HOMOLOGICAL ALGEBRA - math.colorado.edu

HOMOLOGICAL ALGEBRA

Juan Moreno

These notes were taken for a class on homological algebra at the University of Colorado, Boulder.The course was taught in the Spring 2020 semester by Jonathan Wise. The reference textbook is Anintroduction to homological algebra by Charles Weibel. The class met three days a week and one ofthese days was spent presenting solutions to exercises either given in class or from Weibel. Thesenotes were, with the exception of a few arguments either missed or left for the student, taken live,and so there are bound to be errors. If you do happen to find errors or have questions, feel free toemail me at [email protected].

CONTENTSLecture 1: Abelian Categories - 1/13/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2Lecture 2: Abelian Categories (continued) - 1/15/2020 . . . . . . . . . . . . . . . . . . . . . . . . . 4Lecture 3: Chain Complexes - 1/17/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6Lecture 4: The Snake Lemma - 1/24/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8Lecture 5: Chain Homotopy - 1/27/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11Lecture 6: Projective Objects -1/31/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14Lecture 7: Injective Objects & Derived Functors - 02/03/2020 . . . . . . . . . . . . . . . . . . . . . 17Lecture 8: Derived Functors - 02/07/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Lecture 9: Universality of Derived Functors - 02/10/2020 . . . . . . . . . . . . . . . . . . . . . . . 22Lecture 10: Flat Modules - 02/14/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24Lecture 11: The Meaning of Tor and Ext - 02/17/2020 . . . . . . . . . . . . . . . . . . . . . . . . . 26Lecture 12: Limits & Colimits - 02/21/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Lecture 13: Derived Functors of the Inverse Limit - 02/24/2020 . . . . . . . . . . . . . . . . . . . 30Lecture 14: Filtered Comlimits - 02/28/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32Lecture 15: Spectral Sequences - 03/02/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33Lecture 16: Spectral Sequences via Filtrations - 02/06/2020 . . . . . . . . . . . . . . . . . . . . . 36Lecture 17: Spectral Sequence of a Double Complex - 03/09/2020 . . . . . . . . . . . . . . . . . . 38Lecture 18: Grothendieck Spectral Sequences - 03/13/2020 . . . . . . . . . . . . . . . . . . . . . . 40Lecture 19: Derived Categories - 03/16/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43Lecture 20: Existence of Derived Categories - 03/20/2020 . . . . . . . . . . . . . . . . . . . . . . . 44Lecture 21: Derived Categories (continued) 03/30/2020 . . . . . . . . . . . . . . . . . . . . . . . . 46Lecture 22: Sifted Categories - 04/01/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48Lecture 23: Triangulated Categories - 04/03/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50Lecture 24: Back to Derived Functors - 04/06/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . 52Lecture 25: Derived Categories - 04/10/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54Lecture 26: Group Cohomology - 04/13/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55Lecture 27: Group Cohomology (continued) - 04/15/2020 . . . . . . . . . . . . . . . . . . . . . . . . 55Lecture 28: - 04/20/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57Lecture 29: Galois Cohomology - 04/24/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58Lecture 30: ∞-Categories - 04/27/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59Lecture 31: ∞-Categories (continued) - 04/29/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . 61Solutions to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

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mailto: [email protected].

LECTURE 1: ABELIAN CATEGORIES - 1/13/2020

"Homological Algebra is linear algebra in an Abelian category"

The idea is the usual one: linear algebra makes things easier. There are examples of this through-out all of mathematics. We study functions f : R→ R by studying their derivatives, we study man-ifolds by studying their tangent spaces. In each case there is a process of linearization that makesthe problem at hand more tractible. Homological algebra is the toolset that we use to linearize moreabstract problems. The general setting where we apply this toolset is an abelian category.

Definition 1. An abelian category is a category C such thatAB0) C has finite direct sums, that is, it has finite products and coproducts, and these coincide,AB1) C has kernels and cokernels,AB2) images and coimages coincide.

We note that this definition differs from that in Weibel’s text. We will elaborate on what this defini-tion means exactly and in doing so we will reconcile our definition with that of Weibel.

Let C be a category and take X ,Y ∈ Ob(C ). The product of X and Y , denoted in this class byX ×Y , is an object of C together with morphisms pX : X ×Y → X and pY : X ×Y → Y such that thediagram

X ×Y X

Y

pY

pX

is universal. What this means is that if Z is an object of C with morphisms f : Z →Y and g : Z → Xthen there is a unique morphism h : Z → X ×Y such that the following diagram commmutes

Z

X ×Y X

Y

f

g

∃!h

pY

pX

As is always the case in category theory, the co-thing is obtained by reversing the arrows in theoriginal thing. So a coproduct of objects X and Y in a category is an object, denoted X tY , withmorphisms iX : X → X tY and iY : Y → X tY that is universal. To be explicit, the defining propertyof the coproduct is that for any object Z of C with morphisms f : X → Z and g : Y → Z there is aunique morphism h : X tY → Z such that the following diagram commutes.

X

Y X tY

Z

iX fiY

g

∃!h

2

In the former case, we denote the induced map h by a column vector(

fg

): Z → X ×Y and in the

latter case the induced map h will be denoted by a row vector(f g

): X tY → Z. More generally,

suppose we have objects A,B,C,D with maps f1 : A → C, f2 : A → D, and f3 : B → C, f4 : B → D. Then

we have induced maps(f1f2

): A → C×D and

(f3f4

): B → C×D. This gives rise to a unique induced map

(f1 f3f2 f4

): AtB → C×D.

Now note that an initial object ; is the product of an empty collection of objects. Similarly, a terminalobject 1 is the coproduct of an empty collection of objects.It follows that in an Abelian category, theunique morphism ;→ 1 is an isomorphism. We call such an object a zero object and denote it, as wellas any maps into or out of it, by 0. We also obtain, for any two objects X and Y of C a unique zeromap 0 : X → Y which is the composition of the unique maps X → 0 and 0 → Y . It follows that there

are induced maps(idX

0

): X → X ×Y and

(0

idY

): Y → X ×Y . By our discussion above, we get map

(idX 0

0 idY

): X tY → X ×Y

and fits into the commutative diagram

X

X tY X ×Y

Y

iX

∃!

iY

Similarly, we have induced maps from the coproduct(idX 0

): XtY → X and

(0 idY

): XtY →

Y , which by the universal property of the product gives rise to the a unique map in the followingcommutative diagram

X

X tY X ×Y

Y

pX

pY

∃!

What condition AB0 says is that these maps are isomorphisms and inverse to one another. Forcondition AB1, we simply define what we mean by kernel, cokernel. While we are at it, we definewhat we mean by image, and coimage, then elaborate on condition AB2. These are generalizationsof the corresponding notions in the category of vector spaces.

Definition 2. Let f : B → C be a morphism in a category with a zero object, C . A kernel of f ,denoted ker( f ), is an object A of C together with a morphism i : A → B which is universal such that

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f ◦ i = 0. A cokernel of f , denoted coker( f ), is an object D of C with a morphism p : C → D thatis universal such that p ◦ f = 0. The image of f , denoted im( f ), is defined as ker(coker( f )), and thecoimage of f , denoted coim( f ), is defined as coker(ker( f )).

Exercise 1. Verify that, for any ring R, the category of R-modules is an abelian category.Solution

LECTURE 2: ABELIAN CATEGORIES (CONTINUED) - 1/15/2020

We will continue by elaborating on condition AB2. Assuming conditions AB0 and AB1, there is amap coim( f )→ im( f ) for all morphisms f : X →Y in C . Let us see how to construct this map. In thelast definition from last time, we defined the image and coimage of f as the kernel of the cokerneland the cokernel of the kernel, respectively. Here we think of the kernel and cokernel as not justobjects, but objects with the corresponding universal map. We then have the following commutativediagram.

ker( f ) X Y coker( f )

coim( f ) im( f )

0

f

∃!h 0

The dashed map, h comes from the universal property of the triangle-shaped diagram on the right

and the fact that the composition Xf−→Y → coker( f ) is zero. We now use this map and the universal

property of the coimage to construct our desired map. This amounts to showing that the compositionker( f ) → X h−→ im( f ) is in fact the zero map. To see this, note that for any morphism g : A → Band object Z, we have a map Hom(Z,ker(g)) → Hom(Z, A) given by sending t : Z → ker(g) to thecomposition σ◦ t where σ is the universal map ker(g) → X . Now note that g ◦σ◦ t = 0. So the imageof this map contains only morphisms Z → A such that the post-composition with g is 0. By theuniversal property of the kernel, for any morphism Z → A such that the composition with g is 0,there is a unique morphism Z → ker(g). Thus, the map on Hom-sets is injective. It follows that ifany map Z → A post-composes to 0 with g, then ker(g) must be the 0 element. This is indeed thecase with the map im( f )→Y so that the kernel of this map must be the zero element. It follows thatsince the composition

ker( f )→ X h−→ im( f )→Y

is the same as the composition

ker( f )→ Xf−→Y ,

which is zero, we must have that ker( f )→ X h−→ im( f ) is indeed zero, as desired.Condition AB2 is then simply that we require that the map coim( f )→ im( f ) be an isomorphism.

Exercise 2. Construct the map coim( f )→ im( f ) using first the universal property of coim( f ).Solution

That this map is an isomorphism is exactly condition AB2. Now let us look at some examples ofabelian categories.

• vector spaces

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• R-modules, for an associative algebra R

• diagrams in an abelian categories

• presheaves valued in an abelian category (usually)

The following proposition explains the name of an abelian category. It says that an abelian categoryis a category enriched in the category of abelian groups.

Proposition 1. If A is an abelian category and X ,Y objects in A . Then Hom(X ,Y ) has an abeliangroup structure.

Proof. Before we begin, we introduce some notation. Let

∆∗x =

(idXidX

), ∆y =

(idY idY

)Now we define the addition operation. Take morphisms f , g : X →Y . This will be the composition

( f + g) : X∆∗

X−−→ X ⊕ X →Y ⊕Y ∆Y−−→Y ,

where X ⊕ X → Y ⊕Y is(f 00 g

). That this operation is commutatie, assoiative, and unital is left as

an exercise.In order to obtain inverses we need some additional results.

Lemma 1. • ker(idX 0

)= (0

idX

)

• coker(idX

0

)= (

0 idX)

Proof. That the composition

X X ⊕ X X

0

idX

(idX 0

)

is zero follows directly from the definitions of each of the maps since they are the canonical inclusionsand projections to and from the direct sum. Then, if f : X → X ⊕ X is any other map such that(

idX 0)◦ f = 0,

by definition of the projection map, we have that

f =(0g

),

for some g : X → X . Evidently, then

f =(

0idX

)◦ g,

so that the inclusion is indeed the kernel of the projection. The second bullet point follows from asymmetric argument.

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We can use this result to construct an explicit negation map −1 : X → X . Let W = ker(X ⊕ X ∆−→ X ),and σ : W → X the corresponding universal map, where

∆= (1 1

).

We then have the following commutative diagram

X

W X ⊕ X X

X

σ ∆

Lemma 2. In an abelian category, f : X →Y is an isomorphism if and only if ker( f )∼= coker( f )∼= 0.

Proof. (TODO: the argument)

Now we define

−1=((

idX 0)◦σ)

◦((

0 idX)◦σ)−1

.

One can then check that this indeed gives us a negation map so that for any f : X →Y ,

−1( f )+ f = 0.

Exercise 3. Show that the operation defined on the set Hom(X ,Y ) is commutative and associative.Show also that the zero map is the corresponding identity, and the operation is biadditive, that is,for morphisms f : A → B, g,h : B → C, and k : C → D, we have

k ◦ (g+h)◦ f = k ◦ g ◦ f +k ◦h◦ f .

Solution

Here are some additional exercises from the textbook.

Exercise 4. Weibel 1.1.5 Solution


LECTURE 3: CHAIN COMPLEXES - 1/17/2020

Before we talk about chain complexes, we will introduce a category that is not quite abelian, butclose and very important. Let A = {filtered abelian groups}, that is

A = {(A,F ·A)| · · · ⊂ Fn A ⊂ Fn+1 A ⊂ ·· · }A morphism (A,F ·A) → (B,F ·B) is ϕ : A → B such that ϕ(F i A) ⊂ F iB). This satisfies both axiomsAB0 and AB1. The product and coprooduct of such a morphism ϕ ends up being just the product orcoproduct of the restriction to each degree. Similarly,

ker(ϕ)= (ker(ϕ),F ·A∩ker(ϕ))

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andcoker(ϕ)= (B/A,F i(B/A))= im(Fk(B)→ B/A).

this category will be extremely important when we come to talk about spectral sequences. Now forchain complexes.

Definition 3. Let A be an abelian category. The category of chain complexes in A , denoted byCh(A ), is the category whose objects are sequences

· · · dn+1−−−→ Cndn−−→ Cn−1

dn−1−−−→ ·· · ,

where the Ci are objects of A and di morphisms in A . The morphisms in this category are commut-ing diagrams like the following

· · · An+1 An · · ·

· · · Bn+1 Bn · · ·ϕn+1 ϕn

Theorem 1. Ch(A ) is an abelian category if A is an abelian category.

Proof. We use the fact that if I is any category, then the category Hom+(I,A ) is abelian, whereHom+ denotes the category of additive functors with natural transformations as morphisms. Theproof follows from the fact that Hom+(I, A) is isomorphic to the category of chain complexes in thecase that I is the poset category with objects integers and morphisms x → y given by a relationx < y.

Exercise 6. Fill in the details of the last proof.

Suppose A and B are each categories satisfying AB0. A functor F : A → B is called additive if F(X ⊕Y )

∼=−→ F(X )⊕F(Y ).

Exercise 7. If F is additive then F(ϕ+ψ)= F(ϕ)+F(ψ), where ϕ,ψ : X →Y .

Suppose C· is a chain complex in A , an abelian category. Let

Zn = ker(dn), and Bn = im(dn+1).

Now that these are both contained in Cn. We define the n-th homology group of C· as

Hn(C·)= Zn/Bn.

Remark 1. We will switch back and forth from lower to upper indices in chain complexes using theconvention Cn = C−n.

Note that each Zn,Bn and Hn is functorial in Cn, that is, the solid arrows below induce the dashedarrows, which in turn induce a map on homology.

Cn+1 Bn Zn Cn

C′n+1 B′

n Z′n C′

n

7

Exercise 8. Construct the dashed arrows and show that there is an induced map Hn(C·)→ Hn(C′·).

Definition 4. A morphism ϕ : C· → C′· of chain complexes is called a quasi-isomorphism if Hn(C·)→Hn(C′·) is an isomorphism for all n.

Example 1. The diagram to the right is a quasi-isomorphism.

· · · 0 Z Z 0 · · ·

· · · 0 0 0 0 · · ·






LECTURE 4: THE SNAKE LEMMA - 1/24/2020

Here is just a regular Lemma, not of the cold-blooded kind.

Lemma 3. If A is an abelian category and

X →Y → Z →W

is an exact sequence in A . Then for any morphism V →Y ,

X →Y /V → Z/V →W

is also exact.

Proof. First, we give a different characterization of exactness in an abelian category. Using thesequence X →Y → Z as a model, we have

ker(Y → Z)= im(X →Y ) =⇒ coker(ker(Y → Z))= coker(ker(Y → coker(X →Y )))

=⇒ im(Y → Z)= coker(ker(Y → coker(X →Y )))coker(ker(Y → coker(X →Y ))).

However, coker(ker(coker(X → Y )) = coker(X → Y ). Therefore, another way to phrase exactness ofthe given sequence is to say that

coker(X →Y )= im(Y → Z), and coker(Y → Z)= im(Z →W).

Which is true if and only ifcoker(X →Y )

∼=−→ ker(Z →W).

In the above, by equality we mean that there exists a canonical isomorphism.Let us denote by A/B the cokernel object of a map B → A. Now

im(Z/V →W)= im(Z →W)= coker(Y → Z)= coker(Y → Z/V )= coker(Y /V → Z/V ).

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The first equality above follows from the fact that Z → Z/V is an epi. The other equalities are left forthe reader to ponder. This gives exactess at Z/V .Now for exactness at Y /V , we have

coker(X →Y /V )= coker(X ⊕V →Y )= coker(V → coker(X →Y ))= coker(V → im(Y → Z))

= im(Y → Z/V )= im(Y /V → Z/V ).

The reasoning behind the equalities is again left to ponder. We note that there is a rather large stepgoing from the first to the second line above. Nevertheless, these equalities should seem plausible inthe category of R-modules.

Let us denote by A : B the kernel ker(B → A).

Lemma 4. Suppose the following is an exact sequence in an abelian category A

X →Y → Z →W .

Let Z →U be a morphism in A . Then

X →U : Y →U : Z →W

is an exact sequence.

Proof. A op is an abelian category so we can apply the previous lemma.

We will record the actual Snake Lemma as a theorem

Theorem 2. (Snake Lemma) Suppose the diagram bellow commutes with exact rows and columns,then there exists the dashed arrow.

0 0 0

K1 K2 K3

A1 A2 A3 0

0 B1 B2 B3

L1 L2 L3

0 0 0

Furthermore, the following sequence is exact

K2 K3 L1 L2

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Proof. The main step is that the red portions of the diagram can be killed off by taking the corre-sponding cokernels, while the blue portions can be killed off by taking the corresponding kernels.By the preceing lemmas this process preserves exactness. We then obtain the diagram shown belowthis argument. Since every nonzero morphism is in between two zero morphisms, and the rows andcolumns are exact, we have that all the remaining nonzero morphisms must be isomorphisms. Thus

ker(L1 → L2)∼= coker(K2 → K3),

implying the sequence

K2 K3 L1 L2

is exact.

0

0 K3/K2

0 B3 : A2/(A1 +K2) B3 : A3/K2 0

0 L2 : B1/A1 (L2 +B3) : B2/A1 0

L2 : L1 0

0

∼=∼=

∼=∼=

∼=

Now let’s put this Snake Lemma to work. Suppose that I have an exact sequence of chain complexes

0→ C′· → C· → C′′

· → 0.

We want to obtain from this a long exact sequence on the homology of these chain complexes. Firstwe note that ker is a left-exact functor, and coker is a right-exact functor. That is, if we have acommutatuive diagram with exact rows like below

0 A B C 0

0 A′ B′ C′ 0

Then the sequences0→ A′ : A → B′ : B → C′ : C,

andA′/A → B′/B → C′/C → 0

are exact. We then have an exact sequence

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C′n/B′

n Cn/Bn C′′n/B′′

n 0

0 Z′n Zn−1 Z′′

n−1

Applying the snake lemma then gives us the following.

Proposition 2. Given a short exact sequence of chain complexes

0→ C′· → C· → C′′

· → 0,

there is an induced long exact sequence on homology

· · ·H′n → Hn → H′′

n → H′n−1 → Hn−1 → H′′

n−1 →···

We won’t get far today, but let’s move on to chain homotopies.Suppose A·,B· are chain complexes. Define

Homn(A·,B·) := ∏m∈Z

Hom(Am,Bm+n),

and d : Homn(A·,B·)→Homn−1(A·,B·) by

d( f·)m = d ◦ fm + (−1)n−1 fm−1 ◦d : Am → Bm+n−1.

Disclaimer: The d on the right hand side of the equation above corresponds to either the differentialof the complex A· or B·, while the d on the left hand side corresponds to the new differential that weare defining.

Exercise 14. Show that Hom·(A·,B·) forms a chain complex with differentials defined as above.Also, show that Z0 Hom·(A·,B·)=Hom(A·,B·). Solution






LECTURE 5: CHAIN HOMOTOPY - 1/27/2020

Okay, lets talk about math. Recall from last time that if A,B ∈ Ch(A ) we can form a chain complexfrom the Hom-groups between these. Let d be the differential corresponding to this complex. We calla map f· : A· → B· nullhomotopic if f· = d(g·) for some g· ∈Hom1(A·,B·). The following diagram givesa visual representation of the situation.

· · · A1 A0 A−1 · · ·

· · · B1 B0 B−1 · · ·f1 f0g0

f−1g−1

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From the definition of the differential, we see that

d(g·)m = dgm + gm−1d = fm,∀m.

The reason we call such a map nullhomotopic comes from homotopy theory. The essence though, iscaptured in the following result.

Lemma 5. If f· : A· → B· is nullhomotopic then f∗ : Hn(A)→ Hn(B) is 0.

Proof. Let’s assume f is nullhomotopic, then

fn(Zn A·)= (dgn ± gn−1d)(Zn A·)

= dgn(Zn A·)⊂ BnB·.

Since BnB· gets quotiented out in homology, the result follows.

A quick corollary is that since ( f − g)∗ = f∗ − g∗, if f and g are homotopic, that is to say f − g isnullhomotopic, then

f∗ = g∗ : Hn(A·)→ Hn(B·).

For ease of notation, we will write f ' g if f and g are homotopic. It is not too hard to see that thisis indeed an equivalence relation and so the notation is warranted. All of this leads us to a notion of’homotopy equivalence’ of chain complexes.

Definition 5. We call A· and B· chain homotopy equivalent if ∃ f· : A· → B·, g· : B· → A· such that

g· ◦ f· ' idA, f· ◦ g· ' idB.

In this case, we call g a homotopy inverse to f .

Example 2. If f· : A· → B· has a homotopy inverse then it induces an isomorphism on homology,that is f· is a quasi- isomorphism. However, quasi-isomorpihsms do not necessarily come inducedfrom some chain homotopy equivalence. For a concrete example of this, consider the following di-agram. The vertical arrows evidently induce an isomorphism on homology making them a quasi-isomorphism. However, it is also evident that there is no possible sequence of maps from the bottomrow into the top which would compose to anything resembling of the identity.

· · · 0 Z Z 0 · · ·

· · · 0 0 Z/mZ 0 · · ·

m

q

Now let’s talk a bit about cones. Suppose f· : A· → B· is a chain map. Then cone( f )n = An−1 ⊕Bn is achain complex with differential given by

d =(−d 0− f d

): An ⊕Bn+1 → An−1 ⊕Bn.

The cone merges the ideas of taking a kernel and a cokernel simultaneaously. To see this, recall froma previous exercise that we have an associated exact sequence of chain complexes

0→ B· → cone( f )· → A·[−1]→ 0.

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The maps in the middle are given by(01

)and

(−1 0), respectively. According to the Snake Lemma,

this gives us a long exact sequence on homology,

· · ·→ Hn+1(A·[−1])→ Hn(B·)→ Hn(cone( f )·)→ Hn(A·[−1])→ Hn−1(B·)→··· .

So what does this tell us? Well note that if Hn(cone( f )·)= 0 for all n, then the exact sequence impliesthat f· must be a quasi-isomorphism. It follows that cone( f )· is a chain complex that measures thedifference between the homology of A· and B· using the map f between the two. The idea is that oftenwe don’t care about actually computing the homology of an object, we just care about comparing itto another object. The cone construction gives us an object that measures this difference in a preciseway.This is all segway into a larger topic of derived functors which we will start covering in the nextlecture. For now, we provide some motivation. If A is an abelian category and X ∈ Ob(A ), thenHom(X ,_) and Hom(_, X ) are left-exact functors. However, they are not always right-exact dependingon the object X . That is, if

0→Y →W → Z → 0

is exact, then0→Hom(X ,Y )→Hom(X ,W)→Hom(X , Z)→ 0

is not necessarily exact at Hom(X , Z). We would like to measure the extent to which they are notright-exact for a given X . Specifically, taking a hint from the preceding discussion about mappingcones, we might hope to find an object E(X ), which would depend on X , for which the sequence

0→Hom(X ,Y )→Hom(X ,W)→Hom(X , Z)→ E(X )

is exact. Well this is actually not very difficult, we can simply take the cokernel of the last map andwe would be done. However, this does not tell us much. Instead, we might hope to find a sequence ofobjects En(X ), which would also depend on X , for which the sequence

0→Hom(X ,Y )→Hom(X ,W)→Hom(X , Z)→ E1(X )→ E2(X )→···

forms a chain complex. Then the homology of such a chain complex would provide some new infor-mation. The hope is that by choosing the spaces En(X ) wisely, we will get information about thefailure of the functors Hom(X ,_) to be right-exact. It turns out that good choices En exist and thesegive our first examples of derived functors.Before moving on to constructing such functors, let’s talk about objects for which right-exactness ofthese Hom-functors does hold.

Definition 6. An object P ∈A is called projective if the functor Hom(P,_) is exact.

Example 3. Z/mZ is not projective in Ab.

Example 4. All free modules in R−mod are projective.

Proposition 3. The following are equivalent

• P is projective.

• every exact sequence 0→ A → B → P → 0 can be split.

• Given a diagram as the one below, there exists a dahed arrow making the diagram commute.

13

P

B C 0

∃

Additionally, if A is the category of R-modules. Then we also have that the above are equivalent to

• P is a direct summand of a free module.

Exercise 20. Weibel 1.5.3 Solution.



LECTURE 6: PROJECTIVE OBJECTS -1/31/2020

We will pick up where we left off yesterday by proving the following portion of the last proposition.

Proposition 4. Free modules are projective.

Proof. Suppose F is free on a set X . Then

Hom(F, M)=Homset(X , M).

If B → C is injective, then

Hom(F,B) Hom(F,C)

Hom(X ,B) Hom(X ,C)

This morphism is injective by the Axiom of Choice. This is an important thing to notice: projec-tivity is related to the axiom of choice. This is in a sense the key property of projective objects as wewill see later.

Proposition 5. A direct summand of a projective object is projective.

Proof. Suppose F = P⊕Q with F projective. We know that we have a a diagram like the following byprojectivity of F.

F

B C 0∃

However, we can now form the composition with the canonical inclusion

P ,→ F → B,

and we have our desired lift.

Proposition 6. In R−mod, P is projective if and only if P is a direct summand of a free module.

14

Proof. If P is a direct summand of a free module then this follows from the previous proposition. Forthe reverse implication, the key is that for any module, in particular for our P, there is some freemodule F such that P is a quotient of F, that is, there is a surjection F → P.

We will capture this important property of the R-module category used in the last proof into a defi-nition.

Definition 7. A has enough projectives if every X ∈A is a quotient of a projective.

At this point, Weibel moves on to injective objects. However, we will skip ahead a bit to build up somemore motivation. If A has enough projectives, and M ∈A , build a resolution of M by projectives asfollows:

• choose a projective P0 and a surjection P0 → M.

• choose a projective P1 and a surjection P1 → ker(P0 → M)

• proceed in this way to choose Pn.

We then obtain a long exact sequence

· · ·→ P2 → P1 → P0 → M → 0.

What we have accomplished here is to unwind some, possibly complicated, object M into a sequenceof projective objects. We note that this is in fact equivalent to having a chain complex P· of projectiveobjects such that Pn = 0 for n < 0 and a chain map P· → M· where M· is the chain complex withM0 = M and Mn = 0 for all n 6= 0. Lets look at an example.

Example 5. Consider the exact sequence bellow which is an example of a projective resolution.

0 Z Z Z/mZ 0m

Apply the functor Hom(_,Z/mZ) to the sequence above, recalling that this is a left-exact functor, weget an exact sequence

0→Hom(Z/mZ,Z/mZ)→Hom(Z,Z/mZ) m−→Hom(Z,Z/mZ).

This is eqivalent to the sequence

0→Z/mZ 1−→Z/mZ 0−→Z/mZ.

If we consider the problem of extending this to the right as an exact sequence, we can simply addthe cokernel of the last map on the right, namely, Z/mZ. Note, however, that if we consider the chaincomplex of the projective resolution

0→Zm−→Z→ 0

and apply the Hom-functor to get0→Z/mZ 0−→Z/mZ→ 0

and then consider the homology of this complex, we get

Hn ={Z/mZ, if n = 1,00, else

.

That the cokernel shows up in this homology is no coincidence.

15

Now let us talk about injective objects, which are precisely dual to projective objects.

Definition 8. I ∈A is injective if Iop ∈A op is projective.

In this case, we have an equivalent definition given by the following diagram.

0 A B

I∃

We have a similar result for injectives, but this is harder to prove so we will record this as a theorem

Theorem 3. R−mod has enough injectives, that is, every R-module can be embedded in an injective.

Exercise 23. If A = Ab, show that M ∈ A is projective if and only if it is free and injective if andonly if its divisible. Here divisibility means multiplication by an element of R is surjective. Weibeluses this to prove that there are enough injectives in Ab.Solution.

Remark 2. The same actually holds for R-modules over a PID R.

Definition 9. Suppose we have a functor F : A →B of abelian categories. We call F additive if

F(X ⊕Y )→ F(X )⊕F(Y )

is an isomorphism. This implies that the induced map on Hom-sets is a group homomorphism.A functor G : A →B is a left-adjoint to F if there exists an isomorphism

Hom(GX ,Y )→Hom(X ,FY ),

that is natural in X and Y .

If you have not come across the notion of adjoint functors before, the canonical example of such pairsare the free and forgetful functors.

Example 6. LetG : Sets→R-mod

be the functor taking X to the free module generated by X , and let

F : R-mod→Sets

be the functor taking an R-module M to its underlying set M.

Example 7. Define a functorF : R−mod→Ab

by F(M)= M, the underlying abelian group. Then

Hom(A, M)=Hom(R⊗ A, M).

So taking a tensor product with R, A 7→ R ⊗ A, is a left adjoint to F. But F also has a right adjointgiven by

G(A)=HomAb(R, A).


16

LECTURE 7: INJECTIVE OBJECTS & DERIVED FUNCTORS - 02/03/2020

Recall that an injective object is one such that for a given solid arrow diagram below there exists adashed arrow.

0 A B

I

Now we give a sufficient condition for a functor to send injective objects to injective objects.

Proposition 7. Let F : A → B be an additive functor between abelian categories and F has a left-exact left adjoint, G. Then F preserves injectives.

Proof. Assume that I ∈ A is injective. We want to show that HomB(_,F(I)) is exact. Since G is theleft adjoint to F, this is the same as HomB(G(_), I). However, this is now a composition of the exactfunctors HomA (_, I) and G, and so it must be exact.

Let A = Ab and B = R-mod. Let F(A) = HomAb(R, A) and G(M) = M be the underlying abeliangroup. Note that G is exact. We claim that F is right-adjoint to G so F must preserve injectives. TheR-module structure on HomAb(R, A) is given by

(λ ·ϕ)(x)=ϕ(xλ), ∀x,λ ∈ R.

The adjunction comes from the following bijections

HomR-mod(N,HomAb(R, A)) u−→HomAb(N, A)

given byu(ϕ)(x)=ϕ(x)(1),

andv : HomAb(N, A) v−→HomR-mod(N,HomAb(R, A)),

given byv(ψ)(x)(y)=ψ(yx).

We leave it to the reader to verify that these are in fact inverses. You should do this exercise becauseone should enjoy those moments when exact formulas are given!We now have the tools to return to the proof of Theorem 3.

Proof. (Theorem 3) Suppose M is an R-module. Then, by a Exercise 22, ∃ι : M ,→ I with I a divisibleabelian group, and M denoting the underlying abelian group of M. Then ι ∈ HomAb(M, I) so we canuse the bijection v defined above to getn a map

v(ι) : M →HomAb(R, I).

The target of this morphism is injective, so it remains to show that this map is injective. This followssimply from noting that the composition

M →HomAb(R, I)ev1−−→ I,

is nothing but ι. This is injective and so v(ι) must be injective.

17

Now we address our original motivation for these projective and injective objects and define derivedfunctors.

Definition 10. Suppose F : A →B is a right-exact functor. Define LnF(X ) for X ∈A by the followingprocess:

1. choose a projective resolution P·∼=−→ X , where the ∼= symbol means quasi-isomorphism.

2. Set LnF(X )= HnF(P·).

We will show that LnF(X ) is well-defined up to a canonical isomorphism and so the choice of projec-tive resolution will be immaterial.

Proposition 8. If P· is a projective resolution of M and Q· is a resolution (not necessarily projective)of N, then there is an isomorphism H0(Hom·(P·,Q·))

∼=−→Hom(M, N).

Proof. First, recall that Z0 Hom·(P·,Q·)=Hom(P·,Q·) is just the set of chain maps. Let f ∈Hom(M, N).Then we have the following diagram of solid arrows where the rows are the given resolutions in theproposition statement. Then by projectivity of P0, we have the existence of the dashed morphismwhich we denote by ϕ0.

· · · P1 P0 M 0

· · · Q1 Q0 N 0

d

∃ϕ0 f

Now repeat this proccess replacing P0 and Q0 with their cycles to obtain a map ϕ1 : P1 →Q1.

· · · P1 Z(P0) 0

· · · Q1 Z(Q0) 0

∃ϕ0 ϕ0

Continuing in this way gives us a map of chain complexes ϕ· : P· → Q·. We now show that thisinduced chain map is in fact unique up to chain homotopy. Suppose f = 0. We must then show thatϕ· is nullhomotopic. In this case, we have the following diagram, similar to the previous one. Thedashed arrow here exists by the universal property of the kernel ZQ0.

P0 M 0

ZQ0 Q0 N 0

0∃!

ϕ0 0

We then have the following diagram in which the projective property of P0 is again used to obtainthe dashed map σ0.

P0

Q1 ZQ0 0

ϕ0σ0

d

18

Commutativity of the diagram then tells us that

ϕ0 = d ◦σ0 = d ◦σ0 +σ−1 ◦d,

with σ−1 = 0.We now proceed similarly and consider the diagram below, where

α=ϕ1 −σ0 ◦d,

andβ= d ◦ϕ1 +d ◦σ0 ◦d = d ◦ϕ1 +ϕ0 ◦d.

P1

ZQ1 Q1 ZQ0 0

αβ

∃!d d

The unique dashed map again exists by the universal property of the kernel. Repearting the sameprocess as above then gives us a diagram like the following where the dashed map exists by projec-tivity of P1. Then by commutativity of the diagram, we have that d ◦σ1 = α = ϕ1 −σ0 ◦d, implyingthat ϕ1 = d ◦σ1 +σ0 ◦d.

P1

ZQ1 Q1 0

α∃σ1d

Continuing by induction gives us maps σn such that ϕn = d ◦σn +σn−1 ◦d, implying chain map ϕ· isnullhomotopic.

We now use this to show that LnF(M) is well-defined. Suppose P· and Q· are projective resolu-tions of M. We then obtain a chain map ϕ· : P· → Q· which is the image of 1M ∈ Hom(M, M) un-der the isomorphism constructed in Proposition 8. This chain map then induces a homomorphismϕ∗ : HnF(P·)→ HnF(Q·). Furthermore, if ψ· : P· →Q· is another lift of 1M , then the preceding propo-sition implies that ϕ· 'ψ so that ϕ∗ =ψ∗. This shows that this induced homomorphism is canonical.To see that this is in fact an isomorphism, note that since both Q· and P· are projective resolutions,we can also induce a chain map in the opposite direction, η· : Q· → P·. We then have the followingcommutative diagram.

P· M

Q· M

P· M

ϕ·

η·

It follows that η◦ϕ : P· → P· lifts the identity 1M so by Proposition 8 we have that this must in fact bechain homotopic to the identity 1P· , implying that (η◦ϕ)∗ = η∗ ◦ϕ∗ = 1. Similarly, ϕ◦η' 1Q· so thatϕ∗ ◦η∗ = 1. Thus, HnF(P·) and HnF(Q·) are canonically isomorphic, and so the left derived functorsLnF(M) are well-defined up to this canonical isomorphism.

19

LECTURE 8: DERIVED FUNCTORS - 02/07/2020

Recall that projective resolutions are unique up to homotopy. That is, if P· and Q· are projectiveresolutions of M. Then there is a chain map ϕ· : P· → Q· lifting the identity 1M , which is unique upto chain-homotopy. We ended last lecture with showing that this chain map induces an isomorphismon homology. So we have that for any right exact functor F, HnF(P·)∼= HnF(Q·). We also saw that ϕ·is unique up to chain homotopy so this isomorphism is in fact canonical. This allows us to think ofHnF(P·) as being independent of the choice of projective resolution P·, and we call this object the leftderived functor LnF(M).Notice that right exactness was not used at all in the above. What right- exactness gives us is that

L0F(M)= coker(F(P1)→ F(P0))= H0F(P·)= F(coker(P1 → P0))= F(M).

So L0F = F. Now let’s look at some examples.

Example 8. Fix N ∈ R−mod. For M ∈ R−mod, define F(M) = N ⊗R M. This is right-exact (youcan look in Weibel for a verification of this). Assume R = Z, M = Z/nZ. Then we have a projectiveresolution

0 Z Z Z/mZ 0m

Now

LnF(Z/mZ)= HnF([Z m−→Z])= Hn([N m−→ N])=

0, n 6= 0,1N/mN, n = 0ker(m), n−1

.

So we have that LnF(_)=Torn(N,_) computes torsion.

Example 9. Let F(M) = Hom(M, N). Note that this is contravariant, left-exact functor. So we canthink of this as a sort of ’rigth derived functor’

R−mod→Abop.

The reason these derived functors are useful is because they are computable. The reason they arecomputable is the same that ordinary homology is, namely, they have long exact sequences associatedto them.

Theorem 4. Suppose F : A →B is a right-exact, additive functor between abelian categories and A ahas enough projectives. If

0→ M′ → M → M′′ → 0

is exact in A . Then there is a long exact sequence

· · ·→ LnF(M)→ LnF(M′′) δ−→ Ln−1F(M′)→ Ln−1(M)→···where δ is characterized uniquely.

We will make the final part of that statement more precise later. The following result is often calledthe Horseshoe Lemma.

Lemma 6. Suppose0→ M′ → M → M′′ → 0

is exact and P ′· ,P ′′· are projective resolutions of M′ and M′′. Then there is a projective resolution P· ofM· with Pn = P ′

n ⊕P ′′n and a commutative diagram with exact rows

20

0 P ′· P· P ′′· 0

0 M′ M M′′ 0

You can refer to Weibel for a nice proof of this result and also a nice diagram which explains thename of this result. We will now use this to construct the long exact sequence of the theorem above.Choose 0→ P ′· → P· → P ′′· → 0 as in the horeshoe lemma. Then the sequence

0→ F(P ′· )→ F(P·)→ F(P ′′

· )→ 0

is exact since F(Pn) = F(P ′n ⊕P ′′

n) = F(P ′n)⊕F(P ′′

n) and F is a right-exact functor. We now get a longexact sequence on homology

· · ·→ HnF(P·)→ HnF(P ′′· ) δ−→ HnF(P ′

· )→ Hn−1F(P·)→··· .

This is not quite a proof of Theoem 4 above since there are many details to check. We refer the readerto Weibel’s text for those details.Now, as usual in mathematics, we will extract the nice properties of these functors into a definitionof nice objects.

Definition 11. A δ-functor F· : A →B between abelian categories is a sequence of functors Fn : A →B such that if

0→ M′ → M → M′′ → 0

is exact in A then there are natural transformations

Fn(M′′) δ−→ Fn−1(M′)

fitting into a long exact sequence like the one above.

Our hard work above shows that the left derived functors of a right-exact additive functor form aδ-functor.

Definition 12. A morphism of δ-functors F· and G· is a sequence of natural transformations ϕn :Fn →Gn such that the diagram bellow commutes

Fn(M′′) Fn−1(M′)

Gn(M′′) Gn−1(M′)

ϕn

δ

ϕn

δ

Definition 13. A δ-functor F· is called universal if

Homδ(G·,F·)=Hom(G0,F0),

that is every natural transformation G0 → F0 extends uniquely to a morphism of δ-functors.

These universal δ-functors come in handy for showing that the sequence above is independent of themany choices we made along the way.







21

LECTURE 9: UNIVERSALITY OF DERIVED FUNCTORS - 02/10/2020

Last time we started talking about δ-functors as an alternate way of viewing derived functors. Wewill see this to its end and by the end we will have characterized derived functors is by a universalproperty. First note that we can view δ-functors F : A → B as a functor from the category of shortexact sequences in A to the category of long exact seqeuences in B.

Definition 14. A δ-functor F : A →B is called coeffaceable if for all M ∈A there exists P ∈A andan epi P → M such that the induced map Fn(P)→ Fn(M) is the zero map for all n > 0.

Definition 15. A δ-functor F is called universal if

Hom(G·,F·)→Hom(G0,F0)

is a bijection for all δ-functors G·.

Often times it is easy to show that universal objects exist. However, a more difficult thing is todescribe what they look like. The following result, which is actually Exercise 2.4.5 in Weibel, tells uswhat universal δ-functors look like.

Theorem 5. If F· is coeffaceable then it is universal.

Proof. Assume G0 → F0 is a natural transformation, and F· is coeffaceable. We want to construct anatural transformation Gn → Fn by induction on n. So we will assume that Gm → Fm is constructedfor m < n and that it is a morphism of δ-functors where defined. The base case n = 0 is given. We nowhave only one option: to ’coefface the values of M’. Chose an epi P → M such that Fn(P) → Fn(M) isthe zero map for all n > 0. Let

N = ker(P → M),

so that we have a short exact sequence

0→ N → P → M → 0.

We then have the following diagram connecting the long exact sequences associated to F· and G· withthe solid vertical maps coming from our inductive hypothesis.

Gn(P) Gn(M) Gn−1(N) Gn−1(P)

Fn(P) Fn(M) Fn−1(N) Fn−1(P)

un

δ

un−1 un−1

0 δ

The dashed arrow comes from the fact that the composition

Gn(M) δ−→Gn−1(N)un−1−−−→ Fn−1(N)→ Fn−1(P)

is zero and, by exactness of the bottom row, δ : Fn(M)→ Fn−1(N) is the kernel of Fn−1(N)→ Fn−1(P).Now that we have the map un : Gn(M) → Fn(M), we must show three things: (i) it is independent ofP, (ii) it is natural, (iii) it commutes with δ. First we note that this construction is functorial in P,that is given a diagram like the following.

0 N ′ P ′ M′ 0

0 N P M 0

22

we get an induced commutative square

Gn(M′) Gn(M)

Fn(M′) Fn(M)

u′n un

This implies that the construction is independent of P, because then we can choose Q to coeffacethe fiber product P ×M P ′. (Insert diagram and show that naturality follows from this aswell). For (iii), choose coeffacing P → M and assume we have a short exact sequence

0→ M′ → M → M′′ → 0

Then we have

0 N P M′′ 0

0 M′ M M′′ 0

Now the idea is to relate δ for the top row and bottom row to get the following diagram.

Gn(M′′) Gn(N)

Gn(M) Gn(M′)

Fn(M′′) Fn−1(M′)

Fn(M′′) Fn−1(N)

δ

δ

un un−1

δ

δ

(TODO: finish/clear up this proof).

<++>

Example 10. Let N ∈mod−R and M ∈R−mod. Recall that

Tor(N,_)= Ln(N ⊗_).

If M is projective, then M is a projective resolution to itself,

Ln(N ⊗M)={

N ⊗M, n = 00, n 6= 0

.

This implies Tor(M,_) is coeffaceable: for any M choose an epi P → M with P projective, thenTor(N,P)= 0 so the map induced by P → M is 0.

Notice that if F0 is fixed, then there is at most one universal δ-functor (up to a unique isomorphism)F· extending F0.

23

LECTURE 10: FLAT MODULES - 02/14/2020

What we saw last time is that, modulo some details left to the reader, if F· is coeffaceable, then F· isa universal δ-functor. This means that in particular, if A has enough projectives and F : A → cB isadditive, then LnF is a coeffaceable δ-functor, hence universal. From this we get that

Ln(N ⊗ (_)) : R−mod→Ab

andRn Hom(N,_) : R−mod→Ab.

We also haveLn((_)⊗M) : A →Ab

andRn Hom(_, M) : A →Ab

Note that the functor in the last example is contravariant and left-exact. So we obtain the derivedfunctor by taking a projective resolution in A . We will show that there are natural isomorphismsbetween the corresponding derived functors.

Lemma 7. If P is projective then P ⊗_ is an exact functor. We say, in this case, P is flat.

Proof. If P is free then P ∼= R⊕n. Assume

0→ N ′ → N → N ′′ → 0

is exact. Then0→ P ⊗N ′ → P ⊗N → P ⊗N ′′ → 0.

Since, P ⊗ N = N⊗n and the maps in this second sequence are just the corresponding direct sums,this second sequence is also exact. In general, if P is projective then all we have is that there existsa free module F such that F = P⊕Q for some module Q. Then we have the following diagram, wherethe vertical maps are induced by the inclusion P ,→ F and the identity N → N.

0 P ⊗N ′ P ⊗N P ⊗N ′′ 0

0 F ⊗N ′ F ⊗N F ⊗N ′′ 0

Theorem 6.Ln(N ⊗ (_))(M)= Ln((_)⊗M)(N)

Rn Hom(N, (_))= Rn Hom((_), M)(N)

By equality we mean that there is a natural isomorphism.

Proof. We will show that Ln(N⊗_)(M) is a universal δ-functor of N. Since Ln(_⊗M)(N) is a universalδ-functor of N and L0(_⊗M)(N)= N⊗M = L0(N⊗_)(M), once we prove this we will have maps inducedin both directions by universality and so they must be isomorphisms. Suppose P is flat, and choose aprojective resolution Q· → M. Then

Ln(P ⊗_)(M)= Hn(P ⊗Q·)={

P ⊗M, n = 00, n 6= 0

.

24

It follows that all flat modules coefface the functor, so we are done. We leave the proof of the secondisomorphism as an exercise.

Just so we have an idea of what flat modules look like, at least in the category Ab, we state thefollowing proposition.

Proposition 9. Flat abelian groups are torsion free abelian groups.

In general, we have the following hierarchy

free =⇒ projective =⇒ flat .

There are some conditions in which inclusions go both ways. For example, we have seen that if Ris a PID, then a module is projective if and only if it is free. We also have the following interestingtheorem of Lazard.

Theorem 7. All flat modules are filtered colimts of free modules.

Our next step is to obtain as much intuition as we can about what the functors Torn(N, M) andExtn(N, M) are tellings.

Definition 16. If M, N ∈A , an abelian category, an extension of M by N is an exact sequence

0→ N → X → M → 0.

Turns out that extensions for fixed M and N form a category where the morphisms are given bycommutative diagrams like the following

0 N X M 0

0 N Y M 0

Notice that if X =Y = N ⊕M then the options for the vertical map in the center are limited, namelywe must have it be (

1 0f 1

)for some f ∈Hom(M, N).

Definition 17. Ext1(M, N) is the set of extensions of M by N modulo equivalence.

Theorem 8. For all M and N, there is a natural isomorphism

Ext1(M, N)∼=Ext1(M, N)






25

LECTURE 11: THE MEANING OF Tor AND Ext - 02/17/2020

The interpretations of Tor and Ext are not crucial for our voyage into homological algebra. However,these interpretations illustrate the use of homological algebra in mathematics quite well. Addition-ally, they are also quite fun. As such, we will spend some time continuing our discussion from lasttime and discuss the meaning of these particular derived functors. We will start with consequencesof Theorem 8 stated last time. First we note that Ext1(M, N) has a lot of structure. So Theorem 8tells us that all of this structure is transfered to the set of extensions. In particular, Ext1(M, N) is acovariant functor in N and contravariant in M. Therefore, if M′ → M is a morphism of R-modules,then we should have a map

Ext1(M, N)→Ext1(M′, N).

This map is illustrated in the following diagram, where the right-hand square is a pullback.

0 N X ′ = M′×M X M′ 0

0 N X M 0

This is indeed functorial since pullbacks respect composition. Similarly, if N → N ′ is a morphism ofR-modules, then we get a map on Ext1 by the following diagram in which the left-hand square is apushout.

0 N X M 0

0 N ′ X ′ M 0

Furthermore, the square below commutes.

Ext1(M, N) Ext1(M, N ′)

Ext1(M′, N) Ext1(M′, N ′)

We also should have that Ext1(M, N) is an abelian group.

Lemma 8. There is a map

Ext1(M, N)⊕Ext1(M′, N ′)→Ext1(M⊕M′, N ⊕N ′)

Proof. Take the direct sum!

Now let f ⊕ g denote the map (f 00 g

): M⊕M → N ⊕N

of two morphisms f , g : M → N. Let X and Y be two extensions of M by N. Define

X +Y = (1 1

)∗

(11

)∗(X ⊕Y )

26

The zero in Ext1(M, N) is the split extension

0→ N

10

−−−→ N ⊕M

(0 1

)−−−−−→ M → 0.

What else should we have? Well, for an exact sequence

0→ N ′ → N → N ′′ → 0

we should have a mapHom(M, N ′′) δ−→Ext1(M, N ′).

This, again, is given by taking a pullack of a map f ∈Hom(M, N ′′) along N → N ′′. The exercise is nowto verify extactness of the induced long exact sequence in Ext1. One can check that Ext1(M, N) = 0if M is projective or N is injective, implying that Ext1(_,_) is effaceable and coeffaceable in theappropriate slots so that

Ext1(_,_)= R1 Hom(_,_)=Ext1(_,_).

Exercise 36. Show that Ext1(M, N) is effaceable in both M and N. Solution

One can use this to define Extn(M, N) = Extn−1(M, N ′) and try to find the meaning of the highterExt-groups.

Definition 18. (Yoneda Ext.)

Extn(M, N)= {0→ N → X1 →···→ Xn → M → 0| exact }/∼Given X ·,Y· ∈Ext2(M, N), the morphisms are given by butterflies

0 0

0 N X1 X2 M 0

Z

0 N Y1 Y2 M 0

0 0

You can work out Ext3(M, N) yourself for fun, but the diagrams are a bit insane. Moving right along,remember that Tor is the left-derived functor of the tensor product. So Tor should behave similarly tothe tensor product. In particular, it should satisfy a universal property similar to the tensor product.

Definition 19. Suppose M, N,P are abelian groups. A biextension of M, N by P is a family ofextenions

0→ P → X (m)→ N → 0

for all m ∈ M. These come with isomorphisms

X (m+m′)'ϕ X (m)+ X (m′),

that is, a biextension is a homomorphism M →Ext1(N,P).

So for A an abelian group and an injective P,

Hom(TorA(M, N),P)=Biext(M, N,P).

27

LECTURE 12: LIMITS & COLIMITS - 02/21/2020

Today we will introduce limits and colimits. In essence, these are two different ways of forming a newobject out of a collection of objects with some relations (morphisms) between them. The particularlynice thing about these constructions is that they are characterized by universal properties. Afterintroducing these and seeing some examples we will be interested in how these behave with respectto the derived functors Tor and Ext. Long story short, they respect one another. First let us recallthe definitions.

Definition 20. An I-shaped diagram in C is functor from a small category I into C , X : I →C .

For each ξ ∈C there is a constant diagram ∆(ξ) which is just the constant functor sending all objectsof I to ξ and all morphisms to the identity idξ. Now consider the set of all natural transformationsfrom an arbitrary diagram X to ∆(ξ), denoted Hom(X ,∆(ξ)).

Definition 21. A colimit of a diagram X : I →C is an object colimX together with a natural trans-formation η : X =⇒ ∆(colimX ) that is universal. A limit of a diagram X : I → C is an object limXtogether with a natural transformation α :∆(limX ) =⇒ X that is universal.

Remark 3. In the above definition, we used an underline to denote the object underlying the limitand colimit to distinguish from the total data of these constructions which includes a natural trans-formation. This underline will be dropped in what follows for convenience. It should, however, beevident from context whether we are referring to the object or the total data.

By universal we mean, for example if ξ is another object of C with a natural transformation η′ : X =⇒∆(ξ) then there is a morphism colimX → ξ which when viewed as a natural transformation betweenthe corresponding constant functors gives a factorization of η′ in terms of η. We have already seenmany examples of these. Products and coproducts are limits and colimits of the diagrams with nomorphisms. Pullbacks and pushouts are limits and colimits of the diagrams

A B C

and

A B C

respectively. Kernels and cokernels are also examples. Explicitly, the kernel and cokernel of a mapf : A →Y are the limit and colimit, respectively, of the diagram below.

A Bf

0

Such limits and colimits are called equalizers and coequalizers. Yet another nice thing about abeliancategories is that limits and colimits have explicit constructions as kernels and cokernels of particu-lar maps. Explicitly, we have

limX = ker(∏

i∈IX (i)→ ∏

ϕ:i→ jX ( j)

)where the map is given by (ai) 7→ (ϕ(ai)−a j)ϕ:i→ j. Similarly,

colimX = coker(⊕ϕ:i→ j X (i)→⊕ j∈I X ( j)

)where the map is given by ai[ϕ] 7→ϕ(ai)−ai. This implies a nice result.

28

Corollary 1. Abelian categories have finite limits and colimits.

Example 11. Let p ∈Z be a prime. Then the p-adic integers are defined as the colimit

Zp = colim(Z/pZ←Z/p2Z←Z/p3Z←··· ).A category, A , for which all limits (resp. colimits) exist is called complete (resp. cocomplete). In thiscase, we can view the limit and colimit as functors

lim,colim : A I →A ,

where A I denotes the functor category.

Proposition 10. Let I be a small category such that all functors F : I →A have limits and colimits.Then the functor lim is right adjoint to the diagonal functor ∆ : A →A I sending each object ξ ∈A tothe constant functor ∆(ξ), and colim is left adjoint to ∆.

Proof. We want to show that there is a natural isomorphism HomA I (∆(A),F) ∼= HomA (A, limF).Given a natural transformation ∆(A) =⇒ F, the universal property of the limit gives rise to a mor-phism A → limF. It is evident that any morphism A → limF induces a natural transformation∆(A) =⇒ ∆(limF) which one can compose with the universal natural transformation ∆(limF) =⇒ Fto obtain ∆(A) =⇒ F. This isomorphism is natural in both A and F since the construction used onlynatural transformations. The proof of the statement for colimits is nearly identical.

Theorem 2.6.10 in Weibel states that left adjoints preserve colimits and right adjoints preserve limits.Thus, we have that limits commute with limits and colimits commute with colimits.

Example 12. Let’s consider Hom(Z[ 1p ],Q/Z). To determine a homomorphism in this set we must

choose where 1 goes. The set of such choices is Q/Z. Now we must choose where 1p goes, but this is

restricted to lie in Q/pZ depending on the first choice. Similarly, 1p2 has a choice set Q/p2Z and so we

can compute

Hom(Z[ 1

p

],Q/Z)= colim(Q/Z←Q/pZ←Q/p2Z←··· )= Qp

This is used in one of the calculations in Weibel chapter 3 to compute Ext1(Z[p−1]/Z,Z), which wasalso used in one of the exercisies. We end this lecture by introducing particularly interesting limitsand colimits. Let I be the poset category whose objects are the natural numbers 0,1,2. . . , and mor-phisms n → m are given by relations n ≤ m. A functor F : I → A is then a sequence of objects andmorphisms in A :

A0 → A1 → A2 →··· ,

while a contravariant fuctor F : Iop →A is a tower:

· · ·→ A2 → A1 → A0.

The colimit in the former is called the direct limit, is denoted lim−−→A i. The limit in the latter is calledthe inverse limit, and is denote lim←−−A i.





29

LECTURE 13: DERIVED FUNCTORS OF THE INVERSE LIMIT - 02/24/2020

As mentioned last time, under suitable circumstances like A being complete and cocomplete, we canview limits and colimits as functors A I →A for any small category I.

Proposition 11. Suppose I is a small category and A is an abelian category with enough projectivesand all direct sums(i.e A is complete). Then A I has enough projectives. (Note: a projective in A I is afunctor F such that F(i) is projective for all objects i of I).

Proof. For each object i in I, we have an exact evaluation functor

e i : A I →A

sending a functor P to the object P(i). This is right adjoint to the functor Fk : A →A I defined by

Fk(Q)= (i 7→ ⊕Hom(i,k)Q

).

Morphisms are induced from the morphisms Hom(i′,k) → Hom(i,k) for i → i′. Explicitly, if ϕ : i 7→ i′

in I, then the coordinate projection Fk(Q) → Q onto the factor corresponding to ψ : i′ 7→ k is theprojection of Fk(Q) onto to the factor corresponding to ψ◦ϕ. We leave the verification of this to thereader. We then have that Fk is left-adjoint to an exact functor so that Fk preserves projectivesimplying Fk(Q) is projective for any projective Q in A . Suppose M ∈ A I . Let Pi → M(i) be a an epifor each i, with Pi projective. Then

⊕iFi(Pi)→ M

is a surjection from a projective.

Using the fact that the inverse limit and direct limit are left and right exact, respectively, we canthen define the derived functors

Rn lim←−− : A I →A

andLn lim−−→ : A I →A .

We will use the shorthand lim←−−1 for R1 lim←−−. We will focus on understanding the right derived functors

of the inverse limit functor. As we will see, these vanish for n > 1 so lim←−−1 is the only interesting

case. Throughout this section, let I denote the standard poset category of natural numbers. We willconsider the case A =Ab. In this case, lim←−−

1 has a particularly nice description.

Definition 22. Let M ∈A I and take A =Ab.. An M−torsor is a functor P : I →Set such that M(i)acts on P(i) 6= ; simply transitively and the actions commute with the maps in I.

A natural transformation of functors that commutes with M(i) action for all objects i in I is a mor-phism of M-torsors. So these form a category, in fact a groupoid (all morphisms are isomorphisms).We claim that lim←−−

1 M is in fact the set of M-torsors modulo isomorphism. In fact, let us make thefollowing definition, and prove that lim←−−

(1) and lim←−−1 are the same.

Definition 23. If M : I → Ab is a functor, define lim←−−(1)(M(i)) be the set of M-torsors modulo isomor-

phism.

We note that M itself can be considered an M-torsor by composing with the forgetful functor Ab →Set and letting M(i) act on its underlying set by multiplication for each i. If P is an M-torsor, P ∼= Mif and only if lim←−−P(i) 6= ;.

30

Example 13. Let I = 0 ← 1 ← 2 ← ··· and M(n) = pnZ, N(n) = Z/pnZ. Then lim←−−M(n) = 0, and wehave the following Exact sequence where the index n is supressed.

0→ M →Z→ N → 0.

Then for every α ∈ lim←−−N(n)= Zp we get a coset of M in Z. The system P(n)=π−1(αn) is an M-torsor.

This construction in the example is actually a special case of something more general, namely, theexsistence of a connecting homomorphism in the following proposition.

Proposition 12. Let0→ M′ → M → M′′

be exact. Then there is a long exact sequence

0 lim←−−M′ lim←−−M lim←−−M′′ lim←−−(1) M′ lim←−−

(1) M lim←−−(1) M′′δ

Proof. TODO: the proof.

<++>

Proposition 13. lim←−−(1) is effaceable.


<++>

This completes the proof that lim←−−1 = lim←−−

(1). We will end today’s lecture by showing explicitly thatlim←−−

(1) satisfies a particularly nice property that lim←−−1 satisfies.

Definition 24. (Mittag-Leffler Condition) We say that M : I →A satisfies the Mittag-Leffler con-dition if the images M(n+k)→ M(n) stabilize for large enough k.

This condition ends up being very useful as it is a sufficient condition to deduce that lim←−−1 M(i) = 0.

The reader can see Weibel’s text for a proof of this general result. Here we prove explicitly that thisis also the case of lim←−−

(1).

Theorem 9. If M satisfies the Mittag-Lefler condition then lim←−−(1) M = {M}

Proof. Let P be an M-torsor. Let

P ′(n)= ⋂k≥0

im(P(n+k)→ P(n)

).

This is nonempty by assumption. Then P ′(n+ k) → P ′(n) is surjective for all k ≥ 0. So lim←−−nP ′(n) ⊂

lim←−−nP(n) is nonempty. Thus, by our note above, lim←−−n

P(n) 6= ; implying P ∼= M.

31

LECTURE 14: FILTERED COMLIMITS - 02/28/2020

We will finish our discussion on limits and then we will move on to spectral seqeuences.

Definition 25. A (small) category I is called filtered if

1. ∀i, j ∈ I,∃k ∈ I with morphisms i → k and j → k (upper bound)

2. ∀i, j ∈ I with morphisms α : i → j and β : i → j,∃ a coequalizer γ : j → k.

Thinking loosley in terms of homotopy theory, this definition states that I is, in some sense, con-tractible. Colimits of functors with domain a filtered category are called filtered colimits and usuallydenoted the same as a direct limit lim−−→.

Theorem 10. Filtered colimits commute with finite limits in the category of sets.

Consider X : I × J → Set, with I finite and J filtered. The theorem above states that the canonicalmap

lim−−→ jlim←−−i

X (i, j) lim←−−ilim−−→ j

X (i, j)

obtained from the universal properties of the limits involved is an isomorphism

Proof. All (finite) limits are built from equalizers and finite products. It follows that it suffices toprove the theorem for I = {· ·} and I = {·⇒ ·}. We will prove this for the latter case and leave theformer as an exercise. Let α,β be the two nonidentity morphisms in I. Note

lim←−−iX (i, j)= {x0 ∈ X (0, j)|α(x0)=β(x0)}.

In the colimit, x0 inX (0, j) and x′0 ∈ X (0, j′) are equivalent if there exists morphisms j → k and j′ → kthat map x0, x′0 to some x′′0 under X . Thus, we have a map

lim−−→j

lim←−−i

X (i, j)→ lim←−−i

lim−−→j

X (i, j)= {[x0] ∈ lim−−→X (0, j)|[α(x0)]= [β(x0)]},

mapping the image of x0 in the colimit lim−−→ jto the equivalence class [x0]. We first show that this is

surjective. Suppose [x0] ∈ lim−−→ jX (0, j) and [α(x0)] = [β(x0)]. then x0 ∈ X (0, j) for some j. IF α(x0) 6=

β(x0), we still know that ∃ γ : j → j′ such that γ(α(x0)) = γ(β(x0)) because [α(x0)] = [β(x0)]. So γ(x0) ∈lim←−−i

X (i, j′) implying [γ(x0)] ∈ lim−−→ jlim←−−i

X (i, j) is such that ϕ([γ(x0)])= [x0].

Corollary 2. Filtered colimits are exact.

Proof. Colimits commmute with colimits so colimits are right exact. Since kernels are finites limits,filtered colimits preserve kernels and cokernels.

Corollary 3. Homology commutes with filtered colimits, that is

Hn(lim−−→i

Ci)= lim−−→i

Hn(Ci).

Proposition 14. Tor commutes with filtered colimits. Explicitly, M : I → R-mod where I is filtered,and N an R-module,

Torn(lim−−→i

M(i), N)∼= lim−−→i

Torn(M(i), N).

32

Proof. Choose a projective resolution P· → N. then

Torn(lim−−→i

M(i), N)= Hn(lim−−→i

(M(i)⊗N))= lim−−→i

Hn(M(i)⊗N)= lim−−→i

Torn(M(i), N).

Corollary 4. Let M be an abelian group. M is Tor-acyclic if and only if M is torsion-free.

Proof. For any M, M =∪iMi = lim−−→iMi, where Mi are the finitely generated subgroups of M. Thus

Torn(M,_)= lim−−→nTorn(Mi,_).

If M is torsion-free then so are the Mi implying Mi are free, since they are finitely generated. ThusTorn(Mi,_)= 0,n > 0.

Recall that an R-module is called finitely presented if ∃ an exact sequence

Rm → Rn → M → 0,

where m and n are finite.

Proposition 15. M is finitely presented if and only if

Hom(M, lim−−→i

Ni)∼= lim−−→i

Hom(M, Ni).

We leave the proof of this to the reader. However, a nice corollary to this is that if M is finitelypresented and Rn → M is surjective, then the kernel is finitely generated.

LECTURE 15: SPECTRAL SEQUENCES - 03/02/2020

Alright, it’s now time to talk about spectral sequences. Let us start by describing the situation whereone usually meets spectral sequences. Suppose you have a double complex Cpq displayed visuallybelow.

......

...

· · · C22 C12 C02 · · ·

· · · C21 C11 C01 · · ·

· · · C20 C10 C00 · · ·

......

...

33

Where would one encounter such a complex in the first place? One important situation is when youtry to compute a composition of derived functors. These are each computed separately by takingthe homology/cohomology of an injective or projective resolution. When we compose them, we thenhave a double complex to compute the homology/cohomology of. We can combine the data of a doublecomplex into an ordinary chain complex

Tot(C··)n = ∑p+q=n

Cpq,

where the differential is d = dv + dh, where dv is the vertical differential and dh the horizontaldifferential. So we are interested in computing the homology of this double complex, which amountsto computing the homology of Tot(C··). One place to start would be to compute the vertical homologyfirst. With this in mind, set E1

pq = Hq(Cp·). We then have the following sequence of horizontalcomplexes in which the differentials are induced by dh.

...

· · · E122 E1

12 E102 · · ·

· · · E121 E1

11 E101 · · ·

· · · E120 E1

10 E100 · · ·

...

We can then take the horizontal homology of this resulting sequence of complexes. With this in mind,set E2

pq = Hp(E1·q)= Hp(Hq(C··)). We then have the following lattice of homology groups.

...

E222 E2

12 E202

· · · E221 E2

11 E201 · · ·

E220 E2

10 E200

...

34

Now let us see how close this got us to our goal of computing the homology of Tot(C··). For simplicity,suppose that C·· is a first quadrant double complex, meaning Cp,q = 0 if q or p are less than zero.Then at least indegree zero, we have succeeded:

H0H00 = C00

dv(C01)+dh(C10)= H0Tot(C··).

However, in degree one we still have some work to do. First, what should we expect to get? Well arough description of this is the following

H1Tot(C··)= {(x01, x10)|dv(x01)= dh(x10)}⟨(dv(x02),0), (dh(x11),dv(x11)), (0,dh(x02))⟩ .

Now let’s see what we get by following our proceedure above. First we get the vertical homology

H1(C0·)= {x01|dv(x01)= 0}{x01|∃x02 ∈ C02,dv(x02)= x01}

.

Then we take the horizontal homology, this time at degree zero so that the total degree is 1:

H0H1 = {(x01,0)|dv(x01)= 0}{(d(x02),0)|x02 ∈ C02}+ {(dh(x11),0)|dv(x11)= 0}

.

It follows that we have a map H0H1 → H1Tot(C··). This is not quite an isomorphism, but there is anexact sequence

· · ·→ H0H·1 → H1Tot(C··)→ H1H·0 → 0.

So we were able to fit what we wanted to compute into a long exact sequence consisting of simplerthings to compute. This is, in essence, the utitility of spectral sequences. We introduce the formaldefinition of a spectral sequence now so that the reader is aware of the kind of objects we are workingwith.

Definition 26. A (homology) spectral sequence (starting with Ea) in an abelian category A con-sists of the following data:

1. A family {Erpq} of objects of A defined for all integers p, q, and r ≥ a.

2. Morphisms drpq : Er

pq → Erp−r,q+r−1 called differentials, which satisfy drdr = 0.

3. Isomorphisms between Er+1pq and the homology of Er·· at the (p, q)-spot.

We call Er·· the ’Er-page’. Let us move on to re-introduce objects for which spectral sequences are aparticularly useful tool.

Definition 27. A filtration of an object of an abelian category, M ∈A is a sequence of subobjects

· · · ⊂ FnM ⊂ Fn+1M ⊂ ·· · .

(Note: in a general abelian category, subobjects of M are equivalence classes of objects with injectionsinto M.) These define a metric on M

logδ(x, y)=min{n|x− y ∈ FnM}.

Define the associated graded by

gr·M =⊕n

FnMFn−1M

.

35

Example 14.· · · ⊂ p2Z⊂ pZ⊂Z.

The metric completion there is Zp. The n-th associated graded of this filtration is grn = p−nZ/p−(n−1)Z∼=Z/pZ. The total associated graded is gr· = Fp[t].

We can use filtrations do obtain a double complex and hence a spectral sequence. We should note,however, that in this context a spectral sequence does not necessarily compute the homology for us,but it does give us information about how the homology changes as we move along the filtration.Often this combined with some additional information we have about the objects in question will beenough to compute the desired homology.If C· is a complex of filtered objects, then the homology gets an induced filtration:

FpHnC· =FpCn ∩ZnC·FpCn ∩BnC·

.

We want to compute the associated graded, which is given in degree p by

grpHnC· =FpCn ∩ZnC·

FpCn ∩BnC·+Fp−1Cn ∩ZnC·.

We can do this by noting that this is in fact just the n-th homology of the following sequence

d−1(FpCn)→ FpCn

Fp−1Cn→ Cn−1

d(Fp−1Cn).




LECTURE 16: SPECTRAL SEQUENCES VIA FILTRATIONS - 02/06/2020

Definition 28. A filtered complex is a filtered object in the category of chain complexes. This isthe same as having a filtration F·Cn for each n.

We emphasize that this implies d(FpCn) ⊂ FpCn−1. The idea is that we can use the filtrations onCn−1,Cn,Cn+1 to obtain information about Hn(C·). The first thing to notice in this direction is thatthe filtration on C· induces a filtration on Hn(C·), which we saw last time:

FpHnC· =FpCn ∩ZnC·FpCn ∩BnC·

.

The associated graded gr·HpC· is then given by

grpHnC· =FpHnC·

Fp−1HnC·.

We introduce some new notation,

gr]p,q]Cn = FqCn

FpCn,

36

and similarly

gr]p,q]HnC· =FqHnC·FpHnC·

.

Note in this notation p < q. Next we use the filtration to approximate the kernel and the image

HnC·(r, s)= d−1(FsCn−1)d(FrCn+1)

if ∪rFrCn+1 = Cn+1 and ∩sFsCn−1 = 0 then these approximate HnC· as r, s →∞.

Remark 4. Throughout this lecture we use the following notation. If A,B ⊂ C, write

A/B = A/(A∩B)= (A+B)/B = im(A → C/B).

Now let’s combine these two ideas.

gr]p,q]Hn(r, s)= FqCn ∩d−1(FsCn−1)FpCn +d(FrCn+1)

.

What we are doing here is taking approximate cycles and modding out by approximate boundaries.Now what happens when we apply the differential to this object? Well It certainly will land in aquotient of FsCn−1. Further, what we quotient by must contain d(FpCn). Aside from this we havefreedom to choose what this quotient is. In particular, we may choose the image so that d maps intoanother one of our grading quotients as shown below to the right. It turns out we can extend this inthe opposite direction now, and condiser the map induced by the differential into gr]p,q]Hn(r, s). Thisgives us the map on the left below.

gr]r,r′]Hn(p′, q) gr]p,q]Hn(r, s) gr]s′,s]Hn−1(p, q′)d d

We note that here q′, p′, s′, r′ are arbitrarily chosen. Evidently, we now obtain a chain complex, so wemay take its homology and obtain

FqCn ∩d−1Fs′(Cn−1)FpCn +d(Fr′(Cn+1)

.

This is nothing but gr]p, q]Hn(r′, s′). But now we have that r′ < r and s′ < s so this approximationis better than the one we started with! We can then continue in the same manner and improve ourapproximation once more. The important takeaway here is that everytime we take the homology ofthe current complex we get a better approximation of the associated grated of Hn. We note that p′

and q′ played no significant role here. They will be important later when we extend these to longersequences.Now assume that C· is a filtered complex. We then have

Fp−1Cn+1 Fp−1Cn Fp−1Cn−1

FpCn+1 FpCn FpCn−1

· · · grpCn+1 grpCn grpCn−1 · · ·

d d

d d

37

But notice that the last row is the same things as

· · ·→ gr]p−1,p]Hn+1(p−1, p)→ gr]p−1,p]Hn(p−1, p)→ gr]p−1,p]Hn−1(p−1, p)→··· .

LetEr

p,q = gr]p−1,p]Hp+q(p−1+ r, p− r).

Next time we will use this construction to study double complexes.

LECTURE 17: SPECTRAL SEQUENCE OF A DOUBLE COMPLEX - 03/09/2020

Last time we constructed a spectral sequence for a filtered complex. Recall the construction ofgr]p,q]Hn(r, s) as a quotient of ’almost cycles’ by ’almost boundaries’. We also saw that if our orig-inal complex C· is filtered, meaning d(FpC·) ⊂ FpC·, then things become simpler and we get a nicespectral sequence with

Erp,q = gr]p−1,p]Hp+q(p−1+ r, p− r).

Now we move on to a case of particular interest, namely a double complex, C·· of which a singleanticommuting square is shown below.

Cpq Cp−1,q

Cp,q−1 Cp−1,q−1

dh

dv dv

dh

Recall that there is a choice when defining what we mean by the total complex of this double complex

TotΠ(C··)=∏

p+q=nCpq

orTot⊕(C··)=⊕p+q=nCpq.

These are equal if there are only finitely many nonzero Cpq such that p+q = n for each n. We definea filtration on this double complex by letting

FkCpq ={

Cpq p ≥ k0 p < k

.

Now to construct a spectral sequence. The E0-page will simply be the double complex with differen-tial given by dv. Notice that this now has the interpretation of being the associated graded of ourfiltered complex. The E1-page is obtained by taking the homology of the E0 page and the differentialwill simply be induced by dh on the quotients. This leads us to the very important E2-page.Note that grpHn measures the difference between each step of the grading induced on homology.Then if FpCn = 0 for p << 0, FpCn = C−n for p >> 0 and all n. Then

grpHn(a,b)= d−1(FbCn−1)∩FpCn

d(FaCn+1)+Fp−1Cn= FpHnC·.

This implies Erp,n−p = grpHnC· for all r >> 0.

That is quite enough abstraction for now, lets move on to an example.

38

Example 15. Suppose M, N are R-modules. Choose a projective resolution P· of M and Q· of N.Then P·⊗Q· is a double complex. We note that the differentials may need to be adjusted in orderto agree with our sign convention, but we ignore this detail for now. Let’s compute the homologyusing the spectral sequence we constructed. However, note that we made a choice above to takethe filtration along the horizontal direction. Instead, we could start by filtering along the verticaldirection. Let us see how these two approaches differ in this example. In the first case where wefilter horizontally, our E0-page looks like

E0 :...

......

· · · P2 ⊗Q2 P1 ⊗Q2 P0 ⊗Q2

· · · P2 ⊗Q1 P1 ⊗Q1 P0 ⊗Q1

· · · P2 ⊗Q0 P1 ⊗Q0 P0 ⊗Q0

From this, we see that the E1-page, which we obtain by taking homology of E0, is E1pq = Pp ⊗Hq(Q).

However, since Q· is a projective resolution, it is exact for q ≥ 1. For q = 0 we have H0(Q) = B.Visually, we have:

E1 :

0 0 0

0 0 0

· · · P2 ⊗N P1 ⊗N P0 ⊗N

Thus, taking homology gives us

E2pq =

{Lp(_⊗N)(M) q = 00 q 6= 0

.

In the latter case where we filter vertically, we can follow a similar argument to obtain E1pq = Hp(P·)⊗

Qq, and

E2pq =

{Lq(M⊗_)(N) p = 00 p 6= 0

.

Since the complex we started with is a first-quadrant double complex, its filtration is canonicallybounded and so it converges to the homology of the total complex in each case. This, in particular,proves that

Tor(_, N)(M)= L0(_⊗N)(M)= E200 = L0(M⊗_)(N)=Tor(M,_)(N).




39

LECTURE 18: GROTHENDIECK SPECTRAL SEQUENCES - 03/13/2020

This is our fist zoom meeting! Let’s see how it goes.The problem we want to consider today is the following. Suppose

AF−→B

G−→C

is a sequence of right exact functors. How do we relate LnG,LmF,Lp(G ◦F)? Choose a resolution P·of X ∈A . Then LnF(X ) = HnF(P·) and now we need to resolve F(P·) by a double complex Q··. Thenwe can relate G(Q··) to LnG(LmF(X )) and Lp(GF)(X ).

Theorem 11. Suppose C· is a chain complex in A and that A has enough projectives. There exists adouble complex P·· such that the following holds:

1. Each Pm· is a projective resolution of Cm.

2. im(Pm· → Pm−1,·) is a projective resolution of im(Cm → Cm−1).

3. Hm·(P··)→ Hm(C·) is a projective resolution.

4. If Cm = 0 then Pmn = 0 for all n.

Proof. The idea is to start with point 2. Arrange this by choosing a projective resolutions Qm· →d(Cm) and Rm· → Hm(C·). We will combine these two resolutions using the Horseshoe Lemma asfollows.

0 Qm+1,· Sm· Rm· 0

0 d(Cm+1) Zm(C·) Hm(C·) 0

Since the sequence on the bottom is exact, the Horseshoe Lemma gives tells us that Smn =Qm+1,n ⊕Rmn is then a projective resolution of Zm(C·). We play the same game now in the form of the followingdiagram.

0 Sm,· Pm· Qm· 0

0 Zm(C·) Cm d(Cm) 0

Now Pmn = Smn ⊕Qmn =Qm1,n ⊕Rmn ⊕Qmn is a projective resolution of Cm. We then have a doublecomplex of the following form.

......

Qm+1,n ⊕Rmn ⊕Qmn Qmn ⊕Rm−1,n ⊕Qm−1,n

......

Cm Cm−1d

40

The horizontal map above is simply given, in matrix form, by the elementary 3×3 matrix E13. Wenote that signs may be adjusted so that this forms a double complex in the sense of Weibel, i.e. inorder for squares to anticommute.

The construction above is often called the Cartan-Eilenberg resolution of C·. Now we can talkabout the Grothendieck Spectral Sequence. This is a very powerful spectral sequence because manyothers that you will come across, such as the Leray spectral sequence, are special cases of this con-struction. The setting is the one we started with above. We have a composition of right-exact functors

AF−→B

G−→C ,

where A ,B have enough projectives. So we can compute the left-derived functors. Suppose X ∈ A ,and choose a projective resolution P· → X of X . Now choose a Cartan-Eilenberg resolution Q·· →F(P·). Recall that this means that Qm· → F(Pm) is a projective resolution. We now have a doublecomplex G(Q··) on which we can run our spectral sequence from last lecture. The I E0 page then lookslike:

I E0 :

G(Q22) G(Q21) G(Q20)

G(Q12) G(Q11) G(Q10)

G(Q02) G(Q01) G(Q00)

Now we make an additional assumption that LnG(Pm) = 0 for all n > 0. An example of wherethis holds is if F has an exact right-adjoint. Now the homology at the p-th slot of row q is thenHp(G(Qq·)= LpG(F(P·)), and these vanish for projective P, the I E1-page then looks like:

I E1 :

0 0 G(P2)

0 0 G(P1)

0 0 G(P0)

Now let’s move on to the I E2-page. For this, we simply take the vertical homology above and get thefollowing.

41

I E2 :

0 0 L2GF(X )

0 0 L1GF(X )

0 0 L0GF(X )

So we get That Hn(Tot(G(Q··))= LnGF(X ).Now let’s see what happens when we run our double-complex spectral sequence in the opposite di-rection. In this case,

I I E0 :

G(Q22) G(Q21) G(Q20)

G(Q12) G(Q11) G(Q10)

G(Q02) G(Q01) G(Q00)

However, by our construction above, we know that Q·,n is a split complex, and G, being an additivefunctor, preserves this structure. Thus, G(Q·,n) is a split complex, implying Hm(G(Q·n))=G(Hm(Q·n))How?. The E1-page is then:

I I E1 :

G(H2(Q·2)) G(H2(Q·1)) G(H2(Q·0))

G(H1(Q·2)) G(H1(Q·1)) G(H1(Q·0))

G(H0(Q·2)) G(H0(Q·1)) G(H0(Q·0))

Now recall that Hm·(Q··) → Hm(P·) = LmF(X ) is a projecive resolution. So we get the following nicedescription of the E2-page.

42

I I E2 :

L2G(L2F(X )) L1G(L2F(X )) L0G(L2F(X ))



To compute futher, we need more information about the specific functors and categories that we areworking with. In summary, we have created a spectral sequence of the form

E2pq = LpF(LqG(X )) =⇒ Lp+q(GF)(X ),

and so we have succeeded in our goal of relating these derived functors. Next time we’ll see someexamples of this construction.

Exercise 47. Weibel 5.6.2Solution

LECTURE 19: DERIVED CATEGORIES - 03/16/2020

We are going to introduce derived categories today. Weibel takes the standard approach of motivatingderived categories through algebraic topology. This is indeed a very important view and we refer thereader to the textbook for that view. Here we will try to motivate the idea with just homologicalalgebra in mind. The overarching idea of what we have been doing in this class is use complexes toanalyze objects of an abelian category. As we have seen, complexes, as well as beefed-up versionsof complexes (e.g. double complexes, spectral sequences, ect.), are very useful tools. Let’s take theparticular example of Grothendieck spectral sequences from last time.

Remark 5. Following Weibel and Grothendieck, we will now switch to talking about cochain com-plexes and cohomology.

Missed some of this argument.The point is that we want to consider complexes equivalent when they are quasi-isomorphic. So weare in search of a category, D(A ), which is, in a sense, Ch(A ) modulo quasi-isomorphisms.

Proposition 16. If A is an abelian category with enough injectives and C· is a bounded below com-plex (Cn = 0∀n < 0) then there is a bounded below complex of injectives I · and a quasi-isomorphismC· → I ·.

Proof. We will simply hit this problem with a hammer and take the total complex of a Cartan-Eilenberg resolution.

Define RnF(C·)= HnF(I ·). There is a hidden exercise here, which is to check that this is independentof the choice of I ·.

Proposition 17. If A· ϕ−→ B· is a quasi-isomorphism of bounded-below cohcain complexes and α : A· →I · such that I · ∈Ch(inj(A)) then ∃β : B· → I · and a homotopy equivalence h :β◦ϕ'α, i.e. the followingdiagram commutes up to homotopy.

43

A· I ·

B·ϕ

α

β

Proof.

<++>

Corollary 5. If I ·ϕ−→ J· is a quasi-isomorphism of bounded below cochain complexes, then it has an

inverse up to homotopy.

Proof. Consider the diagram bellow.

I · I ·

J·

id

ϕψ

So we get that ψϕ= 1I . Similarly, we can find ψ′ such that ϕψ′ = 1J . Then ψ'ψ◦ϕ◦ψ′ ' π′. So weget a homotopy inverse ψ.

Definition 29.D+(A )= Ch+(inj(A ))

∼Similarly, we can define

D−(A )= Ch−(proj(A ))∼ ,

for cochain complexes which vanish above zero and projective objects in A , assuming A has enoughprojectives. More generally, we have

Db(A )= Chb(inj(A ))∼ = Chb(proj(A ))

∼ ,

where the b denotes complexes with bounded cohomology.

LECTURE 20: EXISTENCE OF DERIVED CATEGORIES - 03/20/2020

Last time we alluded to the universal property of derived categories. To precisely define this propertywill take some work. The reason for this is that the universal property satisfied in this case is ’2-categorical’.

Definition 30. Let A be an abelian category. The derived category of A is the universal cate-gory D(A ) and functor Φ : Ch(A ) → D(A ) such that Φ(quasi-iso) ⊂ (iso). That is if (C ,Ψ) is anothercategory-functor pair such that Ψ(quasi-iso) ⊂ iso, then there exists a functor Γ and a natural iso-morphism γ :Γ◦Φ'Ψ. Diagramatically, we have:

Ch(A ) D(A )

C

Φ

Ψ ∃Γ

∃γ

44

The pair (Γ,γ) is unique up to unique isomorphism, that is, if (Γ′,γ′) is another such datum then thereis a unique natural isomorphism of functors h :Γ

∼=−→Γ′ such that the following diagram commutes

Ch(A )

C D(A )

ΦΨΓ

Γ′

∃!h

Proposition 18. If (D,Φ) satisfies this universal property then (D,Φ) is unique up to an equivalencethat is unique up to unique isomorphism. Explicitly, if (D′,Φ′) is another category-functor pair satisfy-ing the universal property of the derived category then there is an equivalence of categories Γ : D →D′

such that Γ ◦Φ = Φ′. Furthermore, if Γ′ is another such equivalence, then there is a unique naturalisomorphism η :Γ→Γ′ making the diagram below commute.

Ch(A )

D D′

Φ Φ′Γ

Γ′

∃!η

The proof of this is left to the reader. We might come back to it later, but for now we will focus onconstructing the derived category of a large class of abelian categories. We note that in general thederived category of any abelian category need not exist.

Definition 31. An abelian category A has a generator if there is an object X of A such that everyobject of A is a quotient of a direct sum of copies of X .

In general, most everyday categories are going to have a generator. It would take a fairly large andpathological category to not satisfy this.

Theorem 12. If A is an abelian category satisfying:

1. A has all direct sums (is complete)

2. filtered colimits are exact

3. A has a generator

Then D(A ) exists.

Remark 6. These hypotheses also guarantee that A has enough injectives (Grothendieck). The proofis very interesting and the reader is encouraged to read up on it.

Proof. Let D =D(A ). DefineOb(D)=Ob(K(A ))=Ob(Ch(A )),

andHomD(X ,Y ) lim−−→

X ′→XHomK(A )(X ′,Y ),

here the direct limit is over quasi-isomorphisms X ′ → X . There are a couple of things to check here.For one, how do we compose these morphisms? Another thing to check is that the collection of X ′

quasi-isomorphic to X is small, or that there is an equivalent colimit indexed over a small set. Wewill address this latter concern first using the following lemma.

45

Lemma 9. The diagram of quasi-isomorphisms X ′ ∼=−→ X is essentially small.

Proof. We want to find X ′′ ∼=−→ X ′ such that X ′′ has bounded cardinality for any such X ′. (Note: in thecategory of R-modules, this means precisely what it sounds like, in other categories one would needto make sense of this, but we will ignore this for now.) Look at all subcomplexes X ′

µ ⊂ X ′ and notethat

X ′ = lim−−→µ

X ′µ.

Then since filtered colimits in A are exact,

H∗(X ′)= lim−−→µ

H∗(X ′µ).

Now choose λ ∈ N such that every X ′ is a quotient of A⊕λ0 , where A0 is a generator for A . Missed

some stuff here Then X ′µ

∼=−→ X and |X ′µ| is bounded in terms of |H∗(X )|. So complexes of bounded

size are essentially small.

That composition is defined will follow from the following lemma.

Lemma 10. Given the solid arrow diagram below, where t is a quasi-isomorphism, there exists anobject X ′′ and the dashed morphisms where t′ is a quasi-isomorphism.

X ′′

X ′ Y ′

Y

t′f ′

ft

TODO: Complete proof

LECTURE 21: DERIVED CATEGORIES (CONTINUED) 03/30/2020

First day back from spring break!Recall our construction of the derived category of an abelian category. If A is an abelian category, webuilt D(A ) by defining

Ob(D(A ))=Ob(K(A )=Ob(Ch(A )).

Further,HomD(A )(X ·,Y·)= lim−−→

X ′quisoXHomK(A )(X ′

· ,Y·).

Recall that HomK(A )(X ·,Y·) is the chain homotopy classes of chain maps X · →Y·. We proved that thecolimit above makes sense and that composition is well-defined so that we do indeed get a category.We also saw that the derived category satisfies a universal property. Namely, if Ch(A ) Ψ−→ C and Ψtakes quasi-isomorphisms to isomorphisms. Then there exists a factorization Γ : D(A ) → C of thisfunctor throught the derived category and that this factorization Γ is unique up to unique naturalisomorphism.

Proposition 19. D(A ) satisfies this universal property.

46

Proof. Suppose Ch(A ) Ψ−→ C and Ψ takes quasi-isomorphisms to isomorphisms. Let Γ(X ) =Ψ(X ).That was easy, the real work is in defining the functor on morphisms. We need a map

HomD(A )(X ,Y )→HomC (X ,Y ).

This will act by(X ← X ′ →Y ) 7→Ψ( f )◦Ψ(s)−1,

where s is the quasi-isomorphism going to the left and f is the morphism on the right. We need tocheck a couple of things. For one, we need to check that this is well-defined on chain homotopy equiv-alence classes. Additionally, one would have to check that this formula is independent of the choiceof representatives (s, f ). Lastly, one needs to check that (Γ,γ) is unique up to unique isomorphism.

Corollary 6. In D(A ),HomD(A )(X ,Y )= lim−−→

X ′quisoXHomK(A )(X ,Y ).

(TODO: isn’t this â definition?Recall from one of the homework that K(A ) is not an abelian category. This raises the question ofwhat the structure of D(A ) looks like. Is the derived category an abelian category? It turns out thatthe functor Ch(A ) → D(A ) preserves finite direct sums. From this, it follows that D(A ) has finiteproducts, finite coproducts, and that these coincide. To see this, note that

HomD(A )(X ,Y ×Z)= lim−−→X ′quisoX

HomK(A )(X ′,Y ×Z)= lim−−→X ′quisoX

(HomK(A )(X ′,Y )×HomK(A )(X ′, Z)

).

This last step above is where we use the fact that the functor Ch(A ) → D(A ) preserves products.Then because sifted colimits commute with products,

lim−−→X ′quisoX

HomK(A )(X ′,Y )× lim−−→X ′quisoX

HomK(A )(X ′, Z)

=HomD(A )(X ,Y )×HomD(A )(X , Z).

The same argument shows that 0 is an initial object in D(A ). To see this, simply replace Y , Z with 0in the argument. A symmetric argument yields that coproducts are preserved, from which it followsthat these coincide in the derived category. similarly, this symmetric argument modified shows that0 is also a final object in D(A ). Thus, modulo a few details, we have the following result.

Corollary 7. HomD(A )(X ,Y ) has a commutative monoid structure with unit.

We actually have more than this. Recall how we constructed inverses in HomA (X ,Y ). This is dis-played visually below.

X

K X ⊕ X X

X

∼=

∼=∆

47

In A , as in Ch(A ), we have a split exact sequence

0→ X → X ⊕ X → X → 0.

Since, as we just proved, Ch(A ) → D(A ) preseves direct sums, this sequence is split exact in D(A )as well. What do we mean by this exactly? We then get −1 : X → X in D(A ), and

(1 1

)(−11

)= 0

implies that −1+1= 0. Using this, we get − f =−1◦ f .

Corollary 8. HomD(A )(X ,Y ) is an abelian group so that D(A) is an additive category.

But what about kernels and cokernels?

Example 16. Let’s compute coker(Z n−→Z) in D(A ) (or K(A )). Let’s pretend that this exists, so thatwe can write

Q· = coker(Z[0] n−→Z[0].

We then have

HomK(A )(Q·,P·)= {Z[0] α−→ P|α◦n = 0}= {α ∈ P0|d(α)= 0,∃β ∈ P1,d(β)= nα}.

Exercise 48. Compute H∗(Q·) using Pontryagin duality:

Hom(H∗(Q·),Q/Z)= H∗Hom(Q·,Q/Z).

LECTURE 22: SIFTED CATEGORIES - 04/01/2020

Today we will go back and fill in some details from the last lecture. Last time we mentioned siftedcolimits. Let us go through what these are exactly.

Definition 32. A cateory I is called cosifted if for all i, j ∈Ob(I), the category of all diagrams of theform

i

k

j

is connected. Such a diagram is called a cospan. A morphism of cospans is a commutative diagramlike the one below.

i

k′ k

j

48

I is called sifted if the same statement holds with arrows reversed. The diagrams in this case arecalled spans.

Theorem 13. If I is cosifted, then and X ,Y : I →Set. Then

lim−−→(X i ×Yi)→ lim−−→(X i)× lim−−→(Yi)

is a bijection.

Proof. Recall that lim−−→(X i) can be constructed as a quotient of∐

X i by the relation (i,α) ∼ ( j,β) ifthere is a sequence of morphisms in I taking i to j. The image under X of these morphism then givesa sequence of morphisms in Set taking α ∈ X i to β ∈ X j.We construct an inverse. For (k, (u∗(α),v∗(β))) ∈ Xk ×Yk, choose a cospan.

((i,α), ( j,β))→ (k, (u∗(α),v∗(β))

Now let’s go back to the derived category and apply this idea.

Theorem 14. In K(A ), the category of quasi-isomorphisms X ′ → X is sifted.

Proof. We have to show that in the following diagram with dark arrows in which all arrows areisomorphisms, the blue arrows and V exist.

V

W W ′

Y Z

X

fg

g′f ′

pq

h : qg ∼ pf , h′ : qg′ ∼ pf ′.

The idea is to define

V = {w ∈W ,w′ ∈W ′, s : f (w) ∼−→ f ′(w′), t : g(w) ∼−→ g′(w′),H : ps ' qt)}.

Concretely, we letVn =Wn ⊕W ′

n ⊕Yn+1 ⊕Zn+1 ⊕ Xn+2

with differential

d =

d0 −df − f ′ dg g′ 0 d±h ±h′ p −q d

.

We now need to prove that the square

49

W V

Y W ′f

f ′

commutes up to homotopy. So we construct an explicit chain homotopy S : Vn → Yn+1 by just takingit to be the projection of Vn onto the appropriate factor. One easily verifies that dS±Sd = f ′− f .

LECTURE 23: TRIANGULATED CATEGORIES - 04/03/2020

In the derived category, kernels and cokernels are replaced by fibers and cones. These are relatedby fiber = cone [−1]. Note that if X → Y is injective, we can recover X = ker(Y → coker). In generalwe can’t do this. In the cokernel, we identify the image of any element of X with 0, but we forgetwhich element of X gave the identification. (TODO:Fill in some details that were missed dueto technical difficulties).

Definition 33. A triangulated category is an additive category, D, with an automorphism, T :D →D (i.e. a functor to itself that gives an equivalence). The action of this functor is usually denotedby T(X )= X [1], and called the shift. There is also a collection of triples (u,v,w) usually denoted

X Y

Z

u

vw

and called exact triangles. Here w : Z → X [1]. Additionally, this data is subject to the followingaxioms:

1. (a) Every u : X →Y extends to an exact triangle.

(b) X idX−−→ X 0−→ X [1] is an exact triangle

(c) any triangle is isomorphic to an exact triangle.

2. If X u−→Y v−→ Z w−→ X [1] is exact, then so are

Y v−→ Z w−→ X [1] −u−−→Y [1]

andZ[−1] −w−−→ X u−→Y v−→ Z.

3. Given a solid arrow commutative diagram below, where both horizontal rows are exact trian-gles,the dashed arrow exists and preserves commutativity of the diagram. Note: this morphismis not neccessarily unique.

X Y Z X [1]

X ′ Y ′ Z′ X ′[1]

f g ∃ f (1)

4. The octahedral axiom: Given three exact triangles

Xf−→Y u−→ Z′ d−→ X [1],

50

Yg−→ Z v−→ X ′ d′

−→Y [1],

andX

gf−−→ Z w−→Y ′ d′′−→ X [1],

There exists a fourth exact triangle

Z′ φ−→Y ′ ψ−→ X ′ θ−→ Z′[1]

such that the following diagram commutes.

X Z X ′ Z′[1]

Y Y ′ Y [1]

Z′ X [1]

gf

f

v

w

θ

d′g

u d′′

ψ u

φ

d

f

<++>

<++>

Remark 7. In an abelian category, the dashed arrows below exist by the universal property of thecokernel. This suggests that the octahedral axiom is the snake lemma in a triangulated category.

X X Y

Y Z Z

Z′ Y ′ X ′

0 0 0

Theorem 15. If A is abelian, K(A ) and D(A ) are triangulated where X u−→Y → Z → X [1] exact if itis isomorphic to X u−→Y → cone(u)→ X [1].

Proof. Axiom 1(a) follows from the fact that any map has a cone. Axiom 1(b) follows from the factthat cone(idX ) ' 0 for any object X . Axiom 1(c) follows by our definition. Axiom 2 follows from thefact that for any morphism u : X →Y ,

cone(Y → cone(u))' X .

Axiom 3 follows from the fact that the cone construction is functorial.





51

LECTURE 24: BACK TO DERIVED FUNCTORS - 04/06/2020

Now that we have the derived category, today we will spend some time seeing how it fits into theframework of some of the things we’ve done so far.

Definition 34. A triangulated functor between triangulated categories is a functor F : D → D′

such that

1. F(exact triangles)⊂ exact triangles

2. F(X [1])= F(x)[1]

Exercise 53. If F is a triangulated functor D(A )→D(B) then {Hn(F(q(X )))}n∈Z is a δ-functor, whereq : A →D(A ) is the natural functor.

Our goal is to start with F : A → B and extend this to a triangulated functor RF : D(A ) → D(B).The notation here is meant to be suggestive and should remind you of derived functors!First note that F extends automatically to Ch(A ) by just applying F to each object and morphismwhich builds up a chain complex. Further, F(a chain homotopy) is also a chain homotopy (Note:here, as in most of these lectures, we are assuming F is additive). Thus, we automatically have anextension F : K(A ) → K(B) which we will also denote by F. Now the question is, does the dashedarrow below exist?

K(A ) K(B)

D(A ) D(B)

F

q q

∃?RF

The answer to this question is: of course not!

Example 17. Let A be the category of abelian groups and let F(X ) =Z/nZ⊗ X . Then F(Z/nZ[0]) =Z/nZ[0], but Z/nZ' [Z n−→Z] and

F([Z n−→Z])= [Z/nZ 0−→Z/nZ].

So this extension idea was a bust, but maybe we can try to approximate an extension.

Definition 35. A right derived functor of a triangulated funtor

F : K(A )→ K(B),

is a functorRF : D(A )→D(B)

and a natural transformationη : qF → RF ◦ q

such that the pair (RF,η) is universal. A left derived functor of a triangulated functor F as aboveis a functor LF : D(A )→D(B) such with a natural transformation η′ : RF ◦ q → LF.

Example 18. Going back to the example above. . . TODO: missed this.

Now let us see how to actually construct these derived functors in this general setting. The idea is tofind a triangulated subcategory K ′ of K(A )=: K such that

52

1. ∀x ∈ K ,∃ a quasi-isomorphism X → X ′ ∈ K ′

2. If X ′ ∈ K ′ and X ′ is quasi-isomorphic to 0, then F(X ′)' 0.

Remark 8. If X ′ f−→ X ′′ is a quasi-isomorphism, then complete it to an exact triangle

X ′ → X ′′ → cone( f )→ X ′[1].

Note cone( f ) ' 0. Then applying F to this sequence gives us that F(X ′) → F(X ′′) is also a quasi-isomorphism.

Definition 36. Let RF(X )= F(X ′) where X → X ′ is as in 1. above.

We need to show that this definition is independent of our choice of X ′ and quasi-isomorphism.TODO: this.

Theorem 16. Suppose there exists a triangulated K ′ ⊂ K such that 1. and 2. above hold. Then

RF : D ↔D′ F−→D(B)

is a right derived functor of F.

Remark 9. Here we are implicitly using the fact that there is actually an extension F : D′ → D(B).The idea is that we are moving down to this smaller category K ′ where the extension problem actuallyworks out well and then use this extension to obtain our ’approximation’.

Now lets pivot back to the topic of spectral sequences. Let C· be a chain complex in A . The process

C· q(C·) ∈D(A )

looses some information. To see this, first note that C· has two filtrations

· · · ⊂σ≥n+1C· ⊂σ≥nC· ⊂ ·· · ⊂ C·

andτ≤nC· = [· · ·→ Cn−1 → ZnC· → 0→··· ]⊂ τ≤n+1C·.

For historic reasons, we mention here that τ stands for ’truncation’ and σ stands for ’stupidtruncation’. The reason σ gets an insulting name is because it does not behave well with respect tohomology, i.e. it changes the homology in each degree.Suppose now that Γ : A → B is left exact. Then Γ(F·I ·) gives a filtration of Γ(I ·) = RF(C·) so Γ(F·I ·)has a spectral sequence

Hp(grqΓ(I ·)) =⇒ RpΓ(C·).

We will continue with this idea next time.

53

LECTURE 25: DERIVED CATEGORIES - 04/10/2020

This is the last lecture we will spend on the topic of derived categories. We will use this to say somelast minute things that were forgotten in the previous lectures. One thing to note is that not everytriangulated category is the derived category for some abelian category. A prime example of this isthe stable homotopy category. This implies that for a general triangulated category D there is not acanonical homology functor

Hn : D →A .

However, we do have the functors Hom(_, X ) for each object X of D.

Proposition 20. Suppose thatX u−→Y v−→ Z w−→ X [1]

is an exact triangle in a triangulated category D. Then for all objects W of D, if we apply the functorHom(W ,_), we an exact sequence

Hom(W , X )→Hom(W ,Y )→Hom(W , Z).

Proof. Suppose α ∈ ker(Hom(W ,Y ) → Hom(W , Z)). Then we have a diagram like the following inwhich the dashed arrow exists by virtue of axiom TR3 in Weibel.

0 W W 0

Z[−1] X Y Z

1

β

u

TODO: Finish this proof.

Claim:Hom(W , X [n])=Extn(W , X ). How do we see this? Well we know that the functor

Hom : A op ×A →Ab

is left exact in each variable. It extends to

Hom : K(A )op ×K(A )→Ab

and evenHom : D(A )op ×D(A )→Ab.

But we also have an enriched hom:

Hom· : Ch(A )op ×Ch(A )→Ch(Ab)

whereHomn(X ·,Y ·)= ∏

m∈ZHom(X m,Y n+m).

This functor plays nicely with chain homotopies and iduces an enriched triangulated hom

Hom· : K(A )op ×K(A )→ K(Ab).

We can then try to derive this functor. Assuming that A has enough projectives, we get the followingfunctor

R Hom(_,Y ·) : D−(A )op →D(A ),

54

and assuming that A has enough injectives, we also have

R Hom(X ·,_) : D+(A )→D(Ab).

We expect to find a diagram like the following which gives us a derived functor given by the dashedarrow. (TODO: draw diagram). We want to show that

R Hom(_,Y ·)(X ·)

factors as a functor of Y · ∈ K+(A ), throught D+(A ). (TODO: fill in details missed here. Thepoint is to show that Homs in the derived category correspond to ext groups.)We will say one last thing about derived categories in relation to spectral sequences. It appears thatthere is not a great way to obtain a spectral sequence from the derived category. Indeed, this isone of the advantages of the approach of ∞-categories. Let C· ∈ Ch+(A ). This object has canonicalfiltrations, which we have mentioned before as being given by τ or σ truncations. These correspondto maps

· · ·←σ≥nC· ←σ≥n+1C· ←···in D(A). (TODO: fill in triangles above.) Then if Γ : D(A ) →D(B) is a triangulated functor thenRΓ will preserve these triangles so that RΓ(C·) as a filtered object of D(B).

Theorem 17. Assume you have a sequence of functors AF−→ B

G−→ C along with adapted classesK ′(A )⊂ K(A ),K ′(B)⊂ K(B) and F(K ′(A ))⊂ K ′(B), then we get RG ◦RF = R(G ◦F).

LECTURE 26: GROUP COHOMOLOGY - 04/13/2020

LECTURE 27: GROUP COHOMOLOGY (CONTINUED) - 04/15/2020

Today we will continue computing the group cohomology in particular cases of interest. The firstresult of these is the following. As usual, we work within an abelian category A with enough projec-tives and injectives to define derived functors.

Proposition 21. If |G| acts invertibly on M, then Hn(G, M)= Hn(G, M)= 0 and H·(G, M)= H0(G, M)=N(m).

Proof. Let G −A be the category of G-objects of A , and G −A [m−1] the full subcategory wherem = |G| acts invertively. Our strategy will be to compare the group cohomology for these two casesand use the latter to compute the former. We start with the forgetfull functors

F : A [m−1]→A , and F : G−A [m−1]→G−A .

The first of these is just the special case that G = {1}. These admit a right adjoint

Φ(P)= lim←−−k

mkP

and a left adjointΨ(P)= lim−−→

km−kP.

It follows that F is exact. Now Φ, being a right adjoint, preserves injectives. Let Γ : G −A → A beΓ(M)= MG . This is now precisely the data we need to get a Grothendieck spectral sequence.

55

<++> TODO:Fix up this proof.

<++>Recall that we have the norm element N = ∑

g∈G g in ZG. It is not too hard to see that N2 =∑g,h∈G gh = mN, where m = |G|. It follows that e = m−1N is an idempotent so that there is a direct

sum decomposition for any module M = eM⊕ (1− e)M.

Lemma 11. If |G| acts invertibly on M, then eM = MG = MG .


<++>

<++>So we find that in G−A [m−1], every object splits functorially into

M = MG ⊕ (1− e)M = MG ⊕ (1− e)M.

It follows that (_)G and (_)G are exact functors. It follows that the group cohomology in this casevanishes above degree zero.Now lets move on to what the higher cohomology groups actually tell us. Let’s start with cohomologyin degree one. We know that H1(G, M)=Ext1

ZG(Z, M). So the elements of this cohomology group areextensions:

0→ M → Xp−→Z→ 0.

Choose x ∈ X such that p(x) = 1. Now let ϕ : G → M be ϕ(g) = g · x− x. This is not a homomorphism,so why do we care? Well, we have

ϕ(gh)= gh · x− x = g · (hx− x)+ (ax− x)= g ·ϕ(h)+ϕ(g).

We see that this is almost a homomorphism, it is in fact a homomorphism up to this action of g. Wecall such a map a crossed homomorphism. Let Der(G, M) be the set of crossed homomoprphismsG → M. It appears now that we have a map H1(G, M)→Der(G, M). This turns out to almost give usan isomorphism. To see this, first we ask what happens if we replace this x by x′? Write x′ = x+ y fory ∈ M. Let ϕ′(x)= g · x′− x′ and ψ(g)= g · y− y. Then a quick computation shows that

ϕ′(g)=ϕ(g)+ψ(g).

This shows the following.

Theorem 18. There is an isomorphism

H1(G, M) ∼−→Der(G, M)/M.

Proof. The inverse is given by X = M×Z with action given by

g · (x,n)= (nϕ(g)+ g · x,n).

What about homology? Suppose that I is an injective abelian group. Then by the universal coefficienttheorem,

Hom(H1(G,Z), I)= H1(G, I)=Der(G, I)/I.

But the G-action on I is trivial so (TODO: finish the argument here).

Lemma 12. If Hom(X ,_) and Hom(Y ,_ ) are naturally isomorphic functors on the subcategory ofinjective objects of A , and A has enough injectives, then X ∼=Y .

<++>

56

LECTURE 28: - 04/20/2020

First, we go back and fix the result from Proposition 21.

Proposition 22. Suppose G is a group, m ∈Z, m acts invertibly on a G-module A. Then

H∗(G, A)= H∗(G[m−1], A)

andH∗(G, A)= H∗(G[m−1], A).

TODO: what is going on with this proof? Now lets turn to the meaning of H∗(G, A). LetI = ker(Z[G] → Z) be the augmentation ideal which is generated by elements of the form g−1 forg ∈G. From the long exact sequence associated to

0→ I →ZG →Z→ 0

we get thatHn(G, A)=Extn

G-mod(Z, A)=Extn−1(I, A).

Further, we get

H1(G, A)= Hom(I, A)Hom(Z[G], A)

= Hom(I, A)A

.

Proposition 23.HomG(I, A)=Der(G, A)

Proof. The idea is that the map Gϕ−→ I, given by g 7→ g−1 is the universal crossed homomorphism. It

is in fact a crossed homomorphism, as one readily checks. Then for any homomorphism I → A, ψ◦ϕis a crossed homomorphism. For the inverge map, suppose α : G → A is a crossed homomorphism.Let ψ(g−1) = α(g). One can check that this is in fact a homomorphism of G-modules and gives aninverse.

Note that we obtain Theorem 18 as a corollary of this. What about Ext1(I, A)? Let’s define CrExt1(G, A)to be the set of extensions

0→ A → X →G → 0

such that the G-action on A is just the conjugation action. We call such extensions crossed extensionsand claim that H2(G, A)=CrExt1(G, A). Start with an extension

0→ A → Xp−→ I → 0

and the universal map G → I given by g → g−1. Then taking the pullback of this map along p givesY = {(x, g)|p(x) = g−1} with a canonical projection q : Y → G sending (x, g) 7→ g. We define a groupstructure on Y by

(g,u) · (h,v)= (gh, g ·v+u).

Then one can check that this is in fact a crossed extention.The idea is that extensions of G by A is the same as a crossed homomorphism G → BA, which is thesam as a homomorphism I → BA.

57

LECTURE 29: GALOIS COHOMOLOGY - 04/24/2020

Today we will take a look at the particular case of group cohomology in which our group G of interestis the Galois group of a field extension. The hope is that this gives us some information about thefield extensions in question.

Definition 37. Let L be a field, let G be a group acting on L (on the left). A G−L-vector space is anL-vector space with a G-module structure such that

g · (λx)= g · (λ)g · (x), ∀g ∈G,λ ∈ L, x ∈V .

Proposition 24. Let L be a Galois extension of K, G =Gal(L/K). Then H1(G,L∗) is isomorphic to theset of 1-dimensional G−L-vector spaces modulo isomorphism, where L∗ = L\{0} is the multiplicativegroup of L.

Proof. Note that this amounts to procuding an appropriate action on L by G. Recall that H1(G,L∗)=Der(G,L∗)/PDer(G,L∗). So given a crossed homomoprhism σ, we must produce a G-action on L.TODO: missed what this action was.

Theorem 19. Suppose K ⊂ L is a Galois Extension with Galois group G. Then there is an equivalenceof categories

K-Vect→G-L-Vect

sendingV 7→ L⊗K V ,

where the action is given by g · (a⊗ x)= (g ·a)⊗ x.

We will come back to the proof.

Corollary 9.H1(G,L∗)= 0.

Proof. Suppose L′ is a 1-dimensional G −L-vector space, then L′ ∼= L⊗K K since K is the unique1-dimensional K-vector space L′ ∼= L as G−L-vector spaces.

Definition 38. A K-algebra A is L-split if and only if

L⊗K A ∼= Ln

as an L-algebra.

As an example, let K ⊂ L be a Galois extension of the form L = K[x]/ f (x). Then

L⊗K L = L[x]/ f (x)= L[x]/∏

i(x−αi)=

∏i

L[x]/(x−α)=∏i

L

where the correspondence sends x ∈ L[x]/ f (x) to (αi)i.

Remark 10. The theorem above is the fundamental theorem of Galois Theory. The equivalence in thetheorem preserves the tensor product operation since it is given by tensor products and tensor prod-ucts are associative. But the tensor product is used to define algebras, so we also get an equivalenceof categories on the level of algebras:

K-algebras ∼−→G-L-algebras.

58

Similarly, we get a correspondence

(L-split K-algebras)→ (L-split G-L-algebras)→ (Gop −sets)

where the second map above sendsA →HomK (A,L).

Note then thatL 7→HomK (L,L).

We now prove Theorem 19.

Proof. The inverse equivalence will be W 7→WG . We need to show that there are two isomorphisms

V ∼−→ (L⊗K V )G

andL⊗K WG ∼−→W .

To prove the first one, it is sufficient to show that

L⊗V → L⊗ (L⊗V )G

is an isomorphism, where all tensor products are over K unless otherwise decorated.

<++>

LECTURE 30: ∞-CATEGORIES - 04/27/2020

Infinity categories are a bit of a slippery thing. People have different models for what they are.There are methods for talking about them independently of the model, but things get very abstractvery fast. Instead, we will follow Lurie and take an approach via quasicategories. Usually onedescribes a category by specifying objects and morphisms, then checking that composition workscorrectly. The problem comes when we start trying to build categories like the homotopy categoryof chain complexes in an abelian category. Here composition is not necessarily defined on the nose,but defined up to homotopy. A good example to keep in mind visually is the fundamental group of aspace X :

π1(X )= {S1 → X }/∼,

where the quotient is by homotopy equivalence. The group operation is by concatenation. To provetht this is indeed a group one needs to scale the domain.The prototype infinity category is the fundamental groupoid of X , denoted Π(X ). This is a categorywhose objects are the points of X and the morphisms Hom(x, y) for x, y ∈ X are the paths fromx to y. The approach of quasicategories says the following: instead of remembering morphismsand copmosition laws, remember all commutative triangles. More explicitely, given a sequence ofmorphisms below

B

A C

gf

59

Instead of saying there is a single f ∗ g : A → C making the diagram commute, we keep track of allpossible arrows A → C which do this. The way a quasicategory does this is by having, not only a setof objects C0 and a set of morphisms C1, but a set of commuting triangles C2, a set of commutingtetrahedra C3, and so on. If you’ve seen simplicial sets before, then you know where this is going.

Definition 39. ∆ will denote the category whose objects are finite ordered sets [n] = {0, . . . ,n} andmorphisms non-decreasing maps [n]→ [m]. A simplicial set C is a functor C :∆op →Set.

It is usefull to think of C as a recipe for how to glue simplicies together. In fact, this is made explicitby geometric realization which is a functor from the category of simplicial sets to topological spaces.Now it seems like were getting what we want out of this definition, however, we are not quite done.For each n, Let ∆n be the category depicted below

0→ 1→···→ n−1→ n.

If C is a category, let Cn = Hom(∆n,C ). Note then that C0 is the set of objects and C1 the set ofmorphisms in C . Continuing on, we see that C2 is the set of pairs of composable arrows

X1

X0 X2

gf

in C , and C3 is the set of triples of composable arrows

X1 X3

X0 X2

gf h

Remark 11. Note that each Cn is determined by its predecessors. As an example, we can think of C2as a sort of fiber-product:

C2 =C1 ×C0 C1,

where the fiber-product encodes the condition that the target of the first factor in C1 be equal to thesource of the second factor C1.

The key here is to notice that the diagrams above can be filled in for any category. Looking at thef − g diagram above, we can ’fill in the triangle’ by adding g ◦ f : X0 → X2. Similarly, in the f − g−hdiagram above, we can fill in more:

g ◦ f : X0 → X2, h◦ g : X2 → X3, h◦ (g ◦ f ) : X0 → X3

Adding these in amounts to ’filling in the tetrahedron’. This filling condition is precisely what ourdefinition is missing.

Definition 40. The category∆n represents the simplicial set Hom∆(_, [n]). We call this the standardn-simplex. The i-th horn of ∆n, denoted Λn

i is obtained by ommiting the index i in the definition of∆n.

Definition 41. A simplicial set C is called a quasicategory (or ∞-category) if

Hom(∆n,C )→Hom(Λni ,C )

is surjective for all n, 0< i < n.

60

Remark 12. Here we view ∆n both as the category defined above, and the functor that it represents,namely Hom∆(_, [n]).

We note that a slight variant of the definition of a quasicategory above is often of interest. If weadditionally require the same condition to hold for i = 0,n, the resulting notion is called a Kancomplex (∞- groupoid).

Example 19. Infinity category of spaces:

C0 =Ob(Top)

C1 = {X0f01−−→ X1}

C2 ={ X1

X0 X2

gf

h

∣∣∣commutes up to homotopy}

...

LECTURE 31: ∞-CATEGORIES (CONTINUED) - 04/29/2020

Let’s start with a summary of what we learned last time.

Definition 42. An ∞-category is a simplicial set C such that

Hom(∆n,C )→Hom(Λni ,C )

is surjective for all inner horns Λni ⊂∆n.

We called this a quasicategory last time, following Quillen. This definition encodes a composition lawthat is well-defined only up to contractible ambiguity.

Definition 43. An ∞-category C is called stable if

1. C has a zero object

2. C has fibers and cofibers

3. a sequence0→ X →Y → Z → 0

is left exact if and only if it is right exact.

Note the similarities between this and the definition of an abelian category. Note that stability is aproperty of C , not an additional structure. For a comparison, take triangulated categories. Thesecome with an additional structure: distinguished triangles, the shift operator, etc. So if we can dothe same things with ∞-categories that we can do with triangulated categories, then we have madeprogress because the former is more efficient. It remains to say what all the things in the definitionabove mean. In an ∞-category C , X ∈ C0 is initial if HomC (X ,Y ) is contractible for all Y and finalif HomC (Y , X ) is contractible for all Y . Note that here we replace the notion of a single point inordinary categories with the notion of contractibility, being homotopy equivalent to a point. Note by’contractible’ we also mean nonempty. Again, a zero object is an object that is both final and initial.

61

Definition 44. Given (X →Y ) ∈C1, we define the fiber of this morphism by

fiber(X →Y )= X ×Y 0.

That is, it is the universal object

fiber(X →Y ) X

0 Y

In terms of ∞-categories, this means that it is a final object of the slice category C /I, where I is thediagram:

0→Y ← X .

Cofibers are defined dually. The last condition in the definition of an infinity category is equivalentto saying that the square

X Y

0 Z

is a cofiber square if and only if it is a fiber square. Fun fact: Lurie calls this square a triangle.

Theorem 20. Let C be an abelian category. Then the nerve of C , N(C ) is a stable infinity category.

62

SOLUTIONS TO EXERCISES

Solution 1. Exercise 1Let M and N be R-modules. We claim that the standard construction of the direct sum as the setM ⊕ N = {(m,n)|m ∈ M,n ∈ N} with addition and R-action defined component-wise, is in fact theproduct and coproduct of M and N. To show this, we first note that there are canonical projectionsπM : M ⊕ N → M and πN : M ⊕ N → N given by πM(m,n) = m and πN (m,n) = n. Likewise, we havecanonical inclusions ιM : M → M⊕N given by ιM(m)= (m,0) and ιN : N → M⊕N given by ιN (n)= (0,n).

Let Z be an arbitrary R-module with R-module homomorphisms Zf−→ M and Z

g−→ N. We can then amap h : Z → M⊕N by h(z)= ( f (z), g(z)) so that the following diagram commutes

Z

M⊕N N

M

g

f

h

πN

πM

The map h is evidently the unique such map, for if h′ : Z → M⊕N were a map such that h′(z)= (m,n)where f (z) 6= m or g(z) 6= n then either πN ◦ h 6= g or πM ◦ h 6= f . Thus M ⊕ N is the product in thecategory of R-modules.

Now suppose Z is an R-module with maps f : M → Z and g : N → Z. Then we can define a maph : M⊕N → Z by h(m,n)= f (m)+g(n). Now for any m ∈ M, h◦ιM(m)= h(m,0)= f (m)+g(0)= f (m) andfor any n ∈ N,h ◦ ιN (n) = h(0,n) = f (0)+ g(n) = g(n). So the corresponding diagram for the coproductcommutes. Again, the map h is the unique such map with this property. To see this, suppose h′ :M⊕N → Z satisfies h′ ◦ ιM(m)= f (m) and h′ ◦ ιN (n)= g(n) for all m ∈ M and n ∈ N. Then

h′(m,n)= h′(m,0)+h′(0,n)= h′ ◦ ιM(m)+h′ ◦ ιN (n)= f (m)+ g(n).

Thus M⊕N is the coproduct in the category of R-modules.That finite products and coproducts agree follows by a simple induction. This shows that R-modsatisfies condition AB0.

For condition AB1, it is easy to see that the usual constructions ker( f : M → N)= {m ∈ M| f (m)= 0}and coker( f : M → N)= N/ f (M) i satisfy the required properties of the kernel and cokernel as definedin Lecture 1.

Similarly, one can see that im( f )= {n ∈ N|∃m ∈ Ms.t. f (m)= n} and coim( f )= M/ker( f ), satisfy therequired properties of image and coimage. That these coincide is a standard so-called ’isomorphismtheorem’.

Solution 2. Exercise 2Assume conditions AB1 and AB2. Using universality of the coimage and the fact that the composition

ker( f ) → Xf−→ Y is 0 we obtain a unique morphism h : coim( f ) → Y making the following diagram

commute.

ker( f ) X Y coker( f )

coim( f ) im( f )

0

f

∃!h 0

63

Now we want to use the universal property of im( f ), which amounts to showing that the commposi-tion coim( f ) h−→Y → coker( f ) is 0. We proceed in a manner precisely dual to the argument in Lecture2. Let g : A → B be any morphism and Z any object. Let σ : B → coker(g) be the universal map.Then we have a map Hom(coker(g), Z) → Hom(B, Z) given by (t : coker(g) → Z) 7→ t◦σ. Note that theimage of this map consists of maps t′ : B → Z such that t′ ◦ g = 0. Then by the universal property ofcokernels, to each such t′, there is a unique t : coker(g) → Z such that t′ = t ◦σ. This shows that themap on Hom-sets is in fact injective. In partiular, if t◦σ= 0 then t = 0. Viewing the coimage coim( f )above as a cokernel then since the composition

X → coim( f ) h−→Y → coker( f )

is zero, we must have that coim( f ) h−→ Y → coker( f ) is zero, as desired. This gives us our mapcoim( f )→ im( f ).

Solution 3. Exercise 3

First we prove that the composition of the maps we denote by matrices can be computed viamatrix multiplication at least in the case that they are diagonals. Let A i,Bi,Ci, for i = 1,2 be objectsin an abelian category A . Let f i : A i → Bi and g i : Bi → Ci be morphisms. Now consider the followingdiagram.

A1 B1 C1

A1 ⊕ A2 B1 ⊕B2 C1 ⊕C2

A2 B2 C2

f1

( f10 )

g1

(g10)

M f Mg

f2

( 0f2

)

g2

(0g2)

Here

M f =(f1 00 f2

), and Mg =

(g1 00 g2

).

Now the outer maps g i ◦ f i : A i → Ci induce a unique map(g1 ◦ f1 0

0 g2 ◦ f2

): A1 ⊕ A2 → C1 ⊕C2,

which makes the diagram commute. However, the map Mg ◦ M f is seen the make the diagramcommute, so indeed (

g1 00 g2

)(f1 00 f2

)=

(g1 ◦ f1 0

0 g2 ◦ f2

).

Now the proof of associativity follows from a diagram chase of the following diagram. Here we denotea two by two diagonal matrix with f and g along the diagonal, for example, by M f ,g. One can seethat the bottom-most and top-most paths X →Y correspond to ( f + g)+h and f +(g+h), respectively.The two quarter-circle shaped diagrams with blue arrows commute by what was just proven, and thecomplements of each of these within their trapezoidal-shaped diagrams commute by associativity ofthe direct sum. Thus, the entire diagram commutes and the result follows.

64

X ⊕Y ⊕Y Y ⊕Y

X X ⊕ X X ⊕ X ⊕ X Y ⊕Y ⊕Y Y

Y ⊕Y ⊕ X Y ⊕Y

M f ,∆Y

M f ,idY⊕Y∆Y

∆∗X

M∆∗X ,idX

idX 0

0 Mg,h

M f ,g 0

0 idX

f 0 0

0 g 0

0 0 h

MidY⊕Y ,h

M∆Y ,h

∆Y

A similar argument as the one above shows that matrix composition works as it should for off-diagonal matrices as well, that is,(

g1 00 g2

)(0 f1f2 0

)=

(0 g1 ◦ f1

g2 ◦ f2 0

).

Using this, one can see that the following digram commutes, which in turn implies that the additionof functions as defined is commutative.

X ⊕ X Y ⊕Y

X X ⊕ X Y ⊕Y Y

X ⊕ X Y ⊕Y

M f ,g

0 idX

idX 0

∆Y∆∗X

∆∗X

∆∗X

0 f

g 0

∆Y

0 idY

idY 0

Mg, f

∆Y

It remains to show that the zero map X 0−→ Y is the identity with respect to this operation. To seethis, consider the following diagram.

X X ⊕ X Y ⊕Y Y

X Y

∆∗X

idX

f 0

0 0

pX

1

∆Y

fιY1

idY

That the two triangles on the sides commute follows simply from the definition of the diagonal maps∆∗

X and ∆Y . That the center square commutes follows from computing

pY1 ◦ (ιY1 ◦ f ◦ pX

1 )◦ ιX1 = (pY1 ◦ ιY1 )◦ f ◦ (pX

1 ◦ ιX1 )= idY ◦ f ◦ idX = f .

65

Thus, the top path X → Y is the same as the bottom path X → Y , implying f +0 = f . By commuta-tivity, we also have 0+ f = f so the zero map is indeed the identity with respect to this operation onHom(X ,Y ).


That 1 =⇒ 2 follows simply from the fact that exactness, by definition, means ker(dn)= im(dn+1)for all n, so that

Hn(C)= ker(dn)/ im(dn+1)= 0.

Similarly, by definition, for all n, we have

Hn(C)= ker(dn)/ im(dn+1).

So if Hn(C) = 0 for all n, then C must be exact. Thus 2 =⇒ 1. Evidently Hn(0) = 0 for all n,where the arugment, 0, of Hn is the zero complex. Thus, if C is acyclic so that Hn(C) = 0 for all n,then the unique map Hn(0) → Hn(C) must be an isomorphism 0 → 0, so 2 =⇒ 3. Similarly, if theunique induced map Hn(0) → Hn(C) is an isomorphism then since the zero object is unique up toisomorphism, we must have Hn(C)= 0 for all n. Thus 3 =⇒ 2, and the proof is complete.


Let R[x1, . . . , xk] denote the free R-module on the generators {x1, . . . , xk}. Since d0 = 0 we haveker(d0)= C0 = R[v1, . . . ,vV ]. Now for any edge e, let s(e), t(e) ∈ {v1, . . . ,vV } denote the source and targetof the edge. Then the differential acts on e by d(e)= t(e)− s(e). Thus, the image of d is generated bythe differences of adjacent vertices. Since Γ is connected, for any two vertices vi1 and vi l there is apath connecting them, that is, there exists a sequence of vertices {vi1 ,vi2 , . . . ,vi l } such that for eachk = 1, . . . , l −1, there exists a vertex ek with either s(ek) = vik , t(ek) = vik+1 or s(ek) = vik+1 , t(ek) = vik .Let εk =+1 in the former case and εk =−1 in the latter. Then

d( l−1∑

k=1εkek

)=

l−1∑k=1

εkd(ek)= (vi2 −vi1)+ (vi3 −vi2)+·· ·+ (vi l −vi l−1)= vi l −vi1 .

It follows that [vi] = [v j] in H0(C) for any i, j = 1, . . . ,V . Moreover, since im(d) is generated bydifferences of edges, [vi] 6= 0 in H0(C) for any i. Thus

H0(C)= R[v1, . . . ,vV ]/(vi −v j)= R[e1],

as desired.Now since d2 = 0,H1(C)= ker(d). Evidently, if (e1, . . . , e l) is a closed path (i.e. s(e1)= t(e l)), then

d( l∑

k=1εl e l

)= 0,

where εl = ±1 depending on the orientaton of each edge and the orientation on the cycle. Further-more, if {e i}l

i=1 is a collection of disjoint edges, then

d( l∑

i=1r i e i

)6= 0,

for any r i ∈ R. Thus, every element of ker(d), hence H1(C), is a sum of closed paths. Now letT ⊂ {e1, . . . , eE} be a minimal spanning tree for Γ. We claim that |T| =V −1. An elementary proof can

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be given by induction as follows. Suppose V = 1, then ; is a spanning tree so we are done. Supposenow that the result holds for a graph with V −1 vertices. Then for any graph with V vertices, wemay take a subgraph of V −1 vertices, find a spanning tree for this graph, then adding a single edgeconnecting the isolated point gives the result. Now, to each of the E−(V−1) edges not in the spanningtree T of Γ, we assign a closed path as follows. Let e ∈ {e1, . . . , eE}\T. Starting at s(e) we trace a pathwhose first edge is e, followed by the unique path (t1(e), . . . , tl(e)) along the spanning tree back to s(e).We claim that the elements

e := e+l∑

k=1εk tk(e),

for each e ∈ {e1, . . . , eE}\ T, form a free basis for ker(d). It is evident that these are linearly indepen-dent, so it remains to show that any closed path is a sum of such elements. For this, let (e i1 , . . . , e i l )be a closed path. Then for at least one k, e ik ∈ {e1, . . . , eE} \ T. Let e′1, . . . , e′j be the edges in thecomplement of the spanning tree. Then

j∑m=1

εm e′m =l∑

m=1εme im ,

so we are done.


Let f : B → C be a chain map. We may view this as a double complex, D, by using the sign trick.This complex is displayed below, where the extensions upwards and donwards are by zero objectsand maps.

......

...

· · · Bp+1 Bp Bp−1 · · ·

· · · Cp+1 Cp Cp−1 · · ·

......

...

(−1)p+1 f (−1)p f (−1)p−1 f

We can then obtain a map of double chain complexes δ : D → B[0][+1], where the notationB[0][+1] means we shift the q-degree of B by +1 (here B is viewed as a double chain complex concen-trated in degree q = 0), by simply applying the identity in degree q = 1 and the zero map elsewhere.Similarly, we obtain a map ι : C → D given by the identity in degree q = 0 and zero maps elsewhere.The resulting sequence is evidently exact.


We take the liberty to assume the preceding theorem in Weibel, which states that a short exactsequence of chain complexes induces a long exact sequence in homology. It follows that the shortexact sequence

0→ A → B → C → 0

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induces a long exact sequence

· · ·→ Hn+1(C) ∂−→ Hn(A)→ Hn(B)→ Hn(C) ∂−→ Hn−1(A)→···By Exercise 4 above, if two of the three chain complexes are exact, we will have that each Hn for

every n will be zero for those two chain complexes. The exact sequence above then implies that theremaining terms must be zero.


Suppose we are given the following commutative diagram in an abelian category such that thecolumns are exact.

0 0 0

0 A′ B′ C′ 0

0 A B C 0

0 A′′ B′′ C′′ 0

0 0 0

α′

f ′

β′

g′

γ′

α

f

β

g

γ

f ′′ g′′

Suppose the bottom two rows are exact. Then, by commutativity, g◦ f ◦α′ = γ′◦g′◦ f ′. By exactnessof the middle row, g ◦ f = 0, implying 0 = γ′ ◦ g′ ◦ f ′. However, by exactness of the rightmost column,γ′ is monic, and so this implies g′ ◦ f ′ = 0. It follows that the top row is also a chain complex andso, by Exercise 11 above, it must be exact. Similarly, suppose the top two rows are exact. Then, bycommutativity, g′′ ◦ f ′′ ◦α= γ◦ g◦ f . By exactness of the middle row, g◦ f = 0, implying g′′ ◦ f ′′ ◦α= 0.However, by exactness of the leftmost column, α is an epi and so this implies g′′ ◦ f ′′ = 0. It followsthat the bottom row is also a chain complex and so, by Exercise 11 above, it must be exact.Now suppose the top and the bottom row are exact and that g◦ f = 0. Then each row is automaticallya chain complex and so the result, again, follows from the preceding exercise.


Let f· : C· → D· be a chain map. We then have the following two short exact sequences

0→ ker( f·)→ C → coim( f·)→ 0

0→ im( f·)→ D → coker( f·)→ 0.

We then have the following two long exact sequences

· · ·→ Hn(ker( f·))→ Hn(C)→ Hn(coim( f·))→ Hn−1(ker( f·))→···· · ·→ Hn+1(coker( f·))→ Hn(im( f·))→ Hn(D)→ Hn(coker( f·))→··· .

Assuming ker( f·) and coker( f·) are acyclic, these sequences give us isomorphisms Hn(C)∼=−→ Hn(coim( f·)),

Hn(im( f·))∼=−→ Hn(D). Then since we are in an abelian category we have a canonical isomorphism

coim( f·)→ im( f·), the composition

Hn(C)∼=−→ Hn(coim( f·))

∼=−→ Hn(coim( f·))∼=−→ Hn(D)

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is then an isomorphism. Further, by definition of this canonical isomorphism between coimages andimages, the diagram bellow commutes.

ker( f·) C· D· coker( f·)

coim( f·) im( f·)

f

∼=

Therefore, this isomorphism is indeed the map induced by f , and so f is a quasi-isomorphism.The converse is not true in general. To see this, consider the following chain map as a counterexample.

0 0 Z Z 0

0 Z Z 0 0

1

11

The kernel of this chain map is0→ 0→ 0→Z→ 0,

which is not acylic. However, one can easily check that this is indeed a quasi-isomorphism.

Solution 14. Exercise 14That this is a chain complex follows simply from the computation

d(d( f·))m = d ◦d( f·)m + (−1)n−2d( f·)m−1 ◦d

= d ◦ (d ◦ fm + (−1)n−1 fm−1 ◦d)+ (−1)n−2(d ◦ fm−1 + (−1)n−1 fm−2 ◦d)◦d

= (−1)n−1d ◦ fm−1 ◦d+ (−1)n−2d ◦ fm−1 ◦d = 0.

As mentioned in the notes, the tricky part is that we do not notationally distinguish between the dbeing defined on the complex Hom·(A·,B·) and the d of the chain complexes A· and B· themselves.

Now to compute the kernel of d0, compute

d0( f·)m = d ◦ fm − fm−1 ◦d : Am → Bm−1,

but since f ∈ Hom0(A·,B·) and Hom0(A·,B·) is exactly the set of chain maps A· → B·, we have thatthis must be zero. Thus, Z0 =Hom0(A·,B·).

Solution 15. Exercise 15 First we wish to show that acyclic bounded below chain complexes of freeR-modules are always split exact. Let C· be such a complex. Since we are given that C· is acyclic itsuffices to show that it is split. Since it is bounded below, we may assume that the last nonzero termis C0. Then the end of this sequence looks like

· · ·→ C2d2−→ C1

d1−→ C0 → 0.

Since C· is acyclic, this sequence is exact and so d1 must be surjective. Then, since C0 is free wecan choose a generating set {tα} and preimages {tα} under d1 so that we get a map s0 : C0 → C1 bymapping tα 7→ tα and extending linearly. Evidently, d1◦s0 = 1 =⇒ d1◦s0◦d1 = d1 so we have definedthe splitting map at the first stage. To define s1, note that we have an exact sequence

0→ ker(d1)→ C1 → C0 → 0,

69

which splits since all modules in sight are free. Thus C1 ∼= ker(d1)⊕C0. Now since C· is acylic, wehave ker(d1) ∼= im(d2). Furthermore, since this is a direct summand of a projective module we have

the following diagram which shows the existence of a morphism im(d2)s′1−→ C2.

C2 im(d2) 0

im(d2)

d2

1∃s′1

We then have a map s1 =(s′1 0

): im(d2)⊕C0 = C1 → C2. That this is indeed a splitting map follows

from the commutativity of the triangle above, that is,

d2 ◦ s′1 = 1 =⇒ d2 ◦ s′1 ◦d2 = 1.

Continuing in this way, at the n-th step we have the exact sequence

0→ ker(dn)→ Cn → im(dn)→ 0,

which gives the splitting Cn = ker(dn)⊕ im(dn) ∼= im(dn+1)⊕ im(dn). From here we can proceed asabove to obtain a splitting map sn : Cn → Cn+1.

Now let C· is an acyclic chain complex of finitely generated abelian groups, not necessarilybounded below. In this case, we still have the following exact sequence

0→ ker(dn)→ Cn → im(dn)→ 0.

Now since Z is a PID, and im(dn) is a submodule of the free Z-module Cn−1, we have that im(dn)itself is free. So we get that the sequence above splits and the process is the same as above to obtainsplitting maps.

Solution 16. Exercise 16Let C· be a split complex. Consider the associated short exact sequence at the n-th stage

0→ ker(dn)→ Cn → coker(dn)→ 0.

The splitting condition d = dsd implies that this sequence is split so that

Cn = ker(dn)⊕coker(dn)= Zn ⊕B′n,

as desired. Similarly, since d2 = 0, we have that

0→ im(dn+1) σ−→ ker(dn)→ coker(σ)→ 0

is an exact sequence. The splitting condition d = dsd again gives us that

ker(dn)= Zn = im(dn+1)⊕coker(σ)= Bn ⊕H′n,

as desired. Conversely, suppose we have such splittings. Then Cn = Bn ⊕H′n ⊕B′

n. However, B′n∼=

Bn−1. Then Cn+1 = Bn+1 ⊕H′n+1 ⊕Bn, so that we get a map

sn =0 0 0

0 0 01 0 0

: Cn = Bn ⊕H′n ⊕B′

n → Cn+1 = Bn+1 ⊕H′n+1 ⊕Bn.

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In terms of this decomposition, we also have

dn =0 0 1

0 0 00 0 0

so that dn+1 = dn+1sndn+1, implying sn is indeed a splitting map.

Solution 17. Exercise 17Let C· be a split exact chain complex. Then, by the previous exercise, we have that Cn = Bn ⊕Bn−1.The splitting map is then given by

sn =(0 01 0

): Cn → Cn+1

and the differential by

dn =(0 10 0

): Cn → Cn−1.

Then

dn+1 ◦ sn + sn−1 ◦dn =(0 10 0

)(0 01 0

)+

(0 01 0

)(0 10 0

)=

(1 00 1

)= 1Cn .

So the identity map is nullhomotopic.

Solution 18. Exercise 18Let C· be a split complex and H· the complex consisting of Hn(C·) at the n-th stage with zero maps asdifferentials. Now, by Exercise 1.4.2, since C· is split, we have a decomposition Cn = Zn ⊕B′

n whereZn = ker(dn) and B′

n = coker(dn) for all n. Furthermore, we have Zn = Bn ⊕Hn. Thus, we have themaps

fn = (0 1 0

): Cn = Bn ⊕Hn ⊕B′

n → Hn

and

gn =0

10

: Hn → Cn.

Evidently fn ◦ gn = 1 : Hn → Cn, but

gn ◦ fn =0 0 0

0 1 00 0 0

.

TODO:complete solution.

Solution 19. Exercise 19First we show that chain homotopy equivalence defines an equivalence relation on Hom(C·,D·), theset of chain maps. Taking sn = 0 for all n we see that f − f = 0 = d ◦0+0 ◦ d. So this relation isreflexive. Suppose, f − g = ds+sd. Then g− f = d(−s)+(−s)d, so this relation is symmetric. Suppose,additionally, that g−h = ds′+ s′d, then

f −h = ( f − g)+ (g−h)= d(s+ s′)+ (s+ s′)d.

This shows that the relation is transitive so that this is indeed an equivalence relation. Let HomK(C·,D·)denote the set of equivalence classes. This inherits an abelian group structure from the abelian

71

group structure on Hom(C·,D·). To see this, suppose f − g = ds+ sd for some collection of morphismssn : Cn → Dn+1. Then ( f +h)− (g+h) = f − g = ds+ sd so that f +h is in the same equivalence classas g+h.For the next part, suppose f − g = ds+ sd, where f , g : C· → D·. Let u : B → C and v : D → E be chainmaps. Then

vf u−vgu = v( f − g)u = v(ds+ sd)u = vdsu+vsdu = d(vsu)+ (vsu)d.

Thus, vf u and vgu are chain homotopic. It follows that we can compose equivalence classes of chainmaps and so there is a category K whose objects are chain complexes and morphisms are chain-homotopy equivalence classes of chain maps.Now suppose f i, g i : C· → D· for i = 1,2 are chain maps. Let f i − g i = dsi + sid. Then

( f1 + f2)− (g1 + g2)= ( f1 − g1)+ ( f2 − g2)= ds1 + s1d+ds2 + s2d = d(s1 + s2)+ (s1 + s2)d.

So the sums f1 + f2 and g1 + g2 are chain homotopic. It follows that K is an additive category andthat the quotient functor that is the identity on objects and maps a chain map to its chain-homotopyequivalence class, is in fact an additive functor.Now we turn to the final question: is K an abelian category? TODO: complete solution.

Solution 20. Exercise 20Suppose first that f·, g· : C· → D· are chain homotopic. Then we have maps sn : Cn → Dn+1 such that

f − g = ds+ sd.

Now consider the map (f s g

): Cn ⊕Cn−1 ⊕Cn → Dn.

That this is a chain map follows from the following computation

(f s g

)dC 1C 00 −dC 00 −1C dC

= (f ◦dC f − s◦dC − g g ◦dC

)= (dD ◦ f dD ◦ s dC ◦ g

)= dD◦( f s g).

Conversely, suppose (f s g

): Cn ⊕Cn−1 ⊕Cn → Dn

is a chain map. Then (f s g

)dC 1C 00 −dC 00 −1C dC

= dD(f s g

),

implyingf − s◦dC − g = dD ◦ s =⇒ f − g = dD ◦ s+ s◦dC.

So the maps(0s0

): Cn−1 → Dn provide a chain homotopy between f and g.

Solution 21. Exercise 21Let f· : B· → C· be a chain map. Consider the mapping cylinder cyl( f )· of f . Let α· : C· → cyl( f )· bedefined in each degree by αn(c) = (0,0, c). It was shown in Lemma 1.5.6 in Weibel that α is a quasi-isomorphism. We will see in this exercise that this is in fact a chain homotopy equivalence. Let usstart by defining its homotopy-inverse, β· : cyl( f )· → C· which is given in degree n by

βn(b,b′, c)= f (b)+ c.

72

This is indeed a chain map since

d ◦β(b,b′, c)= d( f (b)+ c)= d ◦ f (b)+d(c)= f ◦d(b)+d(c)

= f ◦d(b)+ f (b′)+d(c)− f (b′)= f (d(b)+b′)+d(c)− f (b′)=β(d(b)+b′,−d(b′),d(c)− f (b′))=β◦d(b,b′, c).

Evidently,β◦α(c)=β(0,0, c)= f (0)+ c = c,

so β◦α= 1C. Nowα◦β(b,b′, c)=α( f (b)c)= (0,0, f (b)+ c),

so we dont get an exact invese in this direction. Consider the morphisms sn : cyl( f )n → cyl( f )n+1given by s(b,b′, c)= (0,b,0), and compute

(ds+sd)(b,b′, c)= d(0,b,0)+s(d(b)+b′,−d(b′),d(c)− f (b′))= (b,−d(b),− f (b))+(0,d(b)+b′,0)= (b,b′,− f (b))

= (1−α◦β)(b,b′, c).

So α◦β and 1 : cyl( f )· → cyl( f )· are chain homotopic, implying α is a chain-homotopy equivalence.

Solution 22. Exercise 22Let f· : B· → C· be a chain map. Recall that the cone of f· is itself a chain complex with cone( f·)n =Bn−1 ⊕Cn and differential

d f =(−dB 0− f dC

).

Now let v· : C· → cone( f·)· be the inclusion sending c 7→ (0, c). Then the cone of this inclusion is itselfa chain complex with

cone(v·)= Cn−1 ⊕cone( f·)n = Cn−1 ⊕Bn−1 ⊕Cn,

and differential given by

d =(−dC 0

v d f

)=

−dC 0 00 −dB 0

−1C − f dC

.

Our goal is to exhibit a chain homotopy equivalence between this chain complex and the shiftedcomplex B[−1]. We define maps

hn : cone(v·)n → B[−1]n, by (c,b, c′) 7→ (−1)nb,

andgn : B[−1]n → cone( f·)n, by b 7→ ((−1)n+1 f (b), (−1)nb,0).

Note that hn ◦ gn = 1B so we are halfway to showing that these maps indeed form a chain homotopyequivalence. Note that

(1cone(v·) − gn ◦hn)(c,b, c′)= (c,b, c′)− ((−1)2n+1 f (b),b,0)= (c+ f (b),0, c′).

Now consider the map

sn : Cn−1 ⊕Bn−1 ⊕Cn → Cn ⊕Bn ⊕Cn+1, given by (c,b, c′) 7→ (−c′,0,0).

73

This may be written as a 3×3 matrix with the only nonzero entry being −1Cn on the top right entry.Then

ds+ sd =−dC 0 0

0 −dB 0−1C − f dC

0 0 −10 0 00 0 0

+0 0 −1

0 0 00 0 0

−dC 0 00 −dB 0

−1C − f dC

=0 0 dC

0 0 00 0 1C

+1C f −dC

0 0 00 0 0

=1C f 0

0 0 00 0 1C

.

This implies(ds+ sd)(c,b, c′)= (c+ f (b),0, c′)= (1cone( f·) − gn ◦hn)(c,b, c′),

so gn ◦hn is chain homotopy equivalent to the identity and completing the proof.

Solution 23. Exercise 23We invoke Proposition 6 in these notes which states that an abelian group is projective if and onlyif it is the direct summand of a free abelian group. Then since free subgroups of free abelian groupsare free, it follows that an abelian group is projective if and only if it is free.For this next part we show the result holds in general for R-modules, where R is a PID. First supposeI is an injective R-module. For any nonzero r ∈ R, the multiplication by r map R r−→ R is injectivesince PID’s are integral domains, and hence have no zero divisors. Now fix x ∈ I and let f : R → I begiven by f (s) = sx. Then we have the following diagram where injectivity of I gives us the dashedmap.

0 R R

I

r

f∃ f

Commutativity of the diagram then implies that x = f (1) = f ◦ r(1) = f (r) = r f (1). It follows any x ∈ Iis of the form ry for some r ∈ R and y ∈ I, so I must be divisible.Conversely, suppose I is divisible. By Baer’s Criterion in Weibel page 39, it suffices to check that wecan extend maps from ideals J ⊂ R. Assuming that R is a PID, we have J = (r) for some r ∈ R. Ifr = 0 then we can simply extend by the zero map and be done. Assume, therefore, that r 6= 0. Supposethat f : (r) → I is an R-module homomorphism. Since I is divisible, we have that there exists ay ∈ I such that f (r) = ry. We can then define f : R → I by f (1) = y. This map is an extension sincef (r)= f (r)= r f (1)= ry. Thus, I must be injective.

Solution 24. Exercise 24Let

I(A)= ∏f ∈HomZ(A,Q/Z)

Q/Z.

We want to show that the canonical evaluation map

eA : A → I(A), given by a 7→ ( f (a)) f ∈HomZ(A,Q/Z)

74

is injective. To do this, it suffices to find, for any a ∈ A, a Z-module homomorphism f : A →Q/Z suchthat f (a) 6= 0. First suppose, a ·n = 0 for some n ∈Z. Then we have a map

f : aZ→Q/Z, given by af7−→ [

1n

].

Then since Q/Z is injective, we have the existence of an extension f as in the following diagram.

0 aZ A

Q/Z

f∃ f

Then f : A →Q/Z is a map such that f (a) = f (a) = [ 1n ] 6= 0. Now if a ·n 6= 0 for all n ∈ Z then we can

simply take f : aZ→Q/Z to by the map sending a to any nonzero element of Q/Z and the result stillholds.

Solution 25. Exercise 25Suppose we have the exact sequence below with P projective

0→ M → P → A → 0.

Then, since the derived functors are δ-functors, we obtain from this the long exact sequence below

· · ·→ LnF(P)→ LnF(A)→ Ln−1F(M)→ Ln−1F(P)→···

Since P is projective, we have that LnF(P) = 0 for all n > 0. Thus, LnF(A) ∼= Ln−1F(M) for all n ≥ 2.For n = 1, we have that

L1F(A)→ L0F(M)= F(M)→ L0F(P)= F(P)

is exact so that L1F(A) = ker(F(M) → F(P)). More generally, suppose we have the exact sequencebelow with Pi projective

0→ Mm → Pmfm−−→ Pm−1

fm−1−−−→ ·· · f1−→ P0 → A → 0.

We can break up this into short exact sequences as displayed in the diagram below.

0

0 Mm Pm im( fm) 0

Pm−1

im( fm−1)

0

75

The horizontal row is exact and hence gives rise to a long exact sequence

· · ·→ LnF(Pm)→ LnF(im( fm))→ Ln−1F(Mm)→ Ln−1F(Pm)→···Since Pm is projective, LnF(Pm) = 0 for all n > 0, this implies LnF(im( fm)) ∼= Ln−1F(Mm) for n ≥ 2.The vertical column is also exact and so we also have the following long exact sequence

· · ·→ LnF(Pm−1)→ LnF(im( fm−1))→ Ln−1F(im( fm))→ Ln−1F(Pm−1)→···Again, since Pm−1 is projective, LnF(Pm−1) = 0 for n > 0 so that LnF(im( fm−1)) ∼= Ln−1F(im( fm)).Putting these two together gives us that LnF(im( fm−1)) ∼= Ln−2F(Mm). Continuing in this way, weobtain

LnF(im( f0))= LnF(A)∼= Ln−1−mF(Mm).

We note that this equation only holds for n ≥ 2+m. For if n < 2+m then the first isomorphism thatwe need, namely LnF(im( fm)) ∼= Ln−1F(Mm), no longer holds. At the tail end of the first short exactsequence we have

0→ L1F(im( fm))→ F(Mm)→ F(Pm)→ F(im( fm)),

implying that L1F(im( fm)) = ker(F(Mm) → F(Pm)). However, the above implies that L1F(im( fm)) ∼=Lm+1F(im( f0))= Lm+1F(A).

Solution 26. Exercise 26For this exercise, we will show that the following are equivalent:

1. B is an injective R-module.

2. HomR(_,B) is an exact functor.

3. Exti(A,B) vanishes for all i 6= 0 and all A.

4. Ext1(A,B) vanishes for all A.

(1 =⇒ 2)Suppose

0→ X →Y → Z → 0

is an exact sequence of R-modules. We know that HomR(_,B) is left-exact, and so it suffices to checkthat the map

HomR(Y ,B)→HomR(X ,B)

induced by precomposition with X → Y is surjective. This follows directly from B being injective ascan be seen in the following diagram

B

0 X Y∃

It is evident that the reverse steps give the reverse implication. Explicitly, if HomR(_,B) is an exactfunctor then the induced map HomR(Y ,B) → HomR(X ,B) is surjective, implying that for any solid-arrow diagram as above, the dashed map exists. so we also have that (2 =⇒ 1).(1 =⇒ 3)Assuming B is injective, the following is an injective resolution of B

B → 0→ 0→··· .

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We denote this chain complex by B. Now for any R-module A,

ExtiR(A,B)= R i HomR(A,_)(B)= H i(HomR(A, B)).

The chain complex HomR(A, B) is zero above and below degree zero and so its cohomology H i(HomR(A, B))must vanish for all i 6= 0.(3 =⇒ 4)This is a direct implication.(4 =⇒ 1)Let 0→ A′ → A be an exact sequence of R-modules. Then we have the short exact sequence

0→ A′ → A → A/A′ → 0

Since ExtR is a derived functor, it is a δ-functor, and so there is an associated long exact sequenceshown below.

Ext0R(A/A′,B)→Ext0

R(A,B)→Ext0R(A′,B)→Ext1

R(A/A′,B)→···By assumption, Ext1

R(A/A′,B) = 0 so that Ext0R(A,B) → Ext0

R(A′,B) is surjective. However, sinceExt0

R = HomR , we have that for any solid arrow diagram below, there exists a dashed map. Thiscompletes the proof.

B

0 A′ A∃

Note that here we have used the fact that ExtiR(A,B) is also a δ-functor on A.

Solution 27. Exercise 27For this exercise, we will show that the following are equivalent:

1. A is a projective R-module.

2. HomR(A,_) is an exact functor.

3. Exti(A,B) vanishes for all i 6= 0 and all B.

4. Ext1(A,B) vanishes for all B.

(1 =⇒ 2)Assume A is projective. Suppose

0→ X →Y → Z → 0

is an exact sequence of R-modules. We know that HomR(A,_) is left-exact, so it suffices to show that

HomR(A,Y )→HomR(A, Z)

is surjective. That is, given f : A → Z, we want to exhibit a map f : A →Y such that the composition

Af−→Y → Z is f . It is evident that this is exactly the condition that A be projective as one can see in

the diagram below.

A

Y Z 0

f∃ f

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As before, the logic walked in reverse shows that the implication (2 =⇒ 1) also holds. Explicitely, iffor any given solid-arrow diagram as above there exists the dashed arrow, then the map HomR(A,Y )→HomR(A, Z) is surjective so HomR(A,_) is exact.(2 =⇒ 3)That HomR(A,_) is an exact functor implies that for any R-module B and injective resolution,

B → I0 → I1 →··· ,

the chain complexHomR(A,B)→HomR(A, I0)→HomR(A, I1)→···

is exact. This is because injective resolutions are themselves exact sequences and the functor HomR(A,_)preserves exactness. It follows that the cohomology of this chain complex is zero for any nonzero de-gree. Since Exti

R(A,B) is the i-th cohomology of this complex, we have that ExtiR(A,B) = 0 for all

i 6= 0.(3 =⇒ 4)This is a direct implication.(4 =⇒ 1).Suppose B → B′ → 0 is an exact sequence of R-modules. Then we have the following short exactsequence, where K = ker(B → B′),

0→ K → B → B′ → 0.

Now according to Example 2.5.3 in Weibel, ExtiR(A,B) can also be viewed as a right-derived con-

travariant functor in B. Thus, we have the following long exact sequence

Ext0R(A,K)→Ext0

R(A,B)→Ext0R(A,B′)→Ext1

R(A,K)→···

By assumption, Ext1R(A,K) = 0 so the map Ext0

R(A,B) → Ext0R(A,B′) is surjective. Since Ext0

R =HomR , we have that given a solid arrow diagram below, the dashed map exists. This completes theproof.

B B′ 0

A∃

Solution 28. Exercise 28Let rR = {r ∈R|rs = 0}. Suppose rR 6= 0 and consider the following exact sequence

0→r R → R r−→ R → R/rR → 0.

Note that since R is a projective R-module, this is dimension shift by 1. Then by a direct applicationof Problem 25 above we have that

TorRn (R/rR,B)∼=TorR

n−2(rR,B)

for all n ≥ 3. Now consider breaking up this sequence into two short exact sequences

0→r R → R → rR → 0

and0→ rR → R → R/rR → 0.

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We can now look at the TorR sequence of these sequences. Keeping in mind that R is projective, wehave

· · ·→TorR2 (R,B)= 0→TorR

2 (rR,B)→TorR1 (rR,B)→TorR

1 (R,B)= 0→TorR1 (rR,B)→r B → B → rB

and

· · ·→TorR2 (R,B)= 0→TorR

2 (R/rR,B)→TorR1 (rR,B)→TorR

1 (R,B)= 0→TorR1 (R/rR,B)→ rB → B → B/rB.

Solution 29. Exercise 29Let R be a commutative domain with field of fractions F. Note that the quotient module F/R isgenerated by the elements 1

s +R ∈ F/R for s ∈ R. To see this, simply note that for rs +R ∈ F/R, we

have r(1s +R)= r 1

s +R = rs +R. Thus,

F/R = lim−−→s∈R

⟨1s+R⟩.

Since Tor commutes with colimits, we have that

TorR1 (F/R,B)=TorR

1 (lim−−→s∈R

⟨1s+R⟩,B)= lim−−→

s∈RTorR

1 (⟨1s+R⟩,B).

Now note that ⟨1s +R⟩ ∼= R/sR under the isomorphism 1

s +R 7→ 1. Thus, by example 3.1.7 in Weibelwhich states that if r is not a zero divisor (a condition which is satisfied automatically since R is adomain) then TorR

1 (R/rR,B)∼=r B = {b ∈ B|rb = 0}, we have that

TorR1 (F/R,B)= lim−−→

s∈R.sB = {b ∈ B|∃s ∈ R, such that sb = 0}.


Solution 31. Exercise 31Suppose the following is an exact sequence of R-modules where both B and C are flat

0→ A → B → C → 0.

Then for every left R-module Z, we have a long exact sequence

· · ·→TorR2 (Z,B)= 0→TorR

2 (Z,C)= 0→TorR1 (Z, A)→TorR

1 (Z,B)= 0→··· .

This implies TorR1 (Z, A)= 0 for all left R-modules Z, implying A must also be flat.

Solution 32. Exercise 32Let k be a field, R = k[x, y], and I = (x, y)R. The point of this exercise is to show that I is a torsion-freeideal, but is not flat. Note that k = R/I so we have the following short exact sequence

0→ I → R → k → 0.

Applying Tor to this sequence gives us the following

· · ·→TorR2 (R,k)= 0→TorR

2 (k,k)→TorR1 (I,k)→TorR

1 (R,k)= 0→··· .

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This implies TorR2 (k,k)∼=TorR

1 (I,k). To show that I is not flat, it thus suffices to show that TorR2 (k,k)

is nonzero. To this end, we take Weibel’s hint and consider the projective resolution of k below

0→ R

−yx

−−−−→ R2

(x y

)−−−−−→ R → k → 0.

Tensoring with k and zooming in on degree 2, we find that TorR2 (k,k) is the kernel of the map

k

−yx

−−−−→ k2.

However since the action of R on k is given by multiplication by the constant term, both y and x acton k by the zero map. Thus, the map above is in fact zero implying

TorR1 (I,k)∼=TorR

2 (k,k)= k.

Solution 33. Exercise 33Let B∗ = HomZ(B,Q/Z) denote the Pontryagin dual of an abelian group B. Since Q/Z is injective,HomZ(_,Q/Z) is exact and so if

A → B → C

is an exact sequence of abelian groups then so is

C∗ → B∗ → A∗.

For the converse, we fist note that by a direct consequence of Exercise 24 above we have that B = 0 ifand only if B∗ = 0. Now suppose f : A → B is an abelian group homomorphism such that f ∗ : B∗ → A∗

is the zero map. Consider the following commutative diagram with exact row and column.

0

A∗ im( f )∗ 0

B∗

f ∗

f ∗ i∗

Then f ∗ ◦ i∗ = f ∗ = 0, but i∗ is surjective, implying f ∗ = 0. However, f ∗ is injective so we must haveim( f )∗ = 0 implying im( f )= 0 so that f = 0. Now suppose

C∗ g∗−→ B∗ f ∗

−→ A∗

is exact. Since (g ◦ f )∗ = f ∗ ◦ g∗ = 0, we have that g ◦ f = 0 so that im( f ) ⊂ ker(g). We now showthat the inclusion i : im( f ) → ker(g) is surjective. Let j : ker(g) ,→ B be the canonical inclusion,p : ker(g)→ ker(g)/ im( f ) the projection, and f : A → im( f ) the canonical map. Consider the followingcommutative diagram with exact rows and columns.

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0

0 im( f )∗ A∗

0 ker(g)∗ B∗ C∗

(ker(g)/ im( f ))∗

0

( f ′)∗

i∗

j∗

f ∗

g∗

p∗

We will show that i∗ is injective. To this end, suppose i∗(k)= 0. Since j∗ is surjective, we have b ∈ B∗

such that k = j∗(b). Since the diagram commutes, ( f ′)∗ ◦ i∗ ◦ j∗(b) = f ∗(b) = 0. by exactness of thedual sequence, there is c ∈ C such that g∗(c) = b. Then k = j∗(g∗(c)) = 0 by exactness of the bottomrow above. Thus, i∗ is injective, implying (ker(g)/ im( f ))∗ = 0, which in turn implies ker(g)/ im( f )= 0.

Solution 34. Exercise 34Since Zp∞ =Z[ 1

p ]/Z, there is an exact sequence

0→Z→Z[ 1

p]→Zp∞ → 0.

This gives rise to a long exact sequence in Ext

0→HomZ(Zp∞ ,Z)→HomZ(Z[ 1

p],Z)→HomZ(Z,Z)→Ext1

Z(Zp∞ ,Z)→Ext1Z(Z

[ 1p

],Z)→Ext1

Z(Z,Z)= 0.

We claim that HomZ(Z[ 1p ],Z)= 0. To see this suppose f :Z[ 1

p ]→Z is an abelian group homomorphismand f (1)= z. Then for any n ∈N,

pn f (1pn )= f (

pn

pn )= f (1)= z.

This is only possible if z = 0, proving the claim. Thus,

Ext1Z(Z

[ 1p

],Z)∼=Ext1

Z(Zp∞ ,Z)/HomZ(Z,Z)∼= Zp/Z,

where the last isomorphism follows from Example 3.3.3 in Weibel and the fact that HomZ(Z,Z)∼=Z.

Solution 35. Exercise 35Let p and m be integers such that p|m. We first note that the following sequence is an injectiveresolution for Z/p in the category of Z/m−modules.

0→Z/p ,→Z/mp−→Z/m

mp−→Z/m

p−→Z/mmp−→ ·· · .

Now let A be an arbitrary Z/p-module. We are interested in computing ExtnZ/m(A,Z/m) for all n in

terms of A∗ =Hom(A,Z/m). Applying the Hom-functor to this resolution gives us

0→HomZ/m(A,Z/m)p∗−−→HomZ/m(A,Z/m)

mp∗

−−→HomZ/m(A,Z/m)p∗−−→HomZ/m(A,Z/m)→··· .

81

The cohomology is then computed to be

ExtnZ/m(A,Z/p)=

p A∗ n = 0mp

A∗/(p∗A∗) n odd

p A∗/(( mp )∗A∗) n even

.

Now let A = Z/p and assume p2|m. Then p∗A∗ = 0, and mp = kp for some k ∈ Z, implying ( m

p )∗A∗ =0 as well. Furthermore, since the zeroth derived functor returns the original functor, we haveExt0

Z/m(Z/p,Z/p) = ker(p∗) = HomZ/m(Z/p,Z/m) = HomZ/m(Z/p,Z/p) ∼= Z/p. Since each map in thecomplex above is zero in this case, we have that

ExtnZ/m(Z/p,Z, p)∼=Z/p, ∀n.

Solution 36. Exercise 36Note that Ext1(M, N) is contravariant in M for fixed N. Thus, to show that this is effaceable in bothvariables it suffices to show that Ext1(M, N) = 0 if either N is injective or M is projective. Let X bean extension of M by N. Then the diagram below shows that under either of these assumptions weget a splitting. Thus X is zero in Ext1(M, N).

0 N X M 0

N M∃ ∃

While we’re at it, let’s go ahead and show that Ext1(A,_) is in fact a δ-functor in degrees ≤ 1. Let

0→ B′ → B → B′′ → 0

be exact. Recall that in the notes, we defined the map δ : Hom(A,B′′) → Ext1(A,B′) by pulling backsequences, namely, given ϕ : A → B′′ we pull back along B → B′′ to get an object X and a map X → A.Including B′ into X gives us an extension

0→ B′ → X → A → 0.

Suppose this sequence is zero, then we get a map A → X giving us a map A → B by composingwith the map X → B in the pullback square. This shows that kerδ is contained in the image ofHom(A,B)→Hom(A,B′′). A similar argument shows that the image of this map is also contained inkerδ, implying that the long sequence we get in Extn for n ≤ 1 is exact at that point.

Solution 37. Exercise 37Let {A i} be a tower of abelian groups in which all the maps A i+1 → A i are inclusions. We regardA = A0 as a topological group where the open sets are sets of the form a+ A i for all a ∈ A and i ≥ 0.We wish to show that lim←−−A i = ∩A i = 0 if and only if A is Hausdorff under this topology. Let A beHausdorff. Suppose, by way of contradiction, that ∩A i 6= 0, and take a0 ∈∩A i to be nonzero. We viewa0 as an element of A under the given inclusions. Then any neighborhood of 0 ∈ A is of the formA i for i ≥ 0. But a0 ∈ A i for all i ≥ 0 so a0 and 0 cannot be separated by open sets in this topologycontradicting our Hausdorff assumption. Conversely, suppose ∩A i = 0. Then for any distinct a,b ∈ A,there is an i such that a− b ∉ A i. Now suppose there are ai,a′

i ∈ A i such that a+ai = b+a′i, then

a−b = a′i −ai ∈ A i, a contradiction. Thus, (a+ A i)∩ (b+ A i)=;, implying A is Hausdorff.

Our next goal is to show that (if A is Hausdorff), then lim←−−1 A = 0 if and only if A is complete in the

sense that Cauchy sequences converge. A Cauchy sequence in a general topological abelian group A

82

is a squence ak ∈ A such that for all open neighborhoods U of 0, there exists N such that ak −al ∈Ufor all k, l > N. In our case, since any open set around 0 is of the form A i for some i, a sequence akis Cauchy if for all i > 0 there is some N such that for all k, l > N, ak −al ∈ A i. First, let us see whatthe assumption lim←−−

1 A i = 0 tells us in the long exact sequence associated to

0→ {A i}→ {A}→ {A/A i}→ 0.

From this, we obtain the following exact sequence.

lim←−−A i A lim←−−A/A i lim←−−1 A i 0

ϕ δ

Here we used the Mittag-Leffler condition and the fact that the maps A → A are surjective to obtainlim←−−

1 A = 0 on the right above. Assuming A is Hausdorff, lim←−−A i = 0, as we have shown. In this case,it follows that lim←−−A/A i ∼= A if and only if lim←−−

1 A i = 0. Thus, we are lured into showing that A iscomplete if and only if lim←−−A/A i ∼= A. We turn to this now.

Suppose first that Aϕ−→ lim←−−A/A i is an isomorphism. Then we can pass the topology on A to lim←−−A/A i

making the latter into a topological group isomorphic to A. Let {ak} be a Cauchy sequence. Thenfor all i there is an Ni such that ∀k, l ≥ Ni, ak ≡ al( mod A i). It follows that {ϕ(ak)} converges inlim←−−A/A i and so ak must converge in A, impying A is complete. Conversely, suppose A is complete.Take any a ∈ A, a 6= 0. To show that ϕ is injective, it suffices to show that a ∉ A i for some i. This fol-lows directly from the Hausdorff property of A. Since a 6= 0 and for all i, A i is an open set containing0, there must be some i for which a ∉ A i. Now we prove that ϕ is surjective. For this, we need anexplicit description for lim←−−A/A i. We take the one given in Weibel’s text, which is lim←−−A/A i = ker(∆),where ∞∏

i=0A/A i

∆−→∞∏

i=0A/A i

maps(· · · , ai, · · · ,a0) 7→ (· · · , ai − f i+1(ai+1), · · · , a1 − f2(a2),a0 − f1(a1)).

Here the overline notation denotes the image under the appropriatie quotient, and take f i : A i → A i−1to be the given inclusions. The map ϕ : A → lim←−−A/A i is then just the appropriate quotient map ineach factor. Now take any element (· · · , ai, · · · , a1,0) ∈ ker(∆) and choose representatives ak of ak foreach k. We note that the zeroth coordinate is determined to be 0 since A/A0 = A/A = 0. We then havethat

ai = f i+1(ai+1), ∀i > 0

=⇒ ai ≡ ai+1( mod A i) ∀i > 0.

However since each A i is included in A j for j < i, we also have that

ai ≡ ai+1( mod A j), ∀0< j ≤ i.

It follows that {ai} is in fact a Cauchy sequence in A, so by our completeness assumption, it mustconverge to some a ∈ A. Moreover, for all i > 0 we have for all k ≥ i,

ak −ai ∈ A i =⇒ ak ≡ ai( mod A i) =⇒ a ≡ ai( mod A i).

Thus,ϕ(a)= (· · · , ai, · · · , ai,0),

implying ϕ is surjective, hence an isomorphism.

83

Solution 38. Exercise 38Suppose {A i} is a tower of finite abelian groups. Since each A j → Ak factors through A j → A j−1, wehave that

im(A j → Ak)⊂ im(A j−1 → Ak),

for all j > k. Thus, for each k, {im(A j → Ak)} j>k is a descending chain of subgroups of Ak. Since Akis finite, this chain must stabilize, that is, there must exist j such that im(A i → Ak) = im(A j → Ak)for all i ≥ j. Thus, {A i} satisfies the Mittag-Leffler condition, implying lim←−−

1 A i = 0.Similarly, suppose {A i} is a tower of finite-dimensional vector spaces over a field. Since each A j → Akfactors through A j → A j−1,

im(A j → Ak)⊂ im(A j−1 → Ak),

for all j > k. It follows that

dim(im(A j → Ak))≤ dim(im(A j−1 → Ak)),

for all j > k. Since dimAk is finite for all k this is a descending chain of integers which is boundedbelow by 0 and above by dimAk, hence must stabilize. It follows that the dimension, hence imagemust stabilize for large enough j.

Solution 39. Exercise 39Our first task is to show that Ext1

Z(Z[p−1],Z)∼= Zp/Z using that Z[p−1]=∪i p−iZ. We have the follow-ing short exact sequence

0 lim←−−1 Hom(p−iZ,Z) Ext1

Z(Z[p−1],Z) lim←−−Ext1Z(p−iZ,Z) 0

Since p−iZ ∼= Z for all i, Ext1Z(p−iZ,Z) = 0 for all i implying the rightmost term above is zero. Thus

we have an isomorphism lim←−−1 Hom(p−iZ,Z)∼=Ext1

Z(Z[p−i],Z). Now, we have the sequence

Z→ p−1Z→ p−2Z→···

which we apply the Hom functor to get

Hom(Z,Z)←Hom(p−1Z,Z)←Hom(p−2Z,Z)←··· .

Note that since p−iZ ∼= Z each of the terms above is isomorphic to Z via the isomorphism f 7→ f (1).However, since pi f (p−i) = f (1), we have that for each i, f (1) must be a multiple of pi. So we caninterpret the map f 7→ f (1) instead as an isomorphism Hom(p−iZ,Z)

∼=−→ piZ. The sequence of Hom-groups above is then nothing but

Z← pZ← p2Z←··· ,

where the maps are all multiplication by p. It follows that lim←−−1 Hom(p−iZ,Z)∼= Zp/Z, as desired.


Let →← denote the poset displayed below

y

x z

ψ

ϕ

84

Let A be a →←-shaped diagram of R-modules. Following the construction in Weibel, we let

C0 = Ax × A y × Az, C1 = A(x)z × A(y)

z ,

and define two differentials between them as follows. Let d0 : C0 → C1 be given by the matrix

d0 =(ϕ 0 00 ψ 0

),

and d1 : C0 → C1 be given by

d1 =(0 0 10 0 1

).

Then the derived functors of the lim functor of A are the cohomology of the chain complex

0→ C0d0−d1

−−−−→ C1 → 0.

Evidently, the zeroth cohomology is given by

ker(d0 −d1)= {(a,b, c)|ϕ(a)=ψ(b)= c}⊂ Ax × A y × Az =⇒ H0(C)∼= {(a,b)|ϕ(a)=ψ(b)}⊂ Ax × A y,

which is the pullback, as expected. The lim1 functor in this case is the first cohomology of thiscochain complex, hence is the cokernel coker(d0 − d1) = H1(C). To see that this is the same as thecokernel of the difference map Ax × A y → Az mapping (a,b) 7→ ϕ(a)−ψ(b), we first note that thediagonal ∆ = {(c, c)|c ∈ Az} ⊂ Az × Az is contained in the image of d0 − d1. It follows that the mapAz×Az → H1(C) factors through the map Az×Az

q−→ Az given by q(c,d)= c−d. That these cokernelsdo indeed coincide then follows from commutativity of the following diagram with exact rows.

Ax × A y × Az A(x)z × A(y)

z H1(C) 0

Ax × A y Az coker(ϕ−ψ) 0

d0−d1

q

ϕ−ψ

To see this, note that since the quotient A(x)z × A(x)

z → H1(C) factors through q : A(x)z × A(y)

z → Az, wehave that H1(C) is also the cokernel of the composition

Ax × A y × Az → Ax × A yϕ−ψ−−−→ Az.

However, since the projection Ax × A y × Az → Ax × A y is surjective, this is the same as coker(ϕ−ψ).

Solution 41. Exercise 41Going off of Example 5.2.6 in Weibel, since Er

pq is a spectral sequence converging to H∗, for each nwe have a filtration

0= F−1Hn ⊂ F0Hn ⊂ ·· · ⊂ FpHn ⊂ ·· ·FnHn = Hn

such that

E∞pq

∼= FpHp+q

Fp−1Hp+q.

In this particular case, the only nonzero columns of the E2-page are p = 0,1. Since d2 : E2pq →

E2p−2,q+1, and for p = 0,1 we have p−1 =−2,−1, it follows that all differentials on this E2-page are

zero and thus, E2 = E3 = ·· · = E∞. We then have

E20n

∼= F0Hn

F−1Hn= F0Hn and E2

1,n−1∼= F1Hn

F0Hn,

85

implying there is a short exact sequence

0→ E20n → F1Hn → E2

1,n−1 → 0.

However, for any k > 1 we have

FkHn

Fk−1Hn∼= E2

k,n−k = 0 =⇒ FkHn = Fk−1Hn.

Thus, Hn = FnHn ∼= F1Hn and we get the desired short exact sequence below

0→ E20n → Hn → E2

1,n−1 → 0.

Solution 42. Exercise 42In this exercise, Er

pq is a spectral sequence converging to H∗ such that Erpq = 0 for q 6= 0,1. As in the

preceding exercise, for each n, we have a filtration

0= F−1Hn ⊂ F0Hn ⊂ ·· · ⊂ FpHn ⊂ ·· ·FnHn = Hn,

satisfying

E∞pq

∼= FpHp+q

Fp−1Hp+q.

In this case, however, we do not have the luxury of every differential being zero, since d2 : E2p0 →

E2p−2,1 is not necessarily zero for p > 1. Still, since d3 : E3

pq → E3p−3,q+2, and the E3-page can have no

more than the two rows q = 0,1, we have that all the differentials on this third page are zero, henceE3 = E4 = ·· · = E∞. Thus

ker(d2p)∼= E∞

p0∼= FpHp

Fp−1Hp= Hp

Fp−1Hp,∀p ≥ 2 and

E2p1

im(d2p+2)

∼= E∞p1

∼= FpHp+1

Fp−1Hp+1,∀p ≥ 0.

From the isomorphism on the left above, we have that the following sequence is exact for all p ≥ 2,

0→ Fp−1Hp → Hp → E2p0

d−→ E2p−2,1.

Now note thatFp−1Hp+1

Fp−2Hp+1∼= E∞

p−1,2 = 0 =⇒ Fp−1Hp+1 = Fp−2Hp+1,∀p ≥ 0.

ThusFp−1Hp+1 = F0Hp+1 ∼= E∞

0,p+1 = 0,∀p ≥ 1.

Using the isomorphism to the right above, we can deduce that

E2p1

im(d2p+2)

∼= FpHp+1,∀p ≥ 1.

Thus, we can extend the exact sequence obtained above to the long exact sequence below

· · ·→ E2p+1,0

d−→ E2p−1,1 → Hp → E2

p0 → E2p−2,1 →··· .

Evidently, this will go on to the left indefinitely. On the right we can use the fact that d20 = d2

1 = 0 toget

E200 = E∞

00∼= F0H0

F−1H0= F0H0 = H0 and E2

10 = E∞10

∼= F1H1

F0H1= H1

F0H1.

86

Now, using the isomorphism on the right above in the case p = 0, we have

E201

im(d22)

∼= F0H1,

implying that the following sequence is exact

H2 → E220

d−→ E201 → H1 → E2

10 → 0→ H0∼=−→ E2

00 → 0.

The sequence can then be continued as above for p ≥ 2, so we are done.

Solution 43. Exercise 43In this exercise, E∗∗ is a double complex whose only nonzero columns are at p and p− 1, E1

pq =Hq(Ep∗) is the vertical homology at site (p, q) and E2

pq = Hp(E1∗q) is the horizontal homology of theresulting sequence of complexes E1∗∗. Our goal is to show that there is a short exact sequence

0→ E2p−1,q+1 → Hp+q(T)→ E2

pq → 0,

where T denotes the total complex of E∗∗. The point here is that computing the vertical and hori-zontal homology consecutively is, in a sense, an approximation to the homology of the total complex,at least when there are only two nonzero columns. Here we take the straightforward approach andcompute each of these groups explicitly. After stumbling through some definitions, one finds

Hp+q(T)= {(a,b) ∈ Ep−1,q+1 ×Epq|dv(a)+dh(b)= dv(b)= 0}{(dv(x)+dh(y),dv(y))|(x, y) ∈ Ep−1,q+2 ×Ep,q+1}

,

E2pq = ker(dh : Hq(Ep∗)→ Hq(Ep−1,∗))= {b ∈ Epq|dv(b)= 0,dh(b)= dv(x), x ∈ Ep−1,q+1}

{dv(y)|y ∈ Ep,q+1},

and

E2p−1,q+1 =

Hq+1(Ep−1,∗)

dh(Hq+1(Ep∗))= {a ∈ Ep−1,q+1|dv(a)= 0}

{dv(x)|x ∈ Ep−1,q+2}+ {dh(y)|y ∈ Ep,q+1}.

From this we see automatically that there are maps

Hp+q(T) α−→ E2pq, α([a,b])= [b],

andE2

p−1,q+1β−→ Hp+q(T), β([a])= [a,0].

Evidently α is surjective, β is injective, and im(β) ⊂ ker(α). For the reverse inclusion, note that if[a,b] ∈ ker(α) then dv(y) = b for some y ∈ Ep,q+1. Thus [a,b] = [a,0]+ [0,dv(y)] =⇒ [a,b] = [a,0] ∈Hp+q(T), implying that the sequence

0→ E2p−1,q+1

β−→ Hp+q(T) α−→ E2pq → 0

is exact.



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Solution 46. Exercise 46The task that this exercise bestows upon us is to prove the Universal Coefficient Theorem for coho-mology using a spectral sequence argument. The theorem states that if (P·,d·) is a chain complex ofprojective R-modules such that dn(Pn) is also projective, for all n, then for any other R-module, M,there is a split exact sequence

0→Ext1R(Hn−1(P·), M)→ Hn(HomR(P, M))→HomR(Hn(P·), M)→ 0.

In the following we will surpress the subscript R. Let M → I· be an injective resolution, and considerthe following double (co)chain complex.

......

......

0 Hom(P1, I0) Hom(P1, I1) Hom(P1, I2) · · ·


0 Hom(P−1, I0) Hom(P−1, I1) Hom(P−1, I2) · · ·

......

......

d∗2 d∗

2 d∗2

d∗1 d∗

1 d∗1

d∗0 d∗

0 d∗0

We first show that the spectral sequence for a double complex is in fact computing what we want,namely Hn(Hom(P·, M)). To do this, we filter the complex vertically and run our spectral sequence.In this case, the I I E0 is given by:

I I E0 :...



0 Hom(P−1, I0) Hom(P−1, I1) Hom(P−1, I2) · · ·

...

Since injective resolutions are exact sequences, and Hom(Pk,_) is exact by virtue of Pk being pro-jective, the homology of these rows vanishes above degree zero. We then obtain the following forI I E1.

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I I E1 :...

......

0 Hom(P1, M) 0

· · · 0 Hom(P0, M) 0 · · ·

0 Hom(P−1, M) 0

......

...

d∗2

d∗1

d∗0

Evidently, the cohomology of this cochain complex is precisely what we want to find. Furthermore,once this cohomology is computed to obtain the E2 page, since there is only a single nonzero column,the differentials dr for r ≥ 2 must vanish. Thus I I E∞

pq =I I E2pq = Hn(Hom(P·, M)).

Now we filter this complex horizontally and run the spectral sequence in hopes of gaining a differentdescription of this cohomology. The E0-page is then given by:

E2 :...

......

...

0 Hom(P1, I0) Hom(P1, I1) Hom(P1, I2)


0 Hom(P−1, I0) Hom(P−1, I1) Hom(P−1, I2)

......

......

d∗2 d∗

2 d∗2

d∗1 d∗

1 d∗1

d∗0 d∗

0

Now since each In is injective, the functor Hom(_, In) is exact and so it commutes with homology, i.e.Hn(Hom(P∗, In))=Hom(Hn(P∗), In). The E1-page is then shown below.

E1 :...

0 Hom(H1(P·), I0) Hom(H1(P·), I1) Hom(H1(P·), I2) · · ·

0 Hom(H0(P·), I0) Hom(H0(P·), I1) Hom(H0(P·), I2) · · ·

0 Hom(H−1(P·), I0) Hom(H−1(P·), I1) Hom(H−1(P·), I2) · · ·

...

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It follows that Epq2 = Extp(Hq(P·), M). However, note that since dn(Pn) is also projective for all n,

Zn = ker(dn) must also be projective. To see this, it suffices to notice that the exact sequence

0→ Zn → Pn → dn(Pn)→ 0

splits since dn(Pn) is projective. Thus Zn is a direct summand of Pn, a projective module, hence mustbe projective. It follows that

0→ dn(Pn)→ Zn → Hn(P·)→ 0

is a projective resolution for Hn(P·) for all n. Then, computing Extq(Hp(P·), M) using this resolutionfor Hp(P·) shows that Extp(Hq(P·), M) = 0 for p ≥ 2. This leads us to the following description of theE2-page shown below.

E2 :...

......

...

0 Hom(H1(P·), M) Ext1(H1(P·), M) 0

0 Hom(H0(P·), M) Ext1(H0(P·), M) 0

0 Hom(H−1(P·), M) Ext1(H−1(P·), M) 0

......

......

It now follows, from an argument dual to that of Exercise 41, that for all n there is a short exactsequence

0→Ext1(Hn−1(P·), M)→ Hn(Hom(P·, M))→Hom(Hn(P·), M)→ 0.

It remains to show that this sequence is in fact split.


Let L ∈ mod-R, M ∈ R-mod, M ∈ mod-S, and N ∈ S-mod. Choose projective resolutions P· → L andQ· → N. We are lead to consider the double complex below.

......

...

0 P0 ⊗M⊗Q2 P1 ⊗M⊗Q2 P2 ⊗M⊗Q2 · · ·

0 P0 ⊗M⊗Q1 P1 ⊗M⊗Q1 P2 ⊗M⊗Q1 · · ·

0 P0 ⊗M⊗Q0 P1 ⊗M⊗Q0 P2 ⊗M⊗Q0 · · ·

0 0 0

90

Running our first spectral sequence on this double complex gives us the following I E0-page.

I E0 :...

......

...

0 P0 ⊗M⊗Q2 P1 ⊗M⊗Q2 P2 ⊗M⊗Q2

· · · 0 P0 ⊗M⊗Q1 P1 ⊗M⊗Q1 P2 ⊗M⊗Q1 · · ·

0 P0 ⊗M⊗Q0 P1 ⊗M⊗Q0 P2 ⊗M⊗Q0

0 0 0 0

Taking homology gives us the next page displayed below.

I E1 :...

0 P0 ⊗TorS2 M, N) P1 ⊗TorS

2 (M, N) P2 ⊗TorS2 (M, N) · · ·

0 P0 ⊗TorS1 (M, N) P1 ⊗TorS

1 (M, N) P2 ⊗TorS1 (M, N) · · ·

0 P0 ⊗TorS0 (M, N) P1 ⊗TorS

0 (M, N) P2 ⊗TorS0 (M, N) · · ·

0 0 0 0 · · ·

Notice that we used the fact that since Pp is projective for all p, Pp⊗_ is exact, hence commutes withhomology so that TorS

q (Pp ⊗M, N)= Pp ⊗TorSq (M, N). Taking homology then gives us that

I E2pq =TorR

p (L,TorSq (M, N)).

A symmetric argument shows that if we run our spectral sequence with the opposite filtration we get

I I E2pq =TorS

p (TorRq (L, M), N).

Now note that if M is a flat S-module, then

I E2pq =

{TorR

p (L, M⊗S N) q = 00 q 6= 0

Since this consists of a single row, this spectral sequence collapses at the E2- page and so it convergesto TorR

∗ (L, M⊗S N). Since we started with a first quadrant spectral sequence, both of these spectralsequences converge to the same thing, namely H∗(Tot(P·⊗M⊗Q·). Thus, in this case, we have

I I E2pq =TorS

p (TorRq (L, M), N) =⇒ TorR

∗ (L, M⊗S N).

91

Similarly, if M is a flat R-module, then

I I E2pq =

{TorS

p (L⊗R M, N), q = 00 q 6= 0

.

Again, since there is only one nonzero row, this spectral sequence collapses at E2 and so it convergesto TorS

p (L⊗R M, N). Since the two converge to the same object, we have

I E2pq =TorS

p (TorRq (L, M), N) =⇒ TorS

p (L⊗R M, N).

Solution 48. Exercise 49Suppose, for a contradiction, that there is a map w :Z/2Z→Z/2Z[−1] that makes the sequence

Z/2Z 2−→Z/4Z 1−→Z/2Z w−→Z/2Z[−1]

into an exact triangle. We then have a solid-arrow diagram below in which the square on the leftcommutes.

Z/2Z Z/4Z Z/2Z Z/2Z[−1]

Z/2Z Z/4Z cone(2) Z/2Z[−1]

2

1

1

1

w

∃h 1

2 v δ

The bottom row is, by definition, an exact triangle. Thus, by axiom TR3, there must exist a map hshown above that makes the diagram commute. However, the degree zero part of the center squarelooks like

Z/4Z Z/2Z

Z/4Z Z/4Z

1

1 ∃h1

We have now reached a contradiction since the identity Z/4Z→Z/4Z does not factor through Z/4Z 1−→Z/2Z. Thus, the sequence on top cannot be an exact triangle.

Solution 49. Exercise 50We are given an exact triangle (u,v,w) on (A,B,C) and want to show that vu = wv = (Tu)w = 0. Weknow that (1A,0,0) is an exact triangle on (A, A,0), so by axiom TR3, there is a dashed arrow belowmaking the diagram commute. However, since the source of this arrow is 0, this map can only be thezero map. Thus, by commutativity of the middle square below, vu = 0.

A A 0 T A

A B C T A

1A

1A u ∃h=0 T1u v w

Now apply the same reasoning to the rotated triangle (v,w,−Tu) to obtain wv = 0. Then since w and−Tu are the v and w of this rotated triangle, (−Tu)w = 0 =⇒ (Tu)w = 0.

Solution 50. Exercise 51For this problem, we are asked to prove a form of the five-lemma for triangulated categories. Specifi-cally, given a morphism of exact triangles shown bellow, in which f and g are isomorphisms, we wantto show that h must also be an isomorphism.

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A B C T A

A′ B′ C′ T A′

u

f∼=

v

g∼=

w

h T f∼=u′ v′ w′

One is tempted to simply extend this diagram to the right with T g : TB → TB′ and use the standardfive-lemma, however, this fails because our category is not necessarily abelian. Our approach will beto carry this diagram into an abelian category and apply the five-lemma there to obtain our result.First, we prove an auxiliary result.

Lemma. Let f : X →Y be a morphism in K. If f ∗ =HomK (C, f ) is an isomorphism for all C ∈Ob(K),then so is f .

Proof. We view f ∗ as a natural transformation hX = HomK (_, X ) → HomK (_,Y ) = hY . The YonedaLemma says that there is a bijection

HomK (hX ,hY )'HomK (X ,Y ),

which preserves composition. Thus, natural isomorphisms correspond to isomorphisms and the re-sult follows.

Now for any W ∈ Ob(K), applying HomK (W ,_) to the diagram above yields a similar diagram in thecategory of abelian groups. This is because K is additive, being triangulated, hence

HomK (W ,_) : K →Ab

for all W . The five lemma then applies to show that HomK (W ,h) is an isomorphism for all W . Theresult then follows from the lemma above.

Solution 51. Exercise 52Consider the morphism of chain complexes shown bellow.

· · · 0 Z/2 0 · · ·

· · · Z/4 Z/4 Z/4 · · ·2

2 2

This is not nullhomotopic since finiding a nullhomotopy amounts to finding a lift shown bellow, whichevidently does not exist.

Z/2

Z/4 Z/4

2@

4

Still, the chain complex below is evidently exact and so it is quasi-isomorphic to the zero complex.Composing this morphism of chain complexes with this quasi-isomorphism shows that this morphismis in fact zero in the derived category.

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