Lecture Notes Math 239 (Algebra) Spring 2015 · 2015. 6. 12. · 9 Derived Functors 75 10 The Functors Ext and Tor 82 11 Double Complexes and Tensor Products of Chain Complexes 92

Lecture Notes

Math 239 (Algebra)

Spring 2015(Robert Boltje, UCSC)

Contents

1 Categories and Functors 1

2 Simplicial and Semi-Simplicial Objects in a Category 16

3 Chain Complexes and Homology 23

4 Chain Complexes and Homotopy 33

5 The Long Exact Homology Sequence 43

6 The Mapping Cone 51

7 Extensions of Modules 59

8 Projective Resolutions 68

9 Derived Functors 75

10 The Functors Ext and Tor 82

11 Double Complexes and Tensor Products of Chain Complexes 92

12 Group Homology and Group Cohomology 102

1 Categories and Functors

1.1 Definition A category C consists of

• a class Ob(C), whose elements are called the objects of C,

• for any two objects C,C ′ ∈ Ob(C), a set HomC(C,C ′), called the set of morphismsfrom C to C ′, and,

• for any three objects C,C ′, C ′′ ∈ Ob(C), a composition map

HomC(C,C′)× HomC(C ′, C ′′)→ HomC(C,C ′′) , (f, g) 7→ g ◦ f .

These data are subject to the following axioms:

(C1) If (C1, C′1) and (C2, C

′2) are different pairs of objects of C then HomC(C1, C

′1) and

HomC(C2, C′2) are disjoint sets. In other words, each morphism has a uniquely

determined domain and a uniquely determined codomain.

(C2) Composition of morphisms is associative.

(C3) For each object C of C there exists a morphism 1C ∈ HomC(C,C) such that 1C ◦f =f , for any A ∈ Ob(C) and any f ∈ HomC(A,C), and g ◦ 1C = g for any A ∈ Ob(C)and any morphism g ∈ HomC(C,A)).

We will often write f : C → C ′ to indicate that f ∈ HomC(C,C ′).

1.2 Examples (a) Sets and functions between sets form a category, denoted by Set.There is also a category of finite sets, denoted by set.

(b) There are categories Gr, gr, Ri, Ric, Ab of groups, finite groups, rings, commutativerings, and abelian groups, together with their respective homomorphisms. We assumethat rings are always associative and have a multiplicative identity element, and that ringhomomorphisms preserve identity elements.

(c) For every ring R, one has a category RMod, the category of left R-modules.Similarly, ModR denotes the category of right R-modules. If also S is a ring, we de-note by RModS the category of (R, S)-bimodules. An (R, S)-bimodule is an abeliangroup M which has a left R-module structure and a right S-module structure such that(rm)s = r(ms) for all r ∈ R, s ∈ S, and m ∈ M . Homomorphisms between (R, S)-bimodules are group homomorphism which are both left R-module homomorphisms andright S-module homomorphisms.

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(d) Top denotes the category of topological spaces and continuous maps.

(e) If C and D are categories then their product category C × D is defined as follows:Its objects are pairs (C,D), where C ∈ Ob(C) and D ∈ Ob(D), and the set of morphismsbetween two objects (C,D) and (C ′, D′) consists of all pairs (f, g) with f ∈ HomC(C,C ′)and g ∈ HomD(D,D′). Composition is defined componentwise.

1.3 Definition Let C and D be categories.

(a) A covariant functor F : C→ D consists of functions of the following form:

• A function Ob(C)→ Ob(D), denoted by C 7→ F(C), and,

• for any two objects C,C ′ of C, a function HomC(C,C ′) → HomD(F(C),F(C ′)),again denoted by f 7→ F(f).

These functions are subject to the following axioms:

(F∗1) F(g◦f) = F(g)◦F(f), for all C,C ′, C ′′ ∈ Ob(C) and all f : C → C ′ and g : C ′ → C ′′in C.

(F∗2) F(1C) = 1F(C) for all C ∈ Ob(C).

(b) Acontravariant functor F : C→ D consists of functions of the following form:

• A function Ob(C)→ Ob(D), denoted by C 7→ F(C), and,

• for any two objects C,C ′ of C, a function HomC(C,C ′) → HomD(F(C ′),F(C)),again denoted by f 7→ F(f).

These functions are subject to the following axioms:

(F∗1) F(g◦f) = F(f)◦F(g), for all C,C ′, C ′′ ∈ Ob(C) and all f : C → C ′ and g : C ′ → C ′′in C.

(F∗2) F(1C) = 1F(C) for all C ∈ Ob(C).

We make the convention that the word ”functor” by itself always means ”covariantfunctor”.

1.4 Examples (a) For any category C one has an obvious identity functor IdC : C→ C.(b) If C, D, and E are categories and F : C → D and G : D → E are covariant or

contravariant functors then one can define G ◦ F : C → E on objects by C 7→ G(F(C))and on morphisms by f 7→ G(F(f)). If both F and G are of the same type (covariant

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or contravariant) then G ◦ F is covariant. If they are of different type then G ◦ F iscontravariant.

(c) The unit group functor U : Ri → Gr maps a ring R to its unit group U(R), and aring homomorphism f : R → S to a the group homomorphism f : U(R) → U(S). Moregenerally, for every positive integer n, one has the functor GLn : Ri → Gr which maps aring R to the general linear group GLn(R) and a ring homomorphism f : R → S to thegroup homomorphism GLn(f) : GLn(R)→ GLn(S), (rij) 7→ (f(rij)).

(d) Let R be a ring and let M be in RMod. There is a covariant functor

HomR(M,−) : RMod→ AbN 7→ HomR(M,N)

(f : N → N ′) 7→(HomR(M, f) = f∗ : HomR(M,N)→ HomR(M,N ′)

)g 7→ f ◦ g

Here HomR(M,N) is an abelian group through (f + f′)(m) := f(m) + f ′(m) for f, f ′ ∈

HomR(M,N) and m ∈M .Similarly, for N ∈ RMod, one has a contravariant functor

HomR(−, N) : RMod→ AbM 7→ HomR(M,N)

(f : M →M ′) 7→(HomR(f,N) = f

∗ : HomR(M′, N)→ HomR(M,N)

)g 7→ g ◦ f

If R is commutative, each of these functors can be considered as functor from RModto RMod. In fact, in this case HomR(M,N) is an R-module via (rf)(m) := rf(m) forf ∈ HomR(M,N), r ∈ R and m ∈ M , and HomR(M, f) and HomR(f,N) are R-modulehomomorphisms.

More generally, if also S is a ring and M is an (R, S) − bimodule then HomR(M,−)can be viewed as a functor form RMod to SMod, where HomR(M,N) is viewed as a leftS-module via (sf)(m) := f(ms), for f ∈ HomR(M,N), s ∈ S and m ∈ M . Similarly,if N is an (R, S)-bimodule then HomR(−, N) can be considered as a functor from RModto ModS, where HomR(M,N) is considered as right S-module via (fs)(m) := f(m)s forf ∈ HomR(M,N), s ∈ S and m ∈M .

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(e) Specializing N = R in (d), we obtain a contravariant functor

HomR(−, R) : RMod→ ModRM 7→ HomR(M,R)

(f : M →M ′) 7→(HomR(f,R) = f

∗ : HomR(M′, R)→ HomR(M,R)

)g 7→ g ◦ f

This functor is often called the R-dual functor.

If R is a k-algebra for a commutative ring k, there is a different type of functor, calledthe k-dual functor:

Homk(−, k) : RMod→ ModRM 7→ Homk(M,k)

(f : M →M ′) 7→(Homk(f, k) = f

∗ : Homk(M′, k)→ Homk(M,k)

)g 7→ g ◦ f

Here Homk(M,k) is a right R-module via (fr)(m) = f(rm) for f ∈ Homk(M,k), m ∈Mand r ∈ R. For certain types of algebras, so-called symmetric algebras R, the R-dualand the k-dual functors are isomorphic in a precise sense which will be defined later, see1.24(b). Group algebras for instance are symmetric algebras.

(f) If R is a ring and M ∈ RMod we obtain a covariant functor

−⊗RM : ModR → AbL 7→ L⊗RM

(f : L→ L′) 7→ (f⊗idM : L⊗RM → L′ ⊗RM)

Simiarly, for M ∈ ModR, one obtains a covariant functor

M ⊗R − : RMod→ Ab .

If M ∈ RModS is an (R, S)-bimodule for an additional ring S, then one obtains functors

−⊗RM : ModR → ModS and M ⊗S − : SMod→ RMod .

1.5 Definition Let C be a category, let B,C ∈ Ob(C), and let g ∈ HomC(B,C).(a) The morphism g is called a monomorphism if g ◦ f1 = g ◦ f1 implies f1 = f2 for all

A ∈ Ob(C) and f1, f2 ∈ HomC(A,B).

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(b) The morphism g is a called an epimorphism if h1 ◦ g = h2 ◦ g implies h1 = h2 forall D ∈ Ob(C) and all h1, h2 ∈ HomC(C,D).

(c) g is called an isomorphism, if there exists a morphism f : C → B in C such thatf ◦ g = 1B and g ◦ f = 1C . In this case, f is uniquely determined by g and denoted byg−1.

(d) The objects B and C of C are called isomorphic (notation B ∼= C), if there existsan isomorphism f : B → C in C.

1.6 Remark (a) In the category Set, a morphism is a monomorphism (resp. epimor-phism, resp. isomorphism) if and only if it is injective (resp. surjective, resp. bijective).The same holds for instance in the categories Gr and RMod; but not in Ri.

(b) Every isomorphism is a monomorphism and an epimorphism. But the converse isnot true: Consider the inclusion Z→ Q in Ri.

(c) If F : C→ D is a covariant or contravariant functor and f : C → C ′ is an isomor-phism in C, then F(f) is an isomorphism in D. However, the class of monomorphisms orepimorphisms is not preserved under covariant or contravariant functors.

(d) Every category C has an opposite category C◦. Its objects are the same as thosein C. But when viewed as objects in C◦ we denote them by C◦. For two objects A◦

and B◦ of C◦, one sets HomC◦(A◦, B◦) := HomC(B,A). We denote f ∈ HomC(B,A) by

f ◦ : A◦ → B◦, if it is viewed as a morphism in C◦. Composition in C◦ is defined as follows:For A,B,C ∈ Ob(C) and f ◦ : A◦ → B◦ and g◦ : B◦ → C◦, one has g◦ ◦ f ◦ := (f ◦ g)◦.

Note that a morphism f ◦ : A◦ → B◦ in C◦ is a monomorphim in C◦ if and only iff : B → A is an epimorphism in C. Morover, f ◦ is an epimorphism in C◦ if and only if fis a monomorphism in C. We say that the concepts of ‘monomorphism’ and ‘epimorphism’are dual to each other: The one translates into the other in the opposite category.

There is an obvious contravariant functor C → C◦, sending an object A of C to theobject A◦ of C◦ and a morphism f in C to f ◦. Moreover, there is a contravariant functorC◦ → C defined by A◦ 7→ A and f ◦ 7→ f . The compositions of these two functors yieldthe identity functors on C and on C◦. For any category D, there is a bijection betweenthe collection of contravariant functors C → D and the collection of covariant functorsC◦ → D, given by precomposing with the above functors. This way one can always switchbetween contravariant and covariant functors if necessary. For instance, for any ring R,one can consider HomR(−,−) as a covariant functor

HomR(−,−) : RMod◦ × RMod→ Ab .

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1.7 Definition Let C be a category. An object A of C is called an initial object (resp. finalobject) of C if |HomC(A,B)| = 1 (resp. |HomC(B,A)| = 1) for all B ∈ Ob(C). If A is bothan initial and a final object in C then we call A a zero object.

1.8 Remark (a) If A and A′ are two initial (resp. final) objects of a category C then thereexist unique morphisms in HomC(A,A

′) and in HomC(A′, A). These two maps are inverse

isomorphisms, since HomC(A,A) and HomC(A′, A′) contain only the respective identity

morphism. Thus, initial (resp. final) objects in C are unique up to unique isomorphism,and we may speak of the initial (resp. final) object in C (if it exists). This also impliesthat zero objects are unique up to unique isomorphism. The zero object, if it exists, isusually denoted by 0.

(b) In the category Set, the empty set is an initial object and any set with one elementis a final object. In the category Gr, the trivial group is a zero object. Similarly, the zeromodule is a zero object in the category RMod, for every ring R.

1.9 Definition Let C be a category and let Ai, i ∈ I, be objects of C. A product(P, (πi)i∈I) of Ai, i ∈ I, consists of an object P of C together with morphisms πi : P → Ai,i ∈ I, in C, called projections, such that for any object Q of C and any family of morphismsfi : Q → Ai, i ∈ I, in C, there exists a unique morphism f : Q → P with the propertyπi ◦ f = fi for all i ∈ I. Note that if (P, (πi)i∈I) is a product of Ai, i ∈ I, and Q ∈ Ob(C),then

HomC(Q,P )→ ×i∈I

HomC(Q,Ai) , f 7→ (πi ◦ f)i∈I ,

is a bijection. In particular, for any two morphisms f, g : Q → P in C one has f = g ifand only if πi ◦ f = πi ◦ g for all i ∈ I.

1.10 Proposition Let C be a category and let Ai, i ∈ I, be a family of objects in C.Assume that (P, (πi)i∈I) and (P

′, (π′i)i∈I) are products of Ai, i ∈ I. Then there exists aunique morphism f : P → P ′ in C such that π′i ◦ f = πi for all i ∈ I. Moreover, f isan isomorphism. Thus, products are unique up to unique isomorphism (if they exist).Therefore, we denote any product of Ai, i ∈ I, by

∏i∈I Ai (without using a notation for

the projections).

Proof Since (P ′, (π′i)i∈I) is a product of Ai, i ∈ I, there exists a unique f : P → P ′in C such that π′i ◦ f = πi for all i ∈ I. Similarly, since (P, (πi)i∈I) is a product of Ai,i ∈ I, there exists a unique morphism g : P ′ → P in C such that πi ◦ g = π′i for all i ∈ I.Therefore, πi ◦ (g ◦ f) = (πi ◦ g) ◦ f = π′i ◦ f = πi = πi ◦ 1P for all i ∈ I. This impliesg ◦ f = 1P . Similarly, one shows that f ◦ g = 1P ′ .

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1.11 Definition Let C be a category and let Ai, i ∈ I, be a family of objects of C. Acoproduct (C, (ιi)i∈I) of Ai, i ∈ I, in C consists of an object C of C together with a familyof morphisms ιi : Ai → C in C, called the injections, satisfying the following property. Forany object D in C and any family of morphisms fi : Ai → D, i ∈ I, in C, there exists aunique morphism f : C → D with f ◦ ιi = fi, for all i ∈ I.

1.12 Remark Assume the situation of the previous definition. Note that (C, (ιi)i∈I) isa coproduct of Ai, i ∈ I, in C, if and only if (C◦, (ι◦)i∈I) is a product of A◦i , i ∈ I, inthe opposite category C◦. Thus, coproducts and products are dual concepts. Now Propo-sition 1.10 applied to C◦ implies that coproducts are unique up to unique isomorphism.We denote the coproduct of Ai, i ∈ I, by

∐i∈I Ai (without including a notation for the

injections).

1.13 Examples (a) In Set, the cartesian product ×i∈IAi of sets Ai, i ∈ I, togetherwith the usual projection maps πi : ×i∈I Ai, is a categorical product in the sense ofDefinition 1.9. The disjoint union

⊎i∈I Ai together with the inclusions ιi : Ai →

⊎Ai,

i ∈ I, is a coproduct.(b) In the category Gr, the cartesian product×i∈IAi, together with the usual projection

maps, is again a categorical product. Coproducts also exist but are more complicated todefine.

(c) Let R be a ring and let Mi, i ∈ I, be objects in RMod. The product of Mi, i ∈ I,is given by ×i∈IMi and the usual projection maps. The coproduct of Mi, i ∈ I, is thesubmodule of ×i∈IMi consisting of those elements that have only finitely many non-zerocomponents. The injections are the usual embeddings. It is instructive to check why thecartesian product of Mi together with the usual inclusion maps is not a coproduct. Notethat the images of the inclusion maps Mj → ×i∈IMi do not generate ×i∈IMi. If I is finitethen the underlying modules of the product and coproduct coincide.

1.14 Definition (a) An additive category is a category C together with abelian groupstructures on each morphism set HomC(A,B), A,B ∈ Ob(C) satisfying the followingaxioms:

(Add 1) C has a zero object.

(Add 2) C has finite products and coproducts.

(Add 3) The composition of morphisms is biadditive.

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(b) Let C and D be additive categories. An additive functor from C to D is a covariantfunctor F : C → D with the property that for any two objects A and B of C, the mapF : HomC(A,B)→ HomD(F(A),F(B)), f 7→ F(f), is a group homomorphism. Similarly,one defines contravariant additive functors.

1.15 Remark (a) If C is an additive category then also C◦ is an additive category. Here,the group structure on HomC◦(A

◦, B◦) is the same as the one on HomC(B,A).

(b) For a commutative ring k, there is also the notion of a k-linear category C. Themorphism sets HomC(A,B) have the additional structure of a k-module and compositionis k-bilinear. Moreover, a functor between k-linear categories is said to be k-linear if it isk-linear on morphism sets. The notions of an additive category and additive functors arethe special case were k = Z.

(c) If C is an additive category then also C◦ is an additive category. The groupstructure on morphism sets of C◦ is defined by f ◦ + g◦ := (f + g)◦ for A,B ∈ Ob(C) andf, g ∈ HomC(A,B).

1.16 Examples (a) The category Set can’t be made into an additive category, since ithas no zero object.

(b) The category Ab is additive. More generally, RMod is additive for an ring R. If Ris a k-algebra over a commutative ring k then RMod is a k-linear category.

1.17 Proposition Assume that C is an additive category and that Ai, i ∈ I, is a finitefamily of objects in C.

(a) If (P, (πi)i∈I) is a product of Ai, i ∈ I, in C then there exist unique morphismsιi : Ai → P , i ∈ I, in C such that

πi ◦ ιj =

0 if i 6= j,1Ai if i = j, (1.17.a)for i, j ∈ I. Moreover, ∑

i∈Iιi ◦ πi = 1P (1.17.b)

and (P, (ιi)i∈I) is a coproduct of Ai, i ∈ I, in C.(b) If (C, (ιi)i∈I) is a coproduct of Ai, i ∈ I, in C then there exist unique morphisms

πi : C → Ai, i ∈ I, in C such that Equation (1.17.a) holds. Moreover, Equation (1.17.b)holds and (C, (πi)i∈I) is a product of Ai, i ∈ I, in C.

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Proof (a) Fix j ∈ I. For i ∈ I define fi ∈ HomC(Aj, Ai) to be the identity of Ai if i = jand the 0-morphism if i 6= j. Since (P, (πi)i∈I) is a product of Ai, i ∈ I, there exists aunique morphism ιj : Aj → P such that πi ◦ ιj is the identity on Ai if i = j and the 0morphism if i 6= j. Thus, there exists a unique family of morphisms ιj : Aj → P , j ∈ I,such that Equation (1.17.a) holds. Moreover, for every i ∈ I, the latter equation implies

πi ◦ (∑j∈I

ιj ◦ πj) =∑j∈I

πi ◦ ιj ◦ πj = πi = πi ◦ 1P ,

and we also obtain Equation (1.17.b).Next we show that (P, (ιi)i∈I) is a coproduct of Ai, i ∈ I, in C. So assume that

D ∈ Ob(C) and that hi : Ai → D, i ∈ I, is a family of morphisms in C. We need to showthat there exists a unique morphism f : P → D in C such that f ◦ ιi = hi for all i ∈ I. Setf :=

∑j∈I hj ◦ πj. Then f ◦ ιi =

∑j∈I hj ◦ πj ◦ ιi = hi, by Equation (1.17.a), so that f has

the desired property. If also g : D → P has this property then f ◦ιi◦πi = hi◦πi = g◦ιi◦πifor all i ∈ I. Summing these equations over i ∈ I and using Equation (1.17.b), yieldsf = g.

(b) This is proved in a very similar way.

1.18 Definition Let C be an additive category and let Ai, i ∈ I, be a finite family ofobjects of C. A biproduct of Ai, i ∈ I, is a triple (P, (πi)i∈I , (ιi)i∈I) consisting of anobject P of C and morphism πi : P → Ai, i ∈ I, and ι : Ai → P , i ∈ I, such thatEquations (1.17.a) and (1.17.b) hold.

1.19 Remark Let C be an additive category and let Ai, i ∈ I, be a finite family ofobjects in C. Since C is additive, there exists a product (P, (πi)i∈I) of Ai, i ∈ I, andProposition 1.17(a) implies that this product can be extended in a unique way into abiproduct of Ai, i ∈ I. Similar, a coproduct of Ai, i ∈ I, can be uniquely extended to abiproduct of Ai, i ∈ I. Conversely, if (P, (πi)i∈I , (ιi)i∈I) is a biproduct of Ai, i ∈ I, then(P, (πi)i∈I) is a product of Ai, i ∈ I, and (P, (ιi)i∈I) is a coproduct of Ai, i ∈ I. In fact, therespective universal properties follow quickly from the Equations (1.17.a) and (1.17.b).One can also show easily that these equations imply the uniqueness of biproducts in astrong sense, whose formulation we leave as an exercise. We will denote the biproduct ofAi, i ∈ I, by

⊕i∈I Ai.

1.20 Definition Let C be an additive category and let f : A→ B a morphism in C.(a) A kernel of f is a pair (K, ι) consisting of an object K in C and a morphism

ι : K → A with f ◦ ι = 0 ∈ HomC(K,B) such that (K, ι) is universal with this property:

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For each pair (L, g) consisting of an object L of C and a morphism g : L → A withf ◦ g = 0, there exists a unique morphism h : L→ K in C with ι ◦ h = g:

Kι qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq A

f qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq B

∃!h ppppppppppppqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq��

qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq

g

L

Kernels are unique up to unique isomorphism and if (K, ι) is a kernel of f we often denoteK by ker(f).

(b) Dually, a cokernel of f is a pair (C, π) consisting of an object C in C and amorphism π : B → C such that (C◦, π◦) is a kernel of f ◦ in C◦. If (C, π) is a cokernel off , we often denote C by cok(f).

1.21 Remark (a) Kernels or cokernels do not always exist in an additive category.

(b) Let f : M → N be a morphism in RMod. Then the usual kernel K := {m ∈M | f(m) = 0} together with the inclusion map ι : K → M is a kernel of f in thesense of Definition 1.20. Similarly, C := N/im(f) together with the natural epimorphismπ : N → N/im(f) is a cokernel in the sense of Definition 1.20.

(c) Let f : A→ B be a morphism in an additive category C. If (K, ι) is a kernel of fthen ι is a monomorphism. Dually, if (C, π) is a cokernel of f then π is an epimorphism.

(d) Let f : A → B be a morphism in an additive category. Suppose that (K, ι) is akernel of f and (C, π) be a cokernel of f . Moreover, suppose that (D, ρ) is a cokernel of ιand that (L, κ) is a kernel of π. We claim that there exists a unique morphism h : D → Lin C such that f = κ ◦ h ◦ ρ.

Kι qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq A

f qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Bπ qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C

ρ

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq


κ

D p p p p p p p p p p p p p qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqh LIn order to show the existence of h, note that, since (D, ρ) is a cokernel of ι and f ◦ ι = 0,there exists a unique morphism g : D → B such that g ◦ρ = f . Moreover, since π ◦g ◦ρ =π ◦ f = 0 = 0 ◦ ρ and ρ is an epimorphism by Part (c), we obtain π ◦ g = 0. Thus, since(L, κ) is a kernel of π, there exists a unique morphism h : D → L such that κ ◦ h = g.

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Altogether, the morphism h satisfies κ◦h◦ρ = g◦ρ = f . Now assume that also h′ : D → Lsatisfies κ ◦ h′ ◦ ρ = f . Then κ ◦ h ◦ ρ = f = κ ◦ h′ ◦ ρ. Since ρ is an epimorphism andκ is a monomorphism by Part (c), we obtain h = h′, and the proof of the claim is nowcomplete.

1.22 Definition An abelian category is an additive category C satisfying the followingaxioms:

(Ab 1) Every morphism in C has a kernel and a cokernel.

(Ab 2) For every morphism f in the C, the morphism h in Remark 1.21(d) is an isomor-phism.

1.23 Example For every ring R, the category RMod is abelian. Note that, for an R-module homomorphism f : M → N , the morphism h in Remark 1.21(d), is given by thehomomorphism

M/ ker(f)→ f(M) , m+ ker(f) 7→ f(m) .

That this homomorphism is an isomorphism is the statement of the first isomorphismtheorem. Thus, Axiom (Ab 2) could be interpreted as saying that the first isomorphismaxiom must hold in an abelian category.

1.24 Definition Let F ,G : C→ D be covariant functors between categories C and D.(a) A natural transformation (or functorial morphism) ϕ : F → G is a family of mor-

phisms ϕC : F(C) → G(C), C ∈ Ob(C), satisfying the following condition: For all mor-phisms f : C → C ′ in C, the diagram

F(C) ϕC qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq G(C)

F(f)qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

G(f)

F(C ′) ϕC′ qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq G(C ′)

is commutative. Similarly, one defines natural transformations between contravariantfunctors.

(b) If ϕ : F → G is a natural transformation and every ϕC , C ∈ Ob(C), is an isomor-phism then ϕ is called a natural isomorphism (or functorial isomorphism). In this caseϕ−1 = (ϕ−1C ) : G → F is also a natural transformation. The functors F and G are callednaturally isomorphic (notation F ∼ G) if there exists a natural isomorphism ϕ : F → G.

11

1.25 Examples (a) For R ∈ Ric and n ∈ N, the map detR : GLn(R)→ U(R) is a grouphomorphism and the collection det = (detR) : GLn → U defines a natural transformationbetween the functors GLn, U : Ric → Gr.

(b) Fix a field K. The K-linear maps

iV : V → (V ∗)∗ , v 7→ (λ 7→ λ(v)) , V ∈ Ob(KMod) ,

form a natural transformation i : IdKMod → (−∗)∗ of functors KMod→ KMod. Restricted

to the categories Kmod of finite dimensional K-vector spaces, this is a natural isomor-phism.

1.26 Remark (a) Natural transformations can be composed. If F ,G,H : C → D arethree functors between categories C and D, and if ϕ : F → G and ψ : G → H are naturaltransformations then also ψ ◦ϕ := (ψC ◦ϕC)C∈Ob(C) : F → H is a natural transformation.

(b) Suppose that C and D are categories and that D is additive. If F ,G : C → Dare functors and if ϕ, ψ : F → G are natural transformations then also ϕ + ψ := (ϕC +ψC)C∈Ob(C) : F → G is a natural transformation.

1.27 Definition Let C and D be categories.

(a) A covariant functor F : C→ D is called an isomorphism between C and D if thereexists a functor G : D → C such that G ◦ F = IdC and F ◦ G = IdD. The categories Cand D are called isomorphic (notation C ∼= D) if there exists an isomorphism F : C→ D.Being isomorphic is an equivalence relation on categories.

(b) A covariant functor F : C → D is called an equivalence between C and D if thereexists a functor G : D→ C such that G ◦ F ∼ IdC and F ◦ G ∼ IdD. The categories C andD are called equivalent (notation C ' D) if there exists an equivalence F : C→ D. Beingequivalent is an equivalence relation on categories.

1.28 Examples (a) For any field K, the categories Kmod and Kmod◦ are equivalent via

the functor (−)∗ of taking K-duals.(b) Fix a field K and define C as the category with Ob(C) = N0, HomC(n,m) :=

Matm×n(K) for n,m ∈ N0, and composition given by matrix multiplication. Here, form = 0 or n = 0 we interpret Matm×n(K) as {0}. Then C is equivalent to the categoryKmod. The verification is left as an exercise.

12

Exercises for §1

1. (a) Show that in the category Set a morphism f : A → B is a monomorphism ifand only if it is injective. Show that it is an epimorphism if and only if it is surjective.Show the same for the category RMod.

(b) Show that in the category Ri the inclusion i : Z → Q is a monomorphism and anepimorphism, but not an isomorphism.

2. Let C be a category.

(a) Show that the composition of two monomorphisms is again a monomorphism.

(b) Show that the composition of two epimorphisms is again an epimorphism.

(c) Let C ∈ Ob(C). An automorphism of C is an isomorphism from C to C in C.Show that the set AutC(C) of automorphisms of C is a group under composition.

3. Recall that if R and S are rings then also R ⊗Z S is a ring whose multiplicationis determined by (r1 ⊗ s1)(r2 ⊗ s2) = r1r2 ⊗ s1s2, for r1, r2 ∈ R and s1, s2 ∈ S.

(a) Show that, for any R, S ∈ Ric, the tensor product R ⊗Z S, together with the ringhomomorphisms

ι1 : R→ R⊗Z S, r 7→ r ⊗ 1S, and ι2 : S → R⊗Z S, s 7→ 1R ⊗ s,

is a coproduct in the category Ric.

(b) Show that the same statement would be wrong for the category Ri.

4. Let C be an additive category and let f : A→ B be a morphism in C. Show thatf = 0 if and only if f = h ◦ g, where g : A→ 0 and h : 0→ B are the unique maps to andfrom the zero object.

5. Let Ai, i ∈ I, and Bj, j ∈ J , be two finite families of objects in an additivecategory C and let (A, (πAi ), (ι

Ai )) and (B, (π

Bj ), (ι

Bj )) be their respective biproducts.

(a) Show that

HomC(A,B)→⊕

(j,i)∈J×IHomC(Ai, Bj) , f 7→ (πBj ◦ f ◦ ιAi ) ,

is a group isomorphism.

(b) Assume further that Ck, k ∈ K, is a third finite family of objects in C and that(C, (πCk ), (ι

Ck )) is its biproduct. Let f : A→ B and g : B → C be morphisms in C and set

13

h := g ◦ f . Let (fji), (gkj) and (hki) be the families of morphisms constructed throughthe isomorphism in (a) applied to the various situations. Show that hki =

∑j∈J gkj ◦ fji

for all (k, i) ∈ K × I (matrix multiplication formula).

6. Let C be a category.(a) Consider the following category D. Its objects are diagrams of the form f : A→ B,

the morphisms between two diagrams f : A → B and f ′ : A′ → B′ are given by pairs ofmorphisms (g, h) with g : A→ A′ and h : B → B′ such that the diagram

Af qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq B

g

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqh

A′f ′ qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq B′

commutes, and composition of such morphisms is the obvious composition of such pairs.Show that this is again a category. Would such a construction also work for more generaldiagrams (not just two objects and an arrow between them)?

(b) Show that if C is additive then also D from (a) is additive. Generalize to otherdiagrams.

(c) Show that if C is abelian then also D from (a) is abelian. Generalize to otherdiagrams.

7. Let C be an additive category and let f : A→ B be a morphism in C.(a) Let ι : K → A be a kernel and let π : B → C be a cokernel of f . Show that ι is a

monomorphism and that π is an epimorphism.

(b) Show that f is a monomorphism (resp. epimorphism) if and only if 0 → A is akernel (resp. B → 0 is a cokernel) of f .

8. Let C and D be additive categories and let F : C → D be an additive functor.Suppose thatA1, . . . , An are objects of C and that (P, (πi), (ιi)) is a biproduct ofA1, . . . , Anin C. Show that (F(P ), (F(πi)), (F(ιi))) is a biproduct of F(A1), . . . ,F(An). (Additivefunctors preserve biproducts.)

9. Let C be a category. A subcategory of C is a category D with Ob(D) ⊆ Ob(C)and HomD(A,B) ⊆ HomC(A,B), for any A,B ∈ D, and with composition in D being thesame as in C. If additionally HomD(A,B) = HomC(A,B), for all A,B ∈ Ob(D), then D

14

is called a full subcategory of C. A skeleton of C is a full subcategory S of C such thatevery object of C is isomorphic to precisely one object in S.

(a) Show that ”equivalence of categories” is an equivalence relation.

(b) Show that any two skeletons of a category are isomorphic.

(c) Show that every skeleton of a category C is equivalent to C.

(d) Show that two categories are equivalent if and only if they have isomorphic skele-tons.

(e) Assume that D is a full subcategory of C such that every object of C is isomorphicto an object of D. Show that the inclusion functor D→ C is an equivalence.

10. A functor F : C→ D between categories C and D is called a full embedding if

HomC(A,B)→ HomD(F(A),F(B)) , f 7→ F(f),

is bijective for all A,B ∈ Ob(C). Show that in this case one has A ∼= B in C if and onlyif F(A) ∼= F(B) in D.

15

2 Simplicial and Semi-Simplicial Objects in a Cate-

gory

2.1 Definition Let ∆ denote the category whose objects are the sets

n := {0, . . . , n}

and whose morphisms are given by the increasing functions between them:

Hom∆(m,n) := {α : m→ n | 0 6 i 6 j 6 m⇒ α(i) 6 α(j)}, .

Composition in this category is the usual composition of functions. The subcategory of∆ with the same objects and only strictly increasing functions is denoted by ∆ 1)

(ii) σ(n−1)j ◦ σ

(n)i = σ

(n−1)i ◦ σ

(n)j+1 : n+ 1→ n− 1 (0 6 i 6 j 6 n− 1, n > 1)

(iii) σ(n)j ◦ δ

(n+1)i = δ

(n)i ◦ σ

(n−1)j−1 : n→ n (0 6 i < j 6 n, n > 1)

(iv) σ(n)j ◦ δ

(n+1)i = δ

(n)i−1 ◦ σ

(n−1)j : n→ n (0 6 j < i− 1 6 n, n > 1)

16

(v) σ(n)i ◦ δ

(n+1)i = σ

(n)i ◦ δ

(n+1)i+1 = 1n : n→ n (0 6 i 6 n, n > 0)

Proof This is a straightforward verification.

2.3 Proposition Let m,n ∈ N0 and let α ∈ Hom∆(m,n). Then there exist uniqueintegers s, t ∈ N0, 0 6 i1 < · · · < is 6 n and 0 6 j1 < · · · < jt 6 m− 1 such that

α = δis ◦ · · · ◦ δi1 ◦ σj1 ◦ · · · ◦ σjt . (2.3.a)

Moreover, there exists a unique injective morphism β and a unique surjective morphismγ in ∆ such that α = β ◦ γ. The expression in (2.3.a) is called the canonical factorizationof α. In particular, every α ∈ Hom∆ 1)

s(n)i := S(σ

(n)i ) : Sn → Sn+1 (0 6 i 6 n, n > 0)

In the following we assume the the objects of C are sets (with possibly extra structure)and the morphisms are functions (with possibly extra conditions). The elements of Snare called the n-simplices of S. The morphism d

(n)i = di is called the i− th face map and

the morphism s(n)i is called the i-th degeneracy map. If x ∈ Sn with n > 1 then d

(n)i (x) is

called the i-th face of x. An n-simplex is called degenerate if it is of the form x = σ(n+1)i (y)

for some y ∈ Sn+1 and some i ∈ {0, . . . , n+ 1}.Similarly, one defines a semi-simplicial object in C to be a contravariant functor

S : ∆< → C. One also uses the relevant terms defined above.

17

A morphism between (semi-) simplicial objects S and T in C is by definition a naturaltransformation f : S → T . We usually write

fn := fn : Sn → Tn (n > 0).

This way we obtain categories ∆(C) and ∆ 1 and morphisms s(n)i : Sn → Sn+1 for 0 6 i 6 n and n > 0 which satisfy thefollowing relations:

(i) d(n)i ◦ d

(n+1)j = d

(n)j−1 ◦ d

(n+1)i : Sn+1 → Sn−1 (0 6 i < j 6 n+ 1, n > 1)

(ii) s(n)i ◦ s

(n−1)j = s

(n)j+1 ◦ s

(n−1)i : Sn−1 → Sn+1 (0 6 i 6 j 6 n− 1, n > 1)

(iii) d(n+1)i ◦ s

(n)j = s

(n−1)j−1 ◦ d

(n)i : Sn → Sn (0 6 i < j 6 n, n > 1)

(iv) d(n+1)i ◦ s

(n)j = s

(n−1)j ◦ d

(n)i−1 : Sn → Sn (0 6 j < i− 1 6 n, n > 1)

(v) d(n+1)i ◦ s

(n)i = d

(n+1)i+1 ◦ s

(n)i = 1Sn : Sn → Sn (0 6 i 6 n, n > 0)

In fact, every contravariant functor S : ∆ → C provides such data. And conversely, ifsuch data are given we can define a contravariant functor S : ∆→ C as follows. For eachn ∈ N0 we set S(n) := Sn. And for any morphism α ∈ Hom∆(m,n) we first express α inits canonical factorization as in (2.3.a) and then set

S(α) := sjt ◦ · · · ◦ sj1 ◦ di1 ◦ · · · ◦ dis .

If α′ ∈ ∆(l,m) then S(α ◦ α′) = S(α′) ◦ S(α), since the relations in Proposition 2.2 allowto transform the composition of the canonical factorization of α and α′ into the canonicalfactorization of α ◦ α′, and the above relations allow to transform S(α′) ◦ S(α) into thesame expression.

Similarly, a semi-simplicial object in C amounts to the data of objects Sn ∈ Ob(C),n > 0, and morphisms d(n)i as above such that the relations in (i) above are satisfied.

If S, T : ∆ → C are two simplical objects in C then a morphism f : S → T is equiv-alent to the data of a sequence fn : Sn → Tn of morphisms in C satisfying the followingconditions:

18

(vi) d(n)i ◦ fn = fn−1 ◦ d

(n)i : Sn → Tn−1 (0 6 i 6 n, n > 1)

(vii) s(n)i ◦ fn = fn+1 ◦ s

(n)i : Sn → Tn+1 (0 6 i 6 n, n > 0)

If S and T are semi-simplicial objects in C only the relations (vi) need to hold.

2.6 Example (Posets) Let (X,6) be a partially order set (for short ‘poset’). We denoteby Γ

for all x, x′ ∈ X. Note that the second property implies the first, so that Poset< is asubcategory of Poset. It is now easy to see that the above constructions yield covariantfunctors

Γ< : Poset< → ∆

These maps form a morphism between the simplicial sets S(X) and S(Y ), since, for anyα : m→ n and any ϕ ∈ Sn(X), one has(

S(Y )(α) ◦ Sn(f))(ϕ) = f ◦ ϕ ◦∆(α) =

(Sm(f) ◦ S(X)(α)

)(ϕ)

It is now easy to verify that S(f ◦ g) = S(f) ◦ S(g), if g : Y → Z is another morphism inTop. Thus, we completed the definition of the covariant functor

S : Top→ ∆(Set) .

21

Exercises for §2

1. (a) Show that in the category ∆(Set) products and coproducts exist.

(b) Show that if A is an additive category then also ∆(A) is an additive category.

(c) Show that if A is an abelian category then also ∆(A) is an abelian category.

2. An (abstract) simplicial complex is a pair (K,Σ) consisting of a non-empty set Ktogether with a collection Σ of non-empty finite subsets σ of K such that Σ contains {x}for every element x ∈ K, and if σ ∈ Σ then every non-empty subset τ of σ also belongsto Σ. If σ ∈ Σ has cardinality n then it is called an n-simplex of (K,Σ).

(a) Define a natural notion of morphism of simplicial complexes so that they form acategory Simp.

(b) Discover natural covariant functors Poset → Simp and Simp → Poset. The com-position of these two functors (both ways) is called the barycentric subdivision of a posetor a simplicial complex.

(c) Define a functor Simp→ ∆(Set).

22

3 Chain Complexes and Homology

Throughout this section R denotes a ring.

3.1 Definition A chain complex in RMod is a two-sided infinite sequence

C• : · · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn+1 cn+1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn cn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn−1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

of objects and morphisms in RMod such that cn ◦ cn+1 = 0, for all n ∈ Z. The module Cnis called the term in degree n or the set of n-chains of C•, and the maps cn, n ∈ Z, arecalled the boundary maps of C•.

Suppose that also

D• : · · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dn+1 dn+1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dn dn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dn−1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

is a chain complex in RMod. A chain map f• : C• → D• is a family of R-module homo-morphisms fn : Cn → Dn, n ∈ Z, such that fn−1 ◦ cn = dn ◦ fn for all n ∈ Z, i.e., suchthat the diagram

C• : · · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn+1cn+1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn

cn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn−1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

f•


fn+1


fn


fn−1


D• : · · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dn+1dn+1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dn

dn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dn−1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

is commutative. Chain maps can be composed in the obvious way and we obtain acategory C(RMod). Similarly and more generally, we obtain a category C(A) of chaincomplexes in A, for any additive category A.

3.2 Remark (a) If C• and D• are chain complexes in RMod then the morphism setHomC(RMod)(C•, D•) is an abelian group under (f•+ g•)n := fn + gn, for n ∈ Z. The chaincomplex 0• all of whose terms are 0-modules is a zero object in C(RMod). For two chaincomplexes C• and D• in RMod we define their direct sum as

C• ⊕D• : · · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn+1 ⊕Dn+1(cn+1, dn+1)qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn ⊕Dn

(cn, dn)qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn−1 ⊕Dn−1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

One has a diagram

23

C• C•@@@qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

ιC•��

qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqπC•

C• ⊕D•

��

qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqιD•

@@@qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

πD•D• D•

of chain complexes and chain maps whose n-th degree components are the injectionsand projections related to the direct sum Cn ⊕ Dn. These chain maps clearly definethe biproduct relations from Proposition 1.17. Similarly, one constructs biproducts ofarbitrary finite families of chain complexes. Thus, the category C(RMod) is an additivecategory. More generally, whenever A is an additive category then also C(A) is an additivecategory.

(b) A chain subcomplex D• of a chain complex C• in C(RMod) consists of R-submodulesDn of Cn such that cn(Dn) ⊆ Dn−1, for all n ∈ Z. Then D• is a chain complex in its ownright with the restrictions of the maps cn, n ∈ Z, as boundary maps. The embeddingi• : D• → C• is a chain map. We write D∗ ⊆ C• to indicate that D• is a chain subcomplexof C•. For any D• ⊆ C• one can define the factor chain complex C•/D•

C•/D• : · · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn/Dn cn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn−1/Dn−1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

with cn(x + Dn) := cn(x) + Dn−1 for x ∈ Cn and n ∈ Z. The canonical epimorphismp• : C• → C•/D•, given by pn : Cn → Cn/Dn, x 7→ x + Dn, for x ∈ Cn and n ∈ Z, is achain map.

(c) Next we will establish that C(RMod) is an abelian category. Let f• : C• → D• bea chain map in C(RMod). We define the chain complexes

ker(f•) : · · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq ker(fn) cn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq ker(fn−1) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

im(f•) : · · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq im(fn) dn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq im(fn−1) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

and

cok(f•) : · · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dn/im(fn) dn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dn−1/im(fn−1) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

with the obvious boundary maps. It is not difficult to see that ker(f•) together with theinclusion map is a categorical kernel of f• and that cok(f•) together with the canonicalepimorphism is a categorical cokernel of f•. Thus, kernels and cokernels exist in C(RMod).Moreover, the axiom (Ab 2) holds in C(RMod), since the degree n part of the relevant

24

chain map h• : C•/ ker(f•) → im(f•) is the canonical isomorphism Cn/ ker(fn) → im(fn)of R-modules. Thus, C(RMod) is an abelian category. With the same arguments one canshow that whenever A is an abelian category then also C(A) is an abelian category.

3.3 Definition For a chain complex C• in RMod and n ∈ Z we call

Hn(C•) := ker(cn)/im(cn+1)

the n-th homology module of C• (or the homology of C• in degree n). Note that im(cn+1) ⊆ker(cn) ⊆ Cn for all n ∈ Z so that the above construction makes sense. The elementsin ker(cn) are called the n-cycles of C• and the elements in im(cn+1) are the called then-boundaries of C•.

If f• : C• → D• is a chain map in C(RMod) then we define

Hn(f•) : Hn(C•)→ Hn(D•) , x+ im(cn+1) 7→ fn(x) + im(dn+1) ,

for x ∈ ker(cn) and n ∈ Z. This map is well-defined and an R-module homomorphism.This way we obtain covariant fuctors

Hn : C(RMod)→ RMod (n ∈ Z).

More generally, if A is an abelian category, one obtains covariant functors Hn : C(A)→ A,for n ∈ Z.

3.4 Remark In this remark we will construct covariant functors

C

If Sn is the set of n-simplices of S and d(n)i : Sn → Sn−1 and s

(n)i : Sn → Sn+1 the face

maps and degeneracy maps then R[S] has n-simplices R[Sn] and structure maps R[d(n)i ]

and R[s(n)i ]. Moreover, if f : S → T is a morphism between simplicial sets, given by

fn : Sn → Tn, then R[f ] : R[S] → R[T ] is given by R[fn] : R[Sn] → R[Tn] on the level ofn-simplices, for n ∈ N0.

Similarly, one obtains a functor

R[−] : ∆

Every morphism f : M → N in ∆

(b) Recall from Example 2.6 the construction of the covariant functors

Γ< : Poset< → ∆

3.7 Remark For any additive category A, there is an isomorphism between Coc(A) andC(A) given by just replacing n with −n. Then the n-th cohomology becomes the (−n)-thhomology. Therefore, there would be no real need for this additional category. But it isused widely in the literature.

3.8 Definition Applying the duality functor −∗ = HomR(−, R) : RMod → ModR to achain complex induces a contravariant functor

−∗ = HomR(−, R) : C(RMod)→ Coc(ModR) .

Thus, for a chain complex

C• : · · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn cn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn−1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

its dual is given by

C∗• : · · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq HomR(Cn−1, R)

c∗n qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq HomR(Cn, R) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

f 7→ f ◦ cn

3.9 Remark Composing the covariant functors

C

Exercises for §3

1. (a) Let F be a field and let C• be chain complex in Fmod with non-zero terms inonly finitely many degrees. Show that∑

n∈Z(−1)n dimF Cn =

∑n∈Z

(−1) dimF Hn(C•) .

The above numbers are called the Euler characteristic of C•.

(b) Let C• be a chain complex in the category of finite abelian groups and assumethat Cn = 0 for all but finitely many n ∈ Z. Show that∏

n∈Z|Cn|(−1)

n

=∏n∈Z|Hn(C•)|(−1)

n

.

(c) Let C• be a chain complex in the category of finite generated abelian groups andassue that Cn = 0 for all but finitely many n ∈ Z. Show that∑

n∈Z(−1)nrkZCn =

∑n∈Z

(−1)rkZHn(C•) .

2. Let R be a ring.

(a) Compute the singular chain complex C•(X;R) for the one-point space X = {x} ∈Ob(Top) and compute the homologies.

(b) Compute the chain complexes C

(b) Let X be a poset and assume that a finite group acts via poset automorphismson X. Show again the associated chain complexes C•(X;R) and C

<• (X;R) are chain

complexes in RGMod.

5. Let A be an abelian category, let C• be a chain complex in A, and let n ∈ Z. Showthat the following are equivalent:

(i) Hn(C•) = 0.

(ii) For all diagrams

A

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqa

Cn+1cn+1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn

cn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn−1

bqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

B

one has: cn ◦ a = 0 and b ◦ cn+1 = 0 implies b ◦ a = 0.

6. Let R be a commutative ring, let C• be a chain complex in RMod and let D• be

its dual. Further, let Zn, Bn, Hn denote the sets of n-cycles, n-boundaries, n-homologyclasses fo C• and similarly let Z

n, Bn, Hn denote the sets of n-cocycles, n-coboundaries,n-cohomogy classes of D•.

(a) Show that the bilinear form

〈−,−〉 : Dn × Cn → R , (f, x) 7→ f(x) ,

induces a bilinear form〈−,−〉 : Hn ×Hn → R .

(b) Assume that R is a field. Show that the bilinear form Hn ×Hn → R from (a) isnon-degenerate in both arguments. In particular, if Hn or H

n is finite-dimensional thenHn and H

n have the same dimension.

7. For a finite poset X = {x1, . . . , xn} conisder the set G(X) of elements (a1, . . . , an) ∈[0, 1]n ⊂ Rn with the property that a1 + · · · an = 1 and that the indices of the non-zero components form a totally ordered subset of X. Show that this construction can beextended to a covariant functor G : poset→ Top on the category of finite posets (what does

31

this functor on morphisms?). The functor G is called the geometric realization functor.Take your favorite poset and try to visualize its geometric realization.

8. Show that each of the functors in the sequence

Top S qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq ∆(Set)R[−] qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq RMod C• qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C(RMod) Hn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq RMod

preserves coproducts. In particular, Hn(X∐Y ;R) ∼= Hn(X;R) ⊕ Hn(Y ;R), for any

topological spaces X and Y .

32

4 Chain Complexes and Homotopy

Throughout this section R denotes a ring.

4.1 Definition Two chain maps f•, g• : C• → D• in C(RMod) are called homotopic (no-tation f• ∼ g•) if there exists a family R-module homomorphisms hn : Cn → Dn+1 suchthat

fn − gn = dn+1 ◦ hn + hn−1 ◦ cn ,

for all n ∈ Z. In this case, h• = (hn)n∈Z is called a homotopy from f• to g• (notationf•∼

h•g•).

C• : · · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn+1cn+1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn

cn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn−1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

f•

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

g• fn+1


gn+1

��

��qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

hnfn


gn

��

��qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

hn−1fn−1


gn−1

D• : · · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dn+1dn+1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dn

dn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dn−1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

These definitions carry over without difficulty when RMod is replaced by any additivecategory A.

4.2 Remark (a) ∼ is an equivalence relation on HomC(RMod)(C•, D•). For any chain mapf • : C• → D• we call its equivalence class the homotopy class of f• and denote it by f•.

(b) If f•∼h•g• and f

′•∼h′•g′• in HomC(RMod)(C•, D•) then f•+f

′• ∼h•+h′•

g•+g′•. In particular,

f• ∼ g• if and only if f• − g• ∼ 0•. Thus, the homotopy class 0• of the 0-chain map is asubgroup of HomC(RMod)(C•, D•) and we have

HomC(RMod)(C•, D•)/ ∼= HomC(RMod)(C•, D•)/O• .

(c) If f•∼h•g• : C• → D• and f ′•∼

h′•g′• : D• → E• then f ′• ◦ f• ∼ g′• ◦ f ′• : C• → E• via

the homotopy f ′ ◦ h + h′ ◦ g. Thus, composition of homotopy classes of chain maps iswell-defined: f ′• ◦ f• := f ′• ◦ f•. We obtain a category

K(RMod) ,

33

the homotopy category of RMod, whose objects are the chain complexes in RMod, whosemorphisms are the homotopy classes of chain maps and whose composition is the com-position of homotopy classes. Note that K(RMod) is again an additive category by usingthe same biproducts as in C(RMod). There is an obvious additive functor

· : C(RMod)→ K(RMod) , C• 7→ C• , f• 7→ f• .

(d) More generally, if A is an additive category, we can define the homotopy categoryK(A).

4.3 Proposition Let f• ∼ g• : C• → D• be homotopic chain maps in C(RMod). Then

Hn(f•) = Hn(g•) .

Thus, one obtains a functor Hn : K(RMod)→ RMod such that the diagram

C(RMod)Hn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq RMod

·qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq ��


Hn

K(RMod)

is commutative.

Proof Let h• be a homotopy from f• to g•. Then

fn − gn = dn+1 ◦ hn + hn−1 ◦ cn ,

for all n ∈ Z. Now let n ∈ Z and denote by BCn ⊆ ZCn ⊆ Cn and by BDn ⊆ ZDn ⊆ Dn theset of n-boundaries and n-cocycles in C and D, respectively. For any x ∈ ZCn we have

(Hn(f•))(x+BCn ) = fn(x) +B

Dn = gn(x) + dn+1(hn(x)) + hn−1(cn(x)) = gn(x) +B

Dn

= (Hn(g•))(x+BCn ) ,

since dn1(hn(x)) ∈ BDn and cn(x) = 0.

4.4 Definition A chain map f• : C• → D• in C(RMod) is called a homotopy equivalenceif there exists a chain map g• : D• → C• such that g• ◦ f• ∼ 1C• and f• ◦ g• ∼ 1D• . This

34

is equivalent to saying that f• is an isomorphism in the category K(RMod). Thus, g• isdetermined by f• up to homotopy and it is called a homotopy inverse of f•.

Two chain complexes C• and D• are called homotopy equivalent if there exists a ho-motopy equivalence f• : C• → D•. In this case we write C• ' D•. This is equivalent tosaying that C• ∼= D• in K(RMod).

The same definitions are valid when RMod is replaced by an additive category A.

4.5 Proposition If C• and D• are homotopy equivalent chain complexes in RMod thenHn(C•) and Hn(D•) are isomorphic R-modules for all n ∈ Z.

Proof Let f• : C• → D• be a homotopy equivalence and let g• : D• → C• be a homotopyinverse to f•. Then, using Proposition 4.3, we obtain

Hn(g•) ◦Hn(f•) = Hn(g• ◦ f•) = Hn(1C•) = 1Hn(C•) ,

and similarlyHn(f•)◦Hn(g•) = 1Hn(D•). ThusHn(f•) : Hn(C•)→ Hn(D•) andHn(g•) : Hn(D•)→Hn(C•) are inverse R-module isomorphisms.

The following general lemma will be used to construct homotopies between chain mapsthat arise through the simplicial chain complex construction from Section 3.

4.6 Lemma Let S and T be semi-simplicial sets, i.e., objects in ∆ 1)

(ii) d(n+1)i ◦ h

(n)j = h

(n−1)j ◦ d

(n)i−1 : Sn → Tn (0 6 j < i− 1 6 n, n > 1)

(iii) d(n+1)i ◦ h

(n)i = d

(n+1)i ◦ h

(n)i−1 : Sn → Tn (1 6 i 6 n, n > 1)

For n ∈ N0 set

fn := d(n+1)0 ◦ h

(n)0 : Sn → Tn and gn := d

(n+1)n+1 ◦ h(n)n : Sn → Tn .

Further, for n ∈ N0 define

hn :=n∑i=0

(−1)ih(n)i : R[Sn]→ R[Tn+1] .

35

(a) The maps fn, n ∈ N0, and the maps gn, n ∈ N0, define morphisms f, g : S → T in∆

Proof Let h : I × X → Y be a homotopy between f and g, i.e., h is continuous andh(0, x) = f(x) and h(1, x) = g(x) for all x ∈ X. Here, I = [0, 1] denotes the unit intervallin R. Define continuous maps

i0 : X → I ×X , x 7→ (0, x) , and i1 : X → I ×X , x 7→ (1, x) .

Then f = h ◦ i0 and g = h ◦ i1. By functoriality and by Remark 4.2(c), it suffices to showthat the chain maps C•(i0;R) and C•(i1;R) are homotopic.

In order to show this, define for n > 0 and 0 6 i 6 n the continuous map

u(n)i : ∆n → I , (t0, . . . , tn) 7→ t0 + · · ·+ ti .

Here ∆n denotes the n-dimensional standard simplex in Rn+1, see Example 2.7. We willaslo use the maps

∆(σ(n)i ) : ∆n+1 → ∆n , (t0, . . . , tn+1) 7→ (t0, . . . , ti + ti+1, . . . , tn+1) ,

see Example 2.7. Finally, for n > 1 and 0 6 i 6 n, define

h(n)i : HomTop(∆n, X)→ HomTop(∆n+1, I ×X) , ϕ 7→ (u

(n+1)i , ϕ ◦∆(σ

(n)i )) .

We want to apply Lemma 4.6 with S = S(X) and T := S(I ×X). Note that

(d(n+1)i ◦ h

(n)j )(ϕ) =

(u

(n+1)j ◦∆(δ

(n+1)i ), ϕ ◦∆(σ

(n)j ) ◦∆(δ

(n+1)i )

),

for all n ∈ N0, 0 6 i 6 n+ 1, and 0 6 j 6 n, and that

(h(n+1)k ◦ d

(n)l )(ϕ) =

(u

(n)k , ϕ ◦∆(δ

(n)l ) ◦∆(σ

(n−1)k )

),

for all n ∈ N, 0 6 k 6 n − 1, and 0 6 l 6 n. It is now straightforward to check therelations in Lemma 4.6. In fact, they follow from the relations in Proposition 2.2 and fromrewriting u

(n+1)j ◦∆(δ

(n+1)i ) = u

(n)j′ for appropriate j

′. Using Lemma 4.6 it now suffices toshow that

d(n+1)0 ◦ h

(n)0 = Cn(i0;R) and d

(n+1)n+1 ◦ h(n)n = Cn(i1;R) .

Note that, for ϕ ∈ Sn = HomTop(∆n, X), we have

Cn(i0;R)(ϕ) = i0 ◦ ϕ and Cn(i1;R)(ϕ) = i1 ◦ ϕ

Finally,

(d(n+1)0 ◦ h

(n)0 )(ϕ) =

(u

(n+1)0 ◦∆(δ

(n+1)0 ), ϕ ◦∆(σ

(n)0 ◦ δ

(n+1)0 )

)= (0, ϕ) = i0 ◦ ϕ ,

37

since u(n+1)0 ◦∆(δ

(n+1)0 ) : ∆n → I is the constant map with value 0, and since σ0(n)◦δ

(n+1)0 =

1n by Proposition 2.2(v). And similarly,

(d(n+1)n+1 ◦ h(n)n )(ϕ) =

(u(n+1)n ◦∆(δ

(n+1)n+1 ), ϕ ◦∆(σ(n)n ◦ δ

(n+1)n+1 )

)= (1, ϕ) = i1 ◦ ϕ ,

since u(n+1)n ◦∆(δ(n+1)n+1 ) : ∆n → I is the constant function with value 1 and σ(n)n ◦δ

(n+1)n+1 = 1n

by Proposition 2.2(v). This completes the proof of the theorem.

We turn back to general properties of chain complexes related to the notion of homo-topy. We formulate the next lemma for chain complexes in RMod, but it would also holdfor chain complexes in an abelian category A.

4.8 Lemma Let C• be a chain complex in RMod. Write 0 ⊆ Bn ⊆ Zn ⊆ Cn for thesubmodules of n-boundaries and n-cocycles of C•. The following are equivalent:

(i) For each n ∈ Z, there exist R-submodules Un and Vn of Cn such that Zn⊕Un = Cnand Bn ⊕ Vn = Zn (and hence Cn = Un ⊕ Vn ⊕Bn with Vn ∼= Hn(C•)).

(ii) The chain complex C• is homotopy equivalent to the chain complex

· · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn+1(C•) 0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn(C•) 0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn−1(C•) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

(iii) The chain complex C• is homotopy equivalent to a chain complex whose boundarymaps are 0.

(iv) There exist R-linear maps sn : Cn → Cn+1 such that cn ◦ sn−1 ◦ cn = cn for alln ∈ Z.

(v) For each n ∈ Z, there exist submodules Un and Wn of Cn such that Cn = Zn⊕Un =Bn ⊕Wn.

Proof (i)⇒(ii): The modules Vn and the modules Un ⊕ Bn form subcomplexes of C•whose direct sum is equal to C•. Note that V• is isomorphic to the chain complex in(ii). We will show that the inclusion map i• : V• → C• is a homotopy equivalence withhomotopy inverse p• : C• → V•, the projection map with respect to the decompositionsCn = Un ⊕Bn ⊕ Vn, n ∈ Z. Clearly, p• ◦ i• = 1V• . Next define

hn : Cn = Un ⊕ Vn ⊕Bn → Un+1 ⊕ Vn+1 ⊕Bn+1 = Cn+1

such that hn vanishes on Un⊕Vn and coincides on Bn with the inverse of the isomorphisimUn+1 → Bn. Then, it is easy to verify that

hn−1 ◦ cn + cn+1 ◦ hn = 1Cn − in ◦ pn

38

on each of the three direct summands.

(ii)⇒(iii): This is trivial.(iii)⇒(iv): Let D• be a chain complex with boundary maps equal to 0. and let

f• : C• → D• be a homotopy equivalence with homotopy inverse g• : D• → C•. Moreover,let h• be a omotopy from 1C• to g• ◦ f•. Then

cn ◦ hn−1 ◦ cn = cn(1Cn − gn ◦ fn − cn+1 ◦ hn) = cn − cn ◦ gn ◦ fn = cn − gn ◦ dn ◦ fn = cn ,

since dn = 0. Thus, sn := hn has the required property.

(iv)⇒(v): For n ∈ Z we have

cn ◦ sn−1 ◦ cn ◦ sn−1 = cn ◦ sn−1 and sn−1 ◦ cn ◦ sn−1 ◦ cn = sn−1 ◦ cn .

Thus

Cn−1 = ker(cn ◦ sn−1)⊕ im(cn ◦ sn−1) and Cn = ker(sn−1 ◦ cn)⊕ im(sn−1 ◦ cn) .

Further, we have

im(cn ◦ sn−1) ⊆ im(cn) = im(cn ◦ sn−1 ◦ cn) ⊆ im(cn ◦ sn−1)

andker(sn−1 ◦ cn) ⊆ ker(cn) = ker(cn ◦ sn−1 ◦ cn) ⊆ ker(sn−1 ◦ cn) .

This implies im(cn ◦ sn−1) = Bn−1 and ker(sn−1 ◦ cn) = Zn, and we obtain

Cn−1 = ker(cn ◦ sn−1)⊕Bn−1 and Cn = Zn ⊕ im(sn−1 ◦ cn)

Now, (v) follows with Wn := ker(cn+1 ◦ sn) and Un := im(sn−1 ◦ cn).(v)⇒(i): Set Vn := Wn ∩ Zn. Then Zn = Bn ⊕ Vn.

4.9 Definition A chain complex C• in RMod is called

(a) exact in degree n, if Hn(C•) = 0;

(b) exact (or acyclic), if Hn(C•) = 0, for all n ∈ Z;(c) split, if it satisfies the equivalent conditions in Lemma 4.8; and

(d) contractible, if C• ' 0•.

39

4.10 Lemma For a chain complex C• in RMod the following are equivalent:

(i) The chain complex C• is contractible.

(ii) The chain complex C• is split and exact.

(iii) There existR-module homomorphisms tn : Zn → Cn+1, n ∈ Z, such that cn+1◦tn =1Zn , for all n. Here Zn := ker(cn).

(iv) One has 1C• ∼ 0.

Proof (i)⇒(ii): (i) implies the condition in Lemma 4.8(iii). Thus, C• is split. Since C•is homotopy equivalent to the zero chain complex, we have Hn(C•) ∼= Hn(0•) = 0, for alln ∈ Z. Thus, C• is exact.

(ii)⇒(iii): Since C• is split, there exist an R-module homomorphisms sn : Cn → Cn+1,n ∈ N, with

cn ◦ sn−1 ◦ cn = cn , (4.10.a)

for all n ∈ Z. Let tn denote the restriction of sn to Zn. Since C• is exact, we havecn(Cn) = Zn−1 and Equation (4.10.a) implies that cn ◦ tn−1 is the identity on Zn−1.

(iii)⇒(iv): From (iii) we obtain Cn = im(tn−1) ⊕ Zn, for all n ∈ Z. We definehn : Cn → Cn+1 by hn(x+y) := tn(y), for x ∈ im(tn−1) and y ∈ Zn. Then, for x = tn−1(z),with z ∈ Zn−1, we have

(cn+1 ◦ hn + hn−1 ◦ cn)(x) = hn−1(cn(tn−1(z))) = hn−1(z) = x

and for y ∈ Zn we have

(cn+1 ◦ hn + hn−1 ◦ cn)(y) = cn+1(hn(y)) = cn+1(tn(y)) = y .

Thus, cn+1 ◦ hn + hn−1 ◦ cn) = 1Cn , for all n ∈ Z, and 1C• ∼ 0.(iv)⇒(i): Let f• : C• → 0• and g• : 0• → C• be the zero chain maps. Then g• ◦ f• =

0 ∼ 1C• by (iv), and f• ◦ g• = 0 = 10• . Thus, C• ' 0•.

Based on the equivalence of the conditions (i) and (ii) in the last proposition, one callscontractible chain complexes also split exact chain complexes.

4.11 Example Let M be a left R-module. Then the chain complex

· · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq 0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq M 1M qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq M qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq 0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

is contractible. The verification is left as an exercise.

40

Exercises for §4

1. Prove the statement in Example 4.11.

2. Consider the functor F : RMod → C(RMod) which assigns to every R-module Mthe chain complex F(M) with M in degree 0 and the zero modules in all other degrees,and which assigns to a module homomorphism f : M → N the chain map F(f) : F(M)→F(N) which is defined by F(f)0 = f and F(f)n := 0 if n 6= 0.

(a) Show that the functor F is a full embedding (i.e., bijective on morphism sets).(b) Show that the functor

F : RMod F qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C(RMod) · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq K(RMod)

is a full embedding.

3. Let C• and D• be chain complexes in RMod. Show the following:

(a) For every n ∈ Z one has: C• ⊕ D• is exact in degree n if and only if C• and D•are exact in degree n.

(b) C• ⊕D• is acyclic if and only if C• and D• are acyclic.(c) C• ⊕D• is split if and only if C• and D• are split.(d) C• ⊕D• is contractible if and only if C• and D• are contractible.

4. One defines the reduced singular chain complex C̃(X;R) of a topological space Xas part of the functor

C̃(−;R)Top→ C(RMod)

which assigns to a topological space X the chain complex

· · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn(X;R) ∂n qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn−1(X;R) ∂n−1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · · ∂1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C0(X;R) ∂0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq R qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq 0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

where ∂0(ϕ) := 1 for every singular 0-simplex ϕ : ∆0 → X. On morphisms this functor isdefined the same way as the functor C(X;R), except that the chain map in degree −1 isthe identity on R.

(a) Show that the reduced chain complex of the topological space X = {x} is con-tractible.

(b) Let X be a contractible topological space, i.e., X is homotopy equivalent to aone-point space. Show that the reduced chain complex of X is contractible.

41

5. Let (X,6) be a partially ordered set and let (X◦,6◦) be the opposite poset, i.e.,X◦ = X and x 6◦ y if an only if y 6 x. Show that C(X;R) and C(X◦;R) are isomorphicchain complexes.

6. (a) Let (X 6) be a partially ordered set. Show that the reduced chain complex

· · · qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn(X;R) ∂n qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn−1(X;R) ∂n−1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · · ∂1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C0(X;R) ∂0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq R qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq 0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

of X, with ∂0(x) := 1, for x ∈ X, defines a functor C̃(−;R) : poset→ C(RMod).(b) Let (X,6) be a poset which has a smallest element x∗, i.e., x∗ 6 x for all x ∈ X.

Show that the reduced chain complex C̃(X;R) is contractible.

7. Let (X,6) be a poset and consider the two chain complexes C

5 The Long Exact Homology Sequence

Again, throughout this section R denotes a ring.

5.1 Definition A short exact sequence in RMod is a sequence

C : 0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C ′ i qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cp qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C ′′ qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq 0

which is exact at C ′, C, and C ′′. A morphism between two short exact sequences C andD in RMod is a triple (f

′, f, f ′′) of morphisms in RMod making the following diagramcommutative:

C : 0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C ′

i qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cp qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C ′′ qqqqqqqqqqq

qqqqqqq qqqqqqqqqqqqqqqqqq 0


f ′


f


f ′′

D : 0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq D′

j qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dq qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq D′′ qqqqqqqqqqq


This way one obtains an additive category Ses(RMod). Similarly one obtains an additivecategory Ses(A) for any abelian category A.

5.2 Lemma (Snake Lemma) Let

C ′i qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C

p qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C ′′ qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq 0


f ′


f


f ′′

0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq D′

j qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dq qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq D′′

be a diagram in RMod with exact rows. Define a map ∂ : ker(f′′) → cok(f ′) as follows:

For c′′ ∈ ker(f ′′) choose c ∈ C such that p(c) = c′′. Then g(f(c)) = f ′′(p(c)) = f ′′(c′′) = 0and there exists d′ ∈ D′ such that j(d′) = f(c). Now set ∂(c′′) := d′ + im(f). Then thedotted snake sequence in the commutative diagram

43

ker(f) ĩp p p p p p p p p p pqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq ker(f) p̃p p p p p p p p p pqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq ker(f ′′) ∂p p p p p p p p p p p p p p p p p p p p p p p p p p p qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq






C ′i qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C

p qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C ′′


f ′


f

ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq


f ′′

D′j qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq D

q qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq D′′

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq∂p p p p p p p p p p p p p p p p p p p p p p p p p p p qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq cok(f) j̄p p p p p p p p p p qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq cok(f) q̄p p p p p p p p p qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq cok(f ′′)

is exact.

Proof It is easy to see that i, p, j and q induce the indicated maps between kernels andcokernels. Next we show that ∂ is well-defined. We show that d′ + im(f) ∈ cok(f) doesnot depend on the choice of c. In fact, if also c̃ ∈ C is such that p(c̃) = c′′ = p(c), thenc̃− c ∈ ker(p) = im(i) so that c̃− c = i(c′) for some c′ ∈ C ′. Let d̃′ ∈ D′ be the elementwith j(d̃′) = f((̃c)). Then j(f ′(c′)) = f(i(c′)) = f(c̃) = f(c) = j(d̃′) = j(d′) = j(d̃′ − d′).Since j is injective, d̃′− d′ = f ′(c′). Thus, d̃′+ im(f ′) = d′+ im(f ′), and ∂ is well-defined.It is straightforward to verify that ∂ is an R-module homomorphism.

Exactness at ker(f): Since p̃ ◦ ĩ = 0, we have im(̃i) ⊆ ker(p̃). Assume that c ∈ ker(f)and that p̃(c) = 0. Then p(c) = 0 and there exists c′ ∈ C with i(c′) = c. Moreover,j(f ′(c′)) = f(i(c′)) = f(c) = 0. Since j is injective, this implies f ′(c′) = 0. Thus,c′ ∈ ker(f ′).

Exactness at ker(f ′′): First we show that ∂ ◦ p̃ = 0. Let c ∈ ker(f). Then, by theconstruction of ∂, applied to p̃(c) = p(c), we obtain ∂(p̃(c)) = 0. Thus, im(p̃) ⊆ ker(δ).Conversely, let c′′ ∈ ker(f ′′) with ∂(c′′) = 0. Let c ∈ C be such that p(c) = c′′ andlet d′ ∈ D′ be such that j(d′) = f(c). Then ∂(c′′) = 0 implies that d′ ∈ im(f ′), andd′ = f ′(c′) for some c′ ∈ C ′. Now we have f(i(c′)) = j(f ′(c′)) = j(d′) = f(c), so thatc−i(c′) ∈ ker(f). Further we have p̃(c−i(c′)) = p(c)−p(i(c′)) = p(c) = c′′ and c′′ ∈ im(p̃).This implies ker(∂) ⊆ im(p̃).

We leave the rest of the proof as exercise.

44

5.3 Remark Assume that situation of the Snake Lemma.

(a) If i is injective then also ĩ is injective. If q is surjective then also q̄ is surjective.

(b) The snake lemma holds in an arbitrary abelian category A. The definition of ∂and also the proof is more involved. See for example [K-S, Lemma 12.1.1].

5.4 Theorem (Long Exact Homology Sequence) Let

C : 0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C ′•i• qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C•

p• qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C ′′• qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq 0

be a short exact sequence in C(RMod). Then there exists a long exact sequence

· · · ∂C•n+1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn(C

′•)Hn(i•) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn(C•)

Hn(p•) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn(C′′• )

∂Cn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq

∂C•n qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn−1(C′•)Hn−1(i•) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn−1(C•)

Hn−1(p•) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqHn−1(C′′• )

∂Cn−1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

in RMod. If also

D : 0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq D′•j• qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq D•

q• qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq D′′• qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq 0

is a short exact sequence in C(RMod) and if

C : 0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C ′

i qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cp qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C ′′ qqqqqqqqqqq



f ′


f


f ′′

D : 0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq D′

j qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dq qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq D′′ qqqqqqqqqqq


is a morphism of short exact sequences, then one obtains a commutative diagram

· · · ∂C•n+1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn(C

′•)Hn(i•) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn(C•)

Hn(p•) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn(C′′• )

∂Cn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn−1(C′•)Hn−1(i•) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

Hn(f′•)


Hn(f•)


Hn(f′′• )


Hn−1(f′•)


· · · ∂D•n+1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn(D

′•)Hn(j•) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn(D•)

Hn(q•) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn(D′′•)

∂Dn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn−1(D′•)Hn−1(j•) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq · · ·

Proof For every n ∈ Z, the commutative diagram

45

C ′n/B′n

īn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn/Bnp̄n qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C ′′n/B

′′n

qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq 0

c̄′n


c̄n


c̄′′n


0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Z ′n−1

in−1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Zn−1pn−1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Z ′′n

in RMod has exact rows. Now apply the Snake Lemma. The commutativity of the lastdiagram in the theorem is easily verified.

5.5 Definition Let S be a semi-simplicial (resp. simplicial) set. A semi-simplicial subset(resp. simplicial subset) T of S consists of subsets Tn ⊆ Sn, n ∈ N0 which are stable underthe maps d

(n)i (resp. under the maps d

(n)i and the maps s

(n)i ). In this case we write T ⊆ S.

If T and U are (semi-) simplicial subsets of S. then also the subsets Tn ∩ Un, n ∈ N0,and Tn ∪ Un, n ∈ N0, form (semi-) simplicial subsets of S, denoted by T ∩ U and T ∪ U .

5.6 Proposition Let T and U be semi-simplicial subsets of the semisimplicial set S.Then there is a short exact sequence

0→ C•(T ∩ U ;R) i• qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C•(T ;R)⊕ C•(U ;R) p• qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C•(T ∪ U ;R)→ 0 ,

where

i• =

(C•(iT ;R)C•(iU ;R)

)and p• =

(C•(jT ;R) −C•(jU ;R)

)and iT : T ∩ U → T , iU : T ∩ U → U , jU : U → T ∪ U , and jT : T → T ∪ U denote thecanonical morphisms.

Proof Fix n ∈ N0. We have to show that

0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq R[Tn ∩ Un] in qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq R[Tn]⊕R[Un] pn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq R[Tn ∪ Un] qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq 0

is exact, where in(x) = (x, x) and pn(t, u) = t − u, for x ∈ Tn ∩ Un, t ∈ Tn and u ∈ Un.Clearly, the map in is injective and the map pn is surjective. It is also clear that pn◦in = 0.Thus, we only need to show that ker(pn) is contained in im(in). So let a =

∑t∈Tn αtt and

b =∑u∈Un βuu be such that (a, b) ∈ ker(pn). Then

0 = pn(a, b) =∑t∈Tn

αtt−∑u∈Un

βuu =∑

t∈Tnr(Tn∩Un)αtt−

∑u∈Unr(Tn∩Un)

βuu+∑

s∈Tn∩Un(αs−βs)s .

46

Since Tn ∪ Un =[Tn r (Tn ∩ Un)

]∪[Un r (Tn ∩ Un)

]∪[Tn ∩ Un

]is a disjoint union, we

obtain αt = 0 for all t ∈ Tn r (Tn ∩ Un), βu = 0 for all u ∈ Un r (Tn ∩ Un) and αs = βsfor all s ∈ Tn ∩ Un. Thus, (a, b) = in(c), where c =

∑s∈Tn∩Un αss =

∑s∈Tn∩Un βss.

5.7 Definition Let X be a topological space and let U be an open covering of X, i.e., aset of open subsets of X with the property that ∪U∈UU = X. Define the simplicial setSU(X)

SU(X) :=⋃U∈U

S(U) ⊆ S(X) ,

where we consider S(U) as simplicial subset of S(X) by mapping a singular simplexϕ : ∆n → U to the singular simplex iU ◦ ϕ, where iU : U → X is the inclusion map.Moreover, we define

CU• (X;R) := C•(SU(X);R) and HUn (X;R) := Hn(C

U• (X;R)) ,

for n ∈ N0.

5.8 Theorem Let U be an open covering of a topological spaceX and denote by i : SU(X)→S(X) the canonical map of simplicial sets. Then

C•(i;R) : CU• (X;R)→ C•(X;R)

is a homotopy equivalence. In particular, C•(i;R) induces isomorphisms HUn (X;R) →

Hn(X;R) for all n ∈ N0.

We omit the proof of the above theorem here and refer to [S, Theorem 4.5.14] andthe preceding pages for a detailed proof. The proof uses the compactness of the standardsimplices ∆n and the barycentric subdivisions of the standard simplices.

5.9 Theorem (Mayer-Vietoris Sequence) let X be a topological space and let U andV be two open subsets of X with U ∪ V = X. Then there exists an exact sequence

· · · → Hn(U ∩ V ;R)→ Hn(U ;R)⊕Hn(V ;R)→ Hn(X;R)→→ Hn−1(U ∩ V ;R)→ Hn−1(U ;R)⊕Hn−1(V ;R)→ Hn−1(X;R)→ · · ·

· · · → H0(U ∩ V ;R)→ H0(U ;R) ∩H0(V ;R)→ H0(X;R)→ 0

called the Mayer-Vietoris sequence of X with respect to U and V .

Proof This follows immediately from Proposition 5.6, Theorem 5.4, and Theorem 5.8.

47

5.10 Corollary Let F be a field, let k ∈ N0 and let Sk := {x ∈ Rk+1 | |x| = 1} denotethe k-dimensional sphere. Then

Hn(Sk;F ) ∼=

F ⊕ F if n = k = 0,F if n ∈ {0, k} and 0 6= k,0 otherwise, i.e. if n /∈ {0, k}.

Proof We prove the theorem by induction on k.

k = 0 : We have S0 = {−1, 1} and therefore C•(S0;F ) = C•({−1};F ) ⊕ C•({1};F ).We also know the singular chain complex of a one-point space and obtain H0(S

0;F ) =F ⊕ F and hn(S0;F ) = 0 if n > 0.

k − 1 : Consider the open subsets U := {(x1, x2) ∈ S1 | x2 6= −1} and V := {(x1, x2) |x2 6= 1} of S1. Clearly, U ∪ V = S1, U ∩ V is homotopy equivalent to S0, and U andV are homotopy equivalent to one-point spaces. Thus, by our results for k = 0, theMayer-Vietoris sequence has the form

· · · ∂n+1−→ Hn(U ∩ V ;F )︸︷︷︸=0

in−→ Hn(U ;F )⊕Hn(V ;F )︸︷︷︸=0

pn−→ Hn(S1;F )∂n−→

∂n−→ Hn−1(U ∩ V ;F )︸︷︷︸=0

in−1−→ Hn−1(U ;F )⊕Hn−1(V ;F )︸︷︷︸=0

pn−1−→ Hn−1(S1;F )∂n−1−→ · · ·

· · · ∂2−→ H1(U ∩ V ;F )︸︷︷︸=0

i1−→ H1(U ;F )⊕H1(V ;F )︸︷︷︸=0

p1−→ H1(S1;F )∂1−→

∂1−→ H0(U ∩ V ;F )︸︷︷︸∼=F⊕F

i0−→ H0(U ;F )⊕H0(V ;F )︸︷︷︸∼=F⊕F

p0−→ H0(S1;F ) −→ 0

and it is exact. Since S1 is path-connected, H0(S1;F ) ∼= F . Since p0 is surjective, we

obtain dimF ker(p0) = 1. This implies dimF im(i0) = 1 and further dimF ker(i0) = 1 anddimF im(∂1) = 1. Again, the exactness of the Mayer-Vietoris sequence implies

dimF H1(S1;F ) = dimF ker(∂1) + dimF im(∂1) = dimF im(p1) + 1 = 0 + 1 = 1 .

This implies H1(S1;F ) ∼= F . Moreover, for n > 1, we have

dimF Hn(S1;F ) = dimF ker(∂n)+dimF im(∂n) = dimF im(pn)+dimF im(∂n) = 0+0 = 0 .

k > 1 : Consider the open subsets U := {(x1, . . . , xk+1) ∈ Sk | xk+1 6= −1} andV := {(x1, . . . , xk+1) | xk+1 6= 1} of X. Then, U ∪ V = X, U ∩ V is homotopy equivalent

48

to Sk−1, and U and V are homotopy equivalent to the one-point space. Since Sk is path-connected, we obtain H0(S

k;F ) ∼= F . The last terms of the Mayer-Vietoris sequence aregiven by

· · · i1−→H1(U ;F )⊕H1(V ;F )︸︷︷︸=0

p1−→ H1(Sk;F )∂1−→

∂1−→ H0(Sk−1;F )︸︷︷︸∼=F

i0−→H0(U ;F )⊕H0(V ;F )︸︷︷︸∼=F⊕F

p0−→ H0(Sk;F )︸︷︷︸∼=F

−→ 0 .

The same way as before, this implies that H1(Sk;F ) = 0. For n > 1, we consider the

following part of the Mayer-Vietoris sequence:

Hn(U ;F )⊕Hn(V ;F )︸︷︷︸=0

pn−→ Hn(Sk;F )∂n−→ Hn−1(Sk−1;F )

in−→ Hn−1(U ;F )⊕Hn−1(V ;F )︸︷︷︸=0

.

Its exactness implies that ∂n is an isomorphism and we obtain the desired result.

5.11 Remark (a) There is also a long exact cohomology sequence for every short exactsequence of cochain complexes.

(b) The result in Corollary 5.10 holds for any ring R instead of a field F . One onlyhas to study the maps i0 and p0 more carefully.

49

Exercises for §5

1. (a) Let M be a left R-module and let L be submodule of M . Let ι : L → M bethe inclusion map and let π : M →M/L denote the natural epimorphism. Show that thesequence

0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq L ι qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq M π qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq M/L qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq 0

is a short exact sequence.(b) Show that every short exact sequence in RMod is isomorphic to one that arises

from the construction in (a).

2. Prove the remaining statements of the snake lemma.

3. Let X be a topological space and let R be a ring. Show that H0(X;R) is iso-morphic to the free R-module whose rank is equal to the number of path-connectedcomponents of X. (Here two elements x, y ∈ X are called path-connected if there ex-ists a continuous function f : [0, 1] → X with f(0) = x and f(1) = y. This defines anequivalence relation on X.)

50

6 The Mapping Cone

Throughout this section we assume that R is a ring.

6.1 Definition Let A be an additive category and let f• : C• → D• be a morphism inC(A). We associate to it a new chain complex E• = Mc•(f•) in A, called the mappingcone of f•, by

En := Cn−1 ⊕Dn and en :=(−cn−1 0fn−1 dn

).

This is a chain complex in A, since(−cn−1 0fn−1 dn

)(−cn 0fn dn+1

)=

(cn−1 ◦ cn 0

−fn−1 ◦ cn + dn ◦ fn dn ◦ dn+1

)=

(0 00 0

)The mapping cone of f• can be pictured as

· · · Cn −cn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn−1−cn−1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Cn−2 · · ·

E• : · · · ⊕@@@@@qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

fn ⊕@@@@@qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

fn−1⊕ · · ·

· · · Dn+1dn+1 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dn

dn qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Dn−1 · · ·

degree n+ 1 n n− 1

6.2 Remark Let A be an additive category. One defines the shift functor

−[1] : C(A)→ C(A)

as follows: If C• is a chain complex in A, then C•[1] is the chain complex in A whosedegree n term is given by Cn−1, and whose n-th boundary map is −cn−1. Moreover, achain map f• : C• → D• is mapped under −[1] to the chain map (fn−1)n∈Z : C•[1]→ D•[1].Note that the shift functor is an isomorphism from A to A, i.e., an automorphism of thecategory A.

Now let f : C• → D• be a morphism in C(A). Note that the inclusion maps in : Dn →Mcn(f•) and the projection maps pn : Mcn(f•) → Cn−1 form chain maps and that weobtain a diagram

0 qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq D•i• qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Mc•(f•)

p• qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq C•[1] qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq 0 (6.2.a)

in C(A). If A is an abelian category then this is a short exact sequence in C(A). Note thatin general Mc•(f•) 6= D• ⊕ C•[1]. However, if f• = 0 then Mc•(f•) = D• ⊕ C•[1]. Alsonote that Hn(C•[1]) = Hn−1(C•), for all n ∈ Z.

51

The proof of the following proposition is left as an exercise.

6.3 Proposition Let f• : C• → D• be a morphism in C(RMod). Then the connectinghomomorphism ∂n : Hn(Cn[1]) = Hn−1(C•) → Hn−1(D•) of the short exact sequence(6.2.a) is equal to Hn−1(f•).

6.4 Corollary Let f• : C• → D• be a morphism in C(RMod). Then the following areequivalent:

(i) The morphism f• is a quasi-isomorphism, i.e., Hn(f•) is an isomorphism for alln ∈ Z.

(ii) The chain complex Mc•(f•) is exact.

Proof Consider the long exact homology sequence of the short exact sequence (6.2.a)and use Proposition 6.3 to obtain a commutative diagram

Hn+1(C•[1])∂n+1qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn(D•) qqqqqqqq

qqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn(Mc•) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn(C•[1])

∂n qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Hn−1(D•) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqHn−1(Mc•(f•))

��


Hn(f•)

��


Hn−1(f•)

Hn(C•) Hn−1(C•)

Now the claim follows immediately (see Exercise 1(d) and (e)).

6.5 Proposition Let f• : C• → D• be a morphism in C(RMod). Then the following areequivalent:

(i) The chain map f• is a homotopy equivalence.

(ii) The chain complex Mc•(f•) is contractible.

Proof (i)⇒ (ii): Let g• : D• → C• be such that g• ◦ f• ∼ 1C• and f• ◦ g• ∼ 1D• . Further,let h• be a homotopy from g• ◦ f• to 1C• and let k• be a homotopy from f• ◦ g• to 1D• .For n ∈ Z, we define

ln : Mcn(f•) = Cn−1 ⊕Dn → Cn ⊕Dn+1 = Mcn+1(f•)

by

ln =

(hn−1 + gn ◦ kn−1 ◦ fn−1 − gn ◦ fn ◦ hn−1 gn

kn ◦ fn ◦ hn−1 − kn ◦ kn−1 ◦ fn−1 −kn

).

52

It is now a straighforward calculation that en+1 ◦ ln + ln−1 ◦ en = 1Mcn(f•). Thus, Mc•(f•)is contractible.

(ii) ⇒ (i): Let l• be a homotopy from 1Mc•(f•) to 0. For n ∈ Z, let

in : Cn−1 → Mcn(f•) , jn : Dn → Mcn(f•) , pn : Mcn(f•)→ Cn−1 , qn : Mcn(f•)→ Dn

denote the canonical injections and projections. Note that

cn ◦ pn+1 = −pn ◦ en+1 : Cn ⊕Dn+1 → Cn+1 ,in ◦ cn = −en+1 ◦ in+1 + jn ◦ fn : Cn → Cn−1 ⊕Dn ,

dn−1 ◦ qn−1 = qn ◦ en+1 − fn ◦ pn+1 : Cn ⊕Dn+1 → Dn ,jn−1 ◦ dn = en ◦ jn : Dn → Cn−2 ⊕Dn−1 .

Next, we set

gn := pn+1 ◦ ln ◦ jn : Dn → Cn ,hn := pn+2 ◦ ln+1 ◦ in+1 : Cn → Cn+1 ,kn := −qn+1 ◦ ln ◦ jn : Dn → Dn+1 .

Then, since en+1 ◦ ln + ln−1 ◦ en = 1Mcn(f•), we have

cn ◦ gn = cn ◦ pn+1 ◦ ln ◦ jn = −pn ◦ jn +

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Lecture Notes Math 239 (Algebra) Spring 2015 · 2015. 6. 12. · 9 Derived Functors 75 10 The Functors Ext and Tor 82 11 Double Complexes and Tensor Products of Chain Complexes 92