeed, consider
ηp : A
pq
ηp(A
pq
ηp(d(⋃
r
Ar
p+q,q−r+1
)).
Suppose a ∈ A
pq, so
a + fp−1C ∈ d
(⋃
Ar
p+r,q−r+1
)
+ Fp−1C
Page 95
and we can write a = b + c for
b ∈ d
(⋃
Ar
p+r,q−r+1
)
c ∈ Fp−1C ∩ A
pq = A
p+q−1
which completes the proof.
Example 15.5. Suppose
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−1B ⊆ F0B
︸︷︷︸
=A
⊆ F1B = B.
Then
E0
0q =
F0Bq
F−1Bq
= Aq, E0
1q =
F1Bq+1
F0Bq
=
Bq+1
Aq+1
= Cq+1,
so E0
pq can be represented as
...
...
...
...
0
A2
C3
0
0
A1
C2
0
0
A0
C1
0
0
0
C0
0
0
0
0
0
d0
d0
d0
d0
d0
Hence E1
0q = Hq(A), E1
1q = Hq+1(C) and in general E1
pq can be represented as
...
...
...
...
0
H1(A)
H2(C)
0
0
H0(A)
H1(C)
0
0
0
H0(C)
0
d1
d1
Page 96
Moreover, by definition of E2
1q, we get the following exact sequence
0
E2
1q
Hq+1(C)
Hq(A)
E2
0q
0.
Finally, looking at the diagram for E0
pq, we see that the maps dr : Er
pq → Er
p−r,q+r−1 for r ≥ 2
are all 0, and hence
E2
pq = E
pq .
By the Convergence Theorem 15.4, there is a filtration on H(B) such that
F0Hq(B) =
F0Hq(B)
F−1Hq(B)
= E
0p,
Hq+1(B)
F0Hq+1(B)
=
F1Hq+1(B)
F0Hq+1(B)
= E
1q ,
since F1Hq+1(B) = Hq+1(B) and F−1Hq(B) = 0. We then have
0
F0H1(B)
︸ ︷︷ ︸
=E
0
Hq(B)
Hq(B)
F0Hq(B)
︸ ︷︷ ︸
=H
q,q−1
0.
We then obtain
···
Hq(A)
Hq(B)
Hq(C)
Hq−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 Epq
r , r ≥ a, and the maps are
dpq
r : Epq
r
→ Ep+r,q−r+1
r
and
E
pq
r+1 =
kerdpq
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
Epq
r
−→ Hp+q,
i.e. there exists filtration FtHn ⊆ Ft−1Hn ⊆··· with quotients Epq
.
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.
Definition 15.6. A mapping p: E → B has the homotopy lifting property (HLP) for space
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.
Definition 15.7. We say p is a (Hurewicz) fibration if p has HLP for all Y . We say p is a
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
E2
pq = Hp(B;Hq(F;M) −→ Hp+q(E;M).
Corollary 15.10. Suppose π: E → Sn is a fibration and F is a fiber, with n ≥ 2. Then
there exists a long exact sequence
···
Hq(F)
Hq(E)
Hq−n(F)
Hq−1(F)
Hq−1(E)
···
Proof. We have
E2
pq = Hp(Sn;Hq(F)) =
{ Hq(F) if p = 0,n
0
otherwise
which gives the following diagram of E2
pq
Page 98
...
...
...
...
...
H2(F)
0
···
0
H2(F)
0
···
H1(F)
0
···
0
H1(F)
0
···
H0(F)
0
···
0
H0(F)
0
···
p = 0
p = 1
···
p = n − 1
p = n
p = n + 1
and if n > 2 then d2 = 0 and E3
pq = E2
pq, and in general
E2
pq = E3
pq = ··· = En
pq.
For En
pq, the map dn is non-trivial:
Hn−1(F)
0
...
...
...
...
...
H2(F)
0
···
0
H2(F)
0
···
H1(F)
0
···
0
H1(F)
0
···
H0(F)
0
···
0
H0(F)
0
···
p = 0
p = 1
···
p = n − 1
p = n
p = n + 1
dn
This gives the exact sequence
(∗)
0
En+1
np
Hq(F)
Hq+n−1(F)
En+1
0,q+n−1
0
dn
Moreover, for r>n, we have that dr = 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
Hq(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)
···⊆ Fp Tot(C) ⊆ Fp+1 Tot(C) ⊆···
where
Fk Tot(C)n =
k
p=0
Cp,n−p.
Let dh be the horizontal maps and dv be the vertical maps in the double complex, with
d = dh + dv and dhdv + dvdh = 0. We set
E0
pq =
Fp(Tot(C)p+q
Fp−1(Tot(C))p+q
= Cpq
and d0 = dv. Then
E1
pq = Hq(Cp•) −→ Hp+q(Tot(C••))
by Theorem 15.4. We have
d1 : E1
pq = Hv
q (Cp) → E1
p−1,q = Hv
q (Cp−q,•)
is induced by dh, and so we write d1 = dh. Then also
E2
pq = Hh
p Hv
p (C••) −→ Hp+q(Tot(C••)).
We could also define Dpq = Cqp (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
LpF(B)
≅ LpG(A) = Torp(A, B)
but the proof can be reinterpreted using spectral sequences of double complexes.
Alternative proof of Theorem 9.10. Suppose P→ A, Q→ B are two projective resolutions,
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
Hv
q (P⊗ Q) = P⊗ Hq(Q) =
{ P⊗ B if q = 0,
0
if q = 0.
Then
IE2
pq =
{ Hh
p (P⊗ B) = LpG(A) if q = 0
0
if q = 0
Hence the diagram for IE2
pq is
...
...
...
0
0
0
···
0
0
0
···
L0G(A)
L1G(A)
L2G(A)
···
from which it is clear that dn = 0 for n ≥ 2. Hence
E2
pq = E3
pq = ··· = E
pq .
This shows that
Hn(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
E2
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 Bn = d(Pn+1) is flat for all n. We show that there is an exact
sequence
0
Hq(P) ⊗ M
Hq(P ⊗ M)
TorR
1 (Hq−1(P),M)
0
Let Zn = ker(d: Pn → Pn−1). We showed before that Zn is also flat, and we have a short
exact sequence
0
Bn
Zn
Hn(P)
0
which gives a flat resolution of Hn(P), showing that the tor dimension of Hn(P) is at most 1.
Then
E2
pq = Torp(Hq(P),M)=0,
for p ≥ 2 and p < 0.
The diagram for E2
pq is
...
...
...
...
0
H2(P) ⊗ M
Tor1(H2(P),M)
0
0
H1(P) ⊗ M
Tor1(H1(P),M)
0
0
H0(P) ⊗ M
Tor1(H0(P),M)
0
which shows that d2 = d3 = ··· = 0, and hence E
pq = E2
pq. Thus Hq(P ⊗ M) has a filtration
with quotients E2
0q and E2
1,q−1. This gives the short exact sequence
0
Hq(P) ⊗ M
Hq(P ⊗ M)
TorR
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
E2
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)) =
{ Hp(P ⊗R B) = TorR
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, d2 = d3 = ··· = 0 and so IE2
pq =I E
pq , and hence
Hp(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 Abe a chain complex in A.
Definition 15.12. A Cartan–Eilenberg resolution (CE resolution) is an upper half plane
double complex P•• of projectives together with augmentation ϵ: P•0 → A
...
...
...
...
···
P−11
P01
P11
P21
···
···
P−10
P00
P10
P20
···
···
A−1
A0
A1
A2
···
such that
(1) Pp• → Ap is a resolution and if we define
Bp(P•q,dh) = im(dh : Pp+1,q → Ppq)
Zp(P•q,dh) = ker(dh : Ppq → Pp−1,q)
Page 103
Hp(P•q,dh) = Zp(P•q,dh)/Bp(P•q,dh)
then
Bp(P•q,dh) → Bp(A)
Hp(P•q,dh) → Hp(A)
are projective resolutions (which implies that Zp(P•q,dh) → Zp(A) is a projective
resolution),
(2) if Ap = 0 then Pp• is the zero complex.
Lemma 15.13. Every chain complex has a CE-resolution.
Proof. The proof is omitted but can be found in [Wei94]. It is similar to the proof that any
A ∈ A has a projective resolution 5.21.
Definition 15.14. Suppose F : A→B is a right exact functor. We define the left
hyperderived functor LpF : Ch(A) → B of F as follows: if Ais a chain complex in A, then
LpF(A) = Hp(Tot(F(P)))
where P•• is a CE-resolution of A. Dually, we can define the right hyperderived functor RpG
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
for the double complex FP•• from Section 15.4. We have
IE2
pq = Hh
p Hv
q (F(P)) = Hh
p (LqF(A)),
since Pp• → Ap is a projecitve resolution, so
Hv
q (F(Pp•)) = Lq(FAp)
by definition. For the other spectral sequence, we note that
Hq(F(P••)) = FHv
q (P••)
because the exact sequences
0
Zh
pq
Ppq
Bh
p−1q
0
0
Bpq
Zpq
Hh
pq
0
split. We then have
IIE2
pq = Hv
p Hh
q (F(P••)) = Hv
p (FHh
q (P••)) = (LpF)(Hh
q (A))
because
Hh
q (P••) → Hq(A)
is a projective resolution.
Page 104
Both of these spectral sequences converge to
Hp+q(Tot(F(P••))) = Lp+qF(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, Ais 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 (RpF(A)=0 for p > 0):
I, injective
GI,F-acyclic
A
B
C
RpF(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 → Iis 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
RqF(G(Ip)) =
{ FG(Ip) if q = 0,
0
if q = 0.
This shows that
IE
pq
2 =
{ Rp(FG)(A) if q = 0,
0
if q = 0.
Page 105
Hence
(Rp+qF)(G(I)) = Rp+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
RpF(RqG(A)) −→ Rp+q(FG)(A),
as required.
Example 15.16. Let X, Y be topological spaces and f : X → Y be a continuous map.
Then the functor
f: Sheaves(X) → Sheaves(Y ),
(fF)(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
RpΓ(X,F) = Hp(X;F)
the sheaf cohomology.
Then
(RpΓ)(Rqf)(F) = Hp(Y ;RqfF),
and since
ΓfF = fF(Y ) = F(f−1(Y )) = F(X) = Γ(X),
we get that
Rp+q(Γf)(F) = Hp+q(X;F).
Then the Grothendieck spectral sequence 15.15 gives
Hp(Y ;RqfF) −→ Hp+q(X;F).
In particular, if RqfF = 0 for q > 0, then
Hp(Y ;fF) = Hp(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
An+2
An+1
Bn+1
Bn
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 =
(−dA
0
−f dB
)
.
We then have an exact sequence
0
B
cone(f)
A[−1]
0.
δ
Similarly, we define the cylinder of f:
An+1
An
An+2
An+1
Bn+1
Bn
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 =
dA
idA
0
0 −dA
0
0
−f dB
.
We have exact sequences
0
A
cyl(f)
cone(f)
0
0
B
cyl(f)
cone(−idA)
0
α
β
Page 107
with
α =
0
0
idB
,
β = (idA 0 idB)
and αβ ∼ idB, βα = idB, so α and β are homotopy equivalences.
We construct a quotient category K = K(A) of Ch = Ch(A) by
Obj(K) = Obj(Ch)
HomK(A,B) = HomCh(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 Hn : 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→ Bis 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(idB)
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→ Bis a map, then we have a triangle
cone(u)
A
B
δ
u
v
We will call this a strict exact triangle.
Definition 16.1. For u: A→ B, v: B→ C, w: C→ A[−1], the triangle
C
A
B
w
u
v
is called an exact triangle if there exists ˜u: ˜
A→ ˜
Band 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.
Example 16.2. The diagram
0
A
A
idA
is an exact triangle, because the diagram
A
A
0
A[−1]
A
A
cone(idA)
A[−1]
=
idA
=
=
idA
Page 109
commutes and letting C = cone(idA)
s =
(0 −id
0
0
)
, dC =
(−dA
0
idA
dA
)
we get
sdC + dCs =
(idA
0
0 idA
)
= idC,
so idC ∼ 0.
Example 16.3. Suppose
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
idB
)
, w = δ = (idA 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 idA 0), h =
−u
idA
0
,
and the map C → D is given by the matrix
0
0
idA
0
0 idB
.
We have that
idD − hπ =
id u 0
0 0 0
0 0 id
 = sdD + dDs
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.
Definition 16.4. An additive category K is called a triangulated category if it has an
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−1w, 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
idB
)
: B → C , ∂ = (idA
0) : C → A,
y =
( 0
idC
)
: C → B , δ = (idA
0) : B → A,
Page 112
x =
( 0
idC
)
: C → A , i = (idB
0) : A → B.
Taking
f =
(idA
0
0
v
)
, g =
(u
0
0 idC
)
,
one can easily verify that
δf = (idA
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 idb
u
0
0 0 0 idC
)
: cone(f) → A ,
and claim that this gives an inverse map in the quotient category. We have that
dA =
(−idb
0
−v dC
)
, dD =
dA
0
0
0
u
−dB
0
0
−idA
0
−dA
0
0
−v −vu dC
and dDh = hdA , dA π = πdD, πh = idA,
hπ =
0 0 0 0
0 idB
u
0
0 0 0 0
0 0 0 idC
.
Page 113
Setting
s =
0 0 −idB
0
0 0
0
0
0 0
0
0
0 0
0
0
we get that
sdD + dDs =
idA
0 0 0
0 0 −u 0
0 0 idA
0
0 0 0 0
= idD − hπ.
This shows that idD ∼ hπ, and hence h, π are homotopy equivalences, i.e. isomorphisms
in K. Finally, πp = g, p = hπp = hg (in K).
Similarly, Kb = Chb / ∼, bounded chain complexes, K+, positive chain complexes, K
negative chain complexes, are all triangulated categories.
Definition 16.6. If H : K→A is an additive functor where K is a triangulated category
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(Ti−1C)
H(TiA)
H(TiB)
H(TiC)
H(Ti+1A)
···
H(Tiu)
H(Tiv)
H(Tiw)
We will then write Hi(A) = H(TiA) and Hi(u) = H(Tiu).
Example 16.7. The functor
K(A)
A
A
H0(A)
is a cohomological functor. Indeed, for an exact triangle
cone(u)
A
B
u
we have a long exact sequence
···
Hi(A)
Hi(B)
Hi(cone(u))
Hi+1(A) = Hi(A[−1])
···
Page 114
Proposition 16.8. If K is a triangulated category, X ∈ ObjK, then HomK(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
···
HomK(X, A)
HomK(X, B)
HomK(X, C)
HomK(X, T A)
HomK(X, T B)
HomK(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
HomK(X, B). We can first rotate the triangles, and then shift using T to show exactness
everywhere.
Example 16.9. In K(Ab), we have
Z[0]
···
0
Z
0
0
···
···
A−1
A0
A1
A2
···
d0
and so
HomCh(Z[0],A) = ker(d0)
HomK(Z[0],A) = ker(d0)/ ∼ = ker(d0)/im(d−1) = H0(A).
This shows that the cohomology functor is representable.
17. Derived categories
Suppose C is a category and S is a collection of morphisms.
Definition 17.1. A localization of C with respect to S is a category S−1C together with a
functor q: C → S−1C such that
(1) q(s) is an isomorphism in S−1C for all s ∈ S,
(2) (S−1C,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−1C→D such that the diagram
C
S−1C
D
q
F
˜F
commutes.
If C is a small category, then S−1C exists. Indeed, let S−1C 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
˜s1 and so on.
Example 17.2. If S is the collection of chain homotopy equivalences in Ch(A), then
K = S−1 Ch(A).
Definition 17.3. If Q is the collection of all quasi-isomorphisms in Ch(A), then the derived
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−1K(A) = D(A):
Ch(A)
S−1 Ch(A) = K(A)
R−1K(A)
Q−1 Ch(A)
where we get the dotted arrows from the universal properties, and they are unique so they
are inverses.
Definition 17.4. Let C be any category. A set of morphisms S is a multiplicative system if
(1) S is closed under composition,
(2) idX ∈ 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−1C 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 s3 ∈ S, and
we can write the composition in the same form again
(f1s−1
1 )(f2s−1
2 ) = f1f3s−1
3 s−1
1
= (f1f3)(s1s3)−1.
Define a category D with Obj(D) = Obj(C) and HomD(A, B) as the set of all diagrams
A
C
B
s
f
with s ∈ S and f ∈ C, modulo ≡ where
[A
s1
←− C1
f1
−→ B] ≡ [A
s2
←− C2
f2
−→ B]
if there exist t1,t2 ∈ S with f1t1 = f2t2 and s1t2 = s2t2:
Page 117
C1
A
D
B
C2
s1
f1
t1
t2
s2
f2
The composition is defined by considering the following diagram
D
C2
A3
C1
A2
A1
˜f1
˜s2
s2
f2
f1
s1
h2∈D
h1∈D
and letting
A1
D
A3
˜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] ∈ HomD(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−1K(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 Hn(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 idx ∈ 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 Hn(s) is an isomorphism for all n, we get
Hn(Z)
Hn(Y )
Hn(C)
Hn+1(Z)
Hn+1(Y )
∼=
∼=
so Hn(C) = 0, and hence Hn(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 HomK(X,−) is a
cohomological functor, we have the exact sequence
HomK(X, Z)
HomK(X, Y )
HomK(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,
Hn(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−1K 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−1K(A) is a triangulated category with
T(fs−1) = T(f)T(s)−1.
Proof. First note that T is well-defined: if f1s−1
1
= f2s−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
in [Wei94].
(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 Db(A) from bounded chain complexes and D+(A) from positive
chain complexes.
Proposition 17.7. Suppose Iis a cochain complex of injectives, bounded from below, Z
is a cochain complex. If t: I→ Zis a quasi-isomorphism in K(A), then there exists
s: Z→ Iwith st = id in K(A).
Corollary 17.8. Suppose Iis a cochain complex of injectives, bounded from below, in D(A).
Then HomD(A)(X, I) = HomK(A)(X, I).
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MR 2050440
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