Exact Categories in Functional Analysis

Home , Coimage, Exact category, Pullback

Script Exact Categories in Functional Analysis

Leonhard Frerick and Dennis Sieg

June 22, 2010 ii To Susanne Dierolf.

iii iv Contents

1 Basic Notions 1 1.1 Categories ...... 1 1.2 Morphisms and Objects ...... 5 1.3 Functors ...... 9

2 Additive Categories 25 2.1 Pre-additive Categories ...... 25 2.2 Kernels and Cokernels ...... 27 2.3 Pullback and Pushout ...... 36 2.4 Product, Coproduct and Biproduct ...... 42 2.5 Additive Categories ...... 50 2.6 Semi-abelian Categories ...... 58

3 Exact Categories 65 3.1 Basic Properties ...... 65 3.2 E-strict morphisms ...... 88

4 Maximal Exact Structure 103 4.1 p-strict morphisms ...... 103 4.2 Maximal Exact Structure ...... 109 4.3 Quasi-abelian Categories ...... 115

5 Derived Functors 121 5.1 Exact Functors ...... 121 5.2 Complexes ...... 126 5.3 Resolutions ...... 142 5.4 Derived Functors ...... 157 5.5 Universal δ-Functors ...... 171

6 Yoneda-Ext-Functors 181 6.1 Yoneda-Ext1 ...... 181 6.2 Yoneda-Extn ...... 195

v vi CONTENTS

7 Appendix 1: Projective Spectra 225 7.1 Categories of Projective Spectra ...... 225 7.2 The Projective Limit ...... 231 Preface

These lecture notes have their roots in a seminar organized by the authors in the years 2008 and 2009 at the University of Trier. The present notes contain much more than the content of the original seminar, which was based on the works of Palamodov [17, 18] and the book of Wengenroth [31]. Over the course of the seminar, due to the research of the second named author for his Ph.D.-thesis, it turned out, that the language of exact categories, as introduced by Quillen [22], is a far more flexible and more useful start- ing point for homological algebra in functional analysis, than the setting of semi-abelian categories used in [17, 18, 31]. For example it allows one not only to treat the classical categories of functional analysis, like locally convex spaces, Banach spaches or Fréchet spaces, but also more complex non semi-abelian categories of current research, like the PLS-spaces introduced by Domanski and Vogt. A homological approach to the splitting theory of these spaces, which uses the homological methods presented in this treatise, can be found in the Ph.D.-thesis of the second named author which is avail- able at http://ubt.opus.hbz-nrw.de/volltexte/2010/572/pdf/SiegDiss.pdf. These notes are aimed at readers who are interested in homological methods for functional analysis, who have some knowledge of functional analysis, but who lack experience in classical homological algebra. Therefore we are not assuming any knowledge of category theory or homological algebra in these notes. The classical introductory textbooks of homological algebra, like [30], always assume the categories to be abelian, which is almost never the case in the interesting categories appearing in functional analysis, hence they are only of limited use for someone who wants to learn about the latter ones. In this notes we treat non-abelian homological algebra completely on its own, without refering to the abelian case, and we prove every assumption in full detail. Furthermore, our examples are always taken from functional analysis. It is our philosophy, that the embedding theorems of category theory, like the Freyd-Mitchell full embedding theorem for abelian categories and the Gabriel-Quillen embedding theorem for exact categories, which are often used to transfer assumptions about diagrams into a category of modules and there argue by “diagram chasing”, are better used as an intuition than as a direct proof. Therefore, we proof everything directly from the axioms in this text by using the defining universal properties. We are convinced that this

vii viii PREFACE approach provides more insight into the subject than “diagram chasing”. This text contains part of the work of Wengenroth [31] and also uses much of the content of the survey article about exact categories of B¨uhler[3]. The content of the article [25] of the second named author and Wegner is also completely contained in this text. In addition this text drew much inspira- tion from books on classical homological algebra like Mitchell [16], Weibel [30] and Adamek, Herrlich, Strecker [1]. As a last remark, we want to remind our readers that this text is not written for publication, but only a collection of lecture notes. It is still a work in progress and we are thankful for comments and suggestions.

Leonhard Frerick and Dennis Sieg. Chapter 1

Basic Notions

1.1 Categories

Deﬁnition 1.1. A category is a quadruple C = (Ob(C), HomC, id, ◦) consisting of the following data:

(1) A class Ob(C), whose members are called objects of C.

(2) For each pair (X,Y ) of objects of C, a set HomC(X,Y ), whose elements are called morphisms from X to Y . Morphisms are generally expressed by using arrows; e.g. we will often write “f : X → Y is a morphism” instead of the statement “f ∈ HomC(X,Y )”.

(3) For each object X ∈ Ob(C), a morphism idX : X → X, called the identity on X.

(4) A composition law associating with each morphism f : X → Y and each morphism g : Y → Z a morphism g ◦ f : X → Z, called the composite of f and g.

These data have to have the following properties:

(C1) The composition is associative; i.e. for morphisms f : X → Y , g : Y → Z, and h: Z → W , the equation h ◦ (g ◦ f) = (h ◦ g) ◦ f holds.

(C2) The identities act as neutral elements with respect to the composition; i.e., for a morphism f : X → Y , we have idY ◦f = f and f ◦ idX = f.

(C3) The sets HomC(X,Y ) are pairwise disjoint.

Remark 1.2. Let C = (Ob(C), HomC, id, ◦) be a category.

i) The class of all morphisms of C is denoted by Mor(C) and is deﬁned to be the union of all the sets HomC(X,Y ) in C.

1 2 CHAPTER 1. BASIC NOTIONS

ii) If f : X → Y is a morphism in C, we call X the domain of f and Y the codomain of f. The property (C3) then guarantees that each morphism has a unique domain and a unique codomain. This is given for technical convenience only, because if the other two properties are satisﬁed one can just replace each morphism f ∈ HomC(X,Y ) with the triple (f, X, Y ) so that (C3) is also satisﬁed. Therefore when checking that an entity is a category we will only show (C1) and (C2).

iii) The composition ◦ is a partial binary operation on the class Mor(C) and for a pair (f, g) of morphisms the composite g ◦ f is deﬁned if and only if the domain of g and the codomain of f coincide.

iv) For an object X of C the identity idX : X → X is uniquely determined because of property (C2).

Example 1.3.

i) The category (Set) whose objects are sets and which has as morphisms Hom(Set)(X,Y ) the set of all mappings from X to Y . The identity morphism idX is the identity mapping on X and ◦ is the usual composition of mappings.

ii) An important type of categories are those, whose objects consist of structured sets and whose morphisms are mappings between these sets that preserve the structure. These categories are called constructs. Some examples of constructs are:

a) (Ab) the category of abelian groups and group morphisms. b) (T op) the category of topological spaces and continous mappings.

c) (Ring1) the category of commutative rings with unity together with the ring morphisms that preserve the unity. d) (F − V ec) the category of vector spaces over a fixed field F and F-linear mappings. e) (TVS) the category of topological vector spaces over a fixed field F ∈ {R, C} and continous F-linear mappings. f) (LCS) the category of (not necessarily Hausdorff) locally convex vector spaces over a fixed field F and continous F-linear mappings.

g) (LCS)HD the category of Hausdorff locally convex vector spaces over a fixed field F and continous F-linear mappings. iii) In the case of constructs it is often clear what the morphisms should be once the objects are defined. But this is not always the case:

a) There are, at least, three diﬀerent constructs having metric spaces as objects: 1.1. CATEGORIES 3

• (Met) the category of metric spaces and contractions.

• (Metu) the category of metric spaces and uniformly continous mappings.

• (Metc) the category of metric spaces and continous mappings. b) The following two categories are natural constructs having as objects all Banach spaces: • (Ban) the category of Banach spaces and continous linear mappings.

• (Banc) the category of Banach spaces and linear contractions.

iv) Not all categories consist of structured sets and mappings preserving this structure, as the following examples show:

a) (Mat) which has as objects the natural numbers N and for which Hom(Mat)(m, n) is the set of all real (m×n)-matrices, idn : n → n is the unit matrix, and the composition is defined by A◦B = BA, where BA is the usual multiplication of matrices. b) If (I, ≤) is a preordered set, i.e. I is a set and ≤ is a reflexive and transitive relation on I we can define a category C(I) in the following way:

{(i, j)} if i ≤ j Ob(C(I)) = I, Hom (i, j) = , C(I) ∅ otherwise

as well as idi = {(i, i)} and {(j, k)} ◦ {(i, j)} = {(i, k)}.

Deﬁnition 1.4. Let C = (Ob(C), HomC, id, ◦) be a category. We deﬁne the dual category of C as

op op C = (Ob(C), HomCop , id, ◦ ), where the morphism sets are deﬁned as HomCop (X,Y ) = HomC(Y,X) and the composition as f ◦op g = g ◦ f.

Remark 1.5.

i) If C is a category, then C and Cop have the same objects, only the direction of the arrows is “reversed”: If f : X → Y is a morphism in C, then f ∈ HomC(X,Y ) = HomCop (Y,X), hence f is a morphism from Y to X when considered as a morphism of the dual category. We write f op in place of f when we consider it as a morphism of the dual category. 4 CHAPTER 1. BASIC NOTIONS

ii) Because of the way dual categories are deﬁned, every statement concerning an object X in a category C can be translated into a logically equivalent statement in the dual category Cop. This observation allows one to associate (in two steps) with every property P concerning objects and morphisms in categories, a dual property concerning objects and morphisms in categories, as demonstrated by the following example: Consider the following property of an object X of a category C:

PC(X) ≡ For any Y ∈ Ob(C) there exists a unique λ ∈ HomC(Y,X).

op Step 1: In PC(X) replace all occurrences of C by C , thus obtaining the following property:

op PCop (X) ≡ For any Y ∈ Ob(C ) there exists a unique λ ∈ HomCop (Y,X).

Step 2: Translate PCop (X) into the logically equivalent statment in C:

op PC (X) ≡ For any Y ∈ Ob(C) there exists a unique λ ∈ HomC(X,Y ).

op PC (X) is called the dual property of PC(X). Roughly speaking, op PC (X) is obtained from PC(X) by reversing the direction of each arrow and the order in which morphisms are composed. In general, op PC (X) is not equivalent to PC(X). For example, if C is the category (Set) then the property PC(X) holds if and only if X is a singleton op set, whereas the dual property PC (X) holds if and only if X is the empty set.

iii) The dual property Pop of a property P holds in a category C if and only if P holds in Cop.

iv) Obviously (Cop)op = C.

v) From the above two observations we get the extremely useful Duality Principle for Categories, namely:

Whenever a property P holds for all categories, then the property Pop holds for all categories.

Because of this principle, each result in category theory has two equivalent formulations. However, only one of them needs to be proved, since the other one follows by virtue of the Duality Principle.

Remark 1.6. Morphisms in a category will usually be denoted by lowercase letters, while uppercase letters will be reserved for objects. The morphism 1.2. MORPHISMS AND OBJECTS 5

f g h = g ◦ f will sometimes be denoted by X → Y → Z or by saying that the triangle f X / Y AA AA g h AA A Z commutes. Similarly, the statement that the square

f X / Y

h g W / Z k commutes means that g ◦ f = k ◦ h.

1.2 Morphisms and Objects

Deﬁnition 1.7. Let C be a category and φ: X → Y be a morphism.

i) φ is called an epimorphism, if for every two morphisms f1, f2 : Y → W with f1 ◦ φ = f2 ◦ φ it follows that f1 = f2.

ii) φ is called a monomorphism, if for every two morphisms g1, g2 : Z → X with φ ◦ g1 = φ ◦ g2 it follows that g1 = g2. iii) φ is called a bimorphism, if φ is both a monomorphism and an epimorphism.

Remark 1.8. Monomorphisms are the dual notion of epimorphisms, i.e. a morphism f : X → Y in a category C is an epimorphism if and only if f op is a monomorphism in Cop. The bimorphisms of C and Cop are the same.

Example 1.9.

i) In every category the identities are bimorphisms.

ii) A morphism in the category (Set) is an epimorphism if and only if it is surjective. It is a monomorphism if and only if it is injective and hence a bimorphism if and only if it is bijective.

iii) In many constructs, the monomorphisms are precisely the morphisms that have injective underlying mappings; e.g. this is the case in all the constructs of example 1.3. The epimorphisms in a number of constructs are those having surjective underlying mappings; e.g. this is the case in the constructs of example 1.3.ii). However, this situation occurs less frequently than that of monomorphisms having injective 6 CHAPTER 1. BASIC NOTIONS

underlying mappings and in quite a few familiar constructs, epimorphisms are not surjective: In the category (T op)HD of topological Hausdorﬀ spaces and continous mappings a morphism is an epimorphism if and only if it has dense range. This is also the case in the category (LCS)HD. Proposition 1.10. Let C be a category and let h = g ◦ f be a morphism in C. i) If g and f are epimorphisms, then h is an epimorphism.

ii) If h is an epimorphism, then g is an epimorphism.

iii) If g and f are monomorphisms, then h is a monomorphism.

iv) If h is a monomorphism, then f is a monomorphism. Proof. Because of the duality principle it suﬃces to show i) and ii), since iii) and iv) are the dual statements. More precisely: Suppose i) and ii) have already been shown. If g and f are monomorphisms, then gop and f op are epimorphisms in Cop. Since i) holds for all categories hop = f op ◦op gop is an epimorphism and therefore a monomorphism in C. Analagously one shows iv). i) If g1 ◦ (g ◦ f) = g2 ◦ (g ◦ f) for two morphisms g1 and g2 it follows ﬁrst that g1 ◦ g = g2 ◦ g and then g1 = g2, since both g and f are epimorphisms. This shows i). ii) If g1 ◦ g = g2 ◦ g, then g1 ◦ (g ◦ f) = g2 ◦ (g ◦ f), hence g1 = g2, since h = g ◦ f is an epimorphism.

Definition 1.11. Let C be a category and φ: X → Y be a morphism. i) φ is called a retraction, if there exists a morphism ψ : Y → X with φ ◦ ψ = idY ; i.e. if φ has a right-inverse. ii) φ is called a coretraction, if there exists a morphism ψ : Y → X with ψ ◦ φ = idX ; i.e. if φ has a left-inverse. iii) φ is called an isomorphism, if there exists a morphism ψ : Y → X with φ ◦ ψ = idY and ψ ◦ φ = idX ; i.e. if φ has an inverse. Remark and Definition 1.12. The inverse of an isomorphism is uniquely determined by the defining property. Therefore, given an isomorphism φ : X → Y , we will denote the inverse morphism ψ : Y → X with φ ◦ ψ = −1 idY and ψ ◦ φ = idX by φ := ψ. We will call two objects X, Y of a category C isomorphic, if there exists an isomorphism φ: X → Y . This is an equivalence relation on the class Ob(C) and we will write X =∼ Y if X and Y are isomorphic to each other. The right-inverse of a retraction and the left-inverse of a coretraction are, in general, not uniquely determined. 1.2. MORPHISMS AND OBJECTS 7

Proposition 1.13. Let C be a category and φ: X → Y be a morphism.

i) If φ is retraction, then it is an epimorphism.

ii) If φ is a coretraction, then it is a monomorphism.

iii) If φ is an isomorphism, then it is a bimorphism.

Proof. i) follows from 1.9.i) and 1.10.i), ii) is the dual statement of i) and iii) follows from i) and ii).

Example 1.14.

i) In the category (Set) a morphism is a retraction if and only if it is surjective, it is a coretraction if and only it is injective and it is an isomorphism if and only if it is bijective. The same holds for the category (F − V ec). ii) In the categories (TVS) and (LCS) a morphism is an isomorphism if and only if it is bijective and open unto its range. An isomorphism in the category (LCS)HD is characterized by being injective, open unto its range and having a dense image. The retractions and coretractions of these categories are not so easily characterized. We will come back to them in a later chapter.

Proposition 1.15. Let C be a category and h = g ◦ f be a morphism in C.

i) If g and f are retractions, then h is a retraction.

ii) If h is a retraction, then g is a retraction.

iii) If g and f are coretractions, then h is a coretraction.

iv) If h is a coretraction, then f is a coretraction.

Proof. It suﬃces to show i) and ii), since iii) and iv) are the dual statements of i) and ii). i) If g ◦ r = idZ for a morphism r : Z → Y and f ◦ s = idY for a morphism s: Y → X then

(g ◦ f) ◦ (s ◦ r) = g ◦ (f ◦ s) ◦ r = g ◦ idY ◦r = g ◦ r = idZ , hence h = g ◦ f has a right-inverse. ii) Given a morphism l : Z → X with (g◦f)◦l = idZ we have g◦(f ◦l) = idZ , hence g is a retraction.

Proposition 1.16. Let C be a category and φ: X → Y be a morphism in C. The following are equivalent:

i) φ is a retraction and a monomorphism. 8 CHAPTER 1. BASIC NOTIONS

ii) φ is a coretraction and an epimorphism.

iii) φ is an isomorphism.

iv) φ is a retraction and coretraction.

Proof. i) ⇒ iii) Let ψ : Y → X be a morphism with φ ◦ ψ = idY . Then we have φ ◦ ψ ◦ φ = idy ◦φ = φ ◦ idX and since φ is a monomorphism it follows that ψ ◦ φ = idX . Hence φ is an isomorphism. The assertion ii) ⇒ iii) follows by duality, iii) ⇒ iv) is trivial and iv) ⇒ i) as well as iv) ⇒ ii) follow from proposition 1.13.

Deﬁnition 1.17. Let C be a category.

i) An object I of C is called an initial object, if for every object X ∈ Ob(C) there exists exactly one morphism iX : I → X. ii) An object T of C is called a terminal object, if for every object X ∈ Ob(C) there exists exactly one morphism tX : X → T . iii) An object Z of C is called a zero object, if it is both initial and terminal.

Remark 1.18.

i) The notion of a terminal object is the dual notion of that of an initial object; i.e. if I is an initial object of a category C then I is a terminal object of Cop and vice versa.

ii) If I and Ie are initial (resp. terminal, resp. zero) objects of a category C, then there exists exactly one isomorphism λ: I → Ie. In fact, since I and I are both initial there exists exactly one morphism i : I → I e Ie e and exactly one morphism eiI : Ie → I. Since idI is the only element of Hom (I,I) and id is the only element of Hom (I, I) it follows that C Ie C e e i ◦ i = id and i ◦ i = id . Hence i is an isomorphism. By duality, Ie eI Ie eI Ie I Ie the same holds for terminal objects. This property is characterized by saying that an initial (resp. terminal, resp. zero) object is unique up to a canonical isomorphism.

iii) If I is an initial object of a category C, and f : X → Y is an arbitrary morphism, the morphism f ◦ iX is a morphism from I to Y , hence f ◦ iX = iY . Dually, if T is a terminal object of a category C we have tY ◦ f = tX for every morphism f : X → Y .

Remark and Deﬁnition 1.19. If a category C possesses a zero object Z, then there exists for every two objects X,Y ∈ Ob(C) a morphism

0X,Y : X → Y 1.3. FUNCTORS 9 called the zero morphism from X to Y , given by the composition

tX iX X / Z /7 Y .

0X,Y

If f : W → X and g : Y → Z are morphisms in C it follows from 1.18.iii) that

0X,Y ◦ f = iY ◦ tX ◦ f = iY ◦ tW = 0W,Y

g ◦ 0X,Y = g ◦ iY ◦ tX = iZ ◦ tX = 0X,Z .

Example 1.20.

i) In the category (Set) the empty set ∅ is an initial object and every singleton is a terminal object.

ii) In the category (Ring1) the ring Z is an initial object and the zero ring is a terminal object.

iii) In the category (Ab) the trivial group is a zero object.

iv) The zero vector space is the zero object of the category (F − V ec).

v) The categories (TVS), (LCS) and (LCS)HD all have as a zero object the zero vector space with its unique topology.

1.3 Functors

Deﬁnition 1.21. Let C and D be categories.

i) A covariant functor F : C → D from C to D is a rule that assigns to every X ∈ Ob(C) an object F (X) ∈ Ob(D) and to every morphism f : X → Y in C a morphism F (f): F (X) → F (Y ) with the following properties:

(F 1) F (idX ) = idF (X) (F 2) F (g ◦ f) = F (g) ◦ F (f)

ii) A contravariant functor G: C → D from C to D is a rule that assigns to every X ∈ Ob(C) an object G(X) ∈ Ob(D) and to every morphism f : X → Y in C a morphism G(f): G(Y ) → G(X) with the following properties:

(F 1) G(idX ) = idG(X) (F 2)op G(g ◦ f) = G(f) ◦ G(g) 10 CHAPTER 1. BASIC NOTIONS

Remark and Deﬁnition 1.22.

i) Every contravariant functor G: C → D can also be seen as a covariant functor G: Cop → D.

ii) Every functor (covariant or contravariant), preserves isomorphisms: Let ψ : X → Y be an isomorphism in a category C and F : C → D a covariant functor, then the functor properties yield

−1 −1 F (ψ) ◦ F (ψ ) = F (ψ ◦ ψ ) = F (idY ) = idF (Y ), −1 −1 F (ψ ) ◦ F (ψ) = F (ψ ◦ ψ) = F (idX ) = idF (X),

hence F (ψ) is an isomorphism in D.

iii) If C,D and E are categories, F : C → D, F 0 : D → E are covariant functors and G: C → D, G0 : D → E are contravariant functors one can deﬁne the composite functor

F 0 ◦ F (X) = F 0(F (X)) F 0 ◦ F : C → E , . F 0 ◦ F (f) = F 0(F (f))

The functor F 0 ◦ F is covariant. In a similar way one can deﬁne the composite functors G0 ◦F , F 0 ◦G and G0 ◦G. In these cases the functors G0 ◦F and F 0 ◦G are contravariant and the functor G0 ◦G is covariant.

Example 1.23.

i) For every category C there is the identity functor

Id(X) = X Id: C → C , . Id(f) = f

ii) For every category there is the duality functor

op(X) = X op : C → Cop , . op(f) = f op

op op op is a contravariant functor with ◦ = IdC.

iii) For every two categories C and D and every object Y0 of D there is the constant functor with value Y0 CY0 (X) = Y0 CY0 : C → D , . Id(f) = idY0 1.3. FUNCTORS 11

iv) For every category C and A ∈ Ob(C), we have a covariant functor X 7→ HomC(A, X) HomC(A, −): C → (Set) , , α 7→ HomC(A, α)

where

0 HomC(A, α): HomC(A, X) → HomC(A, X ), f 7→ α ◦ f

and a contravariant functor Z 7→ HomC(Z,A) HomC(−,A): C → (Set) , , γ 7→ HomC(γ, A)

where

0 HomC(γ, A): HomC(Z,A) → HomC(Z ,A), g 7→ g ◦ γ.

Remark 1.24. Let C be a category and let A ∈ Ob(C). Then we have

op HomC(−,A) = HomCop (A, −) ◦ .

Indeed, we have HomC(X,A) = HomCop (A, X) for each object X of C. If f : X → Y is a morphism in C, then

op op op HomCop (A, −) ◦ (f)(g) = f ◦ g.

Since f op ◦op g in Cop is the morphism gop ◦ f in C, it follows that the op morphisms HomC(f, A) and HomCop (A, f ) coinside.

Example 1.25.

i) There is the covariant functor

F (X) = X F :(Ab) → (Set) , . F (f) = f

Functors of this kind, where part of the structure on the objects and morphisms are “forgotten” are called forgetful functors.

ii) There is the contravariant duality functor

L(X) = X0 L:(TVS) → (TVS) , , L(f) = f t

assigning to each topological vector space X its strong dual X0 and to f : X → Y the transposed map f t : Y 0 → X0. 12 CHAPTER 1. BASIC NOTIONS

iii) For a locally convex space X let Xe be the Hausdorﬀ completion of X, i.e. the completion of X/{0} and let jX : X → Xe be the canonical map (which is injective if and only if X is Hausdorﬀ and surjective if and only if X is complete). Then we have the covariant completion functor ( C(X) = Xe C :(LCS) → (LCS)HD , , C(f) = fe

assigning to X the space Xe and to a continous linear map f : X → Y the canonical map fe: Xe → Ye. Remark and Definition 1.26. We have seen above that functors act as morphisms between categories; they are closed under composition, which is associative and the identity functors act as identities with respect to the composition. Because of this, one is tempted to consider the “category of all categories”. However, there are two difficulties that arise when we try to form this entity. First, the “category of all categories” would have objects such as “all sets” or “all vector spaces”, which are not sets but proper classes. We will not concern ourselves with the set-theoretic difficulties arising in category theory. The reader who is interested in this should consult [1, 4, 5] as well as the references therein. Since proper classes cannot be elements of classes, the conglomerate of all categories would not be a class, thus violating the properties of a category. Second, given any categories C and D, it is not generally true that the conglomerate af all functors from C to D forms a set, which is another violation of the properties. However, if we restrict our attention to categories whose class of objects is actually a set, then both problems are eliminated. A category C is called small if Ob(C) is a set. Otherwise its called large.

Example 1.27.

i) The category (Mat) of natural numbers and m × n-matrices is a small category.

ii) Every preordered set is small when considered as a category.

iii) The category (Set) of sets and mappings is not small, since the class of all sets is not a itself a set (Russell’s paradox).

M iv) Since for every set M there is the vector space F , the category (F − V ec) of all vector spaces is not a small category.

v) Every category that contains one of the above is not small. For example, the category (TVS) of topological vector spaces is not small, since it contains all vector spaces (each vector space is a topological vector space, when considered together with the coarsest topology). 1.3. FUNCTORS 13

vi) The category (Ban) of Banach spaces and continous linear mappings is not small, since for every set M there is the Banach space l∞(M).

vii) The categories (LCS) and (LCS)HD are not small, since they contain the Banach spaces.

Remark and Deﬁnition 1.28. The category (Cat) of small categories has as objects all small categories, as morphisms from C to D the functors from C to D, as identities the identity functors, and as composition the composition of functors. Since each small category is a set, the conglomerate of all small categories is a class, and since for each pair (C, D) of small categories the conglomerate of all functors from C to D is a set, (Cat) is indeed a category. However its not a small category.

Deﬁnition 1.29. A quasicategory is a quadruple C = (Ob(C), HomC, id, ◦) consisting of the following data:

(1) A conglomerate Ob(C), whose members are called objects of C.

(2) For each pair (X,Y ) of objects of C, a conglomerate HomC(X,Y ), whose elements are called morphisms from X to Y .

(3) For each object X ∈ Ob(C), an morphism idX : X → X, called the identity on X.

(4) A composition law associating with each morphism f : X → Y and each morphism g : Y → Z a morphism g ◦ f : X → Z.

These data have to have the following properties:

(C1) The composition is associative.

(C2) The identities act as neutral elements with respect to the composition.

(C3) The sets HomC(X,Y ) are pairwise disjoint. Remark and Deﬁnition 1.30.

i) Every category is a quasicategory.

ii) The quasicategory (CAT ) of all categories has as objects all categories, as morphisms from C to D all functors from C to D, as identities the identity functors, and as composition the composition of functors. (CAT ) is a proper quasicategory in the sense that it is not a category.

iii) Virtually every categorical concept has a natural analogue or interpre- tation for quasicategories. The names for such quasicategorical con- cepts will be the same as those of their categorical analogues. Thus we have, for example, the notion of functor between quasicategories. 14 CHAPTER 1. BASIC NOTIONS

Because the main objects of our study are categories, most notions will be specifically formulated only for categories. In fact, the only quasicategories we will introduce at all are the category (CAT ) defined above, the category which has as objects all functors between two categories C and D, the quasicategory of all classes and that of all big abelian groups, defined further below.

Deﬁnition 1.31. Let C and D be categories and let F : C → D be a covariant functor. For X,Y ∈ Ob(C) we have the mapping

φX,Y : HomC(X,Y ) → HomD(F (X),F (Y )), f 7→ F (f).

i) The functor F is called faithful, if φX,Y is injective for all objects X and Y .

ii) The functor F is called full, if φX,Y is surjective for all objects X and Y .

iii) The functor F is called fully faithful if it is full and faithful.

iv) The functor F is called essentially surjective, if for every Y ∈ Ob(D) there exists an X ∈ Ob(C) with F (X) =∼ Y .

Deﬁnition 1.32. Let C be a category. A subcategory of C is a category C0 with the following properties:

(SC1) Every object X ∈ Ob(C0) is an object of C.

(SC2) For every two objects X,Y ∈ Ob(C0) there is the inclusion

HomC0 (X,Y ) ⊆ HomC(X,Y ).

0 0 (SC3) The identity idX in C of an object X ∈ Ob(C ) is also the identity in C.

(SC4) The composition of two morphisms in C0 is the same as the composition in C.

Remark and Deﬁnition 1.33. If C0 is a subcategory of a category C, we have a covariant faithful functor ι(X) = X ι: C0 → C , , ι(f) = f

the so-called inclusion functor. C0 is called a full subcategory of C, if the functor ι is fully faithful, i.e. if 0 HomC0 (X,Y ) = HomC(X,Y ) for all X,Y ∈ Ob(C ). 1.3. FUNCTORS 15

Example 1.34.

i) Let F be a ﬁeld and let (F − V ec) be the category of F-vector spaces and linear mappings. The category (F − V ec)fin of ﬁnite dimensional vector spaces and linear mappings is a full subcategory of (F − V ec). ii) We have the inclusions

(TVS) ⊇ (LCS) ⊇ (LCS)HD ⊇ (Ban)

of categories. Since all these categories have continous linear mappings as morphisms, each of them is a full subcategory of the categories lying above them.

iii) For the category (TVS) the forgetful functor

F (X, T ) = X F :(TVR) → ( − V ec) , X , F F (f) = f

is full. Every vector space V can be made into a topological vector space (V, TV ) by using the coarsest topology on V , hence the functor F is essentially surjective. It is not faithful, since there are linear mappings between topological vector spaces that are not continous.

iv) Let (Mat) be the category, which has as objects the natural numbers N and as morphisms Hom(Mat)(m, n) the set of all real (m×n)-matrices. Then the functor

n F (n) = F F :(Mat) → (F − V ec)fin , m×n , F (A ∈ F ) = Aφ with n m Aφ : F → F , x 7→ Ax, is fully faithful and essentially surjective.

Proposition 1.35. Let C and D be categories and let F : C → D be a fully faithful functor. Then the following are equivalent:

i) f : X → Y is an isomorphism in C.

ii) F (f): F (X) → F (Y ) is an isomorphism in D.

Proof. i) ⇒ ii) is clear from 1.22.ii). ii) ⇒ i) Let u ∈ HomD(F (Y ),F (Y ) be a morphism with u ◦ F (f) = idF (X) and F (f) ◦ u = idF (Y ). Since the mapping g 7→ F (g) is surjective , there exists an h ∈ HomC(Y,X) with F (h) = u. Then

F (h ◦ f) = F (h) ◦ F (f) = idF (X) = F (idX ) 16 CHAPTER 1. BASIC NOTIONS and F (f ◦ h) = F (f) ◦ F (h) = idF (Y ) = F (idY ).

The mapping g 7→ F (g) is also injective, hence it follows that f ◦ h = idY and h ◦ f = idX . Deﬁnition 1.36. Let C and D be categories and let F,G: C → D be covariant functors. A natural transformation

τ : F → G is a rule that assigns to each X ∈ Ob(C) a morphism τ(X): F (X) → F (X) of D in such a way that the following condition holds: For each morphism f : X → Y in C, the square

τ(X) F (X) / G(X)

F (f) G(f) F (Y ) / G(Y ) τ(Y ) is commutative. We will denote natural transformations from F to G by Nat(F,G). A natural transformation between contravariant functors F 0,G0 : C → D is a natural transformation between the covariant functors F 0,G0 : Cop → D Remark and Deﬁnition 1.37. i) If τ : F → G and τ 0 : G → H are natural transformations between functors F, G, H from a category C to a category D then it is clear that τ 0 ◦ τ : F → H deﬁned by τ 0 ◦ τ(X) := τ 0(X) ◦ τ(X) for each X ∈ C is again a natural transformation, since all the diagrams

0 τX τ (X) F (X) / G(X) / H(X)

F (f) G(f) H(f) F (Y ) / G(Y ) / H(Y ) τY τ 0(Y )

are commutative. Therefore we have a composition of natural transformations, which is obviously associative. ii) Let τ : F → G be a natural transformation between functors F,G: C → D. If for every X ∈ Ob(C) the morphism τ(X): F (X) → G(X) is an isomorphism, then τ is called a natural isomorphism. 1.3. FUNCTORS 17

iii) Two Functors F,G: C → D are called naturally isomorphic, often denoted by F =∼ G, if there exists a natural isomorphism φ: F → G. Example 1.38.

i) For every functor F : C → D we have the natural isomorphism

idF : F → F,

called the identity on F , given by idF (X) := idF (X) for every object X of C.

ii) Let C be a category, let A and B be objects of C and let a: A → B be a morphism. Since for every morphism f : X → Y the diagram

HomC(a,X) HomC(B,X) / HomC(A, X)

HomC(B,f) HomC(A,f) HomC(B,Y ) / HomC(A, Y ) HomC(a,Y )

is commutative, a induces a natural transformation

τa := (HomC(a, X))X∈Ob(C) : HomC(B, −) → HomC(A, −).

Deﬁnition 1.39. Let C and D be categories. The functor quasicategory Fun(C, D) has as objects all functors from C to D, as morphisms from a functor F to a functor G all natural transformations from F to G, as identities the identity natural transformation, and as composition the composition of natural transformations.

Remark 1.40.

i) If C and D are small categories, then Fun(C, D) is a category. If C is small and D is large, then Fun(C, D), though being a proper quasicategory, is isomorphic to a category in (Cat). If C and D are both large, then Fun(C, D) will generally fail to be isomorphic to a category.

ii) A natural transformation between functors from C to D is a natural isomorphism if and only if it is an isomorphism in Fun(C, D).

Deﬁnition 1.41. A functor F : C → D is called an equivalence of categories, if there exists a functor G: D → C and natural isomorphisms ∼ ∼ F ◦ G = IdD and G ◦ F = IdC .

The functor G is then called a quasi-inverse of F . 18 CHAPTER 1. BASIC NOTIONS

Proposition 1.42. Let C and D be categories and let F : C → D be a covariant functor, then the following are equivalent:

i) F is an equivalence of categories.

ii) F is fully faithful and essentially surjective. ∼ ∼ Proof. i) ⇒ ii) Let G: D → C be a functor with F ◦G = IdD and G◦F = IdC. Fix natural isomorphisms φ: F ◦ G → IdD and ψ : G ◦ F → IdD. If Y is an object of D, then we have the isomorphism φ(Y ): F ◦ G(Y ) → Y , hence the functor F is essentially surjective. Let f : X → Y be a morphism in C, then the diagram ψ(X) G ◦ F (X) / X

G◦F (f) f G ◦ F (Y ) / Y ψ(Y ) is commutative and we have

f = ψ(Y ) ◦ (G ◦ F (f)) ◦ ψ(X)−1, which means that the morphism f can be recaptured from F (h), hence F is a faithful functor. Analogously one can show that G is a faithfull functor. If l : F (X) → F (Y ) is a morphism in D deﬁne

f := ψ(Y ) ◦ G(l) ◦ ψ(X)−1, which is an element of HomC(X,Y ). In addition we have

f = ψ(Y ) ◦ (G ◦ F (f)) ◦ ψ(X)−1, as was shown above. Since ψ(Y ) and ψ(X)−1 are isomorphisms, it follows that G(l) = G ◦ F (f). The functor G is faithful, hence we have l = F (f), which shows that F is a full functor. ii) ⇒ i) Since F is essentially surjective we can ﬁx an object XY ∈ Ob(C) and an isomorphism φ(Y ): F (XY ) → Y for every object Y ∈ Ob(D). Deﬁne

G(Y ) = X G: D → C , Y , G(g) = fg where fg : XY → XX is the , because the functor F being fully faithful, uniquely determined morphism with

−1 F (fg) = φ(Z) ◦ g ◦ φ(Y ). 1.3. FUNCTORS 19

Then we have −1 F (idXY ) = φ(Y ) ◦ idY ◦φ(Y ) = idY , hence it follows that G(idY ) = idXY = idG(Y ). In addition it follows from F ◦ G(g ◦ g0) = φ(Z)−1 ◦ g ◦ g0 ◦ φ(Y ) = φ(Z)−1 ◦ g ◦ φ(Y ) ◦ φ(Y )−1 ◦ g0 ◦ φ(Y ) = (F ◦ G(g)) ◦ (F ◦ G(g0)) that G(g ◦ g0) = G(g) ◦ G(g0), hence G is functor. For each morphism g : Y → Z in D, the square

φ(Y ) F ◦ G(Y ) / Y

F ◦G(g) g F ◦ G(Z) / Z φ(Z) commutes, therefore the rule

φ: F ◦ G → IdD, φ = (φ(Y ))Y ∈Ob(D) is a natural isomorphism. Additionally the morphisms

φ(F (X)): F ◦ G ◦ F (X) → F (X) is an isomorphism for every object X of C. Since the functor F is fully faithful there exists a unique morphism ψ(X): G ◦ F (X) → X for each X ∈ Ob(C) and by 1.35 the morphism ψ(X) is always an isomorphism. In addition we have for every morphism f : X → X0 in C that

F ◦ G ◦ F (f) = F ◦ G(φ(F (X0))−1 ◦ F (f) ◦ φ(F (X))) = F ◦ G(F (ψ(X0))−1 ◦ F (f) ◦ F (ψ(X))) = F ◦ G ◦ F (ψ(X0)−1 ◦ f ◦ ψ(X)) and since F is faithful it follows that the diagram

ψ(X) G ◦ F (X) / X

G◦F (f) f G ◦ F (X0) / X0 ψ(X0) is commutative. Therefore the rule

ψ : G ◦ F → IdC, ψ = (ψ(X))X∈Ob(C) is a natural isomorphism, which shows that the functor F is an equivalence of categories. 20 CHAPTER 1. BASIC NOTIONS

Deﬁnition 1.43. Let C be a category. A covariant functor F : C → (Set) is called representable, if there exists an A ∈ Ob(C) and a natural isomorphism ∼ F = HomC (A, −). A contravariant functor is called representable , if there exists an B ∈ Ob(C) and a natural isomorphism ∼ F = HomC (−,B). Remark 1.44. Let F : C → (Set) be a representable covariant (resp. con- ∼ ∼ travariant) functor with F = HomC (A, −) (resp. F = HomC (−,A)) for an object A of C. Then the object A is uniquely determined by this property up to a unique isomorphism. In fact, if A and A0 are objects of C with ∼ ∼ 0 HomC (A, −) = F = HomC(A , −) 0 in Fun(C, (Set)), we have isomorphisms φ(A): HomC(A, A) → HomC(A ,A) 0 0 0 0 and φ(A ): HomC(A, A ) → HomC(A ,A ) making the diagram

φ(A) HomC (A, A) / HomC (A, A)

0 HomC(A,f) HomC(A ,f) HomC (A, A) / HomC (A, A) φ(A0) commutative, if we deﬁne f : A → A0 to be the unique morphism with 0 φ(A )(f) = idA0 . Then it follows that 0 idA0 = φ(A )(f) 0 = φ(A )(f) ◦ HomC(A, f)(idA) 0 = HomC(A , f) ◦ φ(A)(idA)

= f ◦ (φ(A)(idA)). Additionally the diagram

0 0 0 HomC(A ,φ(A)(idX )) 0 HomC (A ,A ) / HomC (A ,A)

φ(A0)−1 φ(A)−1 HomC (A, A) / HomC (A, A) HomC(A,φ(A)(idX )) commutes, hence it follows that

φ(A)(idX ) ◦ f = HomC(A, φ(A)(idA))(f) 0 = HomC(A, φ(A)(idA) ◦ φ(A )(idA0 ) −1 0 = φ(A) ◦ HomC(A , φ(A)(idA))(idA0 −1 = φ(A) ◦ φ(A)(idA) = idA 1.3. FUNCTORS 21

This shows that the morphism f : A → A0 is an isomorphism with inverse 0 φ(A)(idA): A → A. Proposition 1.45 (Yoneda Lemma). Let C be a category and let F : C → (Set) be a covariant functor. For every object A of C the map

Y : Nat(HomC(A, −),F ) → F (A) deﬁned by Y(τ) = τ(A)(idA), is bijective. Proof. Deﬁne a mapping

ψ : F (A) → Nat(HomC(A, −),F ) in the following way: For Y ∈ Ob(C) and s ∈ F (A) let ψ(s)(Y ) be given by the composition

HomC(A, Y ) / Hom(Set)(F (A),F (Y ) / F (Y ) f / F (f) / F (f)(s). 0 For a morphism g : Y → Y and every f ∈ HomC(A, Y ) we have 0 0 (ψ(s)(Y ) ◦ HomC(A, g))(f) = ψ(s)(Y )(g ◦ f) = (F (g) ◦ F (f))(s) = F (g)(F (f)(s)) = F (g)(ψ(s)(Y )(f)) = (F (g) ◦ ψ(s)(Y ))(f), hence the diagram

ψ(s)(Y ) HomC(A, Y ) / F (Y )

HomC(A,g) F (g) HomC(A, Y ) / F (Y ) ψ(s)(Y 0) commutes. This shows that ψ(s) is indeed a natural transformation. In addition we have Y ◦ ψ(s) = Y(ψ(s))

= ψ(s)(A)(idA)

= F (idA)(s) = s for all s ∈ F (A), which shows that Y ◦ ψ = idF (A). For τ ∈ Nat(HomC(A, −),F ) and g ∈ HomC(A, Y ) the diagram

τ(X) HomC(A, A) / F (A)

HomC(A,g) F (g) HomC(A, Y ) / F (A) τ(Y ) 22 CHAPTER 1. BASIC NOTIONS commutes, hence we have

ψ ◦ Y(τ)(g) = ψ(τ(A)(idA))(g)

= F (g)(τ(A)(idA))

= τ(Y ) ◦ HomC(A, g)(idA) = τ(Y )(g).

This shows that ψ ◦ Y = idNat(HomC(A,−),F ) and therefore the mapping Y is a bijection.

Remark and Deﬁnition 1.46 (Yoneda Embedding). Let C be a category and let f : A → B be a morphism in C. Deﬁne

E(f)(X): HomC(B,X) → HomC(A, X), h 7→ h ◦ f for each object X of C. If g : X → Y is another morphism in C, the diagram

E(f)(X) HomC(B,X) / HomC(A, X)

HomC(B,g) HomC(A,g) HomC(B,Y ) / HomC(A, Y ) E(f)(Y ) is commutative, since

HomC(A, g) ◦ E(f)(X)(h) = HomC(A, g)(h ◦ f) = g ◦ h ◦ f = E(f)(Y )(g ◦ h)

= E(f)(Y ) ◦ HomC(B, g).

Therefore E(f) := (E(f)(X))X∈Ob(C) is a natural transformation from the functor HomC(B, −) to HomC(A, −). In addition we have

E(idA) = idHomC(A,−) E(f 0 ◦ f) = E(f 0) ◦ E(f), therefore the rule

A 7→ Hom (A, −) E : C → Fun(C, (Set)) , C f 7→ E(f) is a contravariant functor, the so-called Yoneda Embedding.

Proposition 1.47. For every category C the functor E : C → Fun(C, (Set)) deﬁned above is fully faithful. 1.3. FUNCTORS 23

Proof. By 1.45 we have a bijection

Y : Nat(HomC(B, −), HomC(A, −)) → HomC(A, B), with Y(τ) = τ(A)(idA). Then

(Y )(E(f)) = E(f)(A)(idA) = idA ◦f = f, which shows that the mapping f 7→ E(f) is bijective.

Remark 1.48. For every category C one also has the contravariant Yoneda embedding

A 7→ Hom (−,A) E0 : C → Fun(C, (Set)) , C f 7→ E0(f) with E0(f)(X)(h) = f ◦ h. In analogy to the above one can show that E0 is also a fully faithful functor. 24 CHAPTER 1. BASIC NOTIONS Chapter 2

Additive Categories

2.1 Pre-additive Categories

Deﬁnition 2.1. A category C is called preadditive, if it has the following properties: PA1 C possesses a zero object.

PA2 Each of the sets HomC(X,Y ) carries the structure of an abelian group, written additively, and this structure is compatible with the composition of C, in the sense that the distributive laws

(f + g) ◦ h = f ◦ h + g ◦ h f ◦ (g + h) = f ◦ g + f ◦ h

hold on morphisms whenever the terms are deﬁned. Remark 2.2. If C is a preadditive category, then the dual category Cop is also preadditive.

Remark 2.3. Let C be a preadditive category. If eX,Y is the neutral element of the abelian group HomC(X,Y ) and f : X → Y is an arbitrary morphism it follows from

eY,Y ◦ f = (eY,Y + eY,Y ) ◦ f = eY,Y ◦ f + eY,Y ◦ f that eY,Y ◦ f = eX,Y . Especially, if 0X,Y is the zero morphism, we have 0X,Y = eY,Y ◦ 0X,Y by 1.19, hence

0X,Y = eY,Y ◦ 0X,Y = eX,Y and thus the zero morphism 0X,Y is the neutral element of HomC(X,Y ) for all X,Y ∈ Ob(C). From now on we will simply write “0” instead of 0X,Y . Example 2.4.

25 26 CHAPTER 2. ADDITIVE CATEGORIES

i) The categories (Ab), (F − V ec), (TVS), (LCS) and (LCS)HD are all preadditive categories.

ii) If f, g : R → S are two morphisms in the category (Ring1), we have

f(1R) + g(1R) = 1S + 1S

and therefore the elementwise addition of ring morphisms that preserve the unity does not itself preserve the unity. The category (Ring1) is therefore not preadditive with this addition.

Deﬁnition 2.5. Let C be a preadditive category. A full preadditive subcategory C0 of a preadditive category C is a full subcategory that also contains the zero object.

Deﬁnition 2.6. Let C and D be preadditive categories. A covariant functor F : C → D is called additive if

F (f + g) = F (f) + F (g) holds, for all morphisms f, g ∈ HomC(X,Y ) and all X,Y ∈ Ob(C). A contravariant functor is called additive if it is additive as a covariant functor from Cop to D.

Remark 2.7.

i) Let F : C → D be an additive functor between preadditive categories. From F (0) = F (0 + 0) = F (0) + F (0) it follows that F (0) = 0.

ii) If C is a preadditive category, then the duality functor is an additive functor.

iii) If C is a preadditive category and A an object of C, the functors X 7→ HomC(A, X) HomC(A, −): C → (Ab) , , α 7→ HomC(A, α)

and Z 7→ HomC(Z,A) HomC(−,A): C → (Ab) , , γ 7→ HomC(γ, A) are additive functors, because of

HomC(A, f + g)(h) = (f + g) ◦ h = f ◦ h + g ◦ h

= HomC(A, f)(h) + HomC(A, g)(h) 2.2. KERNELS AND COKERNELS 27

and

HomC(f + g, A)(l) = l ◦ (f + g) = l ◦ f + l ◦ g

= HomC(f, A)(l) + HomC(g, A)(l).

2.2 Kernels and Cokernels

Deﬁnition 2.8. Let C be a preadditive category and let f : X → Y be a morphism in C.

i) A kernel of f is a morphism k : K → X with f ◦ k = 0 that satisﬁes the following universal property: For every morphism t: T → X with f ◦ t = 0 there exists a unique morphism λ: T → K with t = k ◦ λ, i.e. making the diagram

k f K / X / Y O > }} λ }} }} t }} T

commutative.

ii) A cokernel of f is a morphism c: Y → C with c ◦ f = 0 that satisﬁes the following universal property: For every morphism s: Y → S with s ◦ f = 0 there exists a unique morphism µ: C → S with s = µ ◦ c, i.e. making the diagram

f c X / Y / C ~ s ~ ~ µ ~ S

commutative.

iii) The domain of a kernel k of f is called a kernel-object of f and the codomain of a cokernel c of f is called a cokernel-object of f.

iv) The category C is said to have kernels if every morphism f : X → Y has a kernel.

v) The category C is said to have cokernels if every morphism f : X → Y has a cokernel.

Remark 2.9. Let C be a preadditive category. 28 CHAPTER 2. ADDITIVE CATEGORIES

i) A cokernel of a morphism is the dual notion of the kernel of a morphism, i.e. if k ∈ HomC(K,X) is the kernel of a morphism f ∈ op HomC(X,Y ), then k ∈ HomCop (X,K) is the cokernel of the mor- op phism f ∈ HomCop (Y,X). ii) Kernel and cokernel of a morphism are uniquely determined up to a unique isomorphism: If k : K → X and k0 : K0 → X are two kernels of a morphism f : X → Y , then the universal property of k gives rise to a unique morphism λ: K0 → K with k0 = k◦λ and the universal property of k0 gives rise to a unique morphism λ0 : K → K0 with k = k0 ◦ λ0. In 0 addition, the universal properties of k and k show that idK and idK0 0 0 are unique with k ◦ idK = k and k ◦ idK0 = k respectively. Since

k ◦ λ ◦ λ0 = k0 ◦ λ0 = k k0 ◦ λ0 ◦ λ = k ◦ λ = k0

0 0 0 it follows that λ ◦ λ = idK and λ ◦ λ = idK0 , hence λ and λ are isomorphisms inverse to each other. The dual argument shows that the cokernel of f is uniquely determined up to a unique isomorphism.

Example 2.10.

i) In (F − V ec) the kernel of a linear mapping f : X → Y is the inclusion f −1({0}) ,→ X, x 7→ x. The cokernel of f is the quotient map Y → Y/f(X), y 7→ y + f(X). Since every morphism f has a kernel and a cokernel, the category (F − V ec) has kernels as well as cokernels. The same is true in the category (Ab) of abelian groups (written additively) and group morphisms.

ii) In the category (TVS) of topological vector spaces the kernel of a continous linear mapping f : X → Y is also the embedding f −1({0}) ,→ X, where f −1({0}) is endowed with the topology induced by X and the cokernel of f is also the quotient map Y → Y/f(X), where Y/f(X) is endowed with the quotient topology. Therefore the category (TVS) has kernels as well as cokernels. All this is the same in the category (LCS).

iii) The kernel of a morphism f : X → Y in the category (LCS)HD is the same as in the category (LCS), whereas the cokernel of f is the mapping Y → Y/f(X), y 7→ y + f(X), where Y/f(X) is endowed with the quotient topology.

Proposition 2.11. Let C be a preadditive category.

i) Kernels are monomorphisms.

ii) Cokernels are epimorphisms. 2.2. KERNELS AND COKERNELS 29

Proof. It suﬃces to show i), since ii) is the dual statement. If f : X → Y is a morphism in a category C, k : K → X a kernel of f and g1, g2 : T → K two morphisms with k ◦ g1 = k ◦ g2, then we have f ◦ k ◦ g1 = f ◦ k ◦ g2 = 0, hence there exists exactly one morphism λ: T → K with k ◦ λ = k ◦ g1 = k ◦ g2 and therefore we have g1 = g2, which shows that k is a monomorphism. Deﬁnition 2.12. Let C be a preadditive category and f : X → Y be a morphism in C. i) An image of f is a kernel of a cokernel of f, i.e. if c: Y → C is a cokernel of f, then an image of f is a morphism i: I → Y with c ◦ i = 0, so that for every other morphism t: T → Y with c ◦ t = 0 there exists a unique morphism λ: T → I making the diagram

i c I / Y / C O ? ~~ λ ~ ~~t ~~ T commutative.

ii) A coimage of f is a cokernel of a kernel of f, i.e. if k : K → X is a kernel of f, then a coimage of f is a morphism q : X → Q with q ◦ k = 0, so that for every other morphism s: X → S with s ◦ k = 0 there exists a unique morphism µ: Q → S making the diagram

k q K / X / Q ~ s ~ ~ µ ~ S commutative.

iii) The domain of an image i of f is called an image-object of f and the codomain of a coimage q of f is called a coimage-object of f. Remark 2.13. i) The coimage of a morphism is the dual notion of the image of a morphism.

ii) Remark 2.9.iii) shows that image and coimage of a morphism are uniquely determined uo to a unique isomorphism. Example 2.14.

i) In (F − V ec) the image of a morphism f : X → Y is the inclusion f(X) ,→ Y, y 7→ y and the coimage of f is the quotient map X → X/f −1({0}), x 7→ x + f −1({0}). 30 CHAPTER 2. ADDITIVE CATEGORIES

ii) In the category (TVS) of topological vector spaces the image of a continous linear mapping f : X → Y is also the inclusion f(X) ,→ Y , where f(X) is endowed with the topology induced by Y and the coimage of f is also the quotient map X → X/f −1({0}), where X/f −1({0}) is endowed with the quotient topology. All this is the same in the category (LCS).

iii) The image of a morphism f : X → Y in the category (LCS)HD is the inclusion f(X) ,→ Y, y 7→ y, where f(X) is endowed with the topology induced by Y . The coimage of f in (LCS)HD is the same as in the category (LCS), since f −1({0}) is a closed subspace of X when endowed with the induced topology. Notation 2.15. Let f : X → Y be a morphism in a category C. By 2.9.iii) we know that kernel and cokernel of f are unique up to a unique isomorphism. From now on kf : ker f → X will denote the kernel of f,

cf : Y → cok f the cokernel of f, if : im f → Y the image of f, and cif : X → coim f will denote the coimage of f. Deﬁnition 2.16. Let C be a preadditive category and let C0 be a full preadditive subcategory C0 of C. i) C0 is said to reﬂect kernels if whenever

f g X / Y / Z

is a sequence in C such that f is a kernel of g and Y and Z are objects of C0, it follows that also X is an object of C0 (and therefore f is also a kernel of g in C0).

ii) C0 is said to reﬂect cokernels if whenever

f g X / Y / Z

is a sequence in C such that g is a cokernel of f and X and Y are objects of C0, it follows that also Z is an object of C0 (and therefore g is also a kernel of f in C0). 2.2. KERNELS AND COKERNELS 31

Example 2.17.

i) (LCS) is a full preadditive subcategory of (TVS) that reﬂects both kernels and cokernels.

ii) (LCS)HD is a full preadditive subcategory of (TVS) that reﬂects kernels, but does not reﬂect cokernels.

iii) Let (BOR) be the category of bornological locally convex spaces and continous linear mappings (cf. [6, §23,1.5] and [6, §11,2.3]). Since quotients of bornological spaces are again bornological (cf. [6, §23,2.93]), the category (BOR) is a full preadditive subcategory of (TVS) that reﬂects cokernels. On the other hand subspaces of bornological spaces need not be bornological (cf. [19, 6.3]) and therefore (BOR) does not reﬂect the kernels of (TVS).

iv) Let (LCS)c be the category of complete locally convex spaces and continous linear mappings. It is a full preadditive subcategory of (TVS). For a morphism f : X → Y in (LCS)c the subspace f −1({0}) is complete with regard to the topology induced by X if and only if it is closed in X, hence (LCS)c does not reﬂect the kernels of (TVS). In addition, the quotient Y/f(X) is in general not a complete space (cf. [20, §31, 6]) and therefore (LCS)c does not reﬂect the cokernels of (TVS).

Remark 2.18. Let C be a preadditive category and let

f X / Y

α β

0 0 X g / Y be a commutative diagram in C. If f and g both possess a kernel, we have

g ◦ α ◦ kf = β ◦ f ◦ kf = 0, hence the universal property of the kernel kg gives rise to a unique morphism λ: ker f → ker g making the diagram

kf f ker f / X / Y λ α β ker g / X0 / Y 0 kg g commutative. By duality we obtain that, if f and g both possess a cokernel, there exists a 32 CHAPTER 2. ADDITIVE CATEGORIES unique morphism µ: cok f → cok g making the diagram

f cf X / Y / cok f α β µ cg 0 0 X g / Y / cok g commutative. Remark 2.19. Since the diagrams

idX X and X / X @ @@ f }} f @@ f }} @@ }} f @ ~}} Y / YY idY commute for every morphism f : X → Y , the identity idY : Y → Y is the kernel of the morphism Y → 0 and the identity idX : X → X is the cokernel of the morphism 0 → X. Lemma 2.20. Let f : X → Y be a morphism in a preadditive category C. Then the following are equivalent: i) f is a monomorphism.

ii) 0 → X is a kernel of f. Proof. i) ⇒ ii) Let t: T → X be a morphism with f ◦t = 0. Then f ◦t = f ◦0 and since f is a momomorphism it follows that t = 0, i.e. the diagram

t f T / X / Y > ~~ ~~ ~~ ~ 0 ~ commutes, which shows that 0 → X is a kernel of f. ii) ⇒ i) Let g1, g2 : T → X be two morphisms with f ◦ g1 = f ◦ g2. Then f ◦ (g1 − g2) = 0 and since 0 → X is a kernel of f the diagram

g1−g2 f T / X / Y > ~~ ~~ ~~ ~ 0 ~ commutes, which shows that g1 = g2. Hence f is a monomorphism. By duality we obtain: 2.2. KERNELS AND COKERNELS 33

Lemma 2.21. Let f : X → Y be a morphism in a preadditive category C. Then the following are equivalent:

i) f is an epimorphism.

ii) Y → 0 is a cokernel of f.

Lemma and Deﬁnition 2.22. Let f : X → Y be a morphism in a preadditive category C that possesses a kernel, a cokernel, an image and a coimage. Then there exists a unique morphism

fe: coim f → im f making the following diagram commutative:

kf f cf ker f / X / Y / cok f O cif if coim f ___ / im f fe

The factorization f = if ◦ fe◦ cif is called the canonical factorization of f. Proof. Consider the following diagram (in which we ﬁrst ignore the dotted arrows)

kf f cf ker f / X / Y / cok f . v: O v cif v if v λ1 v coim f ___ / im f fe

Since f ◦kf = 0 the universal property of cif gives rise to a unique morphism λ1 : coim f → Y with λ1 ◦ cif = f. Because of

cf ◦ f = cf ◦ λ1 ◦ cif = 0 and since cif is an epimorphism, it follows that cf ◦ λ1 = 0. The universal property of if then gives rise to a unique morphism fe: coim f → im f with λ1 = if ◦ fe. It follows that

f = λ1 ◦ cif = if ◦ fe◦ cif .

Deﬁnition 2.23. Let f : X → Y be a morphism in a preadditive category C that possesses a canonical factorization. The morphism f is called strict, if the induced morphism fe: coim f → im f is an isomorphism. 34 CHAPTER 2. ADDITIVE CATEGORIES

Example 2.24.

i) A morphism f : X → Y in the category (TVS) is strict if and only if it is open onto its range. The same is true for the category (LCS).

ii) A morphism f : X → Y in the category (LCS)HD is strict if and only it is open onto its range and has closed range.

Lemma 2.25. Let f : X → Y be a morphism in a preadditive category C that possesses a canonical factorization.

i) The kernel of f is also a kernel of the coimage of f.

ii) The cokernel of f is also a cokernel of the image of f.

Proof. It suﬃces to show i), since ii) is the dual statement of i). Naturally we have cif ◦ kf = 0. Let t: T → X be a morphism with cif ◦ t = 0, then we have also f ◦ t = if ◦ fe◦ cif ◦ t = 0, hence the universal property of the kernel gives rise to a unique morphism λ: T → ker f with t = kf ◦ λ and λ is unique with this property, since kf is a monomorphism. Hence kf is a kernel of cif .

Corollary 2.26. Let f : X → Y be a morphism in a preadditive category C that possesses a canonical factorization and is strict.

i) A morphism k : K → X is a kernel of f if and only if it is the kernel of cif : X → coim f.

ii) A morphism c: Y → C is a cokernel of f if and only if it is the cokernel of if : im f → coim Y .

Proof. It suﬃces to show i), since ii) is the dual statement. The direction ⇒ is just 2.25. If k is a kernel of cif and t: T → X is a morphism with f ◦ t = 0 and therefore if ◦ fe◦ cif ◦ t = 0 it follows that cif ◦ t = 0, since if ◦ fe is a monomorphism by assumption. Therefore the universal property of the kernel k gives rise to a unique morphism λ: T → K with t = k ◦ λ and λ is unique with this property, since k is a monomorphism.

Proposition 2.27. Let f : X → Y be a morphism in a preadditive category C that possesses a canonical factorization.

i) The morphism f is a strict monomorphism if and only if it is the kernel of its cokernel.

ii) The morphism f is a strict epimorphism if and only if it is the cokernel of its kernel. 2.2. KERNELS AND COKERNELS 35

Proof. Since ii) is the dual statement of i) it suﬃces to show i). If f is a strict monomorphism, then we have a canonical isomorphism fe: coim f → im f and by 2.20 we know that 0 → X is a kernel of f. Then idX : X → X is, as the cokernel of 0 → X, a coimage of f and thus the canonical factorization reads f = if ◦ fe. Since fe is an isomorphism, we see that f is its own image. If f is the kernel of its cokernel, then f is a monomorphism and thus cif is an isomorphism, since we have seen above that the identity idX : X → X is also a coimage of f. Then we have

f = f ◦ fe◦ cif and since f is a monomorphism and cif an isomorphism it follows that −1 fe = cif , hence f is a strict monomorphism. Notation 2.28. Because of the above proposition 2.27 a strict epimorphism in a preadditive category C is also called a cokernel and a strict monomorphism is also called a kernel. Corollary 2.29. Let f : X → Y be a morphism in a preadditive category C that possesses a canonical factorization. i) The kernel of f is a strict monomorphism. ii) The cokernel of f is a strict epimorphism. Proof. This follows directly from 2.25 and 2.27.

Proposition 2.30. Let f : X → Y be a morphism in a preadditive category C that possesses a canonical factorization. Then the following are equivalent: i) f is a strict morphism. ii) There exists a strict epimorphism f 0 : X → Z and a strict monomorphism f 00 : Z → Y with f = f 00 ◦ f 0.

Proof. i) ⇒ ii) We have f = if ◦ fe◦ cif , where fe is an isomorphism, and by 2.29 the morphism if is a strict monomorphism and fe◦ cif is a strict epimorphism. ii) ⇒ i) Since f 00 is a monomorphism, the properties f ◦t = f 00 ◦f 0 ◦t = 0 and f 0 ◦t = 0 are equivalent for every morphism t: T → X. Hence f and f 0 have isomorphic kernels. It follows that the coimage of f and f 0 are isomorphic and since f 0 is its own coimage by 2.27, the morphism f 0 is a coimage of f. Dually we obtain that f 00 is an image of f. Since the diagram

kf f cf ker f / X / Y / cok f O f 0 f 00 Z / Z idZ 36 CHAPTER 2. ADDITIVE CATEGORIES commutes and the above shows that this is the canonical factorization of f, it follows that fe = idZ , hence f is a homomorphism. Remark 2.31. The above proposition 2.30 shows that the following are equivalent for a morphism f : X → Y in a preadditive category C that possesses a canonical factorization: i) f is a strict morphism in C. ii) f op is a strict morphism in Cop.

2.3 Pullback and Pushout

Deﬁnition 2.32. Let f : F → E and g : G → E be two morphisms in a category C.A pullback of f and g is a triple (P, pG, pF ) consisting of an object P and morphisms pG : P → G, pF : P → F making the diagram

pF P / F

pG f G g / E commutative and possessing the following universal property: For every triple (Q, qG, qF ), where Q is an object and qG : Q → G, qF : Q → F are morphisms with f ◦ qF = g ◦ qG there exists a unique morphism λ: Q → P with qF = pF ◦ λ and qG = pG ◦ λ, i.e. the diagram

Q q ? F ? λ ? ? pF P / F qG pG f " G g / E is commutative. The category C is said to have pullbacks, if the pullback of every two morphisms f : F → E and g : G → E exists. From now on we say that the diagram

pF P / F

pG f G g / E is a pullback square, if (P, pG, pF ) is a pullback of f and g. 2.3. PULLBACK AND PUSHOUT 37

The dual notion of the above is the following:

Deﬁnition 2.33. Let f : E → F and g : E → G be two morphisms in a category C.A pushout of f and g is a triple (S, sG, sF ) consisting of an object S and morphisms sG : G → S, sF : F → S making the diagram

f E / F

g sF G / S sG commutative and possessing the following universal property: For every triple (T, tG, tF ), where T is an object and tG : G → T , tF : F → T are morphisms with sF ◦f = sG◦g there exists a unique morphism µ: Q → P with tF = µ ◦ sF and tG = µ ◦ sG, i.e. the diagram

f E / F

g sF tF G s / S G @@ @@ µ @@ @ tG / T is commutative. The category C is said to have pushouts, if the pushout of every two morphisms f : E → F and g : E → G exists. From now on we say that the diagram

f E / F

g sF G / S sG is a pushout square, if (S, sG, sF ) is a pullback of f and g.

Remark 2.34. The pullback (P, pG, pF ) of two morphisms f : F → E and g : G → E is uniquely determined up to a unique isomorphism: The proof is more or less analogous to that of the uniqueness of kernels and cokernels in 0 0 0 2.9.iii). If (P , pG, pF ) is another pullback of f and g there exists a unique 0 0 0 morphism λ: P → P with pG = pg ◦ λ and pF = pF ◦ λ, as well as a unique 0 0 0 0 0 0 morphism λ : P → P with pG = pG ◦ λ and pF = pF ◦ λ . The identities idP and idP 0 are the unique morphisms with pG = pG ◦ idP , pF = pF ◦ idP 38 CHAPTER 2. ADDITIVE CATEGORIES

0 0 0 0 and pG = pG ◦ idP 0 , pF = pF ◦ idP 0 respectively, and since we have 0 0 0 pG ◦ λ ◦ λ = pG ◦ λ = pG 0 0 0 pF ◦ λ ◦ λ = pF ◦ λ = pF 0 0 0 pG ◦ λ ◦ λ = pG ◦ λ = pG 0 0 0 pF ◦ λ ◦ λ = pF ◦ λ = pF

0 0 0 it follows that λ◦λ = idP and λ ◦λ = idP 0 , hence λ and λ are isomorphisms inverse to each other. By duality we obtain that the pushout (S, sG, sF ) of two morphisms f : E → F and g : E → G is uniquely determined up to a unique isomorphism. Example 2.35. i) If g : Y → Z and t: T → Z are two morphisms in the category (TVS), the space P := {(y, r) ∈ Y × T | g(y) = t(r)}, endowed with the topology induced by the product topology, together with the restrictions of the projections Y × T → Y and Y × T → T is a pullback of g and t in (TVS). For morphisms f : X → Y and t: X → T in (TVS) the space

S := (Y × T )/L

with L := {(f(x), t(x))| x ∈ X}, endowed with the quotient topology, together with the compositions of the canonical morphisms Y → Y ×T and T → Y × T with the quotient map Y × T → S is a pushout of f and t in (TVS). The category (LCS) has the same pullbacks and pushouts as (TVS).

ii) The pullback of two morphisms in the category (LCS)HD is the same as in the (TVS), whereas the pushout of two morphisms f : X → Y and t: X → T in (LCS)HD is the space

S := (Y × T )/L,

with L deﬁned as in i), together with the compositions of the canonical morphisms Y → Y × T and T → Y × T with the quotient map Y × T → S. Remark 2.36. Every commutative diagram

f X / Y

φ1 (1) φ2 X0 / Y 0 f 0 2.3. PULLBACK AND PUSHOUT 39 with isomorphisms φ1 and φ2 is a pushout as well as a pullback square. 0 In fact, if lX : Y → L and lX0 : X → L are morphisms with lX0 ◦ φ1 = lY ◦ f, then we have

−1 lY = lY ◦ φ2 ◦ φ2 −1 −1 0 lX0 = lY ◦ f ◦ φ1 = lY ◦ φ2 ◦ f −1 and lY ◦ φ2 is unique with this property, therefore (1) is a pushout square. The dual argument shows that (1) is also a pullback square. Proposition 2.37. Let C be category and

α1 α2 X / Y / Z

γ1 (1) γ2 (2) γ3 X0 / Y 0 / Z0 β1 β2 a commutative diagram. i) If (1) and (2) are pullback squares, then the outer rectangle (1) + (2) is a pullback square.

ii) If the outer ractangle (1) + (2) and (2) are pullback squares, then (1) is a pullback square.

iii) If (1) and (2) are pushout squares, then the outer rectangle (1) + (2) is a pushout square.

iv) If the outer ractangle (1) + (2) and (1) are pushout squares, then (2) is a pushout square. Proof. Since iii) and iv) are the dual statements of i) and ii) it suﬃces to show these two. 0 i) Let tX0 : T → X and tZ : T → Z be morphisms with γ3 ◦tZ = β2 ◦β1 ◦tX0 . The universal property of (2) then gives rise to a unique morphism λ1 : T → Y with β1 ◦ tX0 = γ2 ◦ λ1 and tZ = α2 ◦ λ1. Then the universal property of (1) gives rise to a unique morphism λ2 : T → X with tX0 = γ1 ◦ λ2 and λ1 = α1 ◦ λ2, i.e. making the diagram

tZ T ] Z A U P A λ K A 1 E λ2 A > α1 α2 X / Y / Z tX0 γ1 (1) γ2 (2) γ3 " X0 / Y 0 / Z0 β1 β2 40 CHAPTER 2. ADDITIVE CATEGORIES

commutative. Then α2 ◦ α1 ◦ λ2 = α2 ◦ λ1 = tZ and γ1 ◦ λ2 = tX0 . If µ: T → X is another morphism with α2 ◦ α1 ◦ µ = tZ and γ1 ◦ µ = tX0 , then β1 ◦ tX0 = β1 ◦ γ1 ◦ µ = γ2 ◦ α1 ◦ µ and this together with α2 ◦ α1 ◦ µ = tZ shows that α1 ◦ µ = λ1, because of the uniqueness of λ2. Then the uniqueness of λ2 yields λ2 = µ, which shows that the outer recatngle (1) + (2) is a pullback square. 0 ii) Let tX0 : T → X and tY : T → Y be morphisms with γ2 ◦ tY = β1 ◦ tX0 . Then γ3 ◦ α2 ◦ tY = β2 ◦ γ2 ◦ tY = β2 ◦ β1 ◦ tX0 , hence the universal property of the outer rectangle (1) + (2) gives rise to a unique morphism λ1 : T → X with α2 ◦ tY = α2 ◦ α1 ◦ λ1 and γ1 ◦ λ1 = tX0 .

T t A Y A A (3) λ1 A α1 α2 X / Y / Z tX0 γ1 (1) γ2 (2) γ3 " X0 / Y 0 / Z0 β1 β2

We have to show that the triangle (3) is commutative. Since (2) is a pullback square and the diagrams

T α2◦tY and T α2◦tY AA AA AA tY AAα1◦λ1 AA AA AA AA α2 α2 Y / Z Y / Z tX0 ◦β1 tX0 ◦β1 γ2 (2) γ3 γ2 (2) γ3 " " Y 0 / Z0 Y 0 / Z0 β1 β1 are both commutative, it follows that tY = α1 ◦ λ1. The morphism λ1 is unique with tX0 = γ1 ◦ λ1 and tY = α1 ◦ λ1 because of the universal property of the outer rectangle (1)+(2), hence (1) is a pullback square.

Proposition 2.38. Let C be a category and let

pF P / F

pG f G g / E be a pullback square. 2.3. PULLBACK AND PUSHOUT 41

i) If g is a monomorphism, then pF is a monomorphism.

ii) If g is a retraction, then pF is a retraction.

iii) If g is an isomorphism, then pF is an isomorphism.

Proof. i) Let α1, α2 : Q → P be morphisms with pT α1 = pT ◦ α2. Then we have g ◦ pY ◦ α1 = t ◦ pT ◦ α1 = t ◦ pT ◦ α2 = g ◦ pY ◦ α2 (∗) and since g is a monomorphism, it follows that pY ◦ α1 = pY ◦ α2. Therefore the diagram Q ? pF ◦α1 ? α ?? ?? 1 ?? ?? α2 ?? ? ? pF P / F pG◦α2 pG f * G g / E commutes and the universal property of the pullback gives α1 = α2, hence pF is a monomorphism. ii) Let h: Z → Y be a morphism with g ◦ h = idZ . Then we have

f ◦ idF = f = idZ ◦f = g ◦ h ◦ f, hence the universal property of the pullback gives rise to a unique morphism λ: F → P , making the diagram

F id @ F @ λ @ @ pF P / F h◦f pG f " G g / E commutative, which shows that pF is a retraction. iii) Follows immediately from i), ii) and 1.16.

By duality we get: Proposition 2.39. Let C be a category and let

f E / F

g sF G / S sG be a pushout square. 42 CHAPTER 2. ADDITIVE CATEGORIES

i) If f is an epimorphism, then sG is an epimorphism.

ii) If f is a coretraction, then sG is a coretraction.

iii) If f is an isomorphism, then sG is an isomorphism.

2.4 Product, Coproduct and Biproduct

Deﬁnition 2.40. Let C be a category and X and Y be two objects of C.A product of X and Y is a triple (P, πX , πY ), where P is an object of C and πX : P → X,πY : P → Y are morphisms that fulﬁll the following universal property: For every triple (T, tX , tY ) with T ∈ Ob(C) and morphisms tX : T → X, tY : P → Y there exists a unique morphism λ: T → P with tX = πX ◦ λ and tY = πY ◦ λ, i.e. making the diagram

tX / X π ~> X ~~ ~~ λ ~~ T ___ / P @ @@ @@ πY @@ @ tY / Y commutative. The category C is said to have products, if a product of every two objects exists. The dual notion of the above is the following: Deﬁnition 2.41. Let C be a category and X and Y be two objects of C.A coproduct of X and Y is a triple (S, ωX , ωY ), where S is an object of C and ωX : X → S,ωY : Y → S are morphisms that fulﬁll the following universal property: For every triple (R, rX , rY ) with R ∈ Ob(C) and morphisms rX : X → R, rY : Y → R there exists a unique morphism µ: S → R with rX = µ ◦ ωX and rY = µ ◦ ωY , i.e. making the diagram

X r @ X @ ω @@ X @@ @ µ S ___ / R ~? B ~~ ~~ωY ~~ Y rY commutative. 2.4. PRODUCT, COPRODUCT AND BIPRODUCT 43

The category C is said to have coproducts, if the coproduct of every two objects exists.

Remark 2.42. The product (P, πX , πY ) of two objects X and Y in a category C is uniquely determined up to a unique isomorphism: The proof is in strict analogy to the uniqueness of pullback and pushout shown in 2.34. If 0 0 0 (P , πX , πY ) is another product of X and Y there exists a unique morphism 0 0 0 λ: P → P with πX = πX ◦λ and πY = πY ◦λ, as well as a unique morphism 0 0 0 0 0 0 λ : P → P with πX = πX ◦ λ and πY = πY ◦ λ . The identities idP and idP 0 are the unique morphisms with πX = πX ◦ idP , πY = πY ◦ idP and 0 0 0 0 πX = πX ◦ idP 0 , πY = πY ◦ idP 0 respectively, and since we have 0 0 0 πX ◦ λ ◦ λ = πX ◦ λ = πX 0 0 0 πY ◦ λ ◦ λ = πY ◦ λ = πY 0 0 0 πX ◦ λ ◦ λ = πX ◦ λ = πX 0 0 0 πY ◦ λ ◦ λ = πY ◦ λ = πY

0 0 0 it follows that λ◦λ = idP and λ ◦λ = idP 0 , hence λ and λ are isomorphisms inverse to each other. By duality we obtain that the coproduct (S, ωX , ωY ) of two objects X and Y is uniquely determined up to a unique isomorphism. Notation 2.43. Let C be a category. Since by 2.42 the product of two objects X and Y in C is uniquely determined up to a unique isomorphism we will write from now on

(X × Y, πX , πY ) for the product of X and Y , and

(X ⊕ Y, ωX , ωY ) for the coproduct of X and Y . Example 2.44. In the category (TVS) the product of two objects X and Y is the space X × Y = {(x, y)| x ∈ X, y ∈ Y }, endowed with the product topology, together with the canonical maps X × Y →, (x, y) 7→ x and X × Y →, (x, y) 7→ y. The coporoduct of X and Y is the same space X × Y together with the canonical maps X → X × Y, x 7→ (x, 0) and X → X × Y, y 7→ (0, y). Deﬁnition 2.45. Let C be a category and let C0 be a full subcategory of C. i) C0 is said to reﬂect products if for all objects X and Y in C0 the product 0 X × Y in C is also an object of C (and therefore (X × Y, πX , πY ) is also a product of X and Y in C0). 44 CHAPTER 2. ADDITIVE CATEGORIES

ii) C0 is said to reﬂect coproducts if for all objects X and Y in C0 the coproduct X ⊕ Y in C is also an object of C0 (and therefore (X ⊕ 0 Y, ωX , ωY ) is also a coproduct of X and Y in C ). Example 2.46. All full subcategories of (TVS) discussed this far reﬂect the products and coproducts of (TVS).

Remark 2.47. Let C be a category that possesses a zero object 0 and let X and Y be two objects of C. A triple (P, πX , πY ) is a product of X and Y if and only if the commutative diagram

πX P / X

πY Y / 0 is a pullback square.

Dually, a triple (S, ωX , ωY ) is a coproduct of X and Y if and only if the commutative diagram 0 / X

ωX Y / S ωY is a pushout square.

Deﬁnition 2.48. Let C be a preadditive category and let X and Y be two objects of C.A biproduct of X and Y is a tuple (B, πX , πY , ωX , ωY ), with B ∈ Ob(C) and morphisms

ωX πY / / X o B o Y πX ωY that satisfy the following equations:

πX ◦ ωX = idX , πX ◦ ωY = 0

πY ◦ ωY = idX , πY ◦ ωX = 0

idB = ωX ◦ πX + ωY ◦ πY

The category C is said to have biproducts, if the biproduct of every two objects exists.

Remark 2.49. Let C be a preadditive category and let

(B, πX , πY , ωX , ωY ) be a biproduct of two objects X and Y in C. 2.4. PRODUCT, COPRODUCT AND BIPRODUCT 45

i) It is clear that πX and πY are retractions and ωX and ωY are coretractions.

ii) If D is another preadditive category and F : C → D is an additive functor (covariant or contravariant) we have F (0) = 0 by 2.7. This together with the functor properties and the additivity of F shows that (F (B),F (πX ),F (πY ),F (ωX ),F (ωY )) is a biproduct in D.

op op op op op iii) The tuple (B, ωX , ωY , πX , πY ) is a biproduct of X and Y in C , since the duality functor op : C → Cop is additive. Hence the notion of biproduct is self-dual.

Proposition 2.50. Let C be a preadditive category and let

(B, πX , πY , ωX , ωY ) be a biproduct of X,Y ∈ Ob(C).

i) ωX is a kernel of πY and ωY is a kernel of πX .

ii) πY is a cokernel of ωX and πX is a cokernel of ωY .

op op op op Proof. By 2.49.iii) we know that (B, , ωX , ωY , πX , πY ) is a biproduct of X and Y in Cop, hence it suﬃces to show i), since ii) can be obtained by duality. Let t: T → B be a morphism with πY ◦ t = 0, then we also have ωY ◦ πY ◦ t = 0, hence

(idB −ωX ◦ πX ) ◦ t = 0 and therefore the diagram

t T / B > }} πX ◦t } }}ωX }} X is commutative. If λ: T → X is another morphism with t = ωX ◦ λ, then πX ◦t = πX ◦ωX ◦λ = λ. This shows that ωX is a kernel of πY . Analogously one shows that ωY is a kernel of πX .

Proposition 2.51. Let (B, πX , πY , ωX , ωY ) be a biproduct of two objects X and Y in a preadditive category C.

i) (B, πX , πY ) is a product of X and Y .

ii) (B, ωX , ωY ) is a coproduct of X and Y . 46 CHAPTER 2. ADDITIVE CATEGORIES

Proof. It suﬃces to show i), since ii) can be obtained by duality. Let tX : T → X and tY : T → Y be morphisms in C. Deﬁne λ := ωX ◦tX +ωY ◦tY , then we have

πX ◦ λ = πX ◦ (ωX ◦ tX + ωY ◦ tY ) = tX ,

πY ◦ λ = πY ◦ (ωX ◦ tX + ωY ◦ tY ) = tY .

0 0 0 If λ : T → B is another morphism with πX ◦ λ = tX and πY ◦ λ = tY , it follows that

0 0 λ = ωX ◦ tX + ωY ◦ tY = ωX ◦ πX ◦ λ + ωY ◦ πY ◦ λ , hence (B, πX , πY ) is a product of X and Y . Notation 2.52. Let C be a preadditive category and X,Y ∈ Ob(C). The biproduct (B, πX , πY , ωX , ωY ) of X and Y is uniquely determined up to 0 0 0 0 0 a unique isomorphism: Let (B , πX , πY , ωX , ωY ) be another biproduct of 0 0 0 X and Y . Since by 2.51 the two triples (B, πX , πY ) and (B , πX , πY ) are 0 0 products, there exists a unique isomorphism λ: B → B with πX ◦ λ = πX 0 and πY ◦ λ = πY , and since, again by 2.51 the two triples (B, ωX , ωY ) and 0 0 0 0 0 (B , ωX , ωY ) are coproducts, there exists a unique isomorphism λ : B → B 0 0 0 0 with λ ◦ ωX = ωX and λ ◦ ωY = ωY . Then 0 0 λ = λ ◦ idB 0 = λ ◦ (ωX ◦ πX + ωY ◦ πY ) 0 0 = λ ◦ ωX ◦ πX + λ ◦ ωY ◦ πY 0 0 = ωX ◦ πX + ωY ◦ πY 0 0 0 0 = ωX ◦ πX ◦ λ + ωY ◦ πY ◦ λ 0 0 0 0 = (ωX ◦ πX + ωY ◦ πY ) ◦ λ = idB0 ◦λ = λ which shows then that the biproduct itself is uniquely determined up to a unique isomorphism. Because of the above we will write from now on

(X q Y, πX , πY , ωX , ωY ) for the biproduct of X and Y . Example 2.53. The biproduct of two objects X and Y in the category (TVS) is the product space X × Y together with the canonical projections X × Y → X, X × Y → Y and the canonical embeddings X → X × Y , Y → X × Y . Proposition 2.54. Let C be a preadditive category. Then the following are equivalent: 2.4. PRODUCT, COPRODUCT AND BIPRODUCT 47

i) C has products. ii) C has coproducts. iii) C has biproducts.

Proof. i) ⇒ iii) Let X and Y be two objects of C. The morphisms idX : X → X and 0: X → Y induce a unique morphism ωX : X → X × Y making the diagram

idX / X w; πX ww ww ww ωX ww X ___ / X × Y GG GG GG πY GG G# 0 / Y commutative. Analagously the morphisms idY : Y → Y and 0: Y → X induce a unique morphism ωY : Y → X ×Y with πX ◦ωY = 0 and πY ◦ωY = idY . Then we have

πX ◦ (ωX ◦ πX + ωY ◦ πY ) = πX ◦ ωX ◦ πX + πX ◦ ωY ◦ πY = πX

πY ◦ (ωX ◦ πX + ωY ◦ πY ) = πY ◦ ωX ◦ πX + πY ◦ ωY ◦ πY = πY , hence the universal property of the product yields ωX ◦πX +ωY ◦πY = idX×Y , which shows that (X × Y πX , πY , ωX , ωY ) is a biproduct. ii) ⇒ iii) is the dual argument of i) ⇒ iii): If two objects X and Y possess a coproduct in C, they possess a product in Cop and hence by i) ⇒ iii) a biproduct in Cop. By 2.49.ii) this means also that X and Y possess a biproduct in C. iii) ⇒ i) and iii) ⇒ ii) are already clear by 2.51.

Proposition 2.55. Let C be a preadditive category. i) If C has products and cokernels, then C has pushouts. ii) If C has coproducts and kernels, then C has pullbacks. Proof. Since ii) is the dual statement of i), it suﬃces to show i). Let f : E → F and g : E → G be morphisms in C. Then f and −g induce a unique morphism λ: E → F × G making the diagram

f / F x; πF xx xx xx λ xx E ___ / F × G GG GG GG πG GG G# −g / G 48 CHAPTER 2. ADDITIVE CATEGORIES commutative. Let (F × G πF , πG, ωF , ωG) be the biproduct constructed as in the proof of 2.54 and let cλ : F × G → cok λ be the cokernel of λ. Deﬁne then

sF := cλ ◦ ωF : F → cok λ,

sG := cλ ◦ ωG : G → cok λ.

Then we have

−(sG ◦ g) + sF ◦ f = −(cλ ◦ ωG ◦ g) + cλ ◦ ωF ◦ f

= cλ ◦ ωG ◦ (−g) + cλ ◦ ωF ◦ f

= cλ ◦ ωG ◦ πG ◦ λ + cλ ◦ ωF ◦ πF ◦ λ

= cλ ◦ (ωG ◦ πG + ωF ◦ πF ) ◦ λ

= cλ ◦ idF ×G ◦λ = 0 hence the diagram f E / F

g sF G / cok λ sG is commutative. Let then tF : F → T and tG : G → T be two morphisms with tF ◦ f = tG ◦ g. Then

(tG ◦ πG + tF ◦ πF ) ◦ λ = tG ◦ πG ◦ λ + tF ◦ πF ◦ λ

= tG ◦ (−g) + tF ◦ f

= −(tG ◦ g) + tF ◦ f = 0, hence the universal property of the cokernel gives rise to a unique morphism µ: cok λ → T with µ ◦ cλ = tG ◦ πG + tF ◦ πF . This yields

µ ◦ sF = µ ◦ cλ ◦ ωF

= (tG ◦ πG + tF ◦ πF ) ◦ ωF

= tG ◦ πG ◦ ωF + tF ◦ πF ◦ ωF

= tF as well as

µ ◦ sG = µ ◦ cλ ◦ ωG

= (tG ◦ πG + tF ◦ πF ) ◦ ωG

= tG ◦ πG ◦ ωG + tF ◦ πF ◦ ωG

= tG 2.4. PRODUCT, COPRODUCT AND BIPRODUCT 49 and therefore the diagram

f E / F

g sF tF G / cok λ sG E EE EE µ EE E" tG / T

0 0 is commutative. If µ : cok λ → T is another morphism with µ ◦ sG = tG 0 and µ ◦ sF = tF then

0 0 µ ◦ cλ = µ ◦ cλ ◦ (ωG ◦ πG + ωF ◦ πF ) 0 0 = µ ◦ cλ ◦ ωG ◦ πG + µ ◦ cλ ◦ ωF ◦ πF 0 0 = µ ◦ sG ◦ πG + µ ◦ sF ◦ πF

= tG ◦ πG + tF ◦ πF , hence it follows that µ0 = µ because µ is unique with this property. This shows that (cok λ, sF , sG) is a pushout of f and g.

Corollary 2.56. Let C be a preadditive category with kernels and cokernels. Then the following are equivalent:

i) C has products.

ii) C has coproducts.

iii) C has biproducts.

iv) C has pullbacks.

v) C has pushouts.

Deﬁnition 2.57. Let C be a category and let C0 be a full subcategory of C.

i) C0 is said to reﬂect pullbacks if whenever

pT P / T

pY PB t Y g / Z

is a pullback square in C with T,Y,Z ∈ Ob(C0) it follows that also P ∈ Ob(C0) (and therefore the above diagram is a pullback square in C0). 50 CHAPTER 2. ADDITIVE CATEGORIES

ii) C0 is said to reﬂect pushouts if whenever

f X / Y

t PO sY T / S sT

is a pushouts square in C with X,Y,T ∈ Ob(C0) it follows that also S ∈ Ob(C0) (and therefore the above diagram is a pushout square in C0).

Example 2.58.

i) (LCS) is a full subcategory of (TVS) that reﬂects pullbacks and pushouts.

ii) (LCS)HD is a full subcategory of (TVS) that reﬂects pullbacks but does not reﬂect pushouts.

iii) (BOR) is a full subcategory of (TVS) that reﬂects pushouts, but does not reﬂect pullbacks (see the references in 2.17.iii)).

iv) (LCS)c is a full subcategory of (TVS) that does neither reﬂect pullbacks nor pushouts (see the references in 2.17.iv)).

2.5 Additive Categories

Deﬁnition 2.59.

i) A category C is called additive, if it has the following properties:

(A1) The category C is preadditive. (A2) The category C has products.

ii) A full additive subcategory C0 of an additive category C is a full preadditive subcategory that also reﬂects products.

Remark 2.60.

i) By 2.54 a preadditive category C possesses products if and only if it possesses biproducts. Therefore one can equivalently deﬁne an additive category as a preadditive category which possesses biproducts.

ii) A full additive subcategory of an additive subcategory is itself an additive category.

Deﬁnition 2.61. 2.5. ADDITIVE CATEGORIES 51

i) A category C is called preabelian, if it is additive and has kernels and cokernels. ii) A full preabelian subcategory C0 of a preabelian category C is a full additive subcategory that also reﬂects kernels and cokernels. Example 2.62.

i) The categories (TVS), (LCS) and (LCS)HD are preabelian by 2.10.

c ii) Let (LCS)sm be the category of complete semi-metrizable locally convex spaces and continous linear maps. This category is not preabelian, since there are morphisms in this category that do not have a kernel. Consider for example the sequence space l1 and the quotient map q : l1 → l1/φ, where φ is the space of sequences with only ﬁnitely many nonzero values. Since φ is a dense subspace of l1, the quotient l1/φ has the coarsest topology and is therefore a complete semi-metrizable c space, hence q is a morphism in (LCS)sm. Suppose k : K → l1 is c a kernel of q in (LCS)sm. For every y ∈ φ we have the morphism c υy : C → l1, λ 7→ λy, which is a morphism in (LCS)sm with q ◦ υy = 0, hence the universal property of the kernel k gives rise to a unique factorization υy / l1 C @ ? @ ~~ @ ~~ µy ~~ k @ ~~ K c in (LCS)sm, which shows that φ ⊆ k(K). The universal property of the kernel φ ,→ l1 in (LCS) yields a factorization

k K / l1 > Ñ@ > ÑÑ j > ÑÑ > 0 ÑÑ φ

c for an injective morphism j, since k is a monomorphism in (LCS)sm and thus injective. Since φ ⊆ k(K), the map j is even bijective. In addition, K is a Hausdorﬀ space because of k being injective, hence a Fr´echet space and therefore a webbed space. Furthermore, φ = n ∪n∈NC is an ultrabornological space, hence it follows from DeWilde’s open mapping theorem (cf. [15, 24.30]) that j is an isomorphism, in contradiction to K being a complete space. Therefore the morphism c q does not have a kernel in (LCS)sm. Remark 2.63. i) A full preabelian subcategory of a preabelian subcategory is itself a preabelian category. 52 CHAPTER 2. ADDITIVE CATEGORIES

ii) If a category C is preabelian, it has pullbacks and pushouts by 2.56.

iii) It follows from the proof of 2.55 that a full preabelian subcategory of a preabelian category reﬂects pullbacks and pushouts.

iv) Any morphism f : X → Y in a preabelian category C possesses a canonical factorization

kf f cf ker f / X / Y / cok f O cif if coim f ___ / im f fe

by lemma 2.22.

v) Let C be a preabelian category and let C0 be a full preabelian subcategory of C. Then the canonical factorization in C of a morphism f : X → Y with X,Y ∈ Ob(C0) coincides with the canonical factorization of f in C0, hence it follows from 1.35 that f is strict in C0 if and only if it is strict in C.

c Example 2.64. Let (LCS)HD be the category of complete Hausdorﬀ locally convex spaces and continous linear maps. In the chain

c (TVS) ⊇ (LCS) ⊇ (LCS)HD each category is a full additive subcategory of the categories above them c and (LCS) is even a full preabelian subcategory of (TVS), but (LCS)HD is not a full preabelian subcategory of neither (TVS) nor (LCS) since it does c not reﬂect their cokernels. However, (LCS)HD is a preabelian category in its own right: Since closed subspaces of complete spaces are again complete, c (LCS)HD reﬂects the kernels of (LCS) and is therefore a category that has c kernels. For a morphism f : X → Y in (LCS)HD the quotient Y/f(X) is not complete in general (see the references in 2.17.iv)), but if C(Y/f(X)) is the completion of Y/f(X) and j : Y/f(X) → C(Y/f(X)) is the canonical morphism, the composition

q j Y / Y/f(X) / C(Y/f(X))

c c is a cokernel of f in (LCS)HD and therefore (LCS)HD has cokernels, hence it is a preabelian category.

Deﬁnition 2.65. Let C be a preadditive category and let X1,...,Xn be objects of C. A biproduct of (Xk)k=1,...,n is a tuple

(B, (πXk )k=1,...,n, (ωXk )k=1,...,n), 2.5. ADDITIVE CATEGORIES 53 where B is an object of C and

πXk : B → Xk, ωXk : Xk → B are morphisms in C, deﬁned for each k = 1, . . . , n, such that the following equations are satisﬁed:

πXk ◦ ωXk = idXk for k = 1, . . . , n,

πXk ◦ ωXl = 0 for k 6= l, n X ωXk ◦ πXk = idB . k=1 Remark 2.66.

i) Let C be an additive category and let X1,...,Xn be objects of C, then a biproduct of (Xk)k = 1, . . . , n can be constructed inductively from the biproducts of two objects: Suppose that a biproduct

(B , (π0 ) , (ω0 ) ) m−1 Xk k=1,...,m−1 Xk k=1,...,m−1

of (Xk)k=1,...,m−1 has already been constructed. Let

(Bm−1 q Xm, πBm−1 , πXm , ωBm−1 , ωXm )

be the biproduct of Bm−1 and Xm and deﬁne

B := Bm−1 q Xm π0 ◦ π for k = 1, . . . , m − 1 Xk Bm−1 πXk := πXm for k = m ω ◦ ω0 for k = 1, . . . , m − 1 Bm−1 Xk ωXk := . ωXm for k = m

Then (B, (πXk )k=1,...,m, (ωXk )k=1,...,m) is a biproduct of (Xk)k=1,...,m.

ii) If (B, (πXk )k=1,...,n, (ωXk )k=1,...,n) is a biproduct of (Xk)k=1,...,n, then,

in analogy to the case n = 2 (see 2.51), the tuple (B, (πXk )k=1,...,n) has the following universal property:

For any tuple (T, (tXk )k=1,...,n), where T is an object of C and where

tXk : T → Xk is a morphism for k = 1, . . . , n there exists a unique morphism λ: T → B making the diagram

λ T / B BB | BB | t B ||πX Xk }| k Xk 54 CHAPTER 2. ADDITIVE CATEGORIES

commutative for k = 1, . . . , n.

Dually, for any tuple (S, (sXk )k=1,...,n), where S is an object of C and

where sXk : Xk → S is a morphism for k = 1, . . . , n there exists a unique morphism µ: B → S making the diagram

µ B / S aB > BB }} ω B }s Xk B }} Xk Xk

commutative for k = 1, . . . , n.

iii) As in the case of n = 2 (see 2.52), it follows from the two universal properties above that the biproduct of X1,...,Xn ∈ Ob(C) is uniquely determined up to a unique isomorphism. Notation 2.67. Because of 2.66.iii) we will write from now on

n (qk=1Xk, (πXk ), (ωXk )) for the biproduct of objects X1,...,Xn in a preadditive category C.

Remark and Deﬁnition 2.68. Let C be an additive category. If X1,...,Xn and Y1,...,Ym are objects of C and fji : Xi → Yj are morphisms in C for all i = 1, . . . , n and j = 1, . . . , m, then the coproduct universal property of the biproduct (see 2.66.ii)) gives rise to unique morphisms

n λj : qi=1 Xi → Yj with fji = λj ◦ωXi for i = 1, . . . , n and j = 1, . . . , m. In addition the product universal property of the biproduct gives rise to a unique morphism

n m λ: qi=1 Xi → qj=1Yj with λj = πYj ◦ λ for j = 1, . . . , m, therefore we have

fji = πYj ◦ λ ◦ ωXi (?) for i = 1, . . . , n and j = 1, . . . , m. It follows from the uniqueness of λ and the λj that λ is unique with the property (?). In addition it follows that every n m morphism f : qi=1 Xi → qj=1Yj is uniquely determined by the morphisms n πYj ◦ f ◦ ωXi . This allows us to identify the morphisms from qi=1Xi to m qj=1Yj with m × n-matrices

n m (fji)j=1,...,m := f : qi=1 Xi → qj=1Yj, i=1,...,n where fji := πYj ◦ f ◦ ωXi for j = 1, . . . , m and i = 1, . . . , n. This notation yields a multiplication law for matrices: 2.5. ADDITIVE CATEGORIES 55

Claim: For two morphisms

n m f = (fji)j=1,...,m : qi=1 Xi → qj=1Yj, i=1,...,n m l g = (gji) k=1,...,l : qj=1 Yj → qk=1Zk j=1,...,m in C, we have

g ◦ f = (hki) k=1,...,l i=1,...,n Pm with hki := t=1 gkt ◦ fti. Indeed, the morphism g ◦ f is uniquely determined by the morphisms

m X πZk ◦ (g ◦ f) ◦ ωXi = πZk ◦ g ◦ ( ωYt ◦ πYt ) ◦ f ◦ ωXi t=1 m X = πZk ◦ g ◦ ωYt ◦ πYt ◦ f ◦ ωXi t=1 m X = gkt ◦ fti, t=1

n which establishes the claim. If (qk=1Xk, (πXk ), (ωXk )) is a biproduct of X1,...,Xn ∈ Ob(C) we will, because of the above multiplication law for matrices, from now on often identify the morphism ωXk with the k-th unit vector and πXk with the transposed of the k-th unit vector. In addition we will often just write 1 for the identity morphism, if it appears as a coeﬃcient of a matrix, e.g. 1 0 : X q Y → X q Y 0 1 will denote the identity morphism on X q Y .

Notation 2.69. If f : X → Y and f 0 : X0 → Y 0 are morphisms in a category C we set f 0 f ⊕ f 0 := : X q X0 → Y q Y 0. 0 f 0 Lemma 2.70. Let C be an additive category, let f : X → Y be a morphism in C and let A be an object of C.

i) The diagram

idA ⊕f A q X / A q Y

πX πY X / Y f is a pullback and a pushout square. 56 CHAPTER 2. ADDITIVE CATEGORIES

ii) The diagram f X / Y

ωX ωY A q X / A q Y idA ⊕f is a pullback and a pushout square. Proof. It suﬃces to show i), since ii) can be obtained by duality. lA If lX : L → X and : L → A q Y are morphisms in C such that lY lA lY = ( 0 1 ) = f ◦ lX , lY then the diagram ( lA ) lY L GG ( lA ) GG lX GG GG G# idA ⊕f A q X / A q Y lX πX πY % X / Y f

0 α is commutative. If λ = β : L → A q X is another morphism such that 0 0 0 πX ◦ λ = lX and (idA ⊕f) ◦ λ = lAqY , then we have β = πX ◦ λ = lX and lA 0 1 0 α α = (idA ⊕f) ◦ λ = = , lY 0 f lX lY 0 hence it follows that α = lA and thus λ = λ. This shows that the diagram in i) is a pullback square. In addition, if lX : X → L and ( lA lY ): A q Y → L are morphisms in C such that 1 0 ( l l ) = l ◦ π = ( 0 l ), A Y 0 f X X X then the diagram

idA ⊕f A q X / A q Y

πX πY ( l l ) A Y X / Y f FF FF lY FF FF F# lX / L is commutative and lY is unique with this property. This shows that the diagram in i) is also a pushout square. 2.5. ADDITIVE CATEGORIES 57

Lemma 2.71. Let C be an additive category.

i) If g : Y → Z, t: T → Z are morphisms in C such that g has a kernel and (P, pT , pY ) is their pullback, then there is a morphism j : ker g → P making the diagram

j pT ker g / P / T

pY t kerg / Y / Z kg g

commutative and being a kernel of pT .

ii) If f : X → Y , t: X → T are morphisms in C such that f has a cokernel and (S, sT , sY ) is their pushout, then there is a morphism c: S → cok f making the diagram

f cf X / Y / cok f

t sY T / S / cok f sT c

commutative and being a cokernel of sT .

Proof. It is enough to show i), since ii) can be obtained dually. Because of g ◦ kg = 0 = t ◦ 0, the universal property of (P, pT , pY ) gives rise to a unique morphism j : ker g → P making the diagram

ker g 0 D D j D D! pT P / T kG pY t $ Y g / Z commutative. We show that j is a kernel of pT : Let h: H → P be a morphism with pT ◦ h = 0. Then g ◦ pY ◦ h = t ◦ pT ◦ h = 0, therefore the universal property of ker g yields a unique morphism λ: H → ker(g) with pY ◦ h = kG ◦ λ, hence pY ◦ h = pY ◦ j ◦ λ and pT ◦ h = 0 = pT ◦ j ◦ λ. Thus the universal property of (P, pT , pY ) provides that h = j ◦ λ and that λ is unique with this property. This shows that j is a kernel of pT . 58 CHAPTER 2. ADDITIVE CATEGORIES

2.6 Semi-abelian Categories

Deﬁnition 2.72. Let C be a preabelian category. i) The category C is called semi-abelian if for every morphism f : X → Y in C the morphism fe: coim f → im f in the canonical factorization

kf f cf ker f / X / Y / cok f O cif if coim f / im f fe

is a bimorphism.

ii) The category C is called abelian if for every morphism f : X → Y in C the morphism fe: coim f → im f is an isomorphism, i.e. every morphism is a strict morphism. Example 2.73. i) Every abelian category C is semi-abelian.

ii) In the preabelian categories (F − V ec) and (Ab) we have the canonical isomorphism fe: X/f −1({0}) → f(X) for a morphism f : X → Y , which shows that these categories are abelian.

iii) In the preabelian category (TVS) the morphism

fe: X/f −1({0}) → f(X)

is always bijective and therefore a bimorphism, which shows that (TVS) is a semi-abelian category. However it is not an abelian category, since fe need not have a continous inverse.

c iv) Let X be an object of the preabelian category (LCS)HD and let A ⊆ X be a closed subspace of X such that the quotient X/A is not a complete space (see the references in 2.17.iv) for an example). Let Y := C(X/A) be the Hausdorﬀ completion of X/A and let j : X/A → Y be the canonical morphism. Let q : X → X/A denote the quotient map and deﬁne p := j ◦ q. Let y0 ∈ Y \j(X/A) and consider the map

r : X × C → Y, (x, λ) 7→ p(x) − λy0.

Since p(X) is dense in Y , it follows that r(X × C) is also dense in Y , c hence r is an epimorphism in (LCS)HD and therefore the identity on Y 2.6. SEMI-ABELIAN CATEGORIES 59

−1 is an image of r. The kernel of r is the inclusion p (0)×{0} → X ×C, hence the coimage of r is the morphism

−1 cir : X × C → C(X × C/p ({0}) × {0}) = Y × C, (x, λ) 7→ (p(x), λ)

(see 2.64). The canonical factorizatiom of r then has the form

r X × C / Y cir Y × C / Y re

with re(y, λ) = y − λy0. Since re(y0, 1) = 0, the map re is not injective and therefore not a bimorphism. c This shows that (LCS)HD is a preabelian category that is not semi- abelian.

Proposition 2.74. Let C be a semi-abelian (resp. an abelian) category. if C0 is a full preabelian subcategory of C, then C0 is also semi-abelian (resp. abelian).

Proof. The category C0 is again preabelian. Let f : X → Y be a morphism in C0 and let kf f cf ker f / X / Y / cok f O cif if coim f / im f fe be its canonical factorization in C. Since C0 reﬂects kernels and cokernels this is also the canonical factorization of f in C0. Since fe is a bimorphism (resp. an isomorphism) in C it is also a bimorphism in C0 (resp. an isomorphism in C0 by 1.35, since the inclusion functor C0 → C is fully faithful).

Proposition 2.75. Let f : X → Y be a morphism in a semi-abelian category C.

i) A morphism k : K → X is a kernel of f if and only if it is the kernel of cif : X → coim f. ii) A morphism c: Y → C is a cokernel of f if and only if it is the cokernel of if : im f → coim Y . Proof. The proof of 2.26 also works in this case.

Proposition 2.76. Let C be a semi-abelian category.

i) If f = h◦g is a strict monomorphism, then g is a strict monomorphism. 60 CHAPTER 2. ADDITIVE CATEGORIES

ii) If f = h ◦ g is a strict epimorphism, then h is a strict epimorphism. Proof. It suﬃces to show i), since ii) is the dual statement. We know by 1.10 that g is a monomorphism. Then idX is a coimage of g and we have g = ig ◦ge. Analogously we get f = if ◦fe. We show that ge is an isomorphism: Let cf : Y → cok f be the cokernel of f, then

cf ◦ h ◦ ig ◦ ge = cf ◦ h ◦ g = cf ◦ f = 0 and since ge is a bimorphism, it follows cf ◦h◦ig = 0. The universal property of f then gives rise to a unique morphism v : im g → im f with if ◦ h ◦ ig. Then if ◦ v ◦ ge = h ◦ ig ◦ ge = h ◦ g = f = if ◦ f,e hence it follows that v ◦ ge = fe, since if is a monomorphism. Then we have −1 fe ◦ v ◦ ge = idcoim g . −1 In addition we have ge ◦ fe ◦ v ◦ ge = ge = idim g ◦ge, hence it follows that −1 ge ◦ fe ◦ v = idim g since ge is an epimorphism. This shows that ge is an isomorphism and therefore g is a strict monomorphism.

Proposition 2.77. Let C be a semi-abelian category. i) If f : X → Y and g : Y → Z are strict monomorphisms, then g ◦ f is a strict monomorphism. ii) If f : X → Y and g : Y → Z are strict epimorphisms, then g ◦ f is a strict epimorphism. Proof. Since ii) is the dual statement of i) it suﬃces to show i) Since f and g are strict monomorphisms they are their own images by 2.27 and since g ◦ f is a monomorphism, we have

g ◦ f = ig◦f ◦ g]◦ f. We show that g ◦ f is its own image, then it is a strict monomorphism by 2.27. We have 0 = cg ◦ g ◦ f = cg ◦ ig◦f ◦ g]◦ f, therefore it follows that cg ◦ ig◦f = 0, since g]◦ f is a bimorphism. The universal property of the image ig then gives rise to a unique morphism v : im g ◦ f → Y = im g, making the diagram

ig◦f im g ◦ f / ; Z ww ww v ww ww g ww Y 2.6. SEMI-ABELIAN CATEGORIES 61 commutative. Then it follows from

g ◦ f = ig◦f ◦ g]◦ f = g ◦ v ◦ g]◦ f that f = v ◦ g]◦ f, since g is a monomorphism. From

cf ◦ f = cf ◦ v ◦ g]◦ f = 0 it follows that cf ◦v = 0, since g]◦ f is a bimorphism. The universal property of the image if then gives rise to a unique morphism w : im g◦f → X = im f making the diagram v im g ◦ f / ; Y vv vv w vv vv f vv Y commutative. If then t: T → Z is a morphism in C with cg◦f ◦t = 0, the universal property of ig◦f gives rise to a unique morphism λ: T → im g ◦ f with t = ig◦f ◦ λ. Then we have t = ig◦f ◦ λ = g ◦ v ◦ λ = g ◦ f ◦ w ◦ λ, hence the diagram t T / Z > ~~ w◦λ ~ ~~g◦f ~~ X is commutative and since g ◦ f is a monomorphism, the morphism w ◦ λ is unique with this property. This shows that g ◦ f is its own image and hence a strict monomorphism.

Corollary 2.78. Let C be a semi-abelian category and let f : X → Y be a strict morphism in C. i) If j : Y → Z is a strict monomorphism, then the morphism j ◦ f is strict. ii) If p: W → X is a strict epimorphism, then the morphism f ◦p is strict. Proof. Follows directly from 2.77 and 2.30.

Proposition 2.79. Let C be a semi-abelian category and let

f X / Y

t (1) sY T / S sT be a pushout square. 62 CHAPTER 2. ADDITIVE CATEGORIES

i) If f is a strict epimorphism, then sT is also a strict epimorphism. ii) If f or t is a strict monomorphism, then the above diagram is also a pullback square.

Proof. i) We show that sT is its own coimage, then it follows from 2.27 that sT is a strict epimorphism. Let r : T → R be a morphism with r ◦ ksT = 0. Since the diagram (1) is commutative there exists a unique morphism λ: ker f → ker sT making the diagram

kf f ker f / X / Y λ t (1) sY ker s / T / S T sT ksT commutative. Then we have

r ◦ t ◦ ksT ◦ λ = 0. Since f is astrict epimorphism it is its own coimage, hence there exists a unique morphism ε: Y → R with r ◦ t = ε ◦ f. Then the universal property of the pushout gives rise to a unique morphism η : S → R with ε = η ◦ sY and r = η ◦ sT , i.e. making the diagram

ksT sT ker sT / T / S η r R

0 0 commutative. If η : S → R is another morphism with η ◦ sT = r, then we have 0 0 η ◦ sY ◦ f = η ◦ sT ◦ t = r ◦ t = ε ◦ f 0 and since f is an epimorphism it follows that η ◦ sY = ε. The universal 0 property of the pushout then yields η = η. This shows that sT is its own coimage, hence a it is a strict epimorphism. f ii) Consider the morphism −t : X → Y q T . The proof of 2.55 shows that (cok p, cp ◦ωY , cP ◦ωT ) is a pushout of f and t, hence we can assume w.l.o.g.:

S = cok p, sY = cp ◦ ωY , sT = cp ◦ ωT

We claim that (im p, −πT ◦ ip, πY ◦ ip) is a pullback of cp ◦ ωY and cp ◦ ωT . In fact since

−(cp ◦ ωT ◦ (−πT ◦ ip)) + cp ◦ ωY ◦ πY ◦ ip = cp(ωT ◦ πT + ωY ◦ πY ) ◦ ip

= cp ◦ ip = 0 2.6. SEMI-ABELIAN CATEGORIES 63 the diagram −πT ◦ip im p / T

πY ◦ip cp◦ωT Y / cok p cp◦ωY is commutative. Let rT : R → T and rY : R → Y be morphisms with cp ◦ ωY ◦ rY = cp ◦ ωT ◦ rT . Then cp ◦ (ωY ◦ rY − ωT ◦ rT ) = 0, hence the universal property of ip gives rise to a unique morphism h: R → im p with ip ◦ h = ωY ◦ rY − ωT ◦ rT . Then we have

−πT ◦ ip ◦ h = −πT ◦ ωY ◦ rY + πT ◦ ωT ◦ rT = rT and πY ◦ ip ◦ h = πY ◦ ωY ◦ rY − πY ◦ ωT ◦ rT = rY . 0 0 0 If h : R → im p is another morphism with −πT ◦ip ◦h = rT and πY ◦ip ◦h = rY then we also have

0 ωT ◦ rT = ωT ◦ (−πT ◦ ip) ◦ h 0 ωY ◦ rY = ωY ◦ πY ◦ ip ◦ h and it follows that

ip ◦ h = ωY ◦ rY − ωT ◦ rT 0 0 = ωY ◦ πY ◦ ip ◦ h − ωT ◦ (−πT ◦ ip) ◦ h 0 = (ωY ◦ πY + ωT ◦ πT ) ◦ ip ◦ h 0 = ip ◦ h .

0 Since ip is a monomorphism this yield h = h and thus establishes the claim that (im p, −πT ◦ ip, πY ◦ ip) is a pullback of cp ◦ ωY and cp ◦ ωT . Since the category C is semi-abelian the morphism pe ◦ cip : X → im p is an epimorphism. Consider then the diagram

f X 7/ Y . CC ppp CCpe◦cip pp CC ppp C ppπY ◦ip C! ppp t im p sY {{ {{ {{−πT ◦ip }{{ T / S sT We have

−(πT ◦ ip) ◦ (pe◦ cip) = −πT ◦ p = t πY ◦ ip ◦ (pe◦ cip) = πY ◦ p = f 64 CHAPTER 2. ADDITIVE CATEGORIES hence the above diagram is commutative. If f or t is a strict monomorphism it follows from 2.76, that pe◦cip is also a strict monomorphism, hence a strict bimorphism and thus an isomorphism. This establishes ii). By duality we obtain: Proposition 2.80. Let C be a semi-abelian category and let

pT P / T

pY (2) t Y g / Z be a pullback square.

i) If g is a strict monomorphism, then pT is also a strict monomorphism. ii) If g or t is a strict epimorphism, then the above diagram is also a pushout square. Corollary 2.81. Let C be a semi-abelian category. i) If g : Y → Z is a strict epimorphism and

T P / T

pY (1) t Y g / Z

a pullback square, then pT is an epimorphism. ii) If f : X → Y is a strict monomorphism and

f X / Y

t (2) sY T / S sT

a pushout square, then sT is a monomorphism. Proof. It suﬃces to show i), since ii) is the dual statement. Since g is an epimorphism, the diagram (1) is also a pushout square by 2.80.

Let cpT : T → cok pT be a cokernel of pT , then 2.71 yields a commutative diagram T cPT P / T / cok pT

pY (1) t Y g / Z c / cok pT where c is a cokernel of g. Since g is an epimorphism, we have cok pT = 0 by 2.21, hence, again by 2.21, it follows that c is an epimorphism. Chapter 3

Exact Categories

3.1 Basic Properties

Deﬁnition 3.1. Let C be an additive category. A pair of composable morphisms f g X / Y / Z in C is called a kernel-cokernel pair, if f is a kernel of g and g is a cokernel of f. Remark 3.2. Let C be an additive category. i) A pair (f, g) of composable morphisms in C is a kernel-cokernel pair if and only if (gop, f op) is a kernel-cokernel pair in Cop. ii) If (f, g) is a kernel-cokernel pair in C and

f g X / Y / Z

iX iY iZ X0 / Y 0 / Z0 f 0 g0

0 0 is a commutative diagram with isomorphisms iX , iY , iZ , then (f , g ) is also a kernel-cokernel pair. In fact, if t: T → Y 0 is a morphism 0 −1 with g ◦ t = 0, it follows that g ◦ iY ◦ t = 0, hence the universal property of the kernel f gives rise to a unique morphism λ: T → X −1 0 0 with f ◦λ = iY ◦t. Then f ◦iX ◦λ = iY ◦f ◦λ = t and f ◦iX is unique with this property, hence f 0 is a kernel of g0. The dual argument shows that g0 is a cokernel of f 0. Lemma 3.3. Let C be an additive category and let

f g (?) X / Y / Z be a sequence of morphisms in C with g◦f = 0. The following are equivalent:

65 66 CHAPTER 3. EXACT CATEGORIES

i) (f, g) is a kernel-cokernel pair. ii) f is a strict monomorphism, g is a strict epimorphism and the induced morphism λ: im f → ker g with if = kg ◦ λ is an isomorphism. Proof. i) ⇒ ii) If (f, g) is a kernel-cokernel pair, then f is a strict monomorphism and g a strict epimorphism by 2.29. Since g is a cokernel of f, the kernel of g is an image of f, hence the unique morphism with λ: im f → ker g with if = kg ◦ λ is an isomorphism. ii) ⇒ i) It suﬃces to show that f is a kernel of g, since g is a strict epimorphism and therefore by 2.27 a cokernel of its kernel. Since f is a strict monomorphism it is its own image by 2.27 and since the morphism λ is an isomorphism it follows that f is a kernel of g.

Example 3.4. The above shows, that in the category (TVS) a sequence

f g X / Y / Z with g ◦ f = 0 is a kernel-cokernel pair if and only if f is injective and open onto its range, g is surjective and open onto its range and g−1({0}) = f(X). Therefore, using the standard terminology of functional analysis (cf. [31]), the kernel-cokernel pairs in (TVS) are the short exact sequences of (TVS). Definition 3.5. Let C be an additive category. If a class E of kernel-cokernel pairs on C is fixed, an admissible monomorphism is a morphism f such that there exists a morphism g with (f, g) ∈ E. Admissible epimorphisms are defined dually. An exact structure on C is a class E of kernel-cokernel pairs which is closed under isomorphisms and has the following properties:

[E0] For each object X, the identity idX : X → X is an admissible monomorphism.

op [E0 ] For each object X, the identity idX : X → X is an admissible epimorphism. [E1] If f : Y → Z and f 0 : Z → V are admissible monomorphisms, then the composition f 0 ◦ f is an admissible monomorphism. [E1op] If g : Y → Z and g0 : Z → V are admissible epimorphisms, then the composition g0 ◦ g is an admissible epimorphism. [E2] If f : X → Y is an admissible monomorphism and t: X → T is a morphism, then the pushout

f X / Y

t sY T / S sT 3.1. BASIC PROPERTIES 67

of f and t exists and sT is an admissible monomorphism. [E2op] If g : Y → Z is an admissible epimorphism and t: T → Z is a morphism, then the pullback

pT P / T

pY t Y g / Z

of g and t exists and pT is an admissible epimorphism. If E is an exact structure on C, then the pair (C, E) is called an exact category.

Remark 3.6. Let C be an additive category.

i) Since each of the properties of an exact structure has its dual formulation, it follows that E is an exact structure on C if and only if Eop is an exact structure on Cop.

ii) In the deﬁnition of an exact category, the requirement that the class E is closed under isomorphisms means the following: If (f, g) is an element of E and

f g X / Y / Z

iX iY iZ X0 / Y 0 / Z0 f 0 g0

is a commutative diagram in C such that iX , iY , iZ are isomorphisms, then it follows that (f 0, g0) is an element of E.

iii) Isomorphisms are admissible monomorphisms as well as admissible epimorphisms. For an isomorphism f : X → Y this follows from the commutative diagrams

f X / Y / 0

f −1 X X / 0

and f 0 / X / Y

f−1 0 / X X 68 CHAPTER 3. EXACT CATEGORIES

iv) If f : X → Y is an admissible monomorphism and g is a cokernel of f, then the pair (f, g) is in E, since E is closed under isomorphisms. Dually: If g : Y → Z is an admissible epimorphism and f a kernel of g, then (f, g) is in E.

Lemma 3.7. If (C, E) is an exact category and X,Y ∈ Ob(C), then the pair

ωX πY X / X q Y / Y is an element of E.

Proof. By 2.47 the diagramm

0 / Y

ωY X / X q Y ωX is a pushout. It follows from the property (E0op) that the morphism 0 → Y is an admissible monomorphism, since it is a kernel of idY : Y → Y . Then ωX is an admissible monomorphism by (E2). Since πY is a cokernel of ωX the pair (ωX , πY ) is an element of E. Lemma 3.8. Let C be an additive category. For a kernel-cokernel pair (f, g) the following are equivalent:

i) f is a coretraction.

ii) g is a retraction.

iii) There is a commutative diagram

ωX πZ X / X q Z / Z

β X / Y / Z f g

such that β is an isomorphism.

Proof. i) ⇒ ii) Let λ: Y → X be a morphism with λ ◦ f = idX . Deﬁne h := idY −f ◦ λ, then h ◦ f = f − f ◦ λ ◦ f = 0, hence the universal property of g gives rise to a unique morphism µ: Z → Y with h = µ ◦ g. Then

g ◦ µ ◦ g = g ◦ h = g ◦ (idY −f ◦ λ) = g, and since g is an epimorphism it follows that g ◦ µ = idZ , hence g is a retraction. 3.1. BASIC PROPERTIES 69 ii) ⇒ iii) Let µ be a morphism with g ◦ µ = idZ , then the diagram

ωX πZ X / X q Z / Z

( f µ ) X / Y / Z f g is commutative. Deﬁne h := idY −µ ◦ g. Then we have

g ◦ h = g ◦ (idY −µ ◦ g) = g − g ◦ µ ◦ g = g − g = 0, hence the universal property of the kernel f gives rise to a unique morphism λ: Y → X with h = f ◦ λ. Then we have

f ◦ λ ◦ µ = h ◦ µ = (idY −µ ◦ g) ◦ µ = µ − µ ◦ gµ = µ − µ = 0 and therefore λ ◦ µ = 0, since f is a monomorphism. Then we have

λ λ ◦ f λ ◦ µ 1 0 ( f µ ) = = g g ◦ f g ◦ µ 0 1 λ ( f µ ) = f ◦ λ + µ ◦ g = id , g Y which shows that β := ( f µ ) is an isomorphism. −1 iii) ⇒ i) The assumption gives β ◦ f = ωX , hence f is a coretraction by 1.15.

f g Deﬁnition 3.9. We say that a sequence X / Y / Z in an additive category C is split exact, if (f, g) is a kernel-cokernel pair and if it satisﬁes the equivalent properties of 3.8.

Proposition 3.10. If C is an additive category the class Emin = (f, g);(f, g) is a split exact kernel-cokernel pair is an exact structure on C. Moreover, Emin is minimal in the sense that all exact structures on C contain it.

Proof. Consider a commutative diagram

f g X / Y / Z

iX iY iZ X0 / Y 0 / Z0 f 0 g0 70 CHAPTER 3. EXACT CATEGORIES with (f, g) ∈ Emin and isomorphisms iX , iY , iZ . If l : Y → X is a morphism with l ◦ f = idX , we have

−1 0 −1 iX ◦ l ◦ iY ◦ f = iX ◦ l ◦ f ◦ iX = idX0 , therefore f 0 is a coretraction and (f, g) is a kernel-cokernel pair by 3.2.ii). This shows that Emin is closed under isomorphisms. op op Note that (f, g) is an element of Emin if and only if (g , f ) is an element of Eop , therefore it suffices to show that the properties [E0op], [E1op] and opmin [E2 ] are satisfied in order to show that Emin is an exact structure, since the other ones can be obtained by duality. op [E0 ] is satisfied since the pair (0, idX ) is split exact for all X ∈ Ob(C). Let (f, g) and (f 0, g0) be split exact sequences such that g0 ◦ g is defined. For f g the split exact sequence X / Y / Z we can choose morphisms l : Y → X and r : Z → Y with l ◦ f = idX , g ◦ r = idZ and f ◦ l + r ◦ g = idY (in fact, −1 if β is the isomorphism of 3.8, the morphisms l := πX ◦ β and r := β ◦ ωZ have the desired properties). In addition, let l0 :: Z → W be a morphism 0 0 with l ◦ f = idW . Then we have

l ◦ r ◦ g = l ◦ (idY −f ◦ l) = l − l ◦ f ◦ l = l − idX ◦l = 0 and therefore l ◦ r = 0, since g is an epimorphism. With this we get  l   l ◦ f l ◦ r ◦ f 0  1 0 0 0 0 0 0 l ◦ g  f r ◦ f = l ◦ g ◦ f l ◦ g ◦ r ◦ f  = 0 1 g0 ◦ g g0 ◦ g ◦ f g0 ◦ g ◦ r ◦ f 0 0 0 and this shows that the diagram

0 f r ◦ f g0◦g X q W / Y / Z0  l   0   l ◦ g   0  g ◦ g X q W / (X q W ) q Z0 / Z0 ωXqW πZ0 is commutative. It follows that the upper row is split exact, hence [E1op] is satisﬁed. f g For a split exact sequence X / Y / Z choose again morphisms l : Y → X and r : Z → Y with l ◦ f = idX , g ◦ r = idZ and f ◦ l + r ◦ g = idY . We claim that for a morphism t: T → Z the diagram

πT X q T / T ( f ) (1) t r◦t Y g / Z 3.1. BASIC PROPERTIES 71 is a pullback square. Let lY : L → Y and lT : L → T be morphisms with t ◦ lT = g ◦ lY . We have l ◦ lY ( f r ◦ t ) = f ◦ l ◦ lY + r ◦ t ◦ lT lT = (idY −r ◦ g) ◦ lY + r ◦ g ◦ lY

= lY therefore the diagram

lT L G l◦lY G ( l ) G T G G # πT X q T / T

f lY (1) (r◦t) t + Y g / Z is commutative. If λ1 : L → X qT is another morphism making the above λ2 diagram commutative, than it follows that λ2 = lT and λ1 ( f r ◦ t ) = λ1 ◦ lY + r ◦ t ◦ lT , lT hence f ◦λ1 = f ◦l◦lY and therefore λ1 = l◦lY , since f is a monomorphism. This shows that (1) is a pullback square, which in turn shows that [E2op] is satisﬁed, since (ωX , πZ ) is split exact. If E is another exact structure on C, then we know by 3.7 that for all objects X and Z in C the pair (ωX , πZ ) is an element of E. Since E is closed under isomorphisms, 3.8 shows that E contains all split exact sequences, hence Emin ⊆ E. Proposition 3.11. Let (C, E) be an exact category. If (f, g) and (f 0, g0) are in E, then (f ⊕ f 0, g ⊕ g0) is in E.

Proof. By 2.70 the diagrams

0 idZ ⊕g g⊕idY 0 Z q Y 0 / Z q Z0 and Y q Y 0 / Z q Y 0

πY 0 πZ0 πY 0 πZ0 Y 0 / Z0 Y 0 / Z0 g0 g op 0 are pullback squares, hence by the property (E2) the morphisms idZ ⊕g op and g ⊕ idY 0 are admissible epimorphisms. By (E1) the morphism

0 0 g ⊕ g = (idZ ⊕g ) ◦ (g ⊕ idY 0 ) 72 CHAPTER 3. EXACT CATEGORIES is then also an admissible epimorphism. Additionally we show: The morphism f ⊕ f 0 is a kernel of g ⊕ g0. Let t: T → Y q Y 0 be a morphism with (g ⊕ g0) ◦ t = 0. Then we have

0 g ◦ πY ◦ t = πZ ◦ (g ⊕ g ) ◦ t = 0 0 0 g ◦ πY 0 ◦ t = πZ0 ◦ (g ⊕ g ) ◦ t = 0.

The universal properties of the kernels f and f 0 give rise to a unique mor- 0 phism λ1 : T → X with πY ◦t = f ◦λ1 and to a unique morphism λ2 : T → X 0 0 λ1 with π 0 ◦ t = f ◦ λ . Deﬁne λ: T → X q X as λ = , then: Y 2 λ2

0 f 0 λ1 • π ◦ (f ⊕ f ) ◦ λ = ( 1 0 ) 0 = f ◦ λ = π ◦ t Y 0 f λ2 1 Y

0 f 0 λ1 0 • π 0 ◦ (f ⊕ f ) ◦ λ = ( 0 1 ) 0 = f ◦ λ = π 0 ◦ t Y 0 f λ2 2 Y Hence it follows that (f ⊕ f 0) ◦ λ = t, and λ is unique with this property because of the uniqueness of λ1 and λ2. This establishes the claim and thus shows that (f ⊕ f 0, g ⊕ g0) is an element of E.

Proposition 3.12. Let (C, E) be an exact category and let

i X / Y

f (1) f 0 X0 / Y 0 i0 be a commutative diagram in C with admissible monomorphisms i and i0. The following are equivalent:

i) (1) is a pushout square.

i 0 0 ii) The pair ( −f , ( f i )) is an element of E. iii) (1) is a pushout and a pullback square.

iv) There is a commutative diagram

i p X / Y / Z

f f 0 X0 / Y 0 / Z i0 p0

with (i, p), (i0, p0) ∈ E. 3.1. BASIC PROPERTIES 73

Proof. i) ⇒ ii) We claim the following: 0 0 i Claim A: (1) is a pushout square ⇔ ( f i ) is a cokernel of −f . 0 Let lX0 : X → L and lY : Y → L be morphisms with lY ◦ i = lX0 ◦ f. This is equivalent to the two morphisms lX0 and lY satisfying the equation i ( l l 0 ) = l ◦ i − l 0 ◦ f = 0, Y X −f Y X hence the following two properties are equivalent (thus establishing Claim A):

a) There exists a unique morphism λ: Y 0 → L making the following diagram commutative:

i X / Y

f f 0 lY X0 / Y 0 i0 @ @ λ @ @ lX0 / L

b) There exists a unique morphism λ: Y 0 → L making the following diagram commutative:

i 0 0 (−f) ( f i ) X / Y q X / Y 0 w ( lY l 0 ) w X wλ w L {w

The following is obtained by duality: i 0 0 Claim B: (1) is a pullback square ⇔ −f is a kernel of ( f i ). i By Claim A it is enough to show that −f is an admissible monomorphism. For the composition

idX 0 i 0 ωX −f idX0 0 idX0 X / X q X0 / X q X0 / Y q X0 we have 1 i i 0 idX 0 = 0 idX0 −f idX0 0 −f and by 3.7 the morphism ωX is an admissible monomorphism, by 3.11 the morphism i 0 is an admissible monomorphism and idX 0 is an 0 idX0 −f idX0 74 CHAPTER 3. EXACT CATEGORIES isomorphism (with inverse idX 0 ). Therefore it follows from the property f idX0 i (E1) that −f is an admissible monomorphism. ii) ⇒ iii) follows from Claim A and Claim B. iii) ⇒ i) is trivial. i) ⇒ iv) Let p: Y → Z be a cokernel of i. Then (i, p) is an element of E. The universal property of (1) gives rise to a unique morphism p0 : Y 0 → Z with p0 ◦ f 0 = p and p0 ◦ i0 = 0, i.e. making the following diagram commutative:

i X / Y

f f 0 p X0 / Y 0 i0 A A p0 A A 0 / Z Since p0 ◦ f 0 = p, the morphism p0 is an epimorphism. In fact: Claim C: p0 is a cokernel of i0. Let g : Y 0 → T be a morphism with g ◦ i0 = 0. Then g ◦ f 0 ◦ i = g ◦ i0 ◦ f = 0, hence there exists a unique morphism h: Z → T with g ◦ f 0 = h ◦ p. It follows that h ◦ p0 ◦ f 0 = h ◦ p = g ◦ f 0 h ◦ p0 ◦ i0 = 0 = g ◦ i0, hence h◦p0 = g by the universal property of (1). This factorization is unique since p0 is an epimorphism, therefore Claim C is established. The diagram i p X / Y / Z

f f 0 X0 / Y 0 / Z i0 p0 commutes and by Claim C we have (i, p), (i0, p0) ∈ E. iv) ⇒ ii) Let (Q, q0, q) be the pullback of p and p0. By the dual argument of i) ⇒ iv) we have a commutative diagram X X

j i j0 q0 X0 / Q / Y

q PB p X0 / Y 0 / Z i0 p0 3.1. BASIC PROPERTIES 75 whose rows and columns are elements of E. Since the diagram

Y Y

f 0 p Y 0 / Z p0 commutes, the universal property of (Q, q0, q) yields a unique morphism 0 0 k : Y → Q with q ◦ k = idY and q ◦ k = f . Then

0 0 0 0 0 q ◦ (idQ −k ◦ q ) = q − q ◦ k ◦ q = 0, hence the universal property of j0, which is the kernel of q0, there exists a 0 0 0 unique l : Q → X with j ◦ l = idQ −k ◦ q . In addition we have l ◦ k = 0, 0 0 0 since j ◦ l ◦ k = (idQ −k ◦ q ) ◦ k = 0 and j is a monomorphism. From

i0 ◦ l ◦ j = q ◦ (j0 ◦ l) ◦ j 0 = q ◦ (idQ −k ◦ q ) ◦ j = −(q ◦ k) ◦ (q0 ◦ j) = −f 0 ◦ i = −i0 ◦ f it follows that l ◦ j = −f, since i0 is a monorphism, and from

0 0 0 0 j ◦ l ◦ j = (idQ −k ◦ q ) ◦ j = j0 − k ◦ q0 ◦ j0 = j0

0 0 it follows that l ◦ j = idX0 , since j is a monomorphism. Therefore we have q0 ( k j ) = k ◦ q0 + j0 ◦ l = id l Q and 0 0 0 0 q q ◦k q ◦j idY 0 ( k j ) = 0 = l l◦k l◦j o idX0 hence the morphism ( k j ): Y q X0 → Q is an isomorphism with inverse q0 0 0 q0 q0◦j l . Since q ◦ ( k j ) = ( q ◦ k q ◦ j ) = ( f i ) and l ◦ j = l◦j the diagram j q X / Q / Y 0

q0 ( l ) X / Y q X0 / Y 0 i ( f 0 i0 ) (−f)

i 0 0 commutes. It follows that the pair ( −f , ( f i )) is an element of E, since E is closed under isomorphisms. 76 CHAPTER 3. EXACT CATEGORIES

By duality we get:

Proposition 3.13. Let (C, E) be an exact category and let

p0 X0 / Y 0

f (2) f 0 X p / Y be a commutative diagram in C with admissible epimorphisms p and p0. The following are equivalent:

i) (2) is a pullback square.

−f 0 ii) The pair ( p0 , ( p f )) lies in E. iii) (2) is a pushout and a pullback square.

iv) There is a commutative square

i0 p0 W / X0 / Y 0

f (2) f 0 / / W i X p Y

with (i, p), (i0, p0) ∈ E.

Proposition 3.14. Let (C, E) be an exact category, let i: X → Y be an admissible monomorphism and let

i0 X0 / Y 0

e0 PB e / X i Y be a pullback square. If e is an admissible epimorphism, then i0 is also an admissible monomorphism.

Proof. Let p be a cokernel of i. Then (i, p) is an element of E and we have the commutative diagram

i0 p◦e X0 / Y 0 / Z

e0 PB e p / / X i Y Z. 3.1. BASIC PROPERTIES 77

Since p ◦ e is a composition of admissible epimorphisms, it is itself an admissible epimorphism. So it is enough to show that i0 is a kernel of p ◦ e. Let g : W → X0 be a morphism with p ◦ e ◦ g = 0. Then the universal property of i gives rise to a unique morphism h: W → X with e◦g = i◦h and the universal property of the pullback yields a unique morphism h0 : W → X0 with e0 ◦ h0 = h and i0 ◦ h0 = g. Since monomorphisms are stable under pullback, the morphism i0 is a monomorphism which makes h0 unique with the property i0 ◦ h0 = g, hence i0 is a kernel of p ◦ e.

Again we get by duality:

Proposition 3.15. Let (C, E) be an exact category, let d: Y → Z be an admissible epimorphism and let

d Y / Z

f PO f 0 Y 0 / Z0 d0 be a pushout square. If f is an admissible monomorphism, then d0 is an admissible epimorphism.

Deﬁnition 3.16. Let (C, E) be an exact category and let C0 be a full additive subcategory of C. We say that C0 possesses the three space property with regard to (C, E) if for every sequence

f g X / Y / Z in C with (f, g) ∈ E and X,Z ∈ Ob(C0) it follows that Y ∈ Ob(C0).

Proposition 3.17. Let (C, E) be an exact category and let C0 be a full additive subcategory of C that possesses the three space property with regard to (C, E). The restriction

f g E0 := {(f, g) ∈ E| X / Y / Z is a sequence in C0} of E to C0 is an exact structure on C0.

Proof. E0 is closed under isomorphisms in C0, because E is closed under isomorphisms in C and also the properties [E0] and [E0op] are satisﬁed in (C0, E0) because they are satisﬁed in (C, E). f g f 0 g [E1] Let X / Y / Z and Y / Z0 / Z00 be sequences in C0 with (f, g), (f 0, g0) ∈ E0. Since f 0 is an admissible monomorphism in (C, E) we 78 CHAPTER 3. EXACT CATEGORIES can construct the pushout of f 0 and g in C. By 3.12 we get a commutative diagram f g X / Y / Z

f 0 PO sZ Z0 / S sZ0 g0 c Z00 Z00 with (sZ , c) ∈ E. In addition it follows from 3.15 that sZ0 is an admissible 0 0 epimorphism in (C, E). We claim that sZ0 is a cokernel of f ◦f. let t: Z → T be a morphism with t◦f 0◦f = 0. Then the universal property of the cokernel g gives rise to a unique morphism λ: Z → T with λ ◦ g = t ◦ f 0. The universal property of the pushout then yields a unique morphism λ0 : S → T 0 0 with t = λ ◦ sZ0 and λ = λ ◦ sZ . If µ: S → T is another morphism with t = µ ◦ sZ0 , then we have

0 0 µ ◦ sZ ◦ g = µ ◦ sZ0 ◦ f = t ◦ f = λ ◦ g, hence it follows that µ ◦ sZ = λ, since g is an epimorphism and therefore we have µ = λ0, because of the universal property of λ0. This shows that 0 0 sZ0 is a cokernel of f ◦ f. Since C possesses the three space property with 0 0 regard to (C, E) we know that S is an object of C and therefore (f ◦f, sZ0 ) is an element of E0. The dual argument of the above shows that the property [E1op] is also satisﬁed. f g [E2] Let X / Y / Z be an element of E0 and let t: X → T be a morphism. Since f is an admissible monomorphism in (C, E) we can construct the the pushout of f and t in C and by 3.12 we get a commutative diagram

f g X / Y / Z

t (1) sY T / S / Z sT c whose rows are elements of E. Since C0 possesses the three space property with regard to (C, E) the object S lies in C0, therefore the above is a com- 0 0 mutative diagram in C and we have (sT , c) ∈ E . In addition (1) is also a pushout diagram in C0, hence [E2] is satisﬁed. The dual argument shows that [E2op] is satisﬁed.

Remark 3.18. Possessing the three space property with regard to a category is transitive. More precisely, if (C, E) is an exact category and C0 a full additive subcategory of C that possesses the three space property with regard to (C, E) and C00 is a full additive subcategory of C0 that possess the 3.1. BASIC PROPERTIES 79 three space property with regard to (C0, E0), where E0 is the restriction of E to C0, then C00 possesses the three space property with regard to (C, E).

Example 3.19.

i) The following are some examples of full additive subcategories of (TVS), that possess the three space property with regard to (TVS):

• Hausdorﬀ spaces (follows from [6, §23,2.7]), • complete spaces (cf. [23, Prop.1.3]), • pseudometrizable spaces (cf. [23, Prop.2.1]), • locally bounded spaces (cf. [23, Prop.3.2]).

ii) The following are some examples of full additive subcategories of (LCS), that possess the three space property with regard to (LCS):

• metrizable spaces (cf. [19, 2.4.3]), • Fr´echet-spaces (cf. [19, 2.4.4]), • barelled spaces (cf. [23, Thm.2.6]), • normed spaces (follows from [23, Thm.3.2]), • Schwartz spaces (cf. [23, Prop.3.7]), • nuclear spaces (cf. [23, Prop.3.8]).

iii) The following are some examples of full additive subcategories of the category (F ) of Fr´echet-spaces and continous linear maps, that possess the three space property with regard to (F ):

• Montel spaces (cf. [23, Prop.4.4]), • (DN)-spaces (cf. [28], [29]), • (Ω)-spaces (cf. [28], [29]).

iv) We will show in chapter 4 that the categories (TVS), (LCS) and (LCS)HD possess a canonical largest exact structure, namely the class of all their kernel-cokernel pairs. Together with 3.17 the above listing gives a few examples of exact categories in functional analysis.

v) The category (LCS) is a full preabelian subcategory of (TVS) that does not have the three space property with regard to (TVS) (cf. [11]). Nonetheless, the restriction of the maximal exact structure of (TVS) to the subcategory (LCS) is again an exact structure, as we will see in chapter 4.

Proposition 3.20. Let (C, E) be an exact category. 80 CHAPTER 3. EXACT CATEGORIES

i) If i: X → Y is a morphism in C that possesses a cokernel, and if j : Y → Z is a morphism such that j◦i is an admissible monomorphism, then i is an admissible monomorphism.

ii) If d: Y → Z is a morphism in C that possesses a kernel, and if e: X → Y is a morphism such that d ◦ e is an admissible epimorphism, then d is an admissible epimorphism. Proof. ii) is the dual statement of i), so it suﬃces to show i). Let c: Y → C be a cokernel of i and let i X / Y

j◦i PO sZ Z / S sY i be a pushout square. By 3.12 the morphism −j◦i : X → Y q Z is an admissible monomorphism. Since idY 0 : Y q Z → Y q Z is an isomorphism (with inverse idY 0 ), j idZ −j idZ it is an admissible monomorphism, hence the morphism i i = idY 0 : X → Y q Z 0 j idZ −j ◦ i is also an admissible monomorphism. The morphism c 0 is a cokernel 0 idZ of i , hence the pair ( i , c 0 ) is an element of E. 0 0 0 idZ In the commutative diagram

i c X / Y / C

ωY ωC X / Y q Z / C q Z i c 0 ( 0 ) 0 idZ the right hand square is a pullback by 2.70. It follows by 3.13 that (i, c) is an element of E.

Corollary 3.21. Let (C, E) be an exact category and let (i, p) and (i0, p0) be two pairs of composable morphisms. If (i ⊕ i0, p ⊕ p0) is an element of E, then (i, p) and (i0, p0) are also elements of E. Proof. We show ﬁrst, that (i, p) and (i0, p0) are kernel-cokernel pairs: Since (p ⊕ p0) ◦ (i ⊕ i0) = 0 we have p ◦ i = 0. If t: T → Y is another p 0 t 0 morphism with p ◦ t = 0, then 0 p0 0 = 0 . The universal property of the cokernel then gives rise to a unique morphism λ = λ1 : T → X q X0 λ2 i 0 λ1 t with 0 = , hence we have i ◦ λ = t and λ is unique with this 0 i λ2 0 1 1 3.1. BASIC PROPERTIES 81 property because of the uniqueness of λ. Therefore i is a kernel of p. The dual argument shows that p is a cokernel of i, hence (i, p) is a kernel-cokernel pair. Analogously one shows that (i0, p0) is a kernel-cokernel pair. We have i 1 = ω ◦ i = i 0 0 Y 0 i0 0 and by 3.7 the right-hand side is an admissible monomorphism. Since i possesses a cokernel it is an admissible monomorphism by 3.20, hence (i, p) is an element of E. Analogously one shows that (i0, p0) is an element of E.

Proposition 3.22. Let (C, E) be an exact category and let

f 0 g0 X0 / Y 0 / Z0

a b c X / Y / Z f g be a commutative diagram in C with (f 0, g0), (f, g) ∈ E. Then there is a commutative diagram

f 0 g0 X0 / Y 0 / Z0

a (1) b0

0 X m / D e / Z b00 (2) c X / Y / Z f g in C, so that the diagrams (1) and (2) are pullback as well as pushout squares and b00 ◦ b0 = b.

Proof. Let f 0 X0 / Y 0

a PO b0 X m / D be a pushout square. Then m is an admissible monomorphism and we have g0 ◦ f 0 = 0 = 0 ◦ a and b ◦ f 0 = a ◦ f, hence by the universal property of the pushout there exists a unique morphism e: D → Z0 with e ◦ b0 = g0 and e ◦ m = 0, as well as a unique morphism b00 : D → Y with b00 ◦ b0 = b and b00 ◦ m = f. Note that e is an epimorphism, since e ◦ b0 = g0. Moreover, the morphism e is a cokernel of m (and thus (m, e) ∈ E): 82 CHAPTER 3. EXACT CATEGORIES

If t: D → T is another morphism with t ◦ m = 0, then t ◦ b0 ◦ f 0 = t ◦ m ◦ a = 0, hence the universal property of the cokernel g0 gives rise to a unique morphism λ: Z0 → T with λ ◦ g0 = t ◦ b0. Then λ ◦ e ◦ b0 = λ ◦ g0 = t ◦ b0 and λ ◦ e ◦ m = 0 = t ◦ m, hence by the universal property of the pushout it follows that λ ◦ e = t. In addition λ is unique with this property, since e is an epimorphism and thus e is a cokernel of m. From the equations g ◦ b00 ◦ m = g ◦ f = 0 g ◦ b00 ◦ b0 = g ◦ b = c ◦ g0 c ◦ e ◦ m = 0 = g ◦ f c ◦ e ◦ b0 = c ◦ g0 and the uniqueness-property of the pushout it follows that c ◦ e = g ◦ b00. This shows that the large diagram in the proposition is commutative and the remaining claims of the proposition follow from 3.12 and 3.13. Corollary 3.23 (Short Five Lemma). Let (C, E) be an exact category. If f 0 g0 X0 / Y 0 / Z0

a b c X / Y / Z f g is a commutative diagram in C with (f 0, g0), (f, g) ∈ E and if a and c are isomorphisms (resp. admissible monomorphisms, resp. admissible epimorphisms), then b is an isomorphism (resp. admissible monomorphism, resp. admissible epimorphism). Proof. We will use the induced diagram of the preceding proposition:

f 0 g0 X0 / Y 0 / Z0

a (1) b0

0 X m / D e / Z b00 (2) c X / Y / Z f g If a and b are isomorphisms, then b0 and b00 are also isomorphisms, since isomorphisms are stable under pushout and pullback. Thus b = b00 ◦ b0 is an isomorphism. If a and b are admissible monomorphisms, then b0 is an admissible monomorphism by the property (E2) and b00 is an admissible monomorphism by 3.14, hence b = b00 ◦ b0 is an admissible monomorphism. The case of admissible epimorphisms is shown by the dual argument. 3.1. BASIC PROPERTIES 83

Proposition 3.24 (Noether Isomorphism). Let (C, E) be an exact category. If

f g A / B / X a (1) a0 A / C / Y f 0 g0 b b0 Z Z is a commutative diagram in C (ignore a0 and b0 for the moment) with (f, g), (f 0, g0), (a, b) ∈ E, then there exists a pair (a0, b0) ∈ E making the above diagram commutative and it is uniquely determined up to isomorphism by this property. In addition the diagram (1) is a pullback and a pushout square.

Proof. Since g0 ◦ a ◦ f = g0 ◦ f 0 = 0, the universal property of the cokernel g gives rise to a unique morphism a0 : X → Y with a0 ◦ g = g0 ◦ a. In addition 0 we have idZ ◦b ◦ f = b ◦ a ◦ f = 0, hence the universal property of the 0 0 0 0 cokernel g gives rise to a unique morphism b : Y → Z with idZ ◦b = b ◦ g . By 3.12 the square (1) is a pushout as well as a pullback square, hence a0 is an admissible monomorphism and b0 is (again by 3.12) a cokernel of a0. This shows that (a0, b0) is an element of E.

By duality we obtain:

Proposition 3.25. Let (C, E) be an exact category. If

X X a0 a f g Y / A / B b0 (1) b Z / C / B f 0 g0 is a commutative diagram in C with (f, g), (f 0, g0), (a, b) ∈ E, then there is a pair (a0, b0) ∈ E making the above diagram commutative and it is uniquely determined up to isomorphism by this property. In addition the diagram (1) is a pullback and a pushout square. 84 CHAPTER 3. EXACT CATEGORIES

Corollary 3.26 (3 × 3-Lemma). Let (C, E) be an exact category and let

f 0 g0 A0 / B0 / C0

a b c f g A / B / C

a0 b0 c0 A00 / B00 / C00 f 00 g00 be a commutative diagram in which the rows are elements of E. Assume in addition that one of the following conditions holds:

i) The outer two columns are elements of E and b0 ◦ b = 0.

ii) The middle column and either one of the outer columns is an element of E.

Then the remaining column is also an element of E

Proof. Assume that condition i) holds. By 3.22 applied to the morphisms between the ﬁrst two rows of the above diagram we get a commutative diagram f 0 g0 A0 / B0 / C0

a (I) ¯b f¯ g¯ A / D / C0

ˆb (II) c A / B / C f g with exact rows, b = ˆb ◦ ¯b and (I), (II) being pushout as well as pullback squares. Since a and c are admissible monomorphisms it follows from the short ﬁve lemma 3.23 that ¯b and ˆb are admissible monomorphisms, hence so is b. It remains to be shown, that b0 is a cokernel of b. (1) Since a0 ◦ a = 0 = 0 ◦ f 0 the universal property of the pushout (I) gives rise to a unique morphisma ¯: D → A00 witha ¯ ◦ ¯b = 0 anda ¯ ◦ f¯ = a0. We claim thata ¯ is a cokernel of ¯b: Let t: D → T be a morphism with t ◦ ¯b = 0, then we also have

t ◦ f¯◦ a = t ◦ ¯b ◦ f 0 = 0, hence the universal property of the cokernel a0 gives rise to a unique morphism µ: A0 → T with t ◦ f¯ = µ ◦ a0 = µ ◦ a¯ ◦ f¯. In addition we have t ◦ ¯b = µ ◦ a¯ ◦ ¯b = 0, hence it follows from the universal property of the 3.1. BASIC PROPERTIES 85 pushout (I), that t = µ ◦ a¯. If µ0 : A00 → T is another morphism with t = µ0 ◦ a¯ we have also t ◦ f¯ = µ0 ◦ a¯ ◦ f¯ = µ0 ◦ a0 and therefore µ0 = µ, so µ is unique with this property. This shows thata ¯ is a cokernel of ¯b. (2) The morphism c0 ◦ g : B → C00 is a cokernel of ˆb: If t: B → T is a morphism with ˆb ◦ t = 0 = 0 ◦ g¯ the universal property of the pushout (II) gives rise to a unique morphism η : C → T with η ◦ c = 0 and t = η ◦ g. Because of 0 = t ◦ ˆb = η ◦ g ◦ ˆb = η ◦ c ◦ g¯ andg ¯ being an epimorphism it follows that η ◦c = 0. The universal property of the cokernel c0 of c then gives rise to a unique morphism ν : C00 → T with η = ν ◦ c0 and therefore ν ◦ c0 ◦ g = η ◦ g = t. Since c0 ◦ g is an epimorphism the morphism ν is unique with this property which shows that c0 ◦ g is a cokernel of ˆb. (3) The diagram a¯ D / A00

ˆb f 00 B / B00 b0 is commutative. This follows from (f 00 ◦ a¯) ◦ f¯ = f 00 ◦ a0 = b0 ◦ f = (b0 ◦ ˆb) ◦ f¯ (f 00 ◦ a¯) ◦ ¯b = 0 = b0 ◦ b = ( b0 ◦ ˆb) ◦ ¯b and the universal property of the pushout (I). (4) Let e: B → E be the cokernel of the admissible monomorphism b. Then 3.24 yields unique morphisms d: A00 → E and d0 : E → C00 making the diagram ¯b a¯ B0 / D / A00 ˆb (III) d B0 / B / E b e c0◦g d0 C00 C00 commutative, with the rows and columns being elements of E, and (III) being a pullback and a pushout square. Since the diagram in (3) is commutative, the universal property of the pushout (III) gives rise to a unique 86 CHAPTER 3. EXACT CATEGORIES morphisme ˆ: E → B00 with f 00 =e ˆ ◦ d and b0 =e ˆ ◦ e. (5) In addition the diagram

d d0 A00 / E / C00

eˆ A00 / B00 / C00 f 00 g00 is commutative: Since f 00 =e ˆ ◦ d, the left-hand square is commutative and in addition we have

g00 ◦ eˆ ◦ e = g00 ◦ b0 = c0 ◦ g = d0 ◦ e g00 ◦ eˆ ◦ d = g00 ◦ f 00 = 0 = d0 ◦ d hence the universal property of the pushout (III) shows that g00 ◦ eˆ = d0, therefore the right-hand square also commutes. Then the short ﬁve lemma 3.23 yields thate ˆ is an isomorphism, therefore b0 =e ˆ ◦ e is a cokernel of b which shows that (b, b0) ∈ E.

The two possibilities in case ii) are dual to each other, so we only need consider the case that the middle and the right hand column are elements of E. By 3.20 it suﬃces to show that a has a0 as a cokernel, since f ◦ a = b ◦ f 0 is an admissible monomorphism. (6) Because of f 00 ◦ a0 ◦ a = b0 ◦ b ◦ f 0 = 0 it follows that a0 ◦ a = 0, since f 00 is a monomorphism. Consider again the diagram

f 0 g0 A0 / B0 / C0

a (I) ¯b f¯ g¯ A / D / C0

ˆb (II) c A / B / C f g with the rows being exact, b = ˆb ◦ ¯b and (I), (II) being pushout as well as pullback squares. By the short ﬁve lemma 3.23 it follows that ˆb is an admissible monomorphism and one shows as before that c0 ◦ g : B → C00 is a cokernel of ˆb. (7) Again we get the commutative square of step (3) and by 3.25 we obtain 3.1. BASIC PROPERTIES 87 the commutative diagram B0 B0 ¯b b ˆb c0◦g D / B / C00 a¯ b0 A00 / B00 / C00 f 00 g00 whose rows and columns are elements of E. From f 00 ◦ (¯a ◦ f¯) = b0 ◦ ˆb ◦ f¯ = b0 ◦ f = f 00 ◦ a0 it follows thata ¯ ◦ f¯ = a0 since f 00 is a monomorphism. (8) Let then t: A → T be a morphism with t ◦ a = 0 = o ◦ f 0. Then the universal property of the pushout (I) gives rise to a unique morphism γ : D → T with γ ◦ f¯ = t and γ ◦ ¯b = 0. Since (¯b, a¯) ∈ E, the universal property of the cokernela ¯ then gives rise to a unique morphism γ0 : A00 → T with γ = γ0 ◦ a¯. Then γ0 ◦ a0 = γ0 ◦ a¯ ◦ f¯ = γ ◦ f¯ = t. In order to see that γ0 is unique with this property, and therefore a0 is a cokernel of a, we show that a0 is an epimorphism: By 3.22 applied to the second two rows of the initial 3 × 3-diagram we get the diagram f g A / B / C

a (IV ) β fe g A00 / D0 e / C0

β0 (V ) c A00 / B00 / C00 f 00 g00 with exact rows, b0 = β0 ◦ β and (IV ), (V ) being pushout as well as pullback squares. Then β is an admissible epimorphism by the short ﬁve lemma 3.23 and by 3.13 there is a unique morphism k : C0 → D making the diagram

k β0 C0 / D0 / B

00 ge g C0 / C0 / C c c0 commutative and (k, β0) ∈ E. We have β0 ◦ β ◦ b = 0 = β0 ◦ k ◦ g0, 0 0 ge ◦ β ◦ b = g ◦ b = c ◦ g = ge ◦ k ◦ g , 88 CHAPTER 3. EXACT CATEGORIES hence the universal property of the pullback (V ) yields β ◦ b = k ◦ g0 and therefore the diagram b b0 B0 / B / B00

g0 β C0 / D0 / B00 k β0 is commutative. Then the short ﬁve lemma shows that β is an admissible epimorphism. Since the square (IV ) is a pullback it follows that a0 is an admissible epimorphism. This ﬁnishes the proof.

3.2 E-strict morphisms

Deﬁnition 3.27. Let (C, E) be an exact category. A morphism f : X → Y is called an E-strict morphism, if it has a factorization

f X / Y , AA ~> AA ~~ ef AA ~~mf A ~~ V where ef is an admissible epimorphism and mf is an admissible monomorphism. Remark 3.28. If f : X → Y is an E-strict morphism in an exact category (C, E), it follows from (E1) and (E1)op that i) The morphism f has a kernel as well as a cokernel. In fact we have a commutative diagram

k f c K / X / Y / C O ef (1) mf V V

with (k, ef ), (mf , c) ∈ E and in this situation k is a kernel of f and c its cokernel. In addition ef is a coimage of f and mf is its image, i.e. the diagram (1) is a canonical factorization of f. Therefore every E-strict morphism is in particular a strict morphism in the sense of 2.23.

ii) If d: W → X is an admissible epimorphism, then f ◦ d is an E-strict morphism.

iii) If i: Y → T is an admissible monomorphism, then i ◦ f is an E-strict morphism. 3.2. E-STRICT MORPHISMS 89

iv) The composition of two E-strict morphisms is in general not again an E-strict morphism: In the category (LCS) of locally convex vector spaces, the class E of all kernel-cokernel pairs is an exact structure (see also 4.12). The E-strict morphisms are exactly the strict morphisms and the morphisms

α: l1 → l1 q l2, x 7→ (x, x) and β : l1 q l2 → l2, (x, y) 7→ y are open onto their ranges, hence they are E-strict morphisms. But the composition β ◦ α: l1 → l2 is not open onto its range. Lemma 3.29. Let (C, E) be an exact category. Then the factorization of an E-strict morphism is unique up to a unique isomorphism. More speciﬁcally: If e X / V } 0 } m e } v ~} V 0 / Y m0 is a commutative diagram with admissible monomorphisms m, m0 and admissible epimorphisms e, e0, then there exists a unique isomorphism v : V → V 0 with e0 = v ◦ e and m = m0 ◦ v and v. Proof. Let k be a a morphism with (k, e) ∈ E. Since m0 ◦e0 ◦k = m◦e◦k = 0 and since m0 is a monomorphism it follows that e0◦k = 0. Then the universal property of e gives rise to a unique morphism v : V → V 0 with e0 = ◦v ◦ e. Additionally we have m0 ◦ v ◦ e = m0 ◦ e = m ◦ e, hence m0 ◦ v = m since e is an epimorphism. Let c be a morphism with (m, c) ∈ E. Then we have c◦m0 ◦e0 = c◦m◦e = 0, hence c ◦ m0 = 0 since e0 is an epimorphism. The universal property of m then yields a unique morphism v0 : V 0 → V with m0 = m◦v0. In addition we have v0 ◦ e0 = e, since m ◦ v0 ◦ e0 = m0 ◦ e0 = m ◦ e and m is a monomorphism. 0 0 0 0 Furthermore it follows from v ◦ v ◦ e = v ◦ e = e that v ◦ v = idV 0 and 0 0 0 0 from v ◦ v ◦ e = v ◦ e = e that v ◦ v = idV , hence v is an isomorphism with inverse v0.

Lemma 3.30. Let (C, E) be an exact category and let f : X → Y be an E-strict morphism. i) If i: X → V is an admissible monomorphism and

f X / Y

i PO sY X0 / S sX0 90 CHAPTER 3. EXACT CATEGORIES

a pushout square, then sX0 is also an E-strict morphism. ii) If p: W → Y is an admissible epimorphism and

pY 0 P / Y 0

pX PB p X / Y f

a pullback square, then pY 0 is also an E-strict morphism. Proof. ii) is the dual statement of i), so it suﬃces to show i). Let f : X → Y be an E-strict morphism with f = mf ◦ ef for an admissible monomorphism 0 mf and an admissible epimorphism ef . Let (R, rvX , rV ) be the pushout of ef and f and let (T, tR, tY ) be the pushout of mf and rV , then we have the commutative diagram e m X / V / Y

i PO rV PO tY X0 / R / T rX0 tR and by 3.15 the morphism rX0 is an admissible epimorphism. Since the morphism tR is an admissible monomorphism, tR ◦ rX0 is an E-strict morphism and since sX0 = λ ◦ tR ◦ rX0 for an isomorphism λ it is also an E-strict morphism.

Deﬁnition 3.31. Let (C, E) be an exact category. A sequence

fk−1 fk Xk−1 / Xk / Xk+1 of E-strict morphisms is called exact at Xk, if for the factorizations

fk−1 fk Xk−1 / Xk / Xk+1 GG z< AA < GG z A zz GG zz AA zz ek−1 GG zzmk−1 ek AA zzmk G# zz A zz Vk−1 Vk of fk−1 and fk one has (mk−1, ek) ∈ E. Note that this independent of the choice of the factorization because of 3.29 and since E is closed under isomorphisms. A sequence

f1 f2 fn−2 fn−1 X1 / X2 / X3 / ... / Xn−2 / Xn−1 / Xn of E-strict morphisms is called exact if it is exact at every Xk for all k = 2, . . . , n − 1. 3.2. E-STRICT MORPHISMS 91

Remark 3.32. Let (C, E) be an exact category. i) A sequence f 0 / X / Y , where f is an E-strict morphism is exact if and only if f is an admissible monomorphism. To show this, let f = mf ◦ ef be the canonical factorization of f. If the sequence is exact, then (0, ef ) ∈ E, hence ef is an isomorphism and thus f is an admissible monomorphism. If f is an admissible monomorphism, then ef is also a monomorphism, hence its a strict bimorphism and thus an isomorphism. This shows (0, ef ) ∈ E and therefore the above sequence is exact. By duality we obtain that a sequence

g Y / Z / 0 ,

where g is an E-strict morphism is exact if and only if g is an admissible epimorphism.

ii) A sequence f g 0 / X / Y / Z of E-strict morphisms f and g is exact if and only if f is a kernel of g. To show this, let f = mf ◦ ef and g = mg ◦ eg be the canonical factorizations of the E-strict morphisms. If the sequence is exact, then (0, ef ), (mf , eg) ∈ E and therefore mf is kernel of eg, hence it is also a kernel of g. Since ef is a cokernel of 0, it is an isomorphism, which shows that f is a kernel of g. If f is a kernel of g, then it is also a kernel of eg, hence (mf , eg) ∈ E. Since f is a monomorphism, hence ef is a strict bimorphism and thus an isomorphism. This shows (0, ef ) ∈ E, hence the above sequence is exact. By duality we obtain, that a sequence

f g X / Y / Z / 0

of E-strict morphisms f and g is exact if and only if g is a cokernel of f.

iii) A sequence f g 0 / X / Y / Z / 0 of E-strict morphisms f and g is exact if and only if (f, g) ∈ E. Indeed it follows from i) and ii) that the above sequence is exact if and only if (f, g) is a kernel-cokernel pair, f is an admissible monomorphism and g an admissible epimorphism 92 CHAPTER 3. EXACT CATEGORIES

Proposition 3.33 (Five Lemma). Let (C, E) be an exact category. If

f1 f2 f3 f4 X1 / X2 / X3 / X4 / X5

α1 α2 α3 α4 α5 Y / Y / Y / Y / Y 1 g1 2 g2 3 g3 4 g4 5 is a commutative diagram with exact rows and isomorphisms α1, α2, α4 and α5, then α3 is also an isomorphism. Proof. We will start by establishing the following claim: Claim A. Let f g X / Y / Z

α (1) β (2) γ X0 / Y 0 / Z0 f 0 g0 be a commutative diagram in in C with (f, g), (f 0, g0) ∈ E. If two out of the three morphisms α, β, γ are isomorphisms, then the third is also an isomorphism. If α and γ are isomorphisms, then β is an isomorphism by 3.23; if α and β are isomorphisms, the square (1) is a pushout and γ is the unique isomorphism connecting the cokernels of f and f 0. If β and γ are isomorphisms, then

g0 Y 0 / Z0

β−1 γ−1 Y g / Z is a pullback square and this diagram induces a unique isomorphism φ: X0 → X with β−1 ◦f 0 = f ◦φ. Since α is uniquely determined by β and γ it follows that α = φ−1, which establishes the claim. Consider the diagram

f1 f2 f3 f4 X1 / X2 / X3 / X4 / X5 . A > A > A > A > AA }} AA }} AA }} AA }} AA }} AA }} AA }} AA }} e1 AA }}m1 e2 AA }}m2 e3 AA }}m3 e4 AA }}m4 A }} A }} A }} A }} I1 I2 I3 I4 α1 φ1 α2 φ2 α3 φ3 α4 φ4 α5 J1 J2 J3 J4 > A > A > A > A l1 }} AA n1 l2 }} AA n2 l3 }} AA n3 l4 }} AA n4 }} AA }} AA }} AA }} AA }} AA }} AA }} AA }} AA }} A }} A }} A }} A Y / Y / Y / Y / Y 1 g1 2 g2 3 g3 4 g4 5 3.2. E-STRICT MORPHISMS 93

−1 −1 Since f1 = α2 ◦ n1 ◦ l1 ◦ α1 and since α2 ◦ n1 is an admissible monomorphism and l1 ◦α1 an admissible epimorphism, there is a unique isomorphism φ1 : I1 → J1 with φ1 ◦ e1 = l1 ◦ α1 and n1 ◦ φ1 = α2 ◦ m1 by 3.29. Analo- −1 gously f4 = α5 ◦ n4 ◦ l4 ◦ α4 yields a unique isomorphism φ4 : I4 → J4 with φ4 ◦ e4 = l4 ◦ α4 and n4 ◦ φ4 = α5 ◦ m4. Since (m1, e2), (n1, l2) ∈ E there it follows from Claim A that there exists a unique isomorphism φ2 : I2 → J2 with φ2 ◦ e2 = l2 ◦ α2. Since e2 is an epimorphism it follows in addition that n2 ◦φ2 = α3 ◦m2. Similarly one gets an isomorphism φ3 : I3 → J3 with φ3 ◦ e3 = l3 ◦ α3 and n3 ◦ φ3 = α4 ◦ m3. Then α3 is an isomorphism by 3.23.

Proposition 3.34 (Snake Lemma). Let (C, E) be an exact category such that C has kernels and cokernels. For every commutative diagram

k k0 K0 ___ / K ___ / K00

kf kg kh a a0 A0 / A / A00 / 0

f g h b b0 0 / B0 / B / B00

cf cg ch C0 ___ / C ___ / C00 c c0 with exact rows and columns there exist morphisms k, k0, c, c0 and a morphism δ : K00 → C0 such that the sequence

k k0 δ c c0 K0 / K / K00 / C0 / C / C00 is exact.

Proof. By 2.18 the morphisms k, k0, c and c0 in the statement of the proposition do exist and are uniquely determined by the requirement that the resulting diagrams must commute. We ﬁrst write down the canonical factorizations of all the E-strict morphisms in the above diagram. Then the 94 CHAPTER 3. EXACT CATEGORIES diagram has the following form (ignore the dotted arrows for the moment):

k k0 K0 / K / K00

kf (1) kg (2) kh

ea ma a0 A0 / A / A / A00 / 0

ef eg eh ι π I0 ______/ I ______/ I00

mf mg mh

b eb0 mb0 0 / B0 / B / B / B00

cf cg ch C0 / C / C00 c c0

Since the squares (1) and (2) are commutative there exist unique morphisms 0 00 0 ι: I → I and π : I → I with ι ◦ ef = eg ◦ ma ◦ ea and π ◦ eg = eh ◦ a . In addition we have

b ◦ mf ◦ ef = mg ◦ eg ◦ ma ◦ ea

= mg ◦ ι ◦ ef 0 eb0 ◦ mb0 ◦ mg ◦ eg = mh ◦ eh ◦ a

= mh ◦ π ◦ eg

0 and therefore b ◦ mf = mg ◦ ι and eb ◦ mmb0 ◦ mg = mh ◦ π, since ef and eg are epimorphisms. This shows that all the squares in the above diagram commute. Since mg ◦ ι = b ◦ mf is an admissible monomorphism 0 and π ◦ eg = eh ◦ a is an admissible monomorphism it follows from 3.20 that ι is an admissible monomorphism and that π is an admissible epimorphism, since the category C has both kernels and cokernels. We will now gather some morphisms and factorizations which we will need for the proof. Let π0 : I → J be a morphism with (ι, π0) ∈ E and let ι0 : J 0 → I be a morphism with (ι0, π) ∈ E. Since

0 π ◦ ι ◦ ef = eh ◦ a ◦ ma ◦ ea = 0 we also have π ◦ ι = 0, since ef is an epimorphism. The universal property of the cokernel π0 then gives rise to a unique morphism π00 : J 00 → I00 with π − π00 ◦ π0 and the universal property of the kernel ι0 gives rise to a unique morphism ι00 : I0 → J 0 with ι = ι0 ◦ ι00. It follows from 3.20 that ι00 is an admissible monomorphism and that π00 is an admissible epimorphism. Next, we have 0 0 π ◦ eg ◦ ma ◦ ea = π ◦ ι ◦ ef = 0, 3.2. E-STRICT MORPHISMS 95

0 and thus π ◦eg ◦ma = 0, since ea is an epimorphism. The universal property 0 of the kenrel ι then yields a unique morphism α: A → I with ι◦α = eg ◦ma and the universal property of the cokernel a0 gives rise to a unique morphism 00 00 0 0 ω : A → J with ω ◦ a = π ◦ eg. We have

00 0 00 0 0 π ◦ ω ◦ a = π ◦ π ◦ eg = π ◦ eg = eh ◦ a

0 0 and therefore π ◦ ω = eh, since a is an epimorphism. 0 Let ea: Ae → A be a morphism with (ea, ea) ∈ E. Since ef = α ◦ ea, it follows that

e − f ◦ ea = α ◦ ea ◦ ea = 0,

therefore the universal property of the kernel kf gives rise to a unique mor- 0 phism β : Ae → K with ea = kf ◦β, in particular β is an admissible monomorphism by 3.20. Since

0 0 mb0 ◦ eb0 ◦ mg ◦ ι = mh ◦ π ◦ ι = 0

0 and mb0 is a monomorphism it follows that eb0 ◦ mg ◦ ι = 0. The universal property of the kernel b then gives rise to a unique morphism ω0 : J 0 → B0 0 0 with mg ◦ι = b◦ω and the universal property of the cokernel π gives rise to 0 00 0 a unique morphism α : I → B with α ◦π = eb0 ◦mg. By 3.20 the morphism ω0 is an admissible monomorphism. In addition we have

0 mh ◦ π = mb0 ◦ eb0 ◦ mg = mb0 ◦ α ◦ π

0 0 and therefore mh = mb0 ◦ α , since π is an epimorphism. This shows that α is an admissible monomorphism. 00 0 Let eb: B → Be be a morphism with (mb0 ,eb) ∈ E. Since α ◦ mb0 = mh it follows that eb ◦ mh = 0, therefore the universal property of the cokernel ch 0 00 0 gives rise to a unique morphism β : C → Be with β ◦ ch = eb. 96 CHAPTER 3. EXACT CATEGORIES

Alltogether we have shown the commutativity of the following diagram:

k k0 K0 / K / K00 ? β ~ ~ k k k ~ f g h ~ a ea ma a0 e / A0 / A / A / A00 / 0 Ae @ @@ ~ | @@ ef ~ eg | 0 @@ ~ α | ω @ ~ ι π0 }| I0 / I ___ / J 00 ef A C A ι00 C A π00 C A ι0 π C! mf J 0 ___ / I / I00 A ~ | AA 0 ~ mg | mh AA ~ ω0 | α0 AA ~ }| A 0 / 0 / / / 00 / B B e 0 B m 0 B Be b b b eb }> cf cg ch } } β0 } C0 / C / C00 c c0

Let then x0 : X0 → A be a morphism with (x0, α) ∈ E, let x00 : X00 → A00 be a morphism with (x00, ω) ∈ E and let z : Z → J 00 be a morphism with 00 0 0 (z, π ) ∈ E. By 3.25 there are pairs (nm, ne), (nm, ne) ∈ E making the diagrams

A A X00 X00 e e 0 00 nm a nm x e kf ef kh eh K0 / A0 / I0 K00 / A00 / I00 0 ne ea ne ω X0 / / I0 Z / J 00 / I00 x0 A α z π00 commutative. In addition 2.18 shows the existence of unique morphisms 00 0 00 00 nm : X → K and ne : K → X making the diagram

00 00 nm ne X0 ___ / K ___ / X00

x0 kg x00

ma a0 A / A / A00

α eg ω I00 / I / J 00 ι π0

00 00 commutative. It follows then from 3.26 that (nm, ne ) is an element of E. 3.2. E-STRICT MORPHISMS 97

This shows already that the sequence

00 0 00 nm◦ne nm◦ne K0 / K / K00 ~> BB }> BB z= AA ~ B }} BB z A ~~nm ne BB } n00 n00 B zzn0 n0 AA ~ ! 0 } m e 00 z m e Ae X X Z is exact. Since

00 0 kg ◦ nm ◦ ne = ma ◦ x ◦ ne = ma ◦ ea ◦ kf 0 00 00 00 0 kh ◦ nm ◦ ne = x ◦ ne = a ◦ kg

00 0 00 0 0 it follows that nm ◦ ne = k and nm ◦ ne = k , because of k and k being unique with this properties. This is already half the desired exact sequence. 0 00 Next, we again use 3.25 to ﬁnd a morphism eι: J → Z with (ι ,eι) ∈ E, making the diagram I0 I0

ι00 ι ι0 π J 0 / I / I00 ι π0 e Z / J 00 / B z π00 commutative. Now let η0 : B0 → Y 0 be a morphism with (ω0, η0) ∈ E and let η : B → Y 00 be a morphism with (α0, η) ∈ E. By 3.24 we then get pairs 0 (nem, nee), (ˆnm, β ) ∈ E making the diagrams

ι00 ι α0 η I0 / J 0 e / Z I00 / B / Y 00 0 ω nm mb0 nˆm e I0 / B0 / C0 I00 / B00 / C00 mf cf mh ch 0 0 η ne b β e e 0 0 Y Y Be Be commutative. Additionally we ﬁnd by 2.18 y : Y 0 → C and y0 : C → Y 00 making the diagram ι0 π J 0 / I / I00

ω0 mg α0

b eb0 B0 / B / B

η0 cg η Y 0 ___ / C ___ / Y 00 y y0 98 CHAPTER 3. EXACT CATEGORIES commutative and it follows from 3.26, that (y, y0) ∈ E. This shows that the sequence

0 y◦ne nˆm◦y C0 e / C / C00 ? A ? A = A AA ~~ AA {{ AA A ~ 0 A { 0 A nm ne A ~ y y A { nˆm β A e e 0 ~ 00 { Z Y Y Be is exact and since

0 y ◦ nee ◦ cf = y ◦ η = cg ◦ b 0 nˆm ◦ y ◦ cg =n ˆm ◦ η ◦ eb0 = ch ◦ mb0 ◦ eb0

0 0 0 it follows that y ◦ nee = c andn ˆm ◦ y = c , because of c and c being unique with this properties. Therefore we have shown, that the sequence

k0 k0 δ c c0 K0 / K / K00 / C0 / C / C00 > C > B < B ? A ? A = A ~~ CC }} BB yy BB AA ~~ AA zz AA ~nm ne C } 00 00 B y 0 0 B n n A ~y 0 A znˆ 0 A ~ ! 0 } nm ne ! 00 y nm ne emee 0 ~ y 00 z m β Ae X X Z Y Y Be with δ := nem ◦ ne is exact.

Remark 3.35.

i) The assumption that C has kernels and cokernels can be weakened to C being “weakly idempotent complete” as shown in [3]. A category C is said to be weakly idempotent complete if every retraction has a kernel or equivalently, if every coretraction has a cokernel. The proof is the same as the one above, only the conclusions we have drawn from 3.20 are replaced with a similar cancellation statement for categories that are weakly idempotent complete. We do not develop the theory of weakly idempotent complete categories in this treatise, since we are mostly interested in preabelian categories. The interested reader is referred to [3] and the references therein.

ii) The sequence

k k0 δ c c0 K0 / K / K00 / C0 / C / C00

depends naturally on the diagram in the statement of the proposition. 3.2. E-STRICT MORPHISMS 99

More precisely, if

K0 / K / K00 > f ? e = e

K0 / K / K00

Af0 / Ae / Af00 / 0 |> ? z= || zz || zz A0 / A / A00 / 0

0 / Bf0 / Be / Bf00 |> ? z= || zz || zz 0 / B0 / B / B00

C0 / C / C00 > f ? e = f C0 / C / C00

is a commutative diagram whose rows and columns are elements of E, then also the dotted arrows exist, make the corresponding cubes commutative and there is a commutative diagram

δ K0 / K / K00 / C0 / C / C00 .

δe Kf0 / Ke / Kf00 / Cf0 / Ce / Cf00

The dotted arrows exist by 2.18. The rest of the proof one can show by constructing all the diagrams in the proof also for the backside of the three-dimensional diagram above and show then, using the di- verse universal properties, that morphisms exist making all squares into commutative cubes.

Corollary 3.36. Retaining the assumptions of the snake lemma 3.34 as well as the notations of its proof, the connecting morphism δ : K00 → C0 has the following property: Given a commutative square

ν R / K00

u κ00 A / A00 a0 100 CHAPTER 3. EXACT CATEGORIES there exists a unique morphism w : R → B0 making the diagram

v R / K00 u kh w A / A00 a0 g h × B0 / B / B00 b b0 cf C0 commutative and, moreover, we have δ ◦ v = cf ◦ w. Proof. We have 00 0 π ◦ eg ◦ u = π ◦ π ◦ eg ◦ u = π00 ◦ ω ◦ a0 ◦ u

= eh ◦ kh ◦ v =, hence the universal property of the kernel ι0 gives rise to a unique morphism 0 0 0 0 0 0 w : R → J with eg ◦ u = ι ◦ w . Deﬁne w := ω ◦ w , then we have 0 0 0 0 b ◦ w = b ◦ ω ◦ w = mg ◦ ι ◦ w = mg ◦ eg ◦ u = g ◦ u and the morphism w is unique with this property, since b is a monomorphism. In addition we have 0 0 0 0 z ◦ eι ◦ w = π ◦ ι ◦ w 0 = π ◦ eg ◦ u

= ω ◦ kh ◦ v 0 = z ◦ ne ◦ v, 0 0 therefore it follows that eι ◦ w = ne ◦ v, since z is a monomorphism. This then shows 0 0 0 0 cf ◦ w = cf ◦ ω ◦ w = nem ◦ eι ◦ w = nem ◦ ne ◦ v = δ ◦ v.

Remark 3.37. In 3.36 consider the special case that C is the category (R − mod) of (left-)modules over a ring R. The morphism v corresponds to an 00 element v0 ∈ K and the morphism u corresponds to some element u0 with 0 0 a (u0) = kh(v0), since the morphism a is then surjective. Moreover the usual diagram chase in the proof of the classical snake lemma in (R − mod) 0 shows that there is an element w0 ∈ B such that cf (w0) is independent of the choice of u0, hence it makes sense to put δ(v0) = cf (w0). Therefore the above corollary provides the link to the classical proof of the snake lemma. 3.2. E-STRICT MORPHISMS 101

Corollary 3.38. Let (C, E) be an exact category such that C has kernels and cokernels. For every commutative diagram

k k0 K0 ___ / K ___ / K00

kf kg kh a a0 0 / A0 / A / A00 / 0

f g h b b0 0 / B0 / B / B00 / 0

cf cg ch C0 ___ / C ___ / C00 c c0 with exact rows and columns there exist morphisms k, k0, c, c0 and a morphism δ : K00 → C0 such that the sequence

k k0 δ c c0 0 / K0 / K / K00 / C0 / C / C00 / 0 is exact and depends naturally on the above diagram.

Proof. It only remains to be shown that k is an admissible monomorphism 0 0 and that c is an admissible epimorphism. Since kg ◦ k = a ◦ kf is an admissible monomorphism, it follows from 3.20, that k is an admissible 0 0 monomorphism. Dually, since c ◦ cg = ch ◦ b is an admissible epimorphism, it follows from 3.20, that c0 is an admissible epimorphism. 102 CHAPTER 3. EXACT CATEGORIES Chapter 4

The Maximal Exact Structure of a Preabelian Category

4.1 p-strict Epimorphisms and Monomorphisms

Deﬁnition 4.1. Let C be a preabelian category. i) A strict epimorphism g : Y → Z in C is called a semi-stable cokernel, if for every pullback square

pT P / T

pY t Y g / Z

the morphism pT is also a strict epimorphism. ii) A strict monomorphism f : X → Y C is called a semi-stable kernel, if for every pushout square

f X / Y

t sY T / S sT

the morphism sT is also a strict monomorphism. Remark 4.2. Let C be a preabelian category. Because pullbacks and pushouts are transitive, retractions are stable under pullbacks, coretractions are stable under pushouts and isomorphisms are stable under both (see chapter 2), we obtain the following:

103 104 CHAPTER 4. MAXIMAL EXACT STRUCTURE

i) In the situation of 4.1, pT and sT are again semi-stable. ii) Retractions are semi-stable cokernels.

iii) Coretractions are semi-stable kernels.

iv) Isomorphisms are semi-stable cokernels and semi-stable kernels. Example 4.3. i) In the category (TVS) every strict epimorphism is a semi-stable cokernel: Let pX P / X

pY f Y g / Z be a pullback square in (TVS), such that g is a strict epimorphism, i.e. an open mapping and such that (P, pX , pY ) is constructed as in 2.35. Let U be a zero-neighbourhood in X, let V be a zero-neighbourhood in Y and deﬁne W := f −1(g(V )) ∩ U. Since g is open, W is a zero- neighbourhood in X. We claim that W ⊆ pX (U × V ): For w ∈ W we have w ∈ f −1(g(V )), hence there exists a y ∈ V with f(w) = g(y). Then we have (w, y) ∈ P with w ∈ U and y ∈ V , hence w = pX (x, y). This establishes the claim and thus shows that pX is an open mapping, i.e. a strict epimorphism. This shows that g is a semi-stable cokernel. The same is true in the categories (LCS) and (LCS)HD. ii) In the category (TVS) every strict monomorphism is a semi-stable kernel: Let f Z / X

g sX Y / S sY be a pushout square in (TVS), such that f is a strict monomorphism, i.e. injective and open onto its range and such that S is constructed as in 2.35. Since (TVS) is semi-abelian by 2.73.iii), it follows from 2.81.ii), that sY is a monomorphim, i.e. injective. It remains to be shown, that sY is open onto its range. Let W be a zero-neighbourhood of Y and choose a zero-neighbourhood V with V + V ⊆ W . Since f is open onto its range, there exists a zero-neighbourhood U in X with −1 U ∩ f(Z) ⊆ f(g (V )). We claim that qL(u × V ) ∩ sY (Y ) ⊆ sY (W ), where qL : Y × X → Y × X/L = S is the quotient map. Let u ∈ U, v ∈ V , y ∈ Y with qL(u, v) = qL(0, y) = sY (y). Then we have (u, v −y) ∈ ker(qL) = L = {(f(z), −g(z)) | z ∈ Z}, hence there exists a z ∈ Z with u = f(z), y − v = g(Z). Since u ∈ f(Z) ∩ U ⊆ f(g−1(W )) 4.1. P -STRICT MORPHISMS 105

it follows that there also exists a w ∈ Z with u = f(w) and g(w) ∈ V . Since f is injective we have w = z, hence g(z) ∈ V . Then y = v+g(z) ∈ V +V ⊆ W and therefore qL(u, v) = qL(0, y) ⊆ qL({0}×W ) = sY (W ), which establishes the claim. This shows that sY is open onto its range and is therefore a strict monomorphism, hence f is a semi-stable kernel. The same is true in the categories (LCS) and (LCS)HD. Lemma 4.4. Let C be a preabelian category that is a full additive subcategory of the category (TVS) and contains the ground ﬁeld K. i) Every semi-stable cokernel is surjective.

ii) Every semi-stable kernel is injective.

Proof. i) Suppose g is not surjective. Choose z0 ∈ Z with z0 ∈/ g(Y ). Then the mapping φz0 : K → Z, λ 7→ λz0 is linear and continous, hence it is a morphism in C. By 2.35 the pullback of φz0 and g in (TVS) is the square

p −1 K ( g − φz0 ) ({0}) / K , p φ Y (1) z0 Y g / Z where ( g − φz0 ): Y × K → Z is the canonical morphism and where pY and pK are the restrictions of the projections. Since z0 ∈/ g(Y ) we have −1 −1 ( g − φz0 ) ({0}) = g ({0}) × {0}, hence pK = 0. Let

qK Q / K

qY φz0 Y g / Z be a pullback diagram in C, then the universal property of (1) gives rise −1 to a unique morphism µ: Q → ( g − φz0 ) ({0}) with pK ◦ µ = qK and pY ◦ µ = qY . Then qK = 0: Q → K, which is a contradiction to g being a semi-stable cokernel, since 0 is not an epimorphism in C. This shows that g is a surjective mapping. ii) We can assume X 6= 0. Suppose then that f is not injective. Then there exists a x0 ∈ X such that f(x0) = 0 and x0 6= 0. Deﬁne

φ1 : K → X , λ 7→ 0 φ2 : K → X , λ 7→ λx0. These are linear and continous mappings and are therefore morphisms in C. It follows that f ◦ φ1 = f ◦ φ2 = 0, but φ1 6= φ2 in contradiction to f being a monomorphism. This shows that f is injective. 106 CHAPTER 4. MAXIMAL EXACT STRUCTURE

c Example 4.5. We have seen in 2.64 that the category (LCS)HD contains strict epimorphisms that are not surjective, hence the class of semi-stable cokernels is a proper subclass of the strict epimorphisms in (LCS)HD. Lemma 4.6. Let C be a preabelian category.

i) Let f : X → Y and g : Y → Z be semi-stable cokernels. Then the composition h := g ◦ f is a semi-stable cokernel.

ii) Let f : X → Y and g : Y → Z be semi-stable kernels. Then the composition h := g ◦ f is a semi-stable kernel.

Proof. It suﬃces to show i), since ii) is the dual statement. We have

h ◦ kf = g ◦ f ◦ kf = 0, therefore the universal property of the kernel kh gives rise to a unique morphism j : ker f → ker h with kf = kh ◦ j. In addition

g ◦ f ◦ kh = h ◦ kh = 0 gives rise to a unique morphism k : ker h → ker g with f ◦ kh = kg ◦ k. We consider the commutative diagram

k ker h / ker g w; j ww ww kh (1) kg ww ww ker f / X / Y kf GG f GG GG g h GG GG # Z and claim that (1) is a pullback square. Let lX : L → X and lk : L → ker g be morphisms with f ◦ lX = kg ◦ lk. Since

h ◦ lX = g ◦ f ◦ lX = g ◦ kg ◦ lk = 0, the universal property of the kernel kh gives rise to a unique morphism λ: L → ker h with lX = kh ◦ λ. Then

kg ◦ k ◦ λ = f ◦ kh ◦ λ = f ◦ lX = kg ◦ lk.

Since kg is a monomorphism this implies k ◦ λ = lk and λ is unique with lX = kh ◦ λ and k ◦ λ = lk since kh is a monomorphism. This establishes the claim. Since f is a p-strict epimorphism, the above provides that k is a strict epimorphism. We claim that h is a cokernel of its kernel (and therefore a strict epimorphism by 2.27): 4.1. P -STRICT MORPHISMS 107

Let t: X → T be a morphism with t ◦ kh = 0. Then t ◦ kf = t ◦ kh ◦ j = 0. Since f is a strict epimorphism, it is a cokernel of its kernel by 2.27, hence there exists a unique morphism t0 : Y → T with t0 ◦ f = t. Additionally we 0 0 have t ◦kg ◦k = t◦kh = 0 which yields t ◦kg = 0, since k is an epimorphism. Since g is a strict epimorphism, it is the cokernel of its kernel by 2.27, hence there exists a unique morphism t00 : Z → T with t00 ◦ g = t0. Therefore we have t00 ◦ h = t00 ◦ g ◦ f = t0 ◦ f = t. Moreover, t00 is unique with this property, since h is an epimorphism as a composition of two epimorphisms. This establishes the claim and thus ﬁnishes the proof.

Lemma 4.7. Let C be a preabelian category.

i) Let f : X → Y and g : Y → Z be morphisms in C. If h := g ◦ f is a semi-stable cokernel then g is a semi-stable cokernel.

ii) Let f : X → Y and g : Y → Z be morphisms in C. If h := g ◦ f is a semi-stable kernel then f is a semi-stable kernel.

Proof. Again it is enough to show i). We claim that

( f kg ) X q ker g / Y

( 1 0 ) (1) g X / Z h is a pullback square. Let lX : L → X and lY : L → Y be morphisms with g◦lY = h◦lX . Hence g◦(lY −f ◦lX ) = g◦lY −g◦f ◦lX = g◦lY −h◦lX = 0. By the universal property of the kernel of g, we get a morphism w : L → ker g lX with kg ◦w = lY −f ◦lX and may consider the morphism w : L → Xqker g. (1 0) lX (f kg) lX Clearly, w = lX and w = f◦lX +kg◦w = f◦lX +lY −f◦lX = lY . Assume that λ1 satisﬁes the same properties. That is (1 0) λ1 = l , hence λ2 λ2 X l = λ , and (f kg) lX = f◦l +k ◦λ = l , i.e. k ◦λ = l −f◦l = k ◦w. X 1 λ2 X g 2 Y g 2 Y X g Since kg is a monomorphism this implies w = λ2 and we have shown the lX uniqueness of the morphism w which establishes the claim. Since h is a semi-stable cokernel the above shows that ( f kg ): Xqker g → Y is a strict epimorphism. Now we claim that g is a cokernel of kg. Let x: Y → F be a morphism with x ◦ kg = 0. We have

x ◦ f ◦ kh = x ◦ kg ◦ k = 0 ◦ k = 0 and since h is a strict epimorphism, it is a cokernel of its kernel by 2.27. This gives rise to a unique morphism x0 : Z → F with x0 ◦ h = x ◦ f. We 108 CHAPTER 4. MAXIMAL EXACT STRUCTURE consider the diagram of the preceeding proof

k ker h / ker g ; C j ww C 0 ww k CC ww kh g CC ww C w x C! ker f / X / Y / F kf GG f z= GG zz GG g zz h GG zz x0 GG zz # Z and claim that the right lower triangle is commutative: We compute x ◦ f = x0 ◦ h = x0 ◦ g ◦ f that is (x − x0 ◦ g) ◦ f = 0. Moreover, 0 0 (x − x ◦ g) ◦ kg = x ◦ kg − x ◦ g ◦ kg = 0 − 0 = 0. We consider the morphism 0 (x − x ◦ g) ◦ ( f kg ): X q ker g → Z, which is zero by the above: 0 0 0 (x − x ◦ g) ◦ ( f kg ) = ( (x − x ◦ g) ◦ f (x − x ◦ g) ◦ kg ) = 0 Since h is a semi-stable cokernel and (1) is a pullback square, it follows that 0 ( f kg ) is an epimorphism, therefore the above implies x − x ◦ g = 0 that is x = x0 ◦ g. Finally, x0 is unique with this property since g is an epimorphism as by 1.10. This establishes the claim that g is a cokernel of its kernel, hence it is a strict epimorphism by 2.27. Proposition 4.8. Let C be a preabelian category and let f : X → Y and g : Y → Z be morphisms in C. Put h := g ◦ f : X → Z. Then i) If f and g are semi-stable cokernels then h is a semi-stable cokernel. ii) If f and g are semi-stable kernels then h is a semi-stable kernel. iii) If h is a semi-stable cokernel then g is a semi-stable cokernel. iv) If h is a semi-stable kernel then f is a semi-stable kernel. Proof. i) Let h: Z0 → Z be arbitrary. The pullback of h and g is of the form

f 0 g0 X0 / Y 0 / Z0

(2) (1) X / Y / Z f g where (1) and (2) are pullback diagrams By 4.6, the composition g ◦ f is a strict epimorphism. By 4.2, f 0 and g0 are semi-stable cokernels and hence by applying 4.6 again, g0 ◦f 0 is a strict epimorphism that is g ◦f is a semi-stable cokernel. iii) We consider the same diagram as in the proof of i). By 4.7, g is a strict epimorphism. By 4.2.i), g0 ◦ f 0 is a semi-stable cokernel and hence by 4.7, g0 is a strict epimorphism that is, by deﬁnition g is a semi-stable cokernel. ii) and iv) can be obtained from i) and iii) by dualization. 4.2. MAXIMAL EXACT STRUCTURE 109

4.2 The Maximal Exact Structure of a Preabelian Category

In what follows the proof of [E1op] was inspired by that of Keller [12, Propo- sition after A.1].

Deﬁnition 4.9. Let C be a preabelian category. A kernel-cokernel pair

f g X / Y / Z is called stable if f is a semi-stable kernel and g a semi-stable cokernel.

Theorem 4.10. If C is a preabelian category then the class

E = (f, g) | (f, g) is a stable kernel-cokernel pair is an exact structure on C. Moreover, E is maximal in the sense that all exact structures on C are contained within it.

Proof. We show that E is closed under isomorphisms. Let (f, g) ∈ E and let f g X / Y / Z

iX iY iZ X0 / Y 0 / Z0 f 0 g0 be a commutative square in C with isomorphisms iX , iY and iZ . Then (f 0, g0) lies in E. In fact (f 0, g0) is a kernel-cokernel pair by 3.2.ii) and it follows from 2.36 that f 0 is a semi-stable kernel and that g0 is a semi-stable cokernel. Note that a pair (e, f) is an element of E, if and only if (f op, eop) is an element of Eop. Therefore it suﬃces to show [E0op], [E1op] and [E2op] in order to show that E is an exact structure on C. [E0op] are satisﬁed by 4.2.iv). [E2op] Let (f, g) be an element of E and assume that

pT P / T

pY PB t Y g / Z is a pullback square. Let k : K → P be a kernel of pT . Then pT is a semi- stable cokernel and (k, pT ) is a kernel-cokernel pair. Thus it remains to be shown that k is a semi-stable kernel. Lemma 2.71.i) combined with the 110 CHAPTER 4. MAXIMAL EXACT STRUCTURE universal property of k shows that there is an isomorphism ρ: K → X such that py ◦ k = f ◦ ρ. Let r : K → R be arbitrary and let

k K / P

r (1) sP R / S sR be a pushout square. We have to show that sR is a strict monomorphism. Construct pushouts

r sp K / R and P / S

ρ (2) ψ1 pY (3) ψ2 X / Q1 S / Q2. φ1 φ2

Then ψ1 is an isomorphism. We have

ψ2 ◦ sR ◦ r = ψ2 ◦ sP ◦ k = φ2 ◦ pY ◦ k = φ2 ◦ f ◦ ρ, hence by the universal property of the pushout (2) gives rise to a unique morphism ε: Q1 → Q2 with ψ2 ◦ sR = ε ◦ ψ1 and φ2 ◦ f = ε ◦ φ1, making the cube sP P / S > = k } sR {{ }} {{ }} {{ }} r {{ K / R ψ2

pY ψ1 ρ Y / Q2 ? φ2 > f || || ||ε || X / Q1 φ1 commutative. We claim that (Q2, ε, φ2) is a pushout of f and φ1. Let l1 : Q1 → L and lY : Y → L be morphisms such that lY ◦ f = l1 ◦ φ1. Then we have

−1 −1 lY ◦ pY ◦ k ◦ ρ = l1 ◦ ψ1 ◦ r ◦ ρ , hence lY ◦ pY ◦ k = l1 ◦ ψ1 ◦ r. The universal property of the pushout (1) gives rise to a unique morphism µ1 : S → L with lY ◦ pY = µ1 ◦ sP and l1 ◦ ψ1 = µ1 ◦ sR and the universal property of the pushout (3) gives rise to a unique morphism µ2 : Q2 → L with lY = µ2 ◦ φ2 and µ1 = µ2 ◦ ψ2. Then

l1 ◦ ψ1 = µ1 ◦ sR = µ2 ◦ ψ2 ◦ sR = µ2 ◦ ε ◦ ψ1, 4.2. MAXIMAL EXACT STRUCTURE 111 and therefore l1 = µ2 ◦ ε, since ψ1 is an isomorphism. The morphism µ2 is unique with this property because of the universal properties of the pushouts (1) and (3). This establishes the claim. Since f is a semi-stable kernel it follows from the above that ε is a semi- stable kernel. By 4.8.ii) ε ◦ ψ1 = ψ2 ◦ sR is also a semi-stable kernel and by 4.8.iv) it follows that sR is a semi-stable kernel, hence k is a semi-stable kernel which shows that (k, pT ) is an element of E. [E1op] Let (f, g), (f 0, g0) ∈ E be pairs so that g0 ◦g is deﬁned and let k : K → Y be a kernel of g0 ◦ g. Then g0 ◦ g is a semi-stable cokernel by 4.8.i) and (k, g0 ◦ g) is a kernel-cokernel pair. Thus it remains to be shown that k is a semi-stable kernel. Since g0 ◦ g ◦ k = 0, there exists a unique morphism α: K → X0 with f 0 ◦ α = g ◦ k. Claim A. The diagram α K / X0

k (4) f 0 Y g / Z is a pullback square. 0 0 Let lY : L → Y and lX0 : L → X be morphisms with f ◦ lX0 = g ◦ lY . Then 0 0 0 g ◦g ◦lY = g ◦f ◦lX0 = 0, hence there exists a unique morphism η : L → K with lY = k ◦ η. This yields 0 0 f ◦ lX0 = g ◦ lY = g ◦ k ◦ η = f ◦ α ◦ η 0 and from this follows lX0 = α◦η, since f is a monomorphism. The morphism η is unique with this property, since k is a monomorphism, hence (4) is a pullback square. Thus, Claim A is established. Claim B. Let (f, g) ∈ E and

pR P / R

pY PB r Y g / Z be a pullback square. Then ( −pR , ( r g )) ∈ E. pY By 2.71.i) we have a commutative diagram

k pR X / P / R

pY t X / Y / Z f g 112 CHAPTER 4. MAXIMAL EXACT STRUCTURE

op such that k is a kernel of pR and by [E2 ] the pair (k, pR) is an element of E. If (S, sY , sP ) is the pushout of f and k, Lemma 2.71.ii) yields a commutative diagram

f g X / Y / Z

k sY P / S / Z sP c

pR c0 R R

0 such that c is a cokernel of sP and c is a cokernel of sY . It follows from [E2] 0 −pR that (sp, c) and (sy, c ) are elements of E. We have ◦k = ωY ◦f, hence pY the universal property of the pushout yields a unique morphism λ: S → −pR R Π Y with = λ ◦ sP and ωY = λ ◦ sY . Then λ is an isomorphism. pY In fact, because of (sY ◦ pY − sP ) ◦ k = 0 there exists a unique morphism γ : R → S with sY ◦ pY − sP = γ ◦ pR. This in turn gives rise to a unique morphism µ: R Π Y → S with γ = µ ◦ ωR and sY = µ ◦ ωY . We have

λ ◦ µ ◦ ωY = λ ◦ sY

= ωY

λ ◦ µ ◦ ωR ◦ pR = λ ◦ γ ◦ pR

= λ ◦ (sY ◦ pY − sp)

= λ ◦ sY ◦ pY − λ ◦ sP −pR −pR = ωY ◦ πY ◦ − pY pY −pR = (ωY ◦ πY − idRqY ) ◦ pY −pR = −ωR ◦ πR ◦ pY = ωR ◦ pR

therefore the universal property of the coproduct yields λ◦µ = idY qR, since 4.2. MAXIMAL EXACT STRUCTURE 113 pR is an epimorphism. In addition we have

µλ ◦ sY = µ ◦ ωY

= sY −pR µλ ◦ sP = µ ◦ pY −pR = µ ◦ (ωY ◦ πY + ωR ◦ πR) ◦ pY −pR −pR = µ ◦ ωY ◦ πY ◦ + µ ◦ ωR ◦ πR ◦ pY pY = sY ◦ pY − γ ◦ pR

= sY ◦ pY − (sY ◦ pY − sP )

= sP hence the universal property of the product yields µ ◦ λ = idS, which shows that λ is an isomorphism. In addition we have

c ◦ µ ◦ ωY = c ◦ sY = g

c ◦ µ ◦ ωR ◦ pR = c ◦ γ ◦ pR

= c ◦ (sY ◦ pY − sP )

= c ◦ sY ◦ pY − c ◦ sP

= g ◦ pY

= r ◦ pR hence the universal property of the coproduct yields c ◦ µ = ( r g ), since pR is an epimorphism. Therefore the diagram

sP c P / S / Z

λ P / R Π Y / Z ( −pR ) ( r g ) pY is commutative. This establishes Claim B, since E is closed under isomorphisms. −α By Claim B we know that the pair (p, q) of morphisms p := k : K → 0 0 0 f 0 0 X Π Y and q := ( f g ): X Π Y → Z is an element of E. We put r := 0 idY and obtain the commutative diagram

f 0 X0 / Z

ωX0 (5) ωZ

0 X Π Y r / Z Π Y. 114 CHAPTER 4. MAXIMAL EXACT STRUCTURE

Claim C (5) is a pushout square. 0 0 Let lX0Π Y : X Π Y → L and lZ : Z → L be morphisms with lZ ◦ f = lX0Π Y ◦

ωX0 . Denote lY := lX0Π Y ◦ ωY and δ := ( lZ lY ): Z Π Y → L. Then we have

0 lX0Π Y ◦ ωX0 = lZ ◦ f 0 = δ ◦ ωZ ◦ f

= δ ◦ r ◦ ωX0 and

lX0Π Y ◦ ωY = δ ◦ ωY

= δ ◦ ωY ◦ πY ◦ r ◦ ωY

= δ ◦ (idZ Π Y −ωZ ◦ πZ ) ◦ r ◦ ωY

= δ ◦ r ◦ ωY .

The universal property of the coproduct then yields lX0Π Y = δ ◦ r. The uniqueness of δ follows from the universal property of the coproduct, which yields Claim C. Now r is a semi-stable kernel and by 4.8.iii) the composition r ◦ p is also a semi-stable kernel. We put σ := −g and obtain idY

0 0 r ◦ p = f 0 −α = −f ◦α = −g◦k = σ ◦ k 0 idY k k k Since r ◦ p is a semi-stable kernel, it follows from 4.8.iv), that k is a semi- stable kernel, which shows that (k, g0 ◦ g) is an element of E and thus that E is an exact structure on C. It remains to check the maximality of E. Let E be a second exact structure on C. If pT P / T

pY PB t Y g / Z is a pullback square, the morphism pT is an admissible epimorphism by [E2op], hence it is a cokernel of its kernel and thus a strict epimorphism, hence g is a semi-stable cokernel. Analogously, by [E2] the morphism f is a semi-stable kernel, which shows (f, g) ∈ E. Hence we have E ⊆ E.

Notation 4.11. If C is a preabelian category we will denote the maximal exact structure

E = (e, f);(e, f) is a stable kernel-cokernel pair on C from now on as Emax. 4.3. QUASI-ABELIAN CATEGORIES 115

Example 4.12. i) We have shown in 4.3, that in the preabelian categories (TVS), (LCS) and (LCS)HD the maximal exact structure Emax coincides with the class of all kernel-cokernel pairs. Categories of this type are called quasiabelian and are studied in more detail in the next section.

c ii) The category (LCS)HD is a preabelian category by 2.64 and contains cokernels that are not semi-stable by 4.5, hence Emax is not the class c of all kernel-cokernel pairs. However, (LCS)HD possesses the three space property with regard to LCS)HD by 3.19, hence the restriction c of the exact structure of (LCS)HD to (LCS)HD is an exact structure c f g E on (LCS)HD by 3.17 with E ⊆ Emax. Let X / / Z be an element of Emax. Since (f, g) is a kernel-cokernel pair, it follows from 2.64 that there is an isomorphism ψ : C(Y/f(X)) → Z making the diagram p Y Q / C(Y/f(X)) QQQ QQQ QQQ ψ g QQQ QQQ Q( Z commutative, where C(Y/f(X)) is the Hausdorﬀ-completion of Y/f(X) and where p is the composition of the quotient map Y → Y/f(X) and the canonical map Y/f(X) → C(Y/f(X)). Since g is surjective by 4.4, the map p is also surjective, which means that C(Y/f(X)) = Y/f(X). This shows that g is also the cokernel of f in the category (LCS)HD. Furthermore (LCS)HD reﬂects the kernels of (LCS)HD by 2.64, hence (f, g) is an element of E. This shows that

f g Emax = { X / Y / Z | (f, g) is a kernel-cokernel pair in (LCS)HD} c in the category (LCS)HD.

4.3 Quasi-abelian Categories

Deﬁnition 4.13. Let C be a preabelian category. The category C is called quasiabelian if its maximal exact structure Emax is the class of all kernel- cokernel pairs. In other words C is quasiabelian if it has the following properties: (Q1) If g : Y → Z is a strict epimorphism and

T P / T

pY PB t Y g / Z 116 CHAPTER 4. MAXIMAL EXACT STRUCTURE

a pullback square, then pT is also a strict epimorphism, i.e. every strict epimorphism is p-strict.

(Q2) If f : X → Y is a strict monomorphism and

f X / Y

t PO sY T / S sT

a pushout square, then sT is also a strict monomorphism, i.e. every strict monomorphism is p-strict.

Example 4.14. We have shown in 4.3, that in the categories (TVS), (LCS) and (LCS)HD every strict epimorphism is a semi-stable cokernel and that every strict monomorphism is a semi-stable kernel, therefore these categories are quasiabelian.

Proposition 4.15. Every abelian category C is quasiabelian.

Proof. Let g : Y → Z be a strict epimorphism and let

T P / T

pY PB t Y g / Z be a pullback square. Since C is abelian it is also semi-abelian, hence it follows from 2.81.ii) that pT is an epimorphism. Every morphism in C is a strict morphism, therefore pT is a strict epimorphism. By using 2.81.i) instead of 2.81.ii) one sees analogously that the property (Q2) is also satisﬁed.

Remark 4.16. The converse of the above proposition is false, as the examples of 4.14 show.

Proposition 4.17. Every quasiabelian category C is semi-abelian.

Proof. Let f be a morphism in C and let

kf f cf ker f / X / Y / cok f O cif if coim f / im f fe 4.3. QUASI-ABELIAN CATEGORIES 117 be its canonical factorization. We have to show that fe is a bimorphism. Let t1, t2 : T → coim f be morphisms with if ◦ fe◦ (t1 − t2) = 0 and let

pX P / X

pT PB cif T / coim f (t1−t2) be a pullback square. Then

f ◦ pX = id ◦ fe◦ cif ◦ pX = if ◦ fe◦ (t1 − t2) ◦ pT = 0, hence the universal property of the kernel kf gives rise to a unique morphism λ: P → ker f with pX = kf ◦ λ. It follows that

(t1 − t2) ◦ pT = cif ◦ pX = cif ◦ kf ◦ λ = 0, and since C is quasiabelian, pT is an epimorphism, hence t1 = t2. This shows that if ◦ fe , and thus fe, is a monomorphism. The dual argumentation of the above shows, that fe◦ cif , and thus also fe, is an epimorphism, hence fe is a bimorphism. Remark 4.18. The converse of the above proposition is false: Let (BOR) be the full subcategory of (LCS) consisting of bornological spaces (cf. [6, § 23, 1.5 and § 11, 2.]). (BOR) is additive and since quotients of bornological spaces are again bornological, it reﬂects the cokernels of (LCS) (cf. [6, § 23, 2.9]). Let f : E → F be a linear and continuous map in (BOR). We consider the linear space f −1(0). If we consider f as a morphism in (LCS), f −1(0) endowed with the topology induced by E would be a kernel of f. Unfortunately, this space is in general not bornological. However, f −1(0) endowed with the associated bornological topology w.r.t. the induced one, (cf. [6, § 11, 2.2]) is bornological, we will denote this space by f −1(0)BOR. It is easy to check that this space together with the inclusion mapping is a kernel of f in (BOR). From the above it follows that for an arbitrary morphism f : E → F in (BOR), the cokernel of the kernel of f is E/f −1(0) endowed with the quotient topology and that the kernel of the cokernel of f is f(E)BOR w.r.t. the topology induced by F . Thus, the canonical morphism fe: coim(f) → im(f) is bijective and it is easy to see that it thus is both a monomorphism and epimorphism. Thus, (BOR) is a semi-abelian category. However, (BOR) is not quasiabelian; Bonet, Dierolf [2] constructed morphisms f : E → F and g : G → F such that g is a strict epimorphism but in the pullback square pE P / E

pG f G g / F 118 CHAPTER 4. MAXIMAL EXACT STRUCTURE pE fails to be a strict epimorphism, that is g is not a semi-stable cokernel.

Remark 4.19. The relationships between the diﬀerent kinds of additive categories we have deﬁned can be visualized in the following scheme:

Abelian Categories

Quasi-abelian Categories

Semi-abelian Categories

Preabelian Categories

Additive Categories

Each of the above arrows is strict as we have shown in the examples 4.16, 4.18, 2.73.iii) and 2.62.ii).

Proposition 4.20. Let C be a quasiabelian category. If C0 is a full additive subcategory of C that reﬂects kernels and cokernels, then C0 is also quasiabelian.

Proof. The category C0 is again preabelian. Let g : Y → Z be a strict epimorphism in C0, i.e. a cokernel of its kernel by 2.27. Since C0 reﬂects kernels and cokernels, g is also a strict epimorphism in C. Let

pT P / T

pY PB t / Y t Z be a pullback square in C. By 2.63.ii) the subcategory C0 reflects pullbacks, 0 hence P is an object of C . Since C is quasiabelian, pT is a strict epimorphism, i.e. a cokernel of its kernel by 2.27. Hence pT is also a strict epimorphism in C0, since C0 reflects kernels and cokernels. This shows, that the property (Q1) is satsified and the dual argument of the above shows that the property (Q2) is also satisfied.

Example 4.21. The above proposition is useful in checking wether a given full additive subcategory of the quasiabelian categories (TVS), (LCS) or (LCS)HD is quasiabelian. For example, it follows directly from 4.20 that the categories

• (Ban) of Banach spaces,

• (F ) of Fr´echet spaces, 4.3. QUASI-ABELIAN CATEGORIES 119

• (FH) of Fr´echet-Hilbert spaces and

• (FS) of Fr´echet-Schwartz spaces are all quasiabelian categories. 120 CHAPTER 4. MAXIMAL EXACT STRUCTURE Chapter 5

Derived Functors

5.1 Exact Functors

Deﬁnition 5.1. Let (C, E) and (C0, E0) be exact categories and let

F : C → C0 be a covariant additive functor. i) F is called exact if (F (f),F (g)) ∈ E0 for all (f, g) ∈ E.

ii) F is said to reﬂect exactness if (F (f),F (g)) ∈ E0 implies (f, g) ∈ E. Remark 5.2. A contravariant additive functor F : C → C0 between exact categories (C, E) and (C0, E0) is said to be exact (resp. to reﬂect exactness) if it is so as a functor from Cop to C0. Remark 5.3. i) The composition of two exact functors between exact categories is again an exact functor.

ii) For every exact category (C, E) the duality functor

op(X) = X op : C → Cop , . op(f) = f op

is exact by 3.6.i).

iii) We have seen in 4.12.ii), that the inclusion functor

c (LCS)HD ,→ (LCS)HD is an exact functor that reﬂects exactness, when both these preabelian categories are equipped with their maximal exact structures. More generally one can say that given two exact categories (C, E) and

121 122 CHAPTER 5. DERIVED FUNCTORS

(C0, E0) such that C0 is a full additive subcategory of C the inclusion functor C0 ,→ C is an exact functor that reﬂects exactness if and only if C0 is a fully exact subcategory of C. Lemma 5.4. Let C and D be additive categories and let F : C → D be an additive functor. If (f, g) is a split exact sequence in C, then (F (f),F (g)) is split exact in D. Proof. The class of all split exact kernel-cokernel pairs is an exact structure on C, by 3.10. By 3.8 there is a commutative diagram

f g X / Y / Z .

β X / X q Z / Z ωX πZ The functor F transforms this into a commutative diagram

F (f) F (g) F (X) / F (Y ) / F (Z) .

F (β) F (X) / F (X q Z) / F (Z) F (ωX ) F (πZ )

We know by 2.49 that F (X qZ) is a biproduct of F (X) and F (Z), F (ωX ) = ωF (X) and F (πZ ) = πF (Z). In addition β is an isomorphism by 3.23, hence it follows that F (β) is an isomorphism. Since (ωX , πZ ) is a kernel-cokernel pair by 3.7 and the kernel-cokernel pairs in an additive category are closed under isomorphisms, it follows from the above diagram that (F (f),F (g)) is split exact.

Remark 5.5. Let C be an additive category. By 3.10 the class Emin of all split exact kernel-cokernel pairs is an exact structure on C, that is contained in every other exact structure on C. If we regard C as an exact category together with the exact structure Emin, then it follows from the above lemma 5.4 that every additive functor F : C → D into an exact category (D, E) is an exact functor. Proposition 5.6. Let (C, E) and (C0, E0, ) be exact categories and let F : C → C0 be an exact functor. i) If f : X → Y is an admissible monomorphism in C and

f X / Y

t PO sY T / S sT 5.1. EXACT FUNCTORS 123

is a pushout square in C, then the image of the above diagram under F is a pushout in C0.

ii) If g : Y → Z is an admissible epimorphism in C and

pT P / T

pY PB t Y g / Z

is a pullback square in C, then the image of the above diagram under F is a pullback in C0. Proof. i) By 3.12 the pushout diagram in the statement of the proposition is part of a commutative diagram with exact rows, which becomes the diagram

F (f) F (g) F (X) / F (Y ) / F (Z)

F (t) (1) F (sY ) F (T ) / F (S) / F (Z) F (sT ) F (c)

0 0 in C . Since F is exact we have (F (f),F (g)), (F (sT ),F (c)) ∈ E and since the above diagram commutes it follows from 3.12 that the square (1) is a pushout. By substituting 3.13 for 3.12 one can show ii) analogously.

Deﬁnition 5.7. Let (C, E) be an exact category, let A be an abelian category and let F : C → A be a covariant (resp. contravariant) additive functor.

f g i) F is called injective, if for every sequence X / Y / Z in C with (f, g) ∈ E the induced sequence

F (f) F (g) 0 / F (X) / F (Y ) / F (Z)

F (g) F (f) (resp. F (Z) / F (Y ) / F (X) / 0)

is exact in A.

f g ii) F is called semi-injective, if for every sequence X / Y / Z in C with (f, g) ∈ E the induced sequence

F (f) 0 / F (X) / F (Y )

F (f) (resp. F (Y ) / F (X) / 0)

is exact in A. 124 CHAPTER 5. DERIVED FUNCTORS

f g iii) F is called projective, if for every sequence X / Y / Z in C in C with (f, g) ∈ E the induced sequence

F (f) F (g) F (X) / F (Y ) / F (Z) / 0

F (g) F (f) (resp. 0 / F (Z) / F (Y ) / F (X)) is exact in A.

f g iv) F is called semi-projective, if for every sequence X / Y / Z in C in C with (f, g) ∈ E the induced sequence

F (g) F (Y ) / F (Z) / 0

F (g) (resp. 0 / F (Z) / F (Y )) is exact in A. Proposition 5.8. Let (C, E) be an exact category, let A be an abelian category and let F : C → A be a covariant additive functor. i) F is injective ⇔ F transforms the kernel of every E-strict morphism g : Y → Z into the kernel of F (g). ii) F is projective ⇔ F transforms the cokernel of every E-strict morphism f : X → Y into the cokernel of F (f). iii) F is semi-injective ⇔ F transforms admissible monomorphisms into monomorphisms. iv) F is semi-projective ⇔ F transforms admissible epimorphisms into epimorphisms. Proof. This is a direct consequence of 3.32.

Remark 5.9. i) Injective functors are also often called left exact functors and projective functors are often called right exact functors. ii) If (C, E) is an exact category, A an abelian category and F : C → A is a functor that is both injective and projective then it is an exact functor if A is considered as an exact category with its maximal exact structure, the class of all kernel-cokernel pairs. iii) Let (C, E) and (C0, E0) be two exact categories and F : C → C0 an exact functor. If A is an abelian category and I : C0 → A an injective functor, then the functor I ◦ F is also injective. Analogously if P : C0 → A is a projective functor, then the functor P ◦ F is also projective. 5.1. EXACT FUNCTORS 125

Proposition 5.10. Let (C, E) be an exact category and let A be an object of C. The functors

HomC(A, −): C → (Ab) and HomC(−,A): C → (Ab) are both injective functors.

op Proof. We have HomC (−,A) = HomCop (A, −) ◦ by 1.24, hence it suﬃces to show that HomC(A, −) is injective by 5.9.iii). f g The functor HomC(A, −) is additive by 2.7. Let 0 / X / Y / Z / 0 be a sequence in C with (f, g) ∈ E, then we get the sequence

HomC(A,f) HomC(A,g) (?) 0 / HomC(A, X) / HomC(A, Y ) / HomC(A, Z) of abelian groups and group morphisms. For (?) to be exact we have to show:

(1) HomC(A, f) is an injective mapping.

(2) im HomC(A, f) = ker HomC(A, g). Since f is a monomorphism, it follows from

HomC(A, f)(g1) = f ◦ g1 = f ◦ g2 = HomC(A, f)(g2) that g1 = g2, therefore the mapping HomC(A, f) is injective. Because of g ◦ f = 0 we have HomC(A, g) ◦ HomC(A, f) = 0, which shows that im HomC(A, −) ⊆ ker HomC(A, −). If t: A → Y is a morphism with g ◦ t = 0, the universal property of f, which is the kernel of g, gives rise to a unique morphism λ: A → X with t = f ◦ λ = HomC(A, f)(λ). This shows (2). Remark 5.11 (Embedding Theorem for Small Exact Categories). Every small exact category can be considered as a fully exact subcategory of an abelian category. More precisely one has the following embedding theorem for small exact categories (cf. [3, A.1]): Let (C, E) be a small exact category. i) There is an abelian category A and a fully faithful exact functor I : C → A that reﬂects exactness. Moreover, when A is equipped with its maximal exact structure, C possesses the three space property with respect to A.

ii) The category A may canonically be chosen to be the category of injective functors Aop → (Ab) and I to be the contravariant Yoneda embedding I(A) = HomA(−,A). 126 CHAPTER 5. DERIVED FUNCTORS

5.2 Complexes

Deﬁnition 5.12. Let C be an additive category. A complex

X X = (Xn, dn )n∈Z with values in C consists of the following data:

• A family (Xn)n∈Z of objects Xn of C. X X • A family (dn )n∈Z of morphisms dn : Xn → Xn+1 in C, so that

X X dn+1 ◦ dn = 0

for all n ∈ Z.

X A morphism from a complex X = (Xn, dn )n∈Z with values in C to a complex Y Y = (Yn, dn )n∈Z with values in C is a family f = (fn)n∈Z of morphisms fn : Xn → Yn, with Y X dn ◦ fn = fn+1 ◦ dn for all n ∈ Z. Remark and Deﬁnition 5.13. Let C be an additive category.

X i) It is useful to visualize a complex X = (Xn, dn )n∈Z with values in C as a sequence

X X X X dn−2 dn−1 dn dn+1 ... / Xn−2 / Xn−1 / Xn / Xn+1 / Xn+2 / ...

of objects and morphisms and a morphism f : X → Y from a complex X Y X = (Xn, dn )n∈Z to a complex Y = (Yn, dn )n∈Z as a commutative diagram

X X X X dn−2 dn−1 dn dn+1 ... / Xn−2 / Xn−1 / Xn / Xn+1 / Xn+2 / ...

fn−2 fn−1 fn fn+1 fn+2 ... Y Y Y Y Y .... / n−2 Y / n−1 Y / n Y / n+1 Y / n+2 / dn−2 dn−1 dn dn+1

X ii) If f = (fn)n∈Z is a morphism between complexes X = (Xn, dn )n∈Z Y and Y = (Yn, dn )n∈Z with values in C and g = (gn)n∈Z is a morphism Y Z between Y = (Yn, dn )n∈Z and a third complex Z = (Zn, dn )n∈Z with values in C, we can deﬁne the composition

g ◦ f := (gn ◦ fn)n∈Z 5.2. COMPLEXES 127

and it is clear that g ◦ f is a morphism of complexes from X to Z and that this composition is associative. In addition we have for every X ∈ Ob(C) an identity morphism of

complexes, deﬁned by idX := (idXn )n∈Z. Therefore we have the category of complexes with values in C, denoted X by C(C), whose objects are the complexes X = (Xn, dn )n∈Z in C and whose morphisms HomC(C)(X,Y ) are the morphisms of complexes from X to Y .

The complex 0 := (0, id0)n∈Z is a zero object of C(C) and together with the component-wise addition

f + g := (fn + gn)n∈Z

HomC(C)(X,Y ) becomes an abelian group for all complexes X and Y . Additionally we have

(f + g) ◦ h = ((fn + gn) ◦ hn)n∈Z

= (fn ◦ hn + gn ◦ hn)n∈Z = f ◦ h + g ◦ h,

l ◦ (f + g) = (ln ◦ (fn + gn))n∈Z

= (ln ◦ fn + ln ◦ gn)n∈Z = l ◦ f + l ◦ g, hence the category C(C) is preadditive.

X Y iii) Let X = (Xn, dn )n∈Z and Y = (Yn, dn )n∈Z be complexes with values in C and let for each n ∈ Z the tuple (Xn q Yn, πXn , π,n ωXn , ωYn ) be the biproduct of Xn and Yn in C. We deﬁne a new complex X q Y by XqY X q Y := (Xn q Yn, dn )n∈Z, X XqY dn 0 with dn := X : Xn q Yn → Xn+1 q Yn+1. 0 dn In addition we deﬁne:

πX := (πXn )n∈Z : X q Y → X

πY := (πYn )n∈Z : Y q Y → X

ωX := (ωXn )n∈Z : X → X q Y

ωY := (ωYn )n∈Z : Y → X q Y XqY It is clear from the deﬁnition of dn that the above are really morphisms of complexes. In addition these morphisms satisfy the equations

πX ◦ ωX = idX , πX ◦ ωY = 0,

πY ◦ ωY = idX , πY ◦ ωX = 0,

idXqY = ωX ◦ πX + ωY ◦ πY ,

hence the tuple (X q Y, πX , πY , ωX , ωY ) is a biproduct of X and Y in C(C). 128 CHAPTER 5. DERIVED FUNCTORS

iv) As we have seen above, the category C(C) is a preadditive category that has biproducts and is therefore an additive category.

v) Let f : X → Y be a morphism in C(C). If the category C has kernels we can deﬁne a complex

ker f ker f := (ker fn, dn )n∈Z,

ker f where dn : ker fn → ker fn+1 is the unique morphism making the diagram

kfn fn ker fn / Xn / Yn dker f dX dY n n n ker fn+1 / Xn+1 / Yn+1 kfn+1 fn+1 commutative. This is in fact a complex, since

ker f ker f X X kfn+2 ◦ dn+1 ◦ dn = dn+1 ◦ dn ◦ kfn = 0

and because kfn+2 is a monomorphism for all n ∈ Z. The above diagram also shows that

kf := (kfn )n∈Z : ker f → X is a morphism of complexes, in fact it is a kernel of f: Let t: T → X be a morphism of complexes with f◦t = 0, i.e. fn◦tn = 0

for all n ∈ Z. Then the universal properties of the kernels kfn each

give rise to a unique morphism λn : Tn → ker fn with tn = kfn ◦ λn for all n ∈ Z. We have

ker f X kfn+1 ◦ dn ◦ λn = dn ◦ kfn ◦ λn X = dn ◦ tn X = tn+1 ◦ dn T = kfn+1 ◦ λn+1 ◦ dn

for all n ∈ Z, hence the diagram

T dn Tn / Tn+1

λn λn+1 ker f ker f n ker f/ n+1 dn

is commutative for all n ∈ Z. This shows that λ := (λn)n∈Z is the unique morphism of complexes with t = kf ◦ λ, hence kf is a kernel of 5.2. COMPLEXES 129

f. Dually, if C has cokernels, we can deﬁne the complex cok f cok f := (cok fn, dn )n∈Z, cok f where dn : cok fn → cok fn+1 is the unique morphism making the diagram

fn cfn Xn / Yn / cok fn dX dY dcok f n n n Xn+1 / Yn+1 / cok fn+1 fn+1 cfn+1 commutative. Then the morphism of complexes

cf := (cfn )n∈Z : Y → cok f is a cokernel of f in the category C(C). Therefore, if C is a preabelian category, then so is C(C).

X vi) A complex X = (Xn, dn )n∈Z with values in C is called bounded above (resp. bounded below, resp. bounded), if there exists an N0 ∈ Z with Xn = 0 for all n ≥ N0 (resp. for all n ≤ N0, resp. for |n| ≥ N0). This gives the following full subcategories of C(C): • C−(C) whose objects are the bounded above complexes of C(C). • C+(C) whose objects are the bounded below complexes of C(C). • Cb(C) whose objects are the bounded complexes of C(C).

vii) Let C0(C) be the full subcategory of C consisting of those complexes X X = (Xn, dn )n∈Z in C(C) with Xn = 0 for n 6= 0. We have the functor X 7→ ι(X) ι: C → C (C) , , 0 f 7→ ι(f)

where ι(X) is the complex with X0 = X and Xn = 0 for n 6= 0 0 and ι(f): ι(X) → ι(X ) is the morphism of complexes with f0 = f and fn = 0 for n 6= 0. This functor is fully faithful and essentially surjective, hence C is equivalent to the full subcategory C0(C) of C(C). We identify C with the subcategory C0(C) via the functor ι. Remark and Deﬁnition 5.14. Let C be a preabelian category and let X X = (Xn, dn )n∈Z be a complex with values in C. For every n ∈ Z there X X X exists a unique morphism σn : im dn−1 → ker dn making the diagram

σX X n X im dn−1 ______/ ker dn (?) GG x GG x GG xx i X GG xxk X dn−1 G# |xx dn Xn 130 CHAPTER 5. DERIVED FUNCTORS commutative. X X In fact, since dn ◦ dn−1 = 0, the universal property of the cokernel gives rise X to a unique morphism µ: cok dn−1 → Xn+1, making the diagram

X X dn−1 dn Xn−1 / Xn / Xn+1 t: c X t dn−1 t t µ t X cok dn−1

X commutative. If i X is the image of d , we have dn−1 n−1 X d ◦ i = µ ◦ c X ◦ i = 0, n dn−1 hence the universal property of the cokernel gives rise to a unique morphism X X X σ : im d → ker d with i X = k X ◦ σn, i.e. making the diagram (?) n n−1 n dn−1 dn commutative. X X For every n ∈ Z ﬁx a cokernel-object cok(σn ) of σn and deﬁne n X H (X) := cok(σn ). It is called the n-th homology of the complex X. Let f : X → Y be a morphism in C(C). By 2.18 there exists a uniquely determined morphism X Y λn : ker dn → ker dn making the diagram

k X dX d X n n (1) ker dn / Xn / Xn+1 λn fn fn+1 Y ker dn / Yn / Yn+1 k Y dY dn n X Y commutative and a uniquely determined morphism µn : cok dn−1 → cok dn−1 making the diagram c dX dX n−1 n−1 X (2) Xn−1 / Xn / cok dn−1 fn−1 fn (+) µn Y Yn−1 / Y / cok d Y n c Y n−1 dn−1 dn−1 commutative. By applying 2.18 again to the square (+) we get a unique X Y morphism νn : im dn−1 → im dn−1 making the diagram i c dX dX X n−1 n−1 X (3) im dn−1 / Xn / cok dn−1 νn fn µn im dY / Y / cok dY n−1 Y n c Y n−1 idn−1 dn−1 5.2. COMPLEXES 131

X commutative. Since, as seen above, we have i X = k X ◦ σ as well as dn−1 dn n Y i Y = k Y ◦ σ . Therefore we have dn−1 dn n

X X k Y ◦ λ ◦ σ = f ◦ k X ◦ σ dn n n n dn n = fn ◦ i X dn−1 = i Y ◦ νn dn−1 Y = k Y ◦ σ ◦ µ dn n n

X Y and since k Y is a monomorphism it follows that λ ◦ σ = σ ◦ ν , i.e. the dn n n n n diagram σX X n X (4) im dn−1 / ker dn

νn λn im dY Y n−1 Y / ker dn σn is commutative. Then, again by 2.18 we have unique morphism

n X Y H (f): cok σn → cok σn making the diagram

X c σ σX X n X n X (5) im dn−1 / ker dn / cok σn n νn λn H (f) im dY / ker dY / cok σY n−1 Y n c Y n σn σn commutative.

Proposition and Deﬁnition 5.15. Let C be a preabelian category. The rule X 7→ Hn(X) Hn : C(C) → C , , f 7→ Hn(f) is an additive functor from C(C) to C. It is called the homology functor of degree n.

X Proof. We use the same notations as in 5.14. Let X = (Xn, dn )n∈Z be a complex with values in C. Since for any morphism αR → S in C the diagram

kα α cα ker α / R / S / cok α

idker α idR idS idcok α ker α / R / S / cok α kα α cα 132 CHAPTER 5. DERIVED FUNCTORS is commutative it follows from 2.18 that in the case of f = idX we have n λn = id X and µn = id X . Following the construction of H (f) in ker dn cok dn−1 this way, it follows from the above, that

n H (idX ) = idHn(X) .

Let h: X → Y and g : Y → Z be morphisms in C(C). If

kf f cf ker f / X / Y / cok f λ α β µ 0 0 0 0 ker f / X 0 / Y c 0 / cok f kf0 f f and 0 kf0 f cf0 ker f 0 / X0 / Y 0 / cok f 0 λ0 α0 β0 µ0 00 00 00 00 ker f / X 00 / Y c 00 / cok f kf00 f f are commutative diagrams in C it follows from 2.18 that λ0 ◦ λ is the unique 0 0 0 morphism with kf 00 ◦ λ ◦ λ = α ◦ α ◦ kf and µ ◦ µ is the unique morphism 0 0 with µ ◦ µ ◦ cf = cf 00 ◦ β ◦ β. g h g h Therefore in the case of f = g◦h it follows that λn = λn◦λn and µn = µn◦µn + (where λn is the unique morphism making the diagram (1) commute in + the case of f = (+) and where µn is the unique morphism making the diagram (2) commute in the case of f = (+) for (+) ∈ {g, h}). Following the construction of Hn(f) in this way, it follows from the above, that

Hn(g ◦ h) = Hn(g) ◦ Hn(h), hence Hn is a functor. Let h, h0 : X → Y be two morphisms in C(C). For commutative diagrams

kα α cα ker α / R / S / cok α

λ u v µ

0 0 0 0 ker α / R 0 / S c 0 / cok α kα0 α α and kα α cα ker α / R / S / cok α

λ0 u0 v0 µ0

0 0 0 0 ker α / R 0 / S c 0 / cok α kα0 α α 5.2. COMPLEXES 133 in C, the diagram

kα α cα ker α / R / S / cok α

λ+λ0 u+u0 v+v0 µ+µ0

0 0 0 0 ker α / R 0 / S c 0 / cok α kα0 α α also commutes. From this it follows that in the case f = h + h0 we have h h0 h h0 + λn = λn + λn and µn = µn + µn (where λn is the unique morphism making + the diagram (1) commute in the case of f = (+) and where µn is the unique morphism making the diagram (2) commute in the case of f = (+) for (+) ∈ {h, h0}). Following the construction of Hn(f) in this way, it follows from the above, that

Hn(h + h0) = Hn(h) + Hn(h0), hence the functor Hn is additive.

Example 5.16. i) Consider the abelian category (Ab) of abelian groups and group ho- X momorphisms. For a complex X = (Xn, dn )n∈N with values in (Ab) X X X the morphism σn : im dn−1 → ker dn is the inclusion

X X −1 dn−1(Xn−1) ,→ (dn ) ({0}), hence the n-th homolgy of the complex X, i.e. the cokernel of this inclusion is the quotient

n X −1 X X X H (X) = (dn ) ({0})/dn−1(Xn−1) = ker dn / im dn−1.

Y Given a second complex Y = (Yn, dn )n∈Z with values in (Ab) and a morphism of complexes f : X → Y , we have

n X Y H (f)(x + im dn−1) = fn(x) + im dn−1. This is a well-deﬁned morphism because the properties of a morphism X Y X Y of complexes give fn(ker dn ) ⊆ ker dn and fn(im dn−1) ⊆ im dn−1 for each n ∈ N. ii) In the category (TVS) of topological vector spaces and continous linear X maps the n-th homology of a complex X = (Xn, dn )n∈Z with values in (TVS) is the space

X X −1 dn−1(Xn−1) ,→ (dn ) ({0}), equipped with the topology given by the quotient topology of the in- X −1 duced topology on (dn ) ({0}). 134 CHAPTER 5. DERIVED FUNCTORS

Y Given another complex Y = (Yn, dn )n∈Z and a morphism of complexes f : X → Y , the morphism Hn(f) is deﬁned in the same way as in i) and it is a continous linear mapping with regard to the topologies given above. The same is true for the category (LCS) of locally convex spaces and continous linear maps. Deﬁnition 5.17. Let (C, E) be an exact category. A complex X ∈ C(C) is called E-exact, if X i) dn is an E-strict morphism for all n ∈ Z, ii) for the canonical factorizations

X X dn−1 dn ... / Xn−1 / Xn / Xn+1 / ... JJ w; CC ; JJ ww C www X JJ ww X X CC w X en−1 $ w mn−1 en ! ww mn Kn−1 Kn

X X of E-strict morphisms one has (mn−1, en ) ∈ E for all n ∈ Z. Proposition 5.18. Let (C, E) be an exact category such that C is pre- X X abelian and let X = (Xn, dn )n∈Z be a complex in C such that dn is an E-homomorphism for all n ∈ Z. The following are equivalent: i) X is E-exact.

n ii) H (X) = 0 for all n ∈ Z. X X X Proof. Let σ : im d → ker d be the unique morphism with i X = n n−1 n dn X X k X ◦ σ . Then σ is an admissible monomorphism by 3.20, hence its is dn+1 n+1 n an isomorphism if and only if its also an epimorphism. Since Hn(X) is the X n cokernel-object of σn this is the case if and only if H (X) = 0. With this the proposition follows from 3.3. Remark 5.19. Because of the above proposition one can say that the functors Hn measure the exactness of a complex. Proposition 5.20 (Long Homology Exact Sequence). Let A be an f g abelian category and let X / Y / Z be a kernel-cokernel pair in C(A). Then there exist morphisms n n+1 δn : H (Z) → H (X) for all n ∈ Z such that the long sequence

δn ... / Hn(X) / Hn(Y ) / Hn(Z) / Hn+1(X) /

δn+1 Hn+1(Y ) / Hn+1(Z) / Hn+2(X) / Hn+2(Y ) / ... is exact in A. 5.2. COMPLEXES 135

X X X Proof. First let σ : im d → ker d be the unique morphism with i X = n n−1 n dn X X X k X ◦ σ . By 2.18 there exists a unique morphism τ : cok d → dn+1 n+1 n n−1 X coim dn making the diagram

i c dX dX X n−1 n−1 X im dn−1 / Xn / cok dn−1 σX τ X n n X X ker dn / Xn / coim dn k X ci X dn dn commutative. Furthermore the rows of this diagram are kernel-cokernel pairs. It follows then from the snake lemma that there is an isomorphism X X n X X γ : ker τn → cok σn = H (X). Since dn ◦ dn−1 = 0 the universal property X of the cokernel gives rise to a unique morphism αn : cok dn−1 → Xn+1 with X αn ◦ c X = d . Then we also have dn−1 n

X X X d ◦ αn ◦ c X = d ◦ d = 0, n+1 dn−1 n+1 n

X hence also d ◦ αn = 0, since c X is an epimorphism. Then the universal n+1 dn−1 X X X property of the kernel kn+1 gives rise to a unique morphism λn : cok dn−1 → X X ker with k X ◦ λ = αn. The diagram n+1 dn+1 n

X X X dn−1 dn dn+1 ... / Xn−1 / Xn / Xn+1 / Xn+2 / ... O c X X dn−1 kn+1 cok dX ___ ker dX n−1 X / n+1 λn

X is also commutative and λ is also unique with this property, since c X n dn−1 is an epimorphism and k X is a monomorphism. Analogously one ﬁnds dn+1 Y Z morphism λn and λn making the above diagram commutative for the complexes Y and Z. X Since A is abelian, the morphism dfn in the canonical factorization is an X X isomorphism for all n ∈ Z. We identify coim dn and im dn via this isomorphism. Then we have

X X X k X ◦ σ ◦ ci X = i X ◦ ci X ◦ ci X = d = k X ◦ λ ◦ c X dn+1 n+1 dn dn dn dn n dn+1 n dn−1

X X and therefore σ ◦ ci X = λ ◦ c X , since k X is a monomorphism. n+1 dn n dn−1 dn+1 Therefore we have

X X c X ◦ σ ◦ ci X = c X ◦ λ ◦ c X = 0. λn n+1 dn λn n dn−1 136 CHAPTER 5. DERIVED FUNCTORS

X X We claim that c X is a cokernel of σ . Let t: ker d → T be a morphism λn n+1 n+1 X with t ◦ σn+1 = 0. Then we also have

X X t ◦ λ ◦ c X = t ◦ σ ◦ ci X = 0, n dn−1 n+1 dn

X hence t ◦ λ = 0 since c X is an epimorphism. The universal property n dn−1 X of the cokernel c X gives rise to a unique morphism β : cok λ → T with λn n X t = α ◦ c X . This establishes the claim that c X is a cokernel of σ . λn λn n+1 X −1 X n X X Deﬁne k := k X ◦ γ , then k : H (X) → cok d is a cokernel of τ n τn n n−1 n X and therefore also of λ , since k X is a monomorphism. This shows that n dn+1 the sequence

c kX λX σX n n X n X n+1 n+1 H (X) / cok dn−1 / ker dn+1 / H is exact. Analagously one ﬁnds this exact sequences for the complexes Y and Z. f g By 2.18 we ﬁnd morphisms νn and νn making the diagram

fn−1 gn−1 0 / Xn−1 / Yn−1 / Zn−1 / 0

X Y Z dn−1 dn−1 dn−1 fn gn 0 / Xn / Yn / Zn / 0

c X c Y c Z dn−1 dn−1 dn−1 cok dX cok dY cok dZ n−1 f / n−1 g / n−1 νn νn

f g commutative and morphisms λn and λn making the diagram

f g X λn Y λn Z ker dn+1 / ker dn+1 / ker dn+1

k X k Y k Z dn+1 dn+1 dn+1 fn+1 gn+1 0 / Xn+1 / Yn+1 / Zn+1 / 0

X Y Z dn+1 dn+1 dn+1 fn+1 gn+2 0 / Xn+2 / Yn+2 / Zn+2 / 0

f g commutative and since the sequence X / Y / Z is a kernel-cokernel pair it follows from 3.38 that the sequences

f g X νn Y νn X cok dn−1 / cok dn−1 / cok dn−1 / 0 Z λn ker dX ker dY ker dZ 0 / n+1 f / n+1 g / n+1 λn λn 5.2. COMPLEXES 137 are exact. We claim that the diagram

Hn(f) Hn(g) Hn(X) / Hn(Y ) / Hn(Z)

X Y Z kn (K1) kn (K2) kn f g X νn Y νn X cok dn−1 / cok dn−1 / cok dn−1 / 0

X Y Z λn (1) λn (2) λn ker dX ker dY ker dZ 0 / n+1 f / n+1 g / n+1 λn λn c X c Y c Z σn+1 (C1) σn+1 (C2) σn+1 Hn+1(X) / Hn+1(Y ) / Hn+1(X) Hn+1(f) Hn+1(g) is commutative. The commutativity of the diagrams (K1) and (K2) follows from the nat- X X n urality of the isomorphisms γX : ker τn → cok σn = H (X), γY and γZ according to the snake lemma (compare 3.35). In addition we have

f X X k Y ◦ λ ◦ λ ◦ c X = fn+1 ◦ k X ◦ λ ◦ c X dn+1 n n dn−1 dn+1 n dn−1 X = fn+1 ◦ dn Y = dn ◦ fn Y = k Y ◦ λ ◦ c Y ◦ fn dn+1 n dn−1 Y f = k Y ◦ λ ◦ ν ◦ c X , dn+1 n n dn−1

X X Y f hence it folloes that k ◦ λ ◦ c X = λ ◦ νn, since k Y is a monomor- n n dn−1 n dn+1 phism and c X is an epimorphism. This shows that the diagram (1) is dn−1 commutative and the commutativity of (2) can be obtained analogously. The diagrams (C1) and (C2) commute because of the deﬁning properties n+1 n+1 f g of H (f), H g, λn and λn (compare 5.14). Then the snake lemma n n 3.34 applied to the above diagram yields a morphism δn : H (Z) → H (X) making the sequence

n n n+ n+1 H (f) H (g) δn H (f) H (g) Hn(X) / Hn(Y ) / Hn(Z) / Hn+1(X) / Hn+1(Y ) / Hn(Z) exact. This sequence is obtained for every n ∈ Z and they ﬁt together to form the desired long exact sequence. 138 CHAPTER 5. DERIVED FUNCTORS

Remark 5.21. If A is an abelian category and

f g 0 / X / Y / Z / 0

αX αY αZ 0 / X0 / Y 0 / Z0 / 0 f 0 g0 is a commutative diagram with exact rows, it follows from the proof of the above proposition 5.20 and the naturality assertion of the snake lemma (see 3.35), that the diagrams

δn Hn(Z) / Hn+1(X)

n n+1 H (αZ ) H (αX ) n 0 n+1 0 H (Z ) 0 / H (X ) δn are commutative for all n ∈ Z. Deﬁnition 5.22. Let C be an additive category and let f : X → Y be a morphism of complexes in C(C). i) The morphism f is called homotopic to zero if there exist morphisms sn : Xn → Yn−1 in C with X Y fn = sn+1 ◦ dn + dn−1 ◦ sn

for all n ∈ Z. We will denote by Ht(X,Y ) the set of all morphisms of complexes from X to Y that are homotopic to zero. ii) Two morphisms of complexes f, g : X → Y are called homotopic, if f − g is homotopic to zero and a complex X is called null-homotopic if idX is homotopic to zero. iii) A morphism of complexes φ: X → Y is called a homotopy equivalence, if there exists a morphism ψ : Y → X such that the morphisms

φ ◦ ψ − idY and ψ ◦ φ − idX are homotopic to zero. Remark 5.23. If C is an additive category we can visualize a morphism of complexes f, g : X → Y that is homotopic to zero as a commutative diagram

X X X X dn−2 dn−1 dn dn+1 ... / Xn−2 / Xn−1 / Xn / Xn+1 / Xn+2 / ... v y y v v y v fn−2 v fn−1 y fn y fn+1 v fn+2 v sn−1 y sn y sn+1 v sn+2 {v |y |y {v ... Y Y Y Y Y ... / n−2 Y / n−1 Y / n Y / n+1 Y / n+2 / dn−2 dn−1 dn dn+1 5.2. COMPLEXES 139 in which for each n the vertical arrow fn is equal to the sum of the triangles Y X dn−1 ◦ sn and sn+1 ◦ dn .

Lemma 5.24. Let C be an additive category.

i) The set of morphisms of complexes Ht(X,Y ) that are homotopic to zero is a subgroup of the abelian group HomC(C)(X,Y ).

ii) If a morphism of complexes f : X → Y is homotopic to zero, then so 0 0 is v ◦ f ◦ u for all u ∈ HomC(C)(X ,X) and all v ∈ HomC(C)(Y,Y ).

X Y Proof. i) Since we have 0 = 0 ◦ dn + dn−1 ◦ 0 for all n ∈ Z, it is clear that the zero morphism 0: X → Y is an element of Ht(X,Y ). X Y X For f, g ∈ Ht(X,Y ) with fn = sn+1 ◦ dn + dn−1 ◦ sn and gn = tn+1 ◦ dn + Y dn−1 ◦ tn for all n ∈ Z we have

X Y fn + gn = (sn+1 + tn+1) ◦ dn + dn−1 ◦ (sn + tn), and X Y −fn = (−sn+1) ◦ dn + dn−1 ◦ (−sn) hence f + g and −f are also elements of Ht(X,Y ), which is therefore a subgroup of HomC(C)(X,Y ). X Y ii) Let f ∈ HomC(C)(X,Y ) with fn = sn+1 ◦ dn + dn−1 ◦ sn for all n ∈ Z, 0 0 u ∈ HomC(C)(X ,X) and v ∈ HomC(C)(Y,Y ). Then we have

X Y vn ◦ fn ◦ un = vn ◦ (sn+1 ◦ dn + dn−1 ◦ sn) ◦ un X Y = vn ◦ sn+1 ◦ dn ◦ un + vn ◦ dn−1 ◦ sn ◦ un X0 Y 0 = (vn ◦ sn+1 ◦ un+1) ◦ dn + dn−1 ◦ (vn−1 ◦ sn ◦ un) for all n ∈ Z, hence v ◦ f ◦ u is homotopic to zero.

Lemma 5.25. Let C be a preabelian category and let Hn : C(C) → C be the homology functor of degree n. Then Hn(f) = 0 for all f ∈ Ht(X,Y ).

Proof. Let f : X → Y be a morphism in C(C) that is homotopic to zero and let n be an integer. We will use the notations of 5.14 and consider the commutative diagram

X c σ σX X n X n X im dn−1 / ker dn / cok σn

n νn λn H (f) im dY / ker dY / cok σY n−1 Y n c Y n σn σn 140 CHAPTER 5. DERIVED FUNCTORS

n therein. Since c X is an epimorphism it is enough to show H (f) ◦ c X = 0 σn σn n in order to show that H (f) = 0. The morphism λn in the above diagram is the unique morphism making the diagram

k X dX d X n n ker dn / Xn / Xn+1

λn fn fn+1 Y ker dn / Yn / Yn+1 k Y dY dn n commutative. Since the morphism f is homotopic to zero we have fk = X Y sk+1 ◦ dk + dk−1 ◦ sk for all k ∈ Z. Then

k Y ◦ λ = f ◦ k X dn n n dn X Y = (s ◦ d + d ◦ s ) ◦ k X n+1 n n−1 n dn Y = d ◦ s ◦ k X . n−1 n dn

Y Y Let d = i Y ◦d] ◦ci Y be the canonical factorization, then the above n−1 dn−1 n−1 dn−1 yields

Y k Y ◦ λ = d ◦ s ◦ k Y dn n n−1 n dn Y = i Y ◦ d] ◦ ci Y ◦ sn ◦ k Y dn−1 n−1 dn−1 dn Y Y = k Y ◦ σ ◦ d] ◦ ci Y ◦ sn ◦ k Y dn n n−1 dn−1 dn

Y Y and therefore λn = σ ◦d] ◦ci Y ◦sn ◦k Y , since k Y is a monomorphism. n n−1 dn−1 dn dn It follows that

n H (f) ◦ c X = c X ◦ λ σn σn n Y Y = c X ◦ σ ◦ d] ◦ ci Y ◦ sn ◦ k Y σn n n−1 dn−1 dn = 0 which was what we wanted.

Deﬁnition 5.26. Let C be an additive category. The homotopy category K(C) is deﬁned by

Ob(K(C)) := Ob(C(C))

HomK(C)(X,Y ) := HomC(C)(X,Y )/ Ht(X,Y ) together with the composition

(g + Ht(Y,Z)) ◦ (f + Ht(X,Y )) := (g ◦ f + Ht(X,Z)) and the identity idX + Ht(X,X) for every complex X. 5.2. COMPLEXES 141

Remark 5.27. Let C be an additive category. i) The composition

(g + Ht(Y,Z)) ◦ (f + Ht(X,Y )) := (g ◦ f + Ht(X,Z))

is well-deﬁned because of 5.24.ii).

ii) The complex 0 is also a zero object of K(C) and for all X,Y ∈ Ob(C) the set HomK(C)(X,Y ) = HomC(C)(X,Y )/ Ht(X,Y ) inherits a structure of an abelian group from HomC(C)(X,Y ), hence the category K(C) is preadditive.

iii) The rule

P (X) = X P : C(C) → K(C) , , P (f) = f + Ht(X,Y )

is an additive functor from C(C) to K(C).

iv) If f, g : X → Y are homotopic morphisms we have P (f) = P (g) in K(C).

v) If φ: X → Y is a homotopy equivalence, the morphism P (φ) is an isomorphism in K(C).

vi) If (X q Y, πX , πY , ωX , ωY ) is the biproduct of two complexes X and Y with values in C, we know by 2.49.ii) that

(P (X q Y ),P (πX ),P (πY ),P (ωX ),P (ωY )) is a biproduct of X = P (X) and Y = P (Y ) in K(C) since P is an additive functor. Therefore K(C) is a preadditive category that has biproducts, hence it is additive.

vii) We have the following full subcategories of K(C):

• K−(C) with Ob(K−(C)) := Ob(C−(C)). • K+(C) with Ob(K+(C)) := Ob(C+(C)). • Kb(C) with Ob(Kb(C)) := Ob(Cb(C)).

• K0(C) with Ob(K0(C)) := Ob(C0(C)). viii) Let X and Y be objects of C0(C) and let f : X → Y be a morphism of X Y complexes, that is homotopic to zero, i.e. fn = sn+1 ◦ dn + dn−1 ◦ sn X Y for all n ∈ Z. Since dn = 0 and dn = 0 for all n ∈ Z it follows that f is the zero morphism from X to Y . This shows that Ht(X,Y ) = 0 and therefore the functor

P ◦ ι: C → K0(C) 142 CHAPTER 5. DERIVED FUNCTORS

is fully faithful and essentially surjective, hence an equivalence of categories. We identify C with K0(C) via this functor and can therefore regard C as a full subcategory of K(C).

Remark 5.28. Let C be a preabelian category. By 5.25 we know that the functor Hn vanishes on those morphisms f : X → Y that are homotopic to zero. Therefore the functor Hn induces an additive functor from K(C) to C, which we will also denote by Hn, in the following way:

X 7→ Hn(X) Hn : K(C) → C , f + Ht(X,Y ) 7→ Hn(f)

5.3 Injective and Projective Resolutions

Deﬁnition 5.29. Let (C, E) be an exact category.

i) An object P ∈ Ob(C) is called E-projective, if for every admissible epimorphism g : Y → Z and every morphism α: P → Z there exists a morphism β : P → Y making the diagram

P β ~ ~ α ~ ~ Y g / Z

commutative.

ii) An object I ∈ Ob(C) is called E-injective, if for every admissible monomorphism f : X → Y and every morphism α: P → C there exists a morphism β : Y → I making the diagram

f X / Y } α } } β ~} I

commutative.

Remark and Deﬁnition 5.30.

i) The notions of E-projective and E-injective objects are dual to each other, i.e. an object P is E-projective in (C, E) if and only if P is E-injective in (Cop, Eop) and vice versa.

ii) We say that an exact category (C, E) has enough E-projective objects, if for every X ∈ Ob(C) there exists an E-projejective object P and an admissible epimorphism ε: P → X. 5.3. RESOLUTIONS 143

Dually we say that (C, E) has enough E-injective objects, if for every X ∈ Ob(C) there exists an E-injective object I and an admissible monomorphism δ : X → I. Obviously the category C has enough E-projective objects if and only if the dual category Cop has enough Eop-injective objects and vice versa. iii) If C is quasiabelian category and therefore the class of all kernel- cokernel pairs is an exact structure, we will always consider C as an additive category with this largest exact structure and will omit the “E” in the above deﬁnitons. Instead we will only speak of “projective” and “injective” instead of “E-projective” and “E-injective”. Example 5.31. i) If (C, E) is an exact category then the zero object is both an E- projective and an E-injective object of C.

ii) Let F be a field and let (F−V ec) be the category of vector spaces over F and F-linear mappings. Every vector space V is both a projective and an injective object of (F − V ec): Let g : Y → Z be a surjective linear mapping, i.e. a strict epimorphism, between vector spaces Y and Z and let α: V → Z be a linear mapping. Chose a basis (bi)i∈I of V and let (α(bi))i∈I be the image of this basis under α. Since g is surjective, we find a family (yi)i∈I in Y with g)yi) = α(bi) for all i ∈ I. Define a linear mapping β : V → Y by β(bi) := yi, then the diagram V β ~ ~ α ~ ~ Y g / Z is commutative, which shows that V is a projective object. If f : X → Y is an injective linear mapping, i.e. a strict monomorphism, and α: X → V is a linear mapping, choose a basis (xi)i∈I of X. Since f is injective, (f(xi))i∈I is a basis of f(X) and we can, with the help of Zorn’s Lemma, find a basis (bi)i∈J of Y with I ⊆ J and bi = f(xi) for all i ∈ I. Define a linear mapping β : Y → V by α(x ) for i ∈ I β(b ) = i , i 0 otherwise then the diagram f X / Y } α } } β ~} V is commutative, which shows that V is an injective object of (F−V ec). 144 CHAPTER 5. DERIVED FUNCTORS

iii) In the category (LCS) of locally convex spaces the only projective objects are direct sums of the ground ﬁeld K, as was proved by V.A. Geiler in [7]. For every set M it follows from the Hahn-Banach theorem, that the Banach space

∞ M lM := {(xi)i∈M ∈ K : k(xi)i∈M k∞ := supi∈M |xi| < ∞} is an injective object of (LCS) and with this one can show that (LCS) has enough injective objects (cf. [31, 2.2.1]). The same is true for the category (LCS)HD. Proposition 5.32. Let (C, E) be an exact category and let P be an object of C. The following are equivalent: i) P is an E-projective object.

ii) For every admissible epimorphism g : Y → Z the induced morphism

HomC(P, g): HomC(P,Y ) → HomC(P,Z)

of abelian groups is surjective.

iii) The functor HomC(P, −) is exact. iv) Every admissible epimorphism q : X → P is a retraction. Proof. i) ⇒ ii) Let g : X → Y be an admissible epimorphism and let h be an element of HomC(P,Z). Since P is E-projective, there exists a morphism β : P → X making the diagram

P β ~ ~ h ~ ~ Y g / Z commutative. Then we have

HomC(P, g)(β) = g ◦ β = h, hence the mapping HomC(P, g) is surjective.

f g ii) ⇒ iii) Let 0 / X / Y / Z / 0 be an exact sequence in (C, E). By applying the functor HomC(P, −) we get the sequence

HomC(P,f) HomC(P,g) 0 / HomC(P,X) / HomC(P,Y ) / HomC(P,Z) / 0 of abelian groups and group morphisms. This sequence is exact at HomC(P,X) and HomC(P,Y ) by 5.10. Since the 5.3. RESOLUTIONS 145 morphism HomC(P, g) is surjective by hypothesis, it is also exact at the right end and is therefore an exact sequence, therefore HomC(P, −) is an exact functor. iii) ⇒ iv) Let q : X → P be an admissible epimorphism. The mapping

HomC(P, q): HomC(P,X) → HomC(P,P ) is surjective by assumption, hence there exists a morphism p: P → X with

q ◦ p = HomC(P, q)(p) = idP , which shows that q is a retraction. iv) ⇒ i) Let g : Y → Z be an admissible epimorphism and let α: P → Z be a morphism in C. We can form the pullback square

qP Q / P

qY PB α Y g / Z and, since qP is an admissible epimorphism, there exists by assumption a morphism h: P → Q with qP ◦ h = idP . Then we have

g ◦ qY ◦ h = α ◦ qP ◦ h = α, which shows that the object P is E-projective.

By duality we obtain the following:

Proposition 5.33. Let (C, E) be an exact category and let I be an object of C. The following are equivalent:

i) I is an E-injective object.

ii) For every admissible monomorphism f : X → Y the induced morphism

HomC(f, I): HomC(Y,I) → HomC(X,I)

of abelian groups is injective.

iii) The functor HomC(−,I) is exact. iv) Every admissible monomorphism i: I → X is a coretraction.

Proposition 5.34. Let (C, E) be an exact category.

i) If P is an E-projective object of C and r : P → X is a retraction, then the object X is also E-projective. 146 CHAPTER 5. DERIVED FUNCTORS

ii) If I is an E-injective object of C and c: X → I is a coretraction, then the object X is also E-injective.

Proof. Since ii) is the dual statement of i), it suﬃces to show i). Let g : Y → Z be an admissible epimorphism and α: X → Z a morphism in C. Since P is E-projective there exists a morphism β1 : P → Y making the diagram (in which we ﬁrst ignore the dotted arrow)

P β r 1 X β2 v v α {v Y g / Z commutative. Let h: X → P be a morphism with r ◦ h = idP and deﬁne β2 := β1 ◦ h. Then we have

g ◦ β2 = g ◦ β1 ◦ h = α ◦ r ◦ h = α, hence is E-projective.

Proposition 5.35. Let (C, E) be an exact category, P1 and P2 be objects of C and let (P1 q P2, πP1 , πP2 , ωP1 , ωP2 ) be the biproduct of P1 and P2. The following are equivalent:

i) P1 and P2 are E-projective.

ii) P1 q P2 is E-projective.

Proof. i) ⇒ ii) Let g : Y → Z be an admissible epimorphism and let α: P1 q P2 → Z be a morphism. Since P1 and P2 are E-projective, there exists a morphism β1 : P1 → Y with g ◦β1 = α◦P1 and a morphism β2 : P2 → Y with g ◦ β2 = α◦P2 . Then the diagram

P1 q P2 v ( β1 β2 ) v v α v {v Y g / Z is commutative, which shows that P1 q P2 is E-projective. ii) ⇒ i) follows from 5.34, since πP1 and πP2 are retractions. The dual statement of the above is the following:

Proposition 5.36. Let (C, E) be an exact category, I1 and I2 be objects of C and let (I1 q I2, πI1 , πI2 , ωI1 , ωI2 ) be the biproduct of I1 and I2. The following are equivalent: 5.3. RESOLUTIONS 147

i) I1 and I2 are E-projective.

ii) I1 q I2 is E-projective. Deﬁnition 5.37. Let (C, E) be an exact category and X ∈ Ob(C). P i) A left-resolution of X is a pair (P, ε), where P = (Pn, dn )n∈Z is an object of C(C) with Pn = 0 for all n > 0 and where ε: P0 → X is an admissible epimorphism in C, such that the augmented complex

P P P d−3 d−2 d−1 ε ... / P−3 / P−2 / P−1 / P0 / X / 0 is E-exact. A left-resolution (P, ε) of X is called an E-projective resolution, if all the Pn are E-projective objects of C. I ii) A right-resolution of X is a pair (I, δ), where I = (In, dn)n∈Z is an object of C(C) with In = 0 for all n < 0 and where δ : X → I0 is an admissible monomorphism in C, such that the augmented complex

I I I δ d0 d1 d2 0 / X / I0 / I1 / I2 / I3 / ... is E-exact. A right-resolution (I, δ) of X is called an E-injective resolution, if all the In are E-injective objects of C. Proposition 5.38. Let (C, E) be an exact category. i) If C has enough E-projective objects, every X ∈ Ob(C) possesses an E-projective resolution. ii) If C has enough E-injective objects, every X ∈ Ob(C) possesses an E-injective resolution. Proof. It suﬃces to show i), since ii) is the dual statement of i). Let X be an object of C. Since C has enough E-projective objects, there exists an E-projective object P0 and an admissible epimorphism ε: P0 → X. Let f0 : K0 → P0 be a morphism with (f0, ε) ∈ E, then we ﬁnd again an E- projective object P1 and an admissible epimorphism p−1 : P1 → K0. Then the morphism

P d−1 := f0 ◦ p−1 : P−1 → P0 p is an E-strict morphism. Inductively construct morphisms dn : Pn → Pn+1 for n, 1:

P P P d−3 d−2 d−1 ε ... / P−3 / P−2 / P−1 / P0 / X / 0 G ; G ; D > GG ww GG ww DD || p G w p G w p D | −3 G# ww f−2 −2 G# ww f−1 −1 D" || f0 K−2 K−1 K0 148 CHAPTER 5. DERIVED FUNCTORS

Since (fn, pn) ∈ E for all n < −1 the sequence thus deﬁned is exact. Deﬁne Pn := 0 for n > 0 and P P := (Pn, dn )n∈Z, then P is a complex and (P, ε) is an E-projective resolution of X.

Remark 5.39.

i) The deﬁning property of E-projective and E-injective objects has not been used in the proof of the preceding proposition. Therefore we have really shown the following for an exact category (C, E): If for every object X in C there exists an object Q possessing a property (?) and an admissible epimorphism η : Q → X, then every object X in C possesses a left-resolution (P, ε) such that all the Pn have the property (?). Dually: If for every object X in C there exists an object J possessing a property (?) and an admissible monomorphism γ : X → J, then every object X in C possesses a left-resolution (I, δ) such that all the In have the property (?).

ii) Let (C, E) be an exact category and let X be an object of C. The E- injective and E-projective resolutions of X are generally not uniquely determined. For example if V 6= 0 is a vector space, then the sequences

idV 0 / V / V / 0

and ωV πV 0 / V / V q V / V / 0 are both exact and both yield a projective (and an injective) resolution of V , but are not isomorphic as objects of C(C).

Theorem 5.40 (Comparison Lemma). Let (C, E) be an exact category and let X be an object of C.

i) If (P, ε) is an E-projective resolution of X and f : X → Y is a morphism in C, then for every left-resolution (Q, η) of Y there exists a morphism of complexes φ: P → Q with η ◦ φ0 = f ◦ ε, i.e. making the diagram

P P P d−3 d−2 d−1 ε ... / P−3 / P−2 / P−1 / P0 / X / 0

φ−3 φ−2 φ−1 φ0 f ... Q Q Q Q / −3 Q / −2 Q / −1 Q / 0 η / Y / 0 d−3 d−2 d−1 5.3. RESOLUTIONS 149

commutative. Moreover the morphism φ is unique up to homotopy, i.e. if ψ : P → Q is another morphism of complexes with f ◦ ε = η ◦ ψ0 then φ and ψ are homotopic. ii) If (I, δ) is an E-injective resolution of X and f : W → X is a morphism in C, then for every right-resolution (J, γ) of W there exists a morphism of complexes ψ : J → I with ψ ◦ γ = δ ◦ f, i.e. making the diagram

J J J γ d0 d1 d2 0 / W / J0 / J1 / J2 / J3 / ...

f ψ0 ψ1 ψ2 ψ3 0 / X / I0 / I1 / I2 / I3 / ... δ I I I d0 d1 d2 commutative. Moreover the morphism φ is unique up to homotopy. Proof. It suffices to show i), since ii) is the dual statement. Define complexes 0 0 P 0 0 0 Q0 P = (Pn, dn )n∈Z and Q = (Qn, dn )n∈Z by 0 P 0 P1 := X and d0 := ε 0 P 0 P Pn := Pn and dn := dn otherwise and ( 0 Q0 Q1 := Y and d0 := η 0 . 0 Q Q Qn := Qn and dn := dn otherwise We begin with the existence of the morphism φ. For this we construct 0 0 0 0 inductively a morphism of complexes φ : P → Q with φ1 = f, then the 0 morphism φ: P → Q defined by φn := φn for n ≤ 0 and φ1 = 0 has the desired properties. 0 0 Define then φ := 0 for n > 1 and φ1 = f. For n ≤ 0 suppose inductively, 0 0 0 that there are morphisms φi : Pi → Qi with 0 P 0 Q0 0 φi+1 ◦ di = di ◦ φi 0 0 0 0 P 0 0 for i ≥ n. We construct φn−1 : Pn−1 → Qn−1 with φn ◦ d Q0 ◦ φn−1. n−1=dn −1 Consider the following diagram (in which we first ignore the dotted arrow)

0 dP P 0 0 n−1 0 dn 0 Pn−1 / Pn / Pn+1 G < B < GG yy BB xx GG yy B x 0 G y 0 0 BB xx 0 eP GG y mP eP B x mP n−1 G# yy n−1 n B xx n Kn−1 Kn 0 0 λn−1 φ φn+1 n Ln−1 L 0 0 n eQ w; EE mQ Q0 |> FF Q0 n−1 ww E n−1 en | F mn ww EE || FF ww EE || FF ww E | F" 0 " 0 | 0 Qn−1 / Q / Qn+1, Q0 n Q0 dn−1 dn 150 CHAPTER 5. DERIVED FUNCTORS

P 0 P 0 P 0 P 0 P 0 P 0 Q0 Q0 Q0 where dn−1 = mn−1 ◦ en−1, dn = mn ◦ en , dn−1 = mn−1 ◦ en−1 and Q0 Q0 Q0 dn = mn ◦ en are the canonical factorizations of these E-strict morphisms P 0 P 0 Q0 Q0 and where (mn−1, en ), (mn−1, en ) ∈ E, since both complexes are exact. 0 P 0 0 By induction hypothesis we have φn+1 ◦ dn ◦ φn, hence there exists a unique Q0 P 0 morphism λn−1 : Kn−1 → Ln−1 with mn−1 ◦λn−1 = φn ◦mn−1 by 2.18. Since 0 e 0 is an admissible epimorphism and P is E-projective, there exists a Qn−1 n−1 0 0 0 P 0 Q0 0 morphism φn−1 : Pn−1 → Qn−1 with λn−1 ◦ en−1 = en−1 ◦ φ n − 1. Then we have

Q0 0 Q0 Q0 0 dn−1 ◦ φn−1 = mn−1 ◦ en−1 ◦ φn−1 Q0 P 0 = mn−1 ◦ λn−1 ◦ en−1 0 P 0 P 0 0 P 0 = φn ◦ mn−1 ◦ en−1 = φn ◦ dn−1, hence the diagram

0 dP P 0 0 n−1 0 dn 0 Pn−1 / Pn / Pn−1 0 0 0 φn−1 φ φn+1 n 0 0 0 Qn−1 / Q / Qn−1 Q0 n Q0 dn−1 dn commutes. This proofs the existence of the morphism φ0 : X0 → Y 0 and therefore also that of φ: X → Y . Next we will show the uniqueness up to homotopy. Let ψ : P → Q be another morphism of complexes with f ◦ ε = η ◦ ψ0 and deﬁne h := φ − ψ. We show that h is homotopic to zero by inductively constructing morphisms P Q sn : Pn → Qn−1 with hn = sn ◦ dn + dn−1 ◦ sn for all n ∈ Z. For n > 0 deﬁne sn = 0. For n = 0 we have

η ◦ h0 = η ◦ (φ0 − ψ0) = f ◦ ε − f ◦ ε = 0. Q Q Q Hence, if d−1 = m−1 ◦ e−1 is the canonical factorization of the E-strict Q Q morphism d−1, then the universal property of mn−1, which is the kernel 0 of η, gives rise to a unique morphism h0 : P0 → L−1 making the following diagram (in which we ﬁrst ignore the dotted arrow)

e a ^ \ P m 0 z zz s0 zz Ù zzh0 }zz 0 L−1 h0 eQ < C mQ −1 yy CC −1 yy CC yy CC yy C! Q Q −1 Q / 0 d−1 5.3. RESOLUTIONS 151

Q commutative. Since P0 is E-projective and e−1 is an admissible epimorphism, 0 Q there exists a morphism s0 : P0 → Q−1 with h0 = e−1 ◦s0, therefore we have

Q 0 Q Q Q Q P h0 = m−1 ◦ h0 = m−1 ◦ e−1 ◦ s0 = d−1 ◦ s0 = d−1 ◦ s0 + s1 ◦ d0 ,

P since s1 ◦ d0 = 0. For n < 0 we assume that morphisms si : Pi → Qi with

P P hi = si+1 ◦ di + di−1 ◦ si have already been constructed for i ≥ n. Then

Q P Q Q P dn−1 ◦ (hn−1 − sn ◦ dn−1) = dn−1 ◦ hn−1 − dn−1 ◦ sn ◦ dn−1 P P P = hn ◦ dn−1 − (hn − sn+1 ◦ dn ) ◦ dn−1 P P = hn ◦ dn−1 − hn ◦ dn−1 = 0

Q Q Q Hence, if dn−2 = mn−2 ◦ en−2 is the canonical factorization of the E-strict Q Q Q morphism dn−2, the universal property of mn−2, which is the kernel of dn−2, 0 gives rise to a unique morphism hn−1 : Pn−1 → Ln−2 with

P Q 0 hn−1 − sn ◦ dn−1 = mn−2 ◦ hn−1, i.e. making the diagram (in which we ﬁrst ignore the dotted arrows)

dP d ` ^ n−1 k g Pn−1 / Pn v vv sn−1 vv Õ vv 0 vv hn−1 {v sn L hn−1 hn n−2 eQ v; HH mQ n−2 vv HH n−2 vv HH vv HH vv H# Q Q Q n−2 Q / n−1 Q / n dn−2 dn−1

Q commutative. Since Pn−1 is E-projective and en−2 is an admissible epimor- 0 Q phism, there exists a morphism sn−1 : Pn−1 → Qn−2 with hn−1 = en−2◦sn−1. Then we have

P Q 0 hn−1 − sn ◦ dn−1 = mn−2 ◦ hn−1 Q Q = mn−2 ◦ en−2 ◦ sn−1 Q = dn−2 ◦ sn−1

P Q hence hn−1 = sn ◦ dn−1 + dn−2 ◦ sn−1, as desired. Corollary 5.41. Let (C, E) be an exact category and let X be an object of C. 152 CHAPTER 5. DERIVED FUNCTORS

i) If (P, ε) and (P 0, ε0) are two E-projective resolutions of X, then there exists a homotopy equivalence φ: P → P 0. ii) If (I, δ) and (I0, δ0) are two E-injective resolutions of X, then there exists a homotopy equivalence ψ : I → I0. Proof. Since ii) is the dual statement of i) it suﬃces to show i). By 5.40 there exist morphisms of complexes φ: P → P 0 and ψ : P 0 → P making the diagram

P P P d−3 d−2 d−1 ε ... / P−3 / P−2 / P−1 / P0 / X / 0 O O O O ψ−3 φ−3 ψ−2 φ−2 ψ−1 φ−1 ψ0 φ0 ... 0 0 0 0 / P−3 / P−2 / P−1 / P0 / X / 0 P 0 P 0 P 0 ε0 d−3 d−2 d−1 commutative. Then the diagrams

P P P d−3 d−2 d−1 ε ... / P−3 / P−2 / P−1 / P0 / X / 0

ψ−3◦φ−3 ψ−2◦φ−2 ψ−1◦φ−1 ψ0◦φ0 ... P P P P / −3 P / −2 P / −1 P / 0 ε / X / 0 d−3 d−2 d−1 and 0 0 P 0 dP dP ... 0 d −3 0 −2 0 −1 0 ε0 / P−3 / P−2 / P−1 / P0 / X / 0

φ−3◦ψ−3 φ−2◦ψ−2 φ−1◦ψ−1 φ0◦ψ0 ... 0 0 0 0 / P−3 / P−2 / P−1 / P0 / X / 0 P 0 P 0 P 0 ε0 d−3 d−2 d−1 are also commutative, hence the uniqueness part of 5.40 states that ψ ◦ φ − idP and φ ◦ ψ − idQ are homotopic to zero, therefore the morphism φ is a homotopy equivalence. Proposition 5.42 (Horseshoe Lemma). Let (C, E) be an exact category. Suppose we are given a horseshoe diagram

dX dX dX γX X 0 X 1 X 2 X ... X / I0 / I1 / I2 // I3 / f Y

γZ Z Z Z Z ... Z / I0 Z // I1 Z / I2 Z // I3 / d0 d1 d2 5.3. RESOLUTIONS 153 that is to say, the column is an element of E and the horizontal rows are Y X Z E-injective resolutions of X and Z. Then the biproducts In := In q In assemble to an E-injective resolution of Y in such a way that the horseshoe can be embedded into a commutative diagram with exact rows and columns

dX dX dX γX X 0 X 1 X 2 X ... X / I0 / I1 / I2 / I3 /

f φ0 φ1 φ2 φ3 dY dY dY γY Y 0 Y 1 Y 2 Y ... Y / I0 / I1 / I2 // I3 /

g ψ0 ψ1 ψ2 ψ3 γZ Z Z Z Z ... Z / I0 Z // I1 Z // I2 Z / I3 / d0 d1 d2

X Proof. Since f is an admissible monomorphism and I0 is E-injective, there X exists a morphism ι: Y → I0 with γX = ι ◦ f. Deﬁne ι X Z γ := : Y → I0 q I0 = I0, γZ ◦ g then the diagram γX X X / I0

ω X f I0 γ Y Y / I0

g π Z I0 γZ Z Z / I0 is commutative. Since (ω X , π Z ) ∈ E, it follows from the short ﬁve lemma I0 I0 Y Y Y 3.23, that γ is an admissible monomorphism. Let then e0 : I0 → K0 be I X X X Z Z Z a morphism with (γ, e0) ∈ E and let d0 = m0 ◦ e0 and d0 = m0 ◦ e0 be the canonical factorizations of these E-strict morphisms. Then we have the commutative diagram

eX γX X 0 X X / I0 / K0 ω X f I f0 0 eY γ Y 0 Y Y / I0 / K0 g π Z g I0 0 0 eI γZ Z 0 Z Z / I0 / K0 154 CHAPTER 5. DERIVED FUNCTORS where f0 and g0 are the morphisms between the cokernels we get by 2.18. It follows from the 3 × 3-lemma 3.26 that (f0, g0)is an element of E. Ap- plying the above construction to (f0, g0) instead of (f, g) we can proceed by induction and get the desired diagram:

X X d0 d1

eX mX " eX mX " γX X 0 X 0 X 1 X 1 X ... X / I0 / K0 / I1 / K1 / I2 / ω X ω X ω X f I f0 I f1 I 0 1 2 eY mY eY mY γ Y 0 Y 0 Y 1 Y 1 Y ... Y / I0 / K0 / I1 / K1 / I2 / π π π g IZ g0 IZ g1 IZ 0 1 2 eZ mZ eZ mZ γZ Z 0 Z 0 Z 1 Z 1 Z ... Z / I0 / K0 / I1 / K1 / I2 / < <

Z Z d0 d1

By duality we obtain: Proposition 5.43. Let (C, E) be an exact category. Suppose we are given a horseshoe diagram

dX dX dX ... X 3 X 2 X 1 X ηX / P3 / P2 / P1 / P0 / X f Y

g ... / P Z / P Z / P Z / P Z / Z, 3 Z 2 Z 1 Z 0 ηZ d3 d2 d1 that is to say, the column is an element of E and the horizontal rows are Y X Z E-projective resolutions of X and Z. Then the biproducts Pn := Pn q Pn assemble to an E-projective resolution of Y in such a way that the horseshoe can be embedded into a commutative diagram with exact rows and columns

dX dX dX ... X 3 X 2 X 1 X ηX / P3 / P2 / P1 / P0 / X

φ3 φ2 φ1 φ0 f dY dY dY ... Y 3 Y 2 Y 1 Y ηY / P3 / P2 / P1 / P0 / Y

ψ3 ψ2 ψ1 ψ0 g ... / P Z / P Z / P Z / P Z / Z. 3 Z 2 Z 1 Z 0 ηZ d3 d2 d1 5.3. RESOLUTIONS 155

Corollary 5.44 (Shelf Lemma). Let (C, E) be an exact category. Given a commutative diagram

f g X / Y / Z

αX αY αZ X0 / Y 0 / Z0 f 0 g0

with (f, g), (f 0, g0) ∈ E and injective resolutions IX ,IX0 ,IZ ,IZ0 of X,X0,Z,Z0, Y X Z Y 0 X0 Z0 the biproducts In := In q In and In := In q In assemble to E-injective resolutions of Y and Y 0 in such a way that they can be embedded into a commutative diagram

X0 X0 X0 γX0 0 d0 0 d1 0 d2 0 X0 / IX / IX / IX / IX / ... @ 0 1 2 3 αX ÐÐ }> }> }> }> ÐÐ }} }} }} }} Ð }d}X }d}X }d}X }} ÐγX X 0 X 1 X 2 X ... X / I0 / I1 / I2 / I3 / 0 0 0 0 0 f f φ0 φ0 φ1 φ1 φ2 φ2 φ3 φ3 Y 0 Y 0 Y 0 γY 0 0 d0 0 d1 0 d2 0 Y 0 / IY / IY / IY / IY / ... @ 0 1 2 3 αY ÐÐ }> }> }> ÐÐ }} }} }} Ð }d}Y }d}Y }d}Y ÐγY Y 0 Y 1 Y 2 Y ... Y / I0 / I1 / I2 / I3 / 0 0 0 0 0 g g ψ0 ψ0 ψ1 ψ1 ψ2 ψ2 ψ3 ψ3 γZ0 0 0 0 0 0 Z Z Z Z ... Z / I0 0 / I1 0 / I2 0 / I3 / @ dZ dZ dZ αZ Ð }> 0 }> 1 }> 2 }> ÐÐ }} }} }} }} ÐÐ }} }} }} }} ÐγZ Z Z Z Z ... Z / I0 Z / I1 Z / I2 Z / I3 / d0 d1 d2

with E-exact rows and columns.

Proof. The front- and backside of the above diagram can be constructed according to the horseshoe lemma 5.42, we just have to show that one can ﬁll up the diagonal arrows. For this we keep the notations of the proof of the horseshoe lemma 5.42 and consider the following diagram (in which we 156 CHAPTER 5. DERIVED FUNCTORS

ﬁrst ignore the dotted arrows)

X0 γX0 0 e0 0 X0 / IX / KX 0 0 X 0 αX @ αX > k0 < ÐÐ } ω X0 z Ð } I0 z ÐÐ X ÐγX }e0 z X X f 0 X / I0 / K0 0 0 (1) ω X f f I0 (3) f0 Y 0 γY 0 0 e0 0 Y 0 / IY / KY 0 0 Y 0 αY @ αY > k0 < ÐÐ } π Z0 z Ð } I0 z ÐÐ Y ÐγY }e0 z Y Y g0 Y / I0 / K0 0 g 0 (2) π Z g g I0 (4) 0 γ 0 0 Z Z0 Z0 Z / I0 0 / K0 @ α0 eZ kZ αZ Ð Z }> 0 0 z< ÐÐ } z ÐÐ } z ÐγZ Z Z Z / I0 Z / K0 d0 whose rows and columns are E-exact. Since γX is an admissible monomor- X0 0 X phism and I0 is an E-injective object there exists a morphism αX : I0 → X0 0 0 Z I0 with αX ◦ γX = γX0 ◦ αX . Analogously one gets a morphism αZ : I0 → Z0 0 I0 with αZ ◦ γZ = γZ0 ◦ αZ . We have ι ι0 0 γY = and γY = 0 γZ ◦ g γZ0 ◦ g X 0 0 X0 0 0 0 for morphisms ι: Y → I0 and ι : Y → I0 with γX = ι◦f and γX = ι ◦f . Since

0 0 0 0 (ι ◦ αY − αX ◦ ι) ◦ f = ι ◦ αY ◦ f − αX ◦ ι ◦ f 0 0 0 = ι ◦ f ◦ αX − αX ◦ γX = γX0 ◦ αX − γX0 ◦ αX = 0 the universal property of the cokernel g gives rise to a unique morphism X0 0 0 α: Z → I0 with α ◦ g − ι ◦ αY − αX ◦ ι. Then, since γZ is an admissible X0 monomorphism and I0 is an E-injective object there exists a morphism Z X0 β : I0 → I0 with α = β ◦ γZ and therefore also 0 0 β ◦ γZ ◦ g = α ◦ g = ι ◦ αY − αX ◦ ι. Deﬁne a morphism 0 0 αX β X Z Y X0 Z0 Y 0 αY := 0 : I0 q I0 = I0 → I0 q I0 = I0 , 0 αZ then we have 0 0 0 αX β 1 αX 0 α ◦ ω X = = = ω X0 ◦ α , X I0 0 I X 0 αZ 0 0 0 5.4. DERIVED FUNCTORS 157

0 0 αX β 0 0 πIZ0 ◦ αY = ( 0 1 ) 0 = ( 0 αZ ) = αZ ◦ πIZ0 0 0 αZ 0 and 0 0 αX β ι αY ◦ γY = 0 0 αZ γZ ◦ g 0 αX ◦ ι + β ◦ γZ ◦ g = 0 αZ ◦ γZ ◦ g α0 ◦ ι + ι0 ◦ α − α0 ◦ ι = X Y X γZ0 ◦ αZ ◦ g 0 ι ◦ αY 0 = 0 = γY ◦ αY . γZ0 ◦ g ◦ αY This shows that the cubes (1) and (2) are commutative. By 2.18 we get X X X0 Y Y Y 0 Z Z Z0 unique morphisms k0 : K0 → K0 , k0 : K0 → K0 and k0 : K0 → K0 X X X0 0 Y Y Y 0 0 Z Z Z0 0 with k0 ◦ e0 = e0 ◦ αX , k0 ◦ e0 = e0 ◦ αY and k0 ◦ e0 = e0 ◦ αZ . Then we have

Y X Y Y k ◦ f0 ◦ e = k ◦ e ◦ ω X 0 0 0 0 I0 Y 0 0 = e ◦ α ◦ ω X 0 Y I0 Y 0 0 = e0 ◦ ω X0 ◦ αX I0 0 X0 0 = f0 ◦ e0 ◦ αX 0 X X = f0 ◦ k0 ◦ e0

Y 0 X X and therefore k0 ◦ f0 = f0 ◦ k0 , since e0 is an epimorphism. Analogously Z 0 Y one gets k0 ◦ g0 = g0 ◦ k0 , which shows that the cubes (3) and (4) in the above diagram are commutative. Applying the above construction to the commutative diagram

X f0 Y g0 Z K0 / K0 / K0

X Y Z k0 k0 k0 X0 Y 0 Z0 K0 0 / K0 0 / K0 f0 g0 which exact rows, one can proceed inductively to get the desired commutative diagram.

5.4 Right- and Left-derived Functors

Remark and Deﬁnition 5.45. Let (C, E) be an exact category such that C has enough E-injective objects. For each object X in C choose a ﬁxed 158 CHAPTER 5. DERIVED FUNCTORS

X E-injective resolution (I , γX ). If f : X → Y is a morphism in C, there exists a morphism of complexes f I : IX → IY that lifts f and is unique up to homotopy, by 5.40. Deﬁne I(f) := f I + Ht(IX ,IY ), then the rule

X 7→ IX I : C → K(C) , f 7→ I(f) is a well-defined covariant functor (observe that a morphism g ◦ f is lifted I I by g ◦ f and idX is lifted by idIX ). Furthermore, since g + f is lifted by gI + f I , the functor I is additive. If we choose for each object X in C a different (fixed) E-injective resolution X 0 (J , γX ), we get another covariant additive functor J : C → K(C) with J(X) = J X . We claim that there is a natural isomorphism σ : I → J: X X 0 Indeed, if (I , γX ) and (J , γX ) are both E-injective resolutions of X then there exists a homotopy equivalence τ X : IX → J X , by 5.41. Define σX := τ X + Ht(IX ,J X ), then σX is an isomorphism in K(C). If f : X → Y is a morphism in C, the diagrams

γX X γ X X / I0 X / I0

J X Y I f f0 ◦τ0 f τ0 ◦f0 Y Y Y 0 / J0 Y 0 / J0 δY δY are commutative, hence the morphisms of complexes f J ◦ τ X and τ Y ◦ f I both lift the morphism f, and are therefore homotopic by 5.40. It follows that J(f) ◦ σX = σY ◦ I(f) in K(C), i.e. the diagram

I(f) I(X) / I(Y )

σX σY J(X) / J(Y ) J(f)

X is commutative. This shows that σ := (σ )X∈Ob(C) is a natural transformation from I to J and thus a natural isomorphism, since the σX are isomorphisms in K(C).

By duality we obtain:

Remark and Deﬁnition 5.46. Let (C, E) be an exact category such that C has enough E-projective objects. For each object X in C choose a ﬁxed X E-projective resolution (P , εX ). If f : X → Y is a morphism in C, there 5.4. DERIVED FUNCTORS 159 exists a morphism of complexes f P : P X → P Y that lifts f and is unique up to homotopy, by 5.40. Then we get a covariant additive functor

X 7→ P X P : C → K(C) , , f 7→ P (f) with P (f) := f P + Ht(P X ,P Y ). If we choose for each object X in C a diﬀerent (ﬁxed) E-projective resolution X 0 (Q , εX ), we get another covariant additive functor Q: C → K(C) with Q(X) = QX . The functors P and P 0 are then naturally isomorphic. Remark 5.47. Let C and D be additive categories and let F : C → D be X an additive functor. If X = (Xn, dn )n∈Z is a complex in C, then we get a X complex Fe(X) := (F (Xn),F (dn ))n∈Z in D, since X X X X F (dn+1) ◦ F (dn ) = F (dn+1 ◦ dn ) = F (0) = 0 for all n ∈ Z. If f : X → Y is a morphism in C(C), then F (f) := (F (fn)n∈Z is obviously a morphism of complexes between Fe(X) and Fe(Y ). Therefore the functor F induces a covariant additive functor between C(C) and C(D). X Y If f is homotopic to zero, we have fn = sn+1 ◦ dn + dn ◦ dn, and since F is an additive functor, we have also

X Y F (fn) = F (sn+1) ◦ F (dn ) + F (dn ) ◦ F (dn) for all n ∈ Z, hence the morphism F (f) is also homotopic to zero. Therefore F induces a covariant additive functor ( X 7→ Fe(X) Fe : K(C) → K(D) , f 7→ Fe(f) with Fe(f) := F (f) + Ht(Fe(X), Fe(Y )). Deﬁnition 5.48. Let (C, E) be an exact category such that C has enough E-injective objects, let A be an abelian category and let F : C → A be a covariant additive functor. For n ∈ Z the covariant additive functor

RnF := Hn ◦ Fe ◦ I : C → D is called the n-th right derived functor of F . Deﬁnition 5.49. Let (C, E) be an exact category such that C has enough E-projective objects, let A be an abelian category and let F : C → A be a covariant additive functor. For n ∈ Z the covariant additive functor

LnF := Hn ◦ Fe ◦ P : C → D is called the n-th left derived functor of F . 160 CHAPTER 5. DERIVED FUNCTORS

Remark 5.50. i) If (C, E) is an exact category such that C has enough E-injective (resp. enough E-projective) objects, then (Cop, Eop) is an exact category that has enough E-projective (resp. enough E-injective) objects. There- fore the left derived functors (resp. the right derived functors) of a contravariant functor G: C → A are deﬁned. ii) If RnF is the n-th right derived functor of F : C → A, then RnF (X) is a ﬁxed cokernel-object of the unique morphism λ making the diagram

I I F (dn−1) F (dn) ... / F (In−1) / F (In) / F (In+1) / ... |= O || i I | dn−1 || || k I || dn || || || im F (dI ) ___ / ker F (dI ) ___ / cok λ = RnF (X) n−1 λ n cλ commutative. iii) Let f : X → Y be a morphism in C and let RnF be the n-th right derived functor of F : C → A. Following the construction of 5.14 we get for n ∈ Z a commutative diagram

X σn IX IX im F (dn−1) / ker F (dn )

µn λn IY IY im F (dn−1) Y / ker F (dn ), σn

X Y where σn , σn are the unique morphisms between the image and the kernel and where µn and λn are also uniquely determined by certain commutativity relations, as shown in 5.14. Then RkF (f) is the unique morphism making the diagram

X c X σn σn IX IX X k im F (dn−1) / ker F (dn ) / cok σN = R F (X) k µn λn R F (f) c Y σn IY IY Y k im F (dn−1) Y / ker F (dn ) / cok σN = R F (Y ) σn commutative. iv) If LnF is the n-th left derived functor of F : C → A, then the object n L F (X) can be obtained as in ii) by substituting Pk for every Ik and P I n dk for every dk. In the same way L F (f) can be obtained from iii). 5.4. DERIVED FUNCTORS 161

Proposition 5.51. Let (C, E) be an exact category such that C has enough E-injective objects, let A be an abelian category and let F : C → A be a covariant additive functor. i) RnF =∼ 0 for n < 0. ii) If I ∈ Ob(C) is an E-injective object, then RnF (I) = 0 for all n > 0. n Proof. i) is clear, since In = 0 for n < 0 and since F and H are additive functors. ii) If I ∈ Ob(C) is an E-injective objective, then the exact sequence

idI 0 / I / I / 0 induces an E-injective resolution of I with I0 = I and In = 0 for all n > 0. Hence it follows that RnF (I) = 0 for all n > 0.

By duality we obtain: Proposition 5.52. Let (C, E) be an exact category such that C has enough E-projective objects, let A be an abelian category and let F : C → A be a covariant additive functor. i) LnF =∼ 0 for n > 0. ii) If P ∈ Ob(C) is an E-projective object, then LnF (P ) = 0 for all n > 0. Proposition 5.53. Let (C, E) be an exact category such that C has enough E-injective objects, let A be an abelian category and let F : C → A be a covariant additive functor such that F is injective. Then there is a natural isomorphism R0F =∼ F. Proof. Let X be an object of C and let (I, γ) be an injective resolution of X. Since F is injective the sequence

I F (γ) F (d0) 0 / F (X) / F (I0) / F (I1)

I is exact in A, hence F (γ) is a kernel of F (d0). We have 0 I I R F (X) = cok(im(F d−1) → ker F (d0)) I = ker F (d0), I since im(F d−1) = 0. Then the universal property of the kernel F (γ) gives 0 rise to a unique isomorphism λX : R F (X) → F (X), making the diagram

0 kX R F (X) / F (I0) : LL uu LLL uu λX L& uu F (γ) F (X) 162 CHAPTER 5. DERIVED FUNCTORS

0 commutative. We claim that λ: R F → F with λ(X) = λX is a natural transformation and thus a natural isomorphism: Let f : X → Y be a morphism in C and let (J, δ) be an injective resolution of Y , then we have the commutative diagram

I F (γ) F (d0) 0 / F (X) / F (I0) / F (I1)

F (f) F (f0) F (f0) 0 / F (Y ) / F (J0) / F (J1) F (δ) J F (d0 ) with exact rows. Using the notations of 5.50.iii), the morphism RkF (f) is the uniquely determined morphism making the diagram

X c σ σX I n I n X k im F (dn−1) / ker F (dn) / cok σN = R F (X) k µn λn R F (f) c σY im F (dJ ) J n Y k n−1 Y / ker F (dn) / cok σN = R F (Y ) σn

I J commutative. Since im F (d ) = im F (d ) = 0, it follows that c X and −1 −1 σn 0 c Y are isomorphisms and that R F (f) makes the diagram σn

F (dI ) 0 kX 0 R F (X) / F (I0) / F (I1)

0 R F (f) F (f0) F (f1) F (dJ ) 0 kY 0 R F (Y ) / F (J0) / F (J1) commutative (since λ0 is unique with k J ◦ λ0 = F (f0) ◦ k I ). F (d0 ) F (d0) We have F (γ) ◦ λX = kX and F (δ) ◦ λY = kY we have

0 0 F (δ) ◦ λY ◦ R F (f) = kY ◦ R F (f)

= F (f0) ◦ kX

= F (f0) ◦ F (γ) ◦ λX

= F (δ) ◦ F (f) ◦ λX

0 and therefore also λY ◦R F (f) = F (f)◦λX , since F (δ) is a monomorphism. Then the diagram

λX R0F (X) / F (X)

R0F (f) F (f)

λY R0F (Y ) / F (Y ) is commutative, which shows that λ is a natural isomorphism. 5.4. DERIVED FUNCTORS 163

By duality we obtain: Proposition 5.54. Let (C, E) be an exact category category such that C has enough E-projective objects, let A be an abelian category and let F : C → A be a covariant additive functor such that F is projective. Then there is a natural isomorphism L0F =∼ F. Remark and Deﬁnition 5.55. Let (C, E) be an exact category category such that C has enough E-injective objects, let A be an abelian category and let F : C → A be a covariant additive functor such that F is semi-injective. Let X be an object of C and let (I, γ) an injective resolution of X. The proof of 5.53 shows that the morphism

0 kX : R F (X) → F (I0)

I I is a kernel of F (d0). Since F is additive, we have F (d0)◦F (γ) = 0, therefore the universal property of the kernel kX gives rise to a unique morphism 0 µX : F (X) → R F (X) making the diagram

F (γ) F (X) / F (I0) L 9 L rr µ rr X L& rr kX R0F (X) commutative (note that F (γ) is a monomorphism, since F is semi-injective, therefore µX is also a monomorphism). Deﬁne

+ F (X) := cok µX .

If f : X → Y is a morphism in C and (J, δ) an injective resolution of Y , then the proof of 5.53 shows that there is a commutative diagram

µX F (X) / R0F (X)

F (f) (1) R0F (f) F (Y ) / R0F (Y ) µY

−1 where µX = λX in the notation of this proof (note, that here µX is generally not an isomorphism). By 2.18 there exists a unique morphism + F (f): cok µX → cok µY making the diagram

c µX 0 µX + F (X) / R F (X) / cok µX = F (X) F (f) R0F (f) F +(f) cµY F (Y ) / R0F (Y ) / cok µ = F +(Y ) µY Y 164 CHAPTER 5. DERIVED FUNCTORS

commutative. From the above diagram it follows that F (idX ) = idF (X) and F +(g ◦ f) = F +(g) ◦ F +(f), therefore the rule X 7→ F +(X) F + : C → A , f 7→ F +(f) is a covariant functor. F + is called the additional right derived functor of F . The dual notion of the above is the following: Remark and Deﬁnition 5.56. Let (C, E) be an exact category such that C has enough E-projective objects, let A be an abelian category and let F : C → A be a covariant additive functor such that F is semi-projective. Let X be an object of C and let (P, ε) an E-projective resolution of X. The dual argument of the proof of 5.53 shows that the morphism

0 cX : F (P0) → L F (X) P P is a cokernel of F (d0 ). Since F is additive, we have F (ε) ◦ F (d0 ) = 0, therefore the universal property of the cokernel cX gives rise to a unique 0 morphism νX : L F (X) → F (X) making the diagram

cX 0 F (P0) / L F (X) J JJ r JJ r F (ε) J$ yr νX F (X) commutative (note that F (ε) is an epimorphism, since F is semi-projective, therefore νX is also an epimorphism). Deﬁne − F (X) := ker νX . If f : X → Y is a morphism in C and (Q, η) an E-projective resolution of Y , then the dual of the proof of 5.53 shows that there is a commutative diagram νX L0F (X) / F (X)

L0F (f) (2) F (f) L0F (Y ) / F (Y ) νY − By 2.18 there exists a unique morphism F (f): ker νX → ker νY making the diagram

k − νX 0 νX F (X) = ker νX / L F (X) / F (X) F −(f) L0F (f) F (f) F −(Y ) = ker ν / L0F (Y ) / F (Y ) Y νY kνY 5.4. DERIVED FUNCTORS 165 commutative. Then the rule X 7→ F −(X) F − : C → A , f 7→ F −(f) is a covariant functor. F − is called the additional left derived functor of F . Remark 5.57. Let (C, E) be an exact category category, let A be an abelian category and let F : C → A be a covariant additive functor. i) If F is semi-injective and C has enough E-injective objects, the deﬁni- tion of F +(X) yields an exact sequence

µX cµX 0 / F (X) / R0F (X) / F +(X) / 0.

ii) If F is semi-projective and C has enough E-projective objects, the deﬁnition of F −(X) yields an exact sequence

kνX νX 0 / F −(X) / L0F (X) / F (X) / 0.

Corollary 5.58. Let (C, E) be an exact category category, let A be an abelian category and let F : C → A be a covariant additive functor. i) If F is semi-injective and C has enough E-injective objects, then F is injective if and only if there is a natural isomorphism F + =∼ 0. ii) If F is semi-projective and C has enough E-projective objects, then F is projective if and only if there is a natural isomorphism F − =∼ 0. Proof. It suﬃces to show i), since ii) is the dual statement. If F is injective, then it follows from the proof of 5.53, that for each object X of C the 0 morphism µX : F (X) → R F (X) is an isomorphism and therefore the exact sequence in 5.57.i) yields F +(X) = 0. This shows F + =∼ 0. If on the other hand, we have F + =∼ 0, then the exact sequence in 5.57.i), shows that µX is an isomorphism and therefore the diagram (1) in 5.55 shows, that there is a natural isomorphism R0F =∼ F . Since R0F is injective, this is also true for F .

Theorem 5.59 (Long Exact Sequence for Right Derived Functors). Let (C, E) be an exact category category such that C has enough E-injective objects, let A be an abelian category and let F : C → A be a covariant additive functor. If f g X / Y / Z is a sequence in C with (f, g) ∈ E, then there are morphisms

n n n+1 δ(f,g) : R F (Z) → R F (X) 166 CHAPTER 5. DERIVED FUNCTORS such that the long sequence

n δ(f,g) ... / RnF (X) / RnF (Y ) / RnF (Z) / Rn+1F (X) / n+1 δ(f,g) Rn+1F (Y ) / Rn+1F (Z) / Rn+2F (X) / Rn+2F (Y ) / ... is exact in A. Moreover, if

f g X / Y / Z

αX αY αZ X0 / Y 0 / Z0 f 0 g0 is a commutative diagram in C such that also (f 0, g0) is an element of E, then the diagrams n δ(f,g) RnF (Z) / Rn+1F (X)

n n+1 R F (αZ ) R F (αX ) RnF (Z0) / Rn+1F (X0) δn (f0,g0) are commutative for all n ∈ Z. Proof. By the horseshoe lemma 5.42 there are E-injective resolutions IX ,IY ,IZ of X,Y,Z and a sequence

v w IX / IY / IZ in C(C) such that v lifts the morphism f, w lifts the morphism g and with the steps X vn Y wn Z In / In / In / 0 being elements of E for all n ∈ Z. By 5.33 and 3.8 this sequences are even split exact, hence it follows from 5.4 that the sequences

F (v ) F (w ) X n Y n Z 0 / F (In ) / F (In ) / F (In ) / 0 are split exact for all n ∈ Z and assemble therefore to a kernel-cokernel pair in C(A) whose image in K(A) is the sequence

Fe(v) Fe(w) Fe(IX ) / Fe(IY ) / Fe(IZ ). By 5.28 and 5.20 it follows that there are morphisms

n n Z n+1 X δ(f,g) : H (Fe(I )) → H (Fe(I )) 5.4. DERIVED FUNCTORS 167 such that the long sequence

... n n n δn n+1 / H (Fe(IX )) / H (Fe(IY )) / H (Fe(IZ )) / H (Fe(IX )) /

δ n+1 n+1 n+1 n+2 n+2 ... H (Fe(IY )) / H (Fe(IZ )) / H (Fe(IX )) / H (Fe(IY )) / is exact in A. This is just the desired sequence. It remains to show the second assertion. If

f g (1) X / Y / Z

αX αY αZ X0 / Y 0 / Z0 f 0 g0 is a commutative diagram in C such that (f 0, g0) is an element of E, then the shelf lemma 5.44 provides E-injective resolutions IX ,IY ,IZ of X,Y,Z and E-injective resolutions IX0 ,IY 0 ,IZ0 of X0,Y 0,Z0 and a commutative diagram

v w (2) IX / IY / IZ

αeX αeY αeZ X0 / Y 0 / Z0 I v0 I w0 I in C(C) such that the morphisms in the diagram (2) lift the ones in the diagram (1) such that the steps of the commutative diagram (2) have rows that are elements of E. The above construction of the connecting morphisms, coupled with the naturalit of the functor Hn 5.21, then shows that the diagrams n δ(f,g) Hn(Fe(IZ )) / Hn+1(Fe(IX ))

n n+1 H (Fe(αZ )) H (Fe(αX ))

0 0 Hn(F (IZ )) / Hn+1(F (IX )) e δn e (f0,g0) are commutative for all n ∈ Z and this is what we wanted. By duality we obtain:

Theorem 5.60 (Long Exact Sequence for Left Derived Functors). Let (C, E) be an exact category category such that C has enough E-projective objects, let A be an abelian category and let F : C → A be a covariant additive functor. If f g X / Y / Z 168 CHAPTER 5. DERIVED FUNCTORS is a sequence in C with (f, g) ∈ E, then there are morphisms (f,g) n n+1 δn : L F (Z) → L F (X) such that the long sequence

(f,g) δn ... / LnF (X) / LnF (Y ) / LnF (Z) / Ln+1F (X) / (f,g) δn+1 Ln+1F (Y ) / Ln+1F (Z) / Ln+2F (X) / Ln+2F (Y ) / ... is exact in A. Moreover, if

f g X / Y / Z

αX αY αZ X0 / Y 0 / Z0 f 0 g0 is a commutative diagram in C such that also (f 0, g0) is an element of E, then the diagrams (f,g) δn LnF (Z) / Ln+1F (X)

n n+1 L F (αZ ) L F (αX ) LnF (Z0) / Ln+1F (X0) (f0,g0) δn are commutative for all n ∈ Z. Remark 5.61. i) Since RnF =∼ 0 for all n < 0 by 5.51, it follows from 5.59 that the functor R1F is left exact. Dually, since LnF =∼ 0 for all n > 0 by 5.52, it follows from 5.60 that the functor L1F is right exact. ii) If in the situation of 5.59 the functor F is injective and therefore R0F =∼ F by 5.53, then the long exact sequence has the following form:

0 δ(f,g) 0 / F (X) / F (Y ) / F (Z) / R1F (X) / ... n δ(f,g) ... / RnF (X) / RnF (Y ) / RnF (Z) / Rn+1F (X) / ....

iii) If in the situation of 5.60 the functor F is projective and therefore L0F =∼ F by 5.54, then the long exact sequence has the following form:

(f,g) δn ... / RnF (X) / RnF (Y ) / RnF (Z) / Rn+1F (X) / ... (f,g) δ−1 ... / L−1F (Z) / F (X) / F (Y ) / F (Z) / 0. 5.4. DERIVED FUNCTORS 169

Proposition 5.62. Let (C, E) be an exact category category such that C has enough E-injective objects, let A be an abelian category and let F : C → A be a covariant additive functor such that F is injective. The following are equivalent:

i) F is exact.

ii) RnF =∼ 0 for all n > 0.

n Proof. i) ⇒ ii) Let n ∈ Z with n > 0. By 5.50 the object R F (X) is a ﬁxed cokernel-object of the unique morphism λ making the diagram

I I F (dn−1) F (dn) ... / F (In−1) / F (In) / F (In+1) / ... |= O || i I | dn−1 || || k I || dn || || || im F (dI ) ___ / ker F (dI ) ___ / cok λ = RnF (X) n−1 λ n cλ commutative. Since F is exact, the morphism λ is an isomorphism by 3.3 and therefore RnF (X) = 0. ii) ⇒ i) follows immediately from the long exact sequence in 5.61.ii).

By duality we obtain:

Proposition 5.63. Let (C, E) be an exact category category such that C has enough E-projective objects, let A be an abelian category and let F : C → A be a covariant additive functor such that F is projective. The following are equivalent:

i) F is exact.

ii) LnF =∼ 0 for all n < 0.

Remark 5.64. Let (C, E) be an exact category category, let A be an abelian category and let

f g 0 / X / Y / Z / 0 be a sequence in C with (f, g) ∈ E.

i) If C has enough E-injective objects and F : C → A is a contravariant additive functor, then the left derived functors of the covariant functor F : Cop → A are deﬁned by 5.50.i). Then there are morphisms

(f,g) n n+1 δn : L F (X) → L F (Z) 170 CHAPTER 5. DERIVED FUNCTORS

such that the functor F induces a long exact sequence

(f,g) δn ... / LnF (Z) / LnF (Y ) / LnF (X) / Ln+1F (Z) / (f,g) δn+1 Ln+1F (Y ) / Ln+1F (X) / Ln+2F (Z) / Ln+2F (Y ) / ...

by 5.60. Moreover, if

f g X / Y / Z

αX αY αZ X0 / Y 0 / Z0 f 0 g0

is a commutative diagram in C such that also (f 0, g0) is an element of E, then the diagrams

(f,g) δn LnF (X) / Ln+1F (Z)

n n+1 L F (αX ) L F (αZ ) LnF (X0) / Ln+1F (Z0) (f0,g0) δn

are commutative for all n ∈ Z. ii) If C has enough E-projective objects and F : C → A is a contravariant additive functor, then the right derived functors of the covariant functor F : Cop → A are deﬁned by 5.50.i). Then there are morphisms

n n+1 δn : R F (X) → R F (Z)

such that the functor F induces a long exact sequence

n δ(f,g) ... / RnF (Z) / RnF (Y ) / RnF (X) / Rn+1F (Z) / n+1 δ(f,g) Rn+1F (Y ) / Rn+1F (X) / Rn+2F (Z) / Rn+2F (Y ) / ...

by 5.59. Moreover, if

f g X / Y / Z

αX αY αZ X0 / Y 0 / Z0 f 0 g0 5.5. UNIVERSAL δ-FUNCTORS 171

is a commutative diagram in C such that also (f 0, g0) is an element of E, then the diagrams

n δ(f,g) RnF (X) / Rn+1F (Z)

n n+1 R F (αX ) R F (αZ ) RnF (X0) / Rn+1F (Z0) δn (f0,g0)

are commutative for all n ∈ Z.

5.5 Universal δ-Functors

Deﬁnition 5.65. Let (C, E) be an exact category and let A be an abelian category. A covariant homological (resp. cohomological) δE -functor from n C to A is a collection F = (Fn)n≤0 (resp. F = (F )n≥0) of covariant n additive functors Fn : C → A (resp. F : C → A), together with a connecting morphism (f,g) δn : Fn(Z) → Fn+1(X) n n n+1 (resp. δ(f,g) : F (Z) → F (X)) in A that is deﬁned for all n ≤ 0 (resp. for all n ≥ 0) and all sequences f g X / Y / Z in C with (f, g) ∈ E, which has the following properties:

0 (δ0) F0 is injective (resp. F is projective).

f 0 g0 0 0 0 (δ1) If X / Y / Y is another sequence in C with (f, g) ∈ E and

f g X / Y / Z

αX αY αZ X0 / Y 0 / Y 0 f 0 g0

is a commutative diagram, then the diagram

(f,g) δn δn n (f,g) n+1 Fn(Z) / Fn+1(X) resp. F (Z) / F (X)

n n+1 Fn(αZ ) Fn+1(αX ) F (αZ ) F (αX ) F n(Z0) / F n+1(X0) F n(Z0) / F n+1(X0) δn δn (f0,g0) (f0,g0)

is commutative for all n ≤ 0 (resp. for all n ≥ 0). 172 CHAPTER 5. DERIVED FUNCTORS

f g (δ2) For every sequence X / Y / Z in C with (f, g) ∈ E the long sequence

(f,g) (f,g) δn δn+1 ... / Fn(Z) / Fn+1(X) / Fn+1(Y ) / Fn+1(Z) / Fn+2(X) / ...

(resp.

n n+1 δ(f,g) δ(f,g) ... / F n(Z) / F n+1(X) / F n+1(Y ) / F n+1(Z) / F n+2(X) / ... )

is a complex with values in A.

Deﬁnition 5.66. Let (C, E) be an exact category, let A be an abelian cate- n n gory and let F = (Fn)n≤0, G = (Gn)n≤ (resp. F = (F )n≥0, G = (G )n≥0) be two covariant homological (resp. cohomological) δE -functors from C to A.A morphism of homological (resp. cohomological) δE -functors from F to G is a family f = (fn)n≤0 of natural transformations fn : Fn → Gn (resp. n n n n a family f = (f )n≥0 of natural transformations f : F → G ), such that f g in addition for n ≤ 0 (resp n ≥ 0) and every sequence X / Y / Z in C with (f, g) ∈ E the diagrams

n,F δ(f,g) δ n,F n (f,g) n+1 Fn(Z) / Fn+1(X) resp. F (Z) / F (X)

n n+1 fn(Z) fn+1(X) f (Z) f (X) Gn(Z) n+1 Gn(Z) n+1 n,G / G (X) n,G / G (X) δ(f,g) δ(f,g) are commutative.

Remark and Deﬁnition 5.67. Let (C, E) be an exact category and let A be an abelian category.

i) For every covariant homological (resp. cohomological) δE -functor F = n (Fn)n≤0 (resp. F = (F )n≥0) from C to A we have the identity morphism of homological (resp. cohomological) δE -functors

F idF = (idn )n≥0 : F → F

F n (resp. id = (idF )n≤0 : F → F ) F n n with idn := idFn for all n ≤ 0 (resp. idF := idF for all n ≥ 0).

ii) If F = (Fn)n≤0, G = (Gn)n≤0 and H = (Hn)≤0 are three covariant homological δE -functors from C to A and f = (fn)n≤0 : F → G, 5.5. UNIVERSAL δ-FUNCTORS 173

g = (gn)n≤0 : G → H are morphisms of δE -functors, then we have the composite δE -functor

g ◦ f := (gn ◦ fn)n≤0 : F → H.

Analogously one gets a composition for morphisms of cohomological δE -functors.

iii) A morphism of homological (resp. cohomological) δE -functors

λ = (λn)n≤0 : F → G

n (resp. λ = (λ )n≥0 : F → G)

between homological (resp. cohomological) δE -functors is called an isomorphism of δE -functors if there exists a morphism of homological (resp. cohomological) δE -functors

µ = (µn)n≤0 : G → F

n (resp. µ = (µ )n≥0 : G → F )

with λ ◦ µ = idG and µ ◦ λ = idF . Remark and Deﬁnition 5.68. Let (C, E) be an exact category and let A be an abelian category.

i) A contravariant homological (resp. cohomological) δE -functor is a co- op variant homological (resp. cohomological) δE -functor from C to A.

ii) A homological (resp. cohomological) δE -functor F is called an exact homological (resp. cohomological) δE -functor, if for every sequence f g X / Y / Z in C with (f, g) ∈ E the induced long sequence of the property (δ2) is exact in A.

Example 5.69. Let A be an abelian category and let Cn≤0(A) and Cn≥0(A) be the full subcategories of C(A) with

X Ob(Cn≤0(A)) := {X = (Xn, dn )n∈Z ∈ C(C)|Xn = 0 for n > 0}, X Ob(Cn≥0(A)) := {X = (Xn, dn )n∈Z ∈ C(C)|Xn = 0 for n < 0}. Then it follows from 5.20 and 5.21 that the homology functors

k H : Cn≤0(A) → A restricted to Cn≥0(A) form an exact homological δE -functor and that the homology functors k H : Cn≥0(A) → A restricted to Cn≥0(A) form an exact cohomological δE -functor. 174 CHAPTER 5. DERIVED FUNCTORS

Example 5.70. Let (C, E) be an exact category, let A be an abelian category and let F : C → A be a covariant additive functor. i) If C has enough E-injective objects, it follows from 5.59 and 5.61.i) that n RF := (R F )n≥0

is an exact cohomological δE -functor from C to A. ii) If C has enough E-projective objects, it follows from 5.60 and 5.61.i) that n LF := (L F )n≤0

is an exact homological δE -functor from C to A. Deﬁnition 5.71. Let (C, E) be an exact category, let A be an abelian cat- n egory and let F = (Fn)n≤0 (resp. F = (F )n≥0) be a covariant homological (resp. cohomological) exact δE -functor from C to A. F is called a universal homological (resp. cohomological) δE -functor, if for every homological (resp. n cohomological) δE -functor G = (Gn)n≤0 (resp. G = (G )n≥0) and every nat- 0 0 0 ural transformation f0 : F0 → G0 (resp. f : F → G ) there exists a unique morphism λ: F → G of homological (resp. cohomological) δE -functors with 0 0 λ0 = f0 (resp. λ = f ). Proposition 5.72. Let (C, E) be an exact category and let A be an abelian n category. If F = (Fn)n≤0 and G = (Gn)n≤0 (resp. F = (F )n≥0 and n G = (G )n≥0) are universal homological (resp. cohomological) δE -functors ∼ 0 ∼ 0 with F0 = G0 (resp. F = G ) in Fun(C, A), then there is an isomorphism of δE -functors φ: F → G.

Proof. We show only the case of homological δE -functors, the other case is shown analogously. Let λ: F0 → G0 and µ: G0 → F0 be natural isomorphisms with λ ◦ µ = idG0 and µ ◦ λ = idF0 . The universal property of the δE -functors F = (Fn)n≤0 and G = (Gn)n≤0 yields morphisms of δE -functors

(λn)n≤0 : F → G

(µn)n≤0 : G → F with λ0 = λ and µ0 = µ. Since the identities idF and idG are the unique F G morphisms of δE -functors with id0 = idF0 and id0 = idG0 we have

λn ◦ µn = idFn

µn ◦ λn = idGn for all n ≤ 0. Then (λn)n≤0 is an isomorphism of δE -functors. Deﬁnition 5.73. Let (C, E) be an exact category and let A be an abelian category. 5.5. UNIVERSAL δ-FUNCTORS 175

i) If C has enough E-injective objects we call a covariant additive functor F : C → A eﬀacable if F (I) = 0 for every E-injective object I of C.

ii) If C has enough E-projective objects we call a covariant additive functor F : C → A coeﬀacable if F (P ) = 0 for every E-projective object P of C.

Proposition 5.74. Let (C, E) be an exact category such that C has enough n E-injective objects, let A be an abelian category and let F = (F )n≥0 be a n covariant exact cohomological δE -functor from C to A. If F is eﬀacable for n > 0 then F is a universal cohomological δE -functor.

n Proof. Let G = (G )n≥0 be a covariant cohomological δE -functor and let f 0 : F 0 → G0 be a natural transformation. We will construct f 1 : F 1 → G1 and inductively f n : F n → Gn. Let A be an object of C. Since C has enough E-injective objects we ﬁnd a sequence u v A / I / A0 with an E-injective object I and (u, v) ∈ E. For the construction of f 1 (and inductively for f n), consider the following commutative diagram:

0,F 0 1 F (v) δ(u,v) F (u) F 0(I) / F 0(A0) / F 1(A) / F 1(I) = 0 f 0(I) f 0(A0) f 1(A) G0(I) / G0(A0) / G1(A) G0(v) 0,G δ(u,v)

0,F 0 Since the upper row is exact, the morphism δ(u,v) is a cokernel of F (v). We have 0,G 0 0 0 0,G 0 0 δ(u,v) ◦ f (A ) ◦ F (v) = δ(u,v) ◦ G (v) ◦ f (I) = 0, since the lower row is a complex. Therefore the universal property of the 0,F 1 1 1 cokernel δ(u,v) gives rise to a unique morphism f (A): F (A) → G (A) with 1 0,F 0,G 0 0 f (A) ◦ δ(u,v) = δ(u,v) ◦ f (A ). We have thus deﬁned f 1(A) for every object A of C. It remains to be shown:

(1) The morphism f 1(A) does not depend on the choice of the sequence u v A / I / A0 .

1 1 (2) f := (f (A))A∈Ob(C) is a natural transformation. 176 CHAPTER 5. DERIVED FUNCTORS

f g (3) For every sequence X / Y / Z with (f, g) ∈ E the diagram

0,F δ(f,g) F 0(Z) / F 1(X)

f 0(Z) f 1(X) 0 1 G (Z) 0,G / G (X) δ(f,g)

is commutative.

u0 v0 (1) Let A / J / A00 be another sequence with an E-injective object J and (u0, v0) ∈ E. Since J is E-injective and u is an admissible monomorphism, there exists a morphism s: I → J with u0 = s ◦ u. Then we also have the commutative diagram u v A / I / A0 s t A / J / A00 u0 v0 by 2.18. As before we ﬁnd a unique morphism fe1(A): F 1(A) → G1(A) making the diagram

0,F 0 0 δ 1 F (v ) (u0,v0) F (u) F 0(J) / F 0(A00) / F 1(A) / F 1(J) = 0 f 0(J) f 0(A00) fe1(A) G0(J) / G0(A00) / G1(A) G0(v0) δ0,G (u0,v0) commutative. Consider then the diagram

0,F 0 F (v) δ(u,v) F 0(I) / F 0(A0) / F 1(A) / 0 r tt qq 0,F rrr t 0 0 0 qq 0 δ rrr yttF (s) F (v ) xq F (t) (u0,v0) rrr F 0(J) / F 0(A00) / F 1(A) / 0 f 0(I) f 0(A00) f 0(A0) f 1(A) f 1(A) e f 0(J) G0(I) / G0(A0) / G1(A) G0(u) δ0,G t q (u,v) r tt qq rrrrr yttG0(s) xqqG0(t) rrrr G0(J) / G0(A00) / G1(A). G0(v0) δ0,G (u0,v0)

Because of the properties the involved cohomological δE -functors and of the natural transformation f 0 the commutativity of all squares in the above 5.5. UNIVERSAL δ-FUNCTORS 177 diagram follows, except that of the square

F 1(A) F 1(A) .

fe1(A) f 1(A) G1(A) G1(A)

Here we have

1 0,F 1 0,F 0 fe (A) ◦ idF 1(A) ◦δ(u,v) = fe (A) ◦ δ(u0,v0) ◦ F (t) 0,G 0 00 0 = δ(u0,v0) ◦ f (A ) ◦ F (t) 0,G 0 0 0 = δ(u0,v0) ◦ G (t) ◦ f (A ) 0,G 0 0 = idG1(A) ◦δ(u,v) ◦ f (A ) 1 0,F = idG1(A) ◦f (A) ◦ δ(u,v),

1 1 0,F hence f (A) = fe (A), since δ(u,v) is an epimorphism. This shows (1). (2) Let r : A → B be a morphism in C and let

u v A / I / A0 be a sequence with I an E-injective object and (u, v) ∈ E, let

u0 v0 B / J / B0 be a sequence with J an E-injective object and (u0, v0) ∈ E. Since J is injective and u is an admissible monomorphism there exists a morphism s: I → J with u0 ◦ r = s ◦ u, hence we have a commutative diagram

u v A / I / A0 r s t B / J / B0 u0 v0 by 2.18. Consider then the diagram

0,F 0 F (v) δ(u,v) F 0(I) / F 0(A0) / F 1(A)

tt rr 0,F rr t 0 0 0 r 0 δ r 1 yttF (s) F (v ) xrrF (t) (u0,v0) yrrF (r) 0 0 0 1 1 F (J) / F (B ) / F (B) fA f 0(I) f 0(B0) f 0(A0) f 1(B) f 0(J) G0(I) / G0(A0) / G1(A) G0(u) δ0,G ttt rrr (u,v) rrr yttG0(s) xrrG0(t) yrrG1(r) G0(J) / G0(B0) / G1(B). G0(v0) δ0,G (u0,v0) 178 CHAPTER 5. DERIVED FUNCTORS

As before the properties of the cohomological δE -functors and of the natural transformation f 0 yield the commutativity of all squares, except that of

F 1(r) F 1(A) / F 1(B)

f 1(A) f 1(B) G1(A) / G1(B). G1(r)

The commutativity of this square follows as before modulo the epimorphism 0,F δ(u,v). This shows (2).

f g (3) Let X / Y / Z be a sequence with (f, g) ∈ E. Choose then a sequence u0 v0 X / I / X0 , where I is an E-injective object and (r, s) ∈ E. Since I is E-injective, we ﬁnd a commutative diagram

f g X / Y / Z s t X / I / X0, u0 v0 which induces the following diagram:

0,F 0 F (g) δ(f,g) F 0(Y ) / F 0(Z) / F 1(X) r ss rr 0,F rrr s 0 0 0 r 0 δ rrr yssF (s) F (u ) xrrF (t) (u0,v0) rrr F 0(I) / F 0(X0) / F 1(X) f 1(X) f 0(Y ) f 0(A00) f 0(Z) f 1(X) f 0(I) G0(Y ) / G0(Z) / G1(X) G0(g) δ0,G s r (f,g) r ss rr rrrrr yssG0(s) xrrG0(t) rrrr G0(I) / G0(X0) / G1(X) G0(v0) δ0,G (u0,v0)

Note that, because we do not assume Y to be E-injective, it does not necessarily follow that F 1(Y ) = 0. Therefore we have to show the commutativity of the diagram 0,F δ(f,g) F 0(Z) / F 1(X)

f 0(Z) f 1(X) 0 1 G (Z) 0,G / G (X). δ(f,g) 5.5. UNIVERSAL δ-FUNCTORS 179

The properties of the cohomological δE -functors and of the natural transformation f 0 yield again the commutativity of all other squares. Here we have 1 0,F 1 0,F 0 f (X) ◦ δ(f,g) = f (X) ◦ idF 1(X) ◦δ(u0,v0) ◦ F (t) 0,G 0 0 0 = δ(u0,v0) ◦ f (X ) ◦ F (t) 0,G 0 0 = δ(u0,v0) ◦ G (t) ◦ f (Z) 0,G 0 = δ(f,g) ◦ f (Z) which shows that the above diagram is commutative and thus (3) holds. n n Inductively one shows (1) − (3) for the f . Then (f )n≥0 : F → G is a n morphism of cohomological δE -functors and since the f are uniquely con- 0 structed up from f it follows that F is a universal δE -functor. By duality we obtain: Proposition 5.75. Let (C, E) be an exact category such that C has enough E-projective objects, let A be an abelian category and let F = (Fn)n≤0 be a covariant exact homological δE -functor from C to A. If Fn is coeﬀacable for n < 0 then F is a universal homological δE -functor. Proposition 5.76. Let (C, E) be an exact category let A be an abelian category and let F : C → A be a covariant additive functor.

i) If C has enough E-injective objects, the exact cohomological δE -functor n RF := (R F )n≥0 is universal.

ii) If C has enough E-projective objects, the exact homological δE -functor n LF := (L F )n≤0 is universal. Proof. This is a direct consequence of 5.74 and 5.75, since the functors RnF are eﬀacable for n > 0 by 5.51 and the functors LnF are coeﬀacable for n < 0 by 5.52.

Corollary 5.77. Let (C, E) be an exact category, let A be an abelian category and let F : C → A be a covariant additive functor. i) If C has enough E-injective objects, the functor F is injective and n G = (G )n≥0 is a universal covariant cohomological δE -functor from C to A with G0 =∼ F , then there is an isomorphism RF =∼ G

of δE -functors. 180 CHAPTER 5. DERIVED FUNCTORS

ii) If C has enough E-projective objects, the functor F is projective and G = (Gn)n≤0 is a universal covariant homological δE -functor from C to A with G0 =∼ F , then there is an isomorphism

LF =∼ G

of δE -functors. Proof. It suﬃces to show i), since ii) can be obtained by duality. Since the functor F is injective, we have R0F =∼ F by 5.53. Then i) follows from 5.72, n since RF = (R F )n≥0 is a universal cohomological δE -functor by 5.76. Corollary 5.78. Let (C, E) be an exact category, let A be an abelian category and let F : C → A be a covariant additive functor.

n i) If C has enough E-injective objects and F = (F )n≥0 is a universal n cohomological δE -functor from C to A, then F is eﬀacable for n > 0.

ii) If C has enough E-projective objects and G = (Gn)n≤0 is a universal homological δE -functor from C to A, then Gn is coeffacable for n < 0. Proof. Again it suffices to show i). The functor F 0 : C → A is injective and 0 n 0 therefore we have the universal cohomological δE -functor RF = (R F )n≥0 with R0F 0 =∼ F 0. By 5.77 it follows that RnF 0 =∼ F n for all n ≥ 0, hence F n is effacable for n > 0 by 5.51. Chapter 6

Yoneda-Ext-Functors

6.1 Yoneda-Ext1

Deﬁnition 6.1. Let (C, E) be an exact category and X,Z ∈ Ob(C). Deﬁne NZ,X to be the class of all short exact sequences of the following form:

f g X / Y / Z

On the class NZ,X deﬁne the following equivalence relation:

(f, g) ∼ (f 0, g0):⇔ There is a commutative diagram f g X / Y / Z

β X / Y 0 / Z f 0 g0

The relation ∼ is really an equivalence relation, since β is an isomorphism by 3.23. Deﬁne then 1 ExtE (Z,X) := NZ,X / ∼ . 1 We will write [f, g] for the equivalence class of (f, g) in ExtE (Z,X). Remark 6.2. If (C, E) is an exact category, a logical diﬃculty (apart from 1 the commonplace one that the members of ExtE (Z,X) may not be sets) 1 arises from the fact that ExtE (Z,X) may not be a set. Of course, if C is 1 a small category, then ExtE (Z,X) will be a set. Likewise it can be shown 1 that ExtE (Z,X) is a set if C has enough E-projective or E-injective objects (see 6.44 at the end of this chapter), or if the category C has a generator or a cogenerator (a generator of a category C is an object G of C such that for every pair of distinct morphisms α, β : X → Z in C there exists a morphism g : G → X such that α ◦ g 6= β ◦ g, a cogenerator is the dual notion; cf. [16].II.15 and VII).

181 182 CHAPTER 6. YONEDA-EXT-FUNCTORS

In the examples arising from functional analysis, the second one will almost always be the case, since in every full subcategory of (TVS) that contains 1 the ground ﬁeld K, the ground ﬁeld is a generator, hence ExtE (Z,X) will always be a set in this case.

Remark and Deﬁnition 6.3. Let (C, E) be an exact category, let X and 1 0 Z be objects of C and [f, g] ∈ ExtE (Z,X). If γ : Z → Z is a morphism, we have a commutative diagram

k pZ0 X / P / Z0

pY PB γ X / Y / Z f g with (k, pZ0 ) ∈ E by proposition 3.13. If additionally

f g X / Y / Z

β X / Y 0 / Z f 0 g0 is a commutative diagram, i.e. [f, g] = [f 0, g0], then we get a commutative diagram k0 qZ0 X / Q / Z0 ~~ ~? } ~~~ λ }}}} ~~~~ ~ }}}} ~~~~ ~ }}}} ~~~ k ~ pZ0 }}} X / P / Z0 γ

pY qY 0 γ

0 X 0 / Y 0 / Z } f > g } }}}} }} }}}} }}} }} }}} }}}} } β }}}} }}}} }} }}}} X / Y / Z f g

0 0 where (Q, qY 0 , qZ0 ) is the pullback of γ and g and (k , qZ0 ) ∈ E. Then 0 g ◦ β ◦ pY = g ◦ pY = γ ◦ idZ ◦pZ0 , hence by the universal property of (Q, qY 0 , qZ0 ) there exists a unique morphism λ: P → Q with β ◦ pY = qY 0 ◦ λ 0 0 and pZ0 = qZ0 ◦ λ. The morphism k is uniquely determined by qZ0 ◦ k = 0 0 0 0 and qY 0 ◦ k = f and in addition we have q ◦ λ ◦ k = pZ0 ◦ k = 0 and 0 0 qY 0 ◦ λ ◦ h = β ◦ pY ◦ k = β ◦ f = f , hence k = λ ◦ k, i.e. all the squares in 0 the above diagram commute. It follows that [k, pZ0 ] = [k , qZ0 ] and therefore the mapping

1 1 0 ExtE (Z,X) → ExtE (Z ,X), [f, g] 7→ [k, pZ0 ] 6.1. YONEDA-EXT1 183 is well-deﬁned. Deﬁne then

[f, g]γ := [k, pZ0 ].

By duality we ﬁnd for a morphism α: X → X0 a commutative diagram

f g X / Y / Z

α PO sY X0 / S / Z sX0 c with (sX0 , c) ∈ E and the dual argument of the above shows that the mapping

1 1 0 ExtE (Z,X) → ExtE (Z,X ), [f, g] 7→ [sX0 , c] is well-deﬁned. Deﬁne α[f, g] := [sX0 , c]. Proposition 6.4. Let (C, E) be an exact category, let X and Z be objects of C, let α: X → X0, α0 : X0 → X00, γ : Z0 → Z, γ0 : Z00 → Z0 be morphisms 1 in C and let [f, g] be an element of ExtE (Z,X).

i) idX [f, g] = [f, g] ii) (α0 ◦ α)[f, g] = α0(α[f, g])

iii) [f, g] idZ = [f, g] iv) [f, g](γ ◦ γ0) = ([f, g]γ)γ0

v) (α[f, g])γ = α([f, g]γ) Proof. i) and ii) follow from the diagrams

f g X / Y / Z

PO X / Y / Z f 0 g0 and f g X / Y / Z

α PO sY sX0 c X0 / S / Z

α0 PO rS X00 / R / Z rX00 c0 184 CHAPTER 6. YONEDA-EXT-FUNCTORS and from the transitivity of the pushout. iii) and iv) are the dual statements of i) and ii). To show v) we construct a commutative cube

sX0 c X0 / S / Z0 > ? } α }} sP }}} }} }}}} }} }}}} } p 0 }}} } k Z 0 } X / P / Z γ pY λ γ 0 / / X t T 0 Z |> X0 ? c }} || }}}} |α }}} || tY }}} || }}}} X / Y / Z f g where (P, pY 0 , pZ0 ) is the pullback of γ and g,(S, sX0 , sp) is the pushout of 0 α and k,(T, tX0 , tY ) is the pushout of α and f and k, c, c are the induced morphisms. Then tY ◦ pY ◦ k = tY ◦ f = tX0 ◦ idX0 , hence the universal property of (S, sX0 , sp) gives rise to a unique morphism λ : S → T with λ ◦ sP = tY ◦ pY 0 and λ ◦ sX0 = tX0 ◦ idX0 . The morphism c ◦ λ is uniquely determined by 0 0 c ◦ λ ◦ sX0 and c ◦ λ ◦ sP and we have

0 0 c ◦ λ ◦ sP = c ◦ tY ◦ pY = g ◦ pY = γ ◦ pZ0 = γ ◦ c ◦ sP 0 0 c ◦ λ ◦ sX0 = c ◦ tX0 ◦ idX0 = 0 = γ ◦ c ◦ sX0 , hence it follows that γ ◦ c = c0 ◦ λ. Therefore the two squares

sX0 c X0 / S / Z0 (1) λ (2) γ 0 X / T 0 / Z tX0 c created by the morphism λ commute and in addition (2) is a pullback square by 3.13, hence

0 (α[f, g])γ = [tX0 , c ]γ = [sX0 , c] = α[k, pZ0 ] = α([f, g]γ).

Notation 6.5. The above proposition 6.4 allows us to write

α0 ◦ α[f, g] := (α0 ◦ α)[f, g] = α0(α[f, g]) [f, g]γ ◦ γ0 := [f, g](γ ◦ γ0) = ([f, g]γ)γ0 α[f, g]γ := (α[f, g])γ = α([f, g]γ) 6.1. YONEDA-EXT1 185

1 Remark and Deﬁnition 6.6. ExtE (Z,X) need not always be a set for a 1 given exact category (C, E) (see 6.2). In order to formulate ExtE -functors we therefore introduce the quasicategory (Class) (see 1.29), whose objects are classes and whose morphisms are mappings between classes.

Remark 6.7. Let (C, E) be an exact category and let A be an object of C. By 6.4 we have a covariant functor

1 1 X 7→ ExtE (A, X) ExtE (A, −): C → (Class) , 1 , α 7→ ExtE (A, α) where

1 1 1 0 ExtE (A, α): ExtE (A, X) → ExtE (A, X ), [f, g] 7→ α[f, g] and a contravariant functor 1 1 Z 7→ ExtE (Z,A) ExtE (−,A): C → (Class) , 1 , γ 7→ ExtE (γ, A) where

1 1 1 0 ExtE (γ, A): ExtE (Z,A) → ExtE (Z ,A), [f, g] 7→ [f, g]γ. Additionally 6.4.v) shows that

1 op ExtE (−, −): C q C → (Class) is a bifunctor, i.e. for morphisms α: A → A0 and β : B0 → B the following diagram commutes:

Ext1 (B,α) 1 E 1 0 ExtE (B,A) / ExtE (B,A )

1 1 0 ExtE (β,A) ExtE (β,A ) 1 0 1 0 0 ExtE (B ,A) 1 0 / ExtE (B ,A ) ExtE (B ,α)

Remark and Deﬁnition 6.8.

i) For X ∈ Ob(C), deﬁne idX ∆X := : X → X q X, ∇X := ( idX idX ): X q X → X idX

Since ωX ◦ ∆X = idX the morphism ∆X is a coretraction and since ∇X ◦ ωX = idX the morphism ∇X is a retraction. For two morphisms α, α0 : A → B simple matrix multiplactions yield:

I) α ◦ ∇A = ∇B ◦ (α ⊕ α) 186 CHAPTER 6. YONEDA-EXT-FUNCTORS

0 0 II) α = α = ∇B ◦ (α ⊕ α ) ◦ ∆A

ii) The mapping

1 1 1 ⊕: ExtE (Z,X) q ExtE (Z,X) → ExtE (Z q Z,X q X) ([f, g], [f 0, g0]) 7→ [f ⊕ f 0, g ⊕ g0] =: [f, g] ⊕ [f 0, g0]

is well-deﬁned. In fact, if the diagrams

f g f 0 g0 X / Y / Z and X / Y 0 / Z

α β / / / / X v1 V v2 ZX w1 W w2 Z

commute, then the diagram

f⊕f 0 g⊕g0 X q X / Y q Y 0 / Z q Z

α⊕β X q X / V q W / Z q Z v1⊕w1 v2⊕w2

also commutes.

Deﬁnition 6.9 (Baer-Sum). Let (C, E) be an exact category and let X and Z be objects of C. From 6.8 and 6.4 it follows that the mapping

1 1 1 +: ExtE (Z,X) q ExtE (Z,X) → ExtE (Z,X) 0 0 0 0 ([f, g], [f , g ]) 7→ ∆X ([f, g] ⊕ [f , g ])∇Z is well-deﬁned. [f, g] + [f 0, g0] is called the Baer-sum of [f, g] and [f 0, g0].

Proposition 6.10. Let (C, E) be an exact category, let X and Z be objects 0 0 1 0 0 0 0 of C,[f, g], [f , g ] ∈ ExtE (Z,X) and let α, α : X → X and γ, γ : Z → Z be morphisms. Then:

i) (α ⊕ α0)([f, g] ⊕ [f 0, g0]) = (α[f, g]) ⊕ (α0[f 0, g0])

ii) (α + α0)[f, g] = α[f, g] + α0[f, g]

iii) α([f, g] + [f 0, g0]) = α[f, g] + α[f 0, g0]

iv) ([f, g] ⊕ [f 0, g0])(γ ⊕ γ0) = ([f, g]γ) ⊕ ([f 0, g0]γ0)

v) [f, g](γ + γ0) = [f, g]γ + [f, g]γ0

vi) ([f, g] + [f 0, g0])γ = [f, g]γ + [f 0, g0]γ 6.1. YONEDA-EXT1 187

Proof. iv) − vi) are the dual statements of i) − iii), so it suﬃces to show these. To show i) we ﬁrst construct the pushout (S, sX0 , sY ) of f and α and 0 0 the pushout (T, tX0 , tY 0 ) of f and α . Then we have a commutative diagram

f⊕f 0 g⊕g0 X q X / Y q Y 0 / Z q Z

0 α⊕α (1) sY ⊕tY 0 X0 q X0 / S q T / Z q Z sX0 ⊕tX0 c1⊕c2 where c1 is uniquely determined by g = c1 ◦sY and c2 is uniquely determined 0 by g = c2 ◦ tY 0 . The square (1) is a pushout by 3.12. Then i) follows from the above diagram. ii) Since the diagram

d g X / Y / Z

∆X ∆Y ∆Z X q X / Y q Y / Z q Z f⊕f g⊕g commutes we have the commutative diagram

d g X / Y / Z

∆X (1) λ1 e m X q X / D / Z

λ2 (2) ∆Z X q X / Y q Y / Z q Z f⊕f g⊕g by 3.22, where (1) and (2) are pullback as well as pushout squares and λ2 ◦ λ1 = ∆Y . Then ∆X [f, g] = [e, m] = [f ⊕ f, g ⊕ g]∆Z and it follows with the help of 6.8 and i) that 0 0 (α + α )[f, g] = ∇X0 ◦ (α ⊕ α ) ◦ ∆X [f, g] 0 = ∇X0 ◦ (α ⊕ α )[f ⊕ f, g ⊕ g]∆Z 0 = ∇X0 ((α[f, g]) ⊕ (α [f, g]))∆Z = α[f, g] + α0[f, g], which shows ii). iii) Again with the help of 6.8 and i) we see that 0 0 0 0 α([f, g] + [f , g ]) = α∇X ([f, g] ⊕ [f , g ])∆Z 0 0 = ∇X0 ◦ (α ⊕ α)([f, g] ⊕ [f , g ])∆Z 0 0 = ∇X0 ((α[f, g]) ⊕ (α[f , g ]))∆Z = α[f, g] + α[f 0, g0]. 188 CHAPTER 6. YONEDA-EXT-FUNCTORS

Remark and Definition 6.11. We introduce the notion of big abelian 1 group at this point in order to deal with the ExtE -functors. This is defined in the same way as an ordinary abelian group, except that the underlying class need not be a set. We get the quasicategory (AB) of big abelian groups and group morphisms. The notions of kernels, cokernels, images, kernel-cokernel pairs, etc. can be defined in (AB) as for ordinary abelian groups, in fact every result stated in this treatise for the category (Ab) has an equivalent formulation for the quasicategory (AB). Theorem 6.12. Let (C, E) be an exact category and X,Z ∈ Ob(C). The 1 Baer-sum makes ExtE (Z,X) a big abelian group. Proof. We begin with the associativity of the Baer-sum. We have

[f, g] + ([f 0, g0] + [f 00, g00]) 0 0 00 00 = [f, g] + ∇X ([f , g ] ⊕ [f , g ])∆Z 0 0 00 00 = ∇X ([f, g] ⊕ ∇X ([f , g ] ⊕ [f , g ])∆Z )∆Z 0 0 00 00 = ∇X ((idX ⊕∇X )([f, g] ⊕ ([f , g ] ⊕ [f , g ]))(idZ ⊕∆Z ))∆Z 0 0 00 00 = ∇X ◦ (idX ⊕∇X )([f, g] ⊕ ([f , g ] ⊕ [f , g ])(idZ ⊕∆Z ) ◦ ∆Z and

([f, g] + [f 0, g0]) + [f 00, g00] 0 0 00 00 = ∇X ◦ (∇X ⊕ idX )(([f, g] ⊕ [f , g ]) ⊕ [f , g ])(∆Z ⊕ idZ ) ◦ ∆Z .

Since in addition we have

[f, g] ⊕ ([f 0, g0] ⊕ [f 00, g00]) = ([f, g] ⊕ [f 0, g0]) ⊕ [f 00, g00], and simple matrix multiplication shows

∇X ◦ (∇ ⊕ idX ) = ∇X ◦ (idX ⊕∇X ) and (idZ ⊕∆Z ) ◦ ∆Z = (∆Z ⊕ idZ )∆Z the associativity follows. To show the commutativity we take a look at the commutative diagram

f⊕f 0 g⊕g0 X q X / Y q Y 0 / Z q Z 0 0 idX 0 idY 0 idZ idX 0 idY 0 idZ 0 X q X / Y 0 q Y / Z q Z f 0⊕f g0⊕g 6.1. YONEDA-EXT1 189 and the induced diagram

f⊕f 0 g⊕g0 X q X / Y q Y 0 / Z q Z

0 idX (1) λ1 idX 0 e m X q X / D / Z q Z

0 idZ λ2 (2) idZ 0 X q X / Y 0 q Y / Z q Z f 0⊕f g0⊕g of 3.22. Then (1) and (2) are pullback as well as pushout squares, which shows that

([f 0, g0] ⊕ [f, g]) 0 idZ = 0 idX ([f, g] ⊕ [f 0, g0]). idZ 0 idX 0

0 idX 0 idZ Additionally we have ∇X ◦ = ∇X and ◦ ∆Z = ∆Z , hence idX 0 idZ 0

0 0 0 0 [f, g] + [f , g ] = ∇X ([f, g] ⊕ [f , g ])∆Z 0 0 0 idZ = ∇X ([f, g] ⊕ [f , g ]) ◦ ∆Z idZ 0 0 idX 0 0 = ∇X ◦ ([f , g ] ⊕ [f, g])∆Z idX 0 0 0 = ∇X ([f , g ] ⊕ [f, g])∆Z = [f 0, g0] + [f, g].

We show that [ωX , πZ ] is the neutral element for the Baer-sum: Let [f, g] 1 be an element of ExtE (Z,X) and construct the commutative diagram

f g X / Y / Z . g (1) 0 idZ (id ) 0 Y idZ 0 ( f ) idZ ⊕g X / Z q Y / Z q Z O O 0 idX (2) 0 idZ 0 idX 0 f 0 idY X q X / (X q Z) q Y / Z q Z ωX ⊕f ( 0 idZ )⊕g

0 By 3.11 the pairs ( f , idZ ⊕g) and (ωX ⊕f, πZ ⊕g) are elements of E, hence by 3.12 and 3.13 the squares (1) and (2) are pullback as well as pushout squares. It follows that

[ωX , πZ ] + [f, g] = ∇X ([ωX , πZ ] ⊕ [f, g])∆Z 0 = [ , id ⊕g]∆ f Z Z = [f, g]. 190 CHAPTER 6. YONEDA-EXT-FUNCTORS

1 To show that every [f, g] ∈ ExtE (Z,X) possesses an inverse element with regard to the Baer-sum we construct the commutative diagram

f g X / Y / Z . 0 0 (1) ( g ) X / X q Z / Z ωX πZ

By 3.12 the square (1) is a pushout, hence 0[f, g] = [ωX , πZ ]. With the help of 6.10 we may write

idX [f, g] + (− idX )[f, g] + (idX +(− idX ))[f, g] = 0[f, g] = [ωX , πZ ], hence (− idX )[f, g] is an inverse element of idX [f, g = [f, g]. Proposition 6.13. Let (C, E) be an exact category. For X,Z ∈ Ob(C) the following are equivalent:

1 i) ExtE (Z,X) = 0

f g ii) Every sequence X / Y / Z with (f, g) ∈ E is split exact. Proof. This follows immediately from 3.8 and 6.12.

Remark and Deﬁnition 6.14. Let (C, E) be an exact category and let

f g X / Y / Z be an element of E. For A ∈ Ob(C) we have the mapping

1 δA : Hom(A, Z) → ExtE (A, X) γ 7→ [f, g]γ and by 6.10 we have

0 0 0 0 δA(γ + γ ) = [f, g](γ + γ ) = [f, g]γ + [f, g]γ = δA(γ) + δA(γ ), hence δA is a group morphism. If β : B → A is a morphism in C, then the diagram δA 1 HomC(A, Z) / ExtE (A, X)

1 HomC(β,Z) ExtE (β,X) 1 HomC(B,Z) / ExtE (B,X) δB commutes, i.e.

1 δ := (δA)A∈Ob(C) : Hom(−,Z) → ExtE (−,X) 6.1. YONEDA-EXT1 191 is a natural transformation. Additionally the commutative diagram

f g X / Y / Z

αX αY αZ X0 / Y 0 / Z0 f 0 g0 with (f 0, g0) ∈ E, induces a commutative diagram

δA 1 HomC(A, Z) / ExtE (A, X)

1 HomC(A,αZ ) ExtE (A,αX ) 0 1 0 HomC(A, Z ) 0 / ExtE (A, X ) δA by 3.22. δA is called the covariant connecting morphism in A relative to (f, g). By duality we get for B ∈ Ob(C) a group homomorphism

1 δB : Hom(X,B) → ExtE (Z,B) α 7→ α[f, g] so that 1 δ := (δB)B∈Ob(C) : Hom(X, −) → ExtE (Z, −) is a natural transformation and for every commutative diagram

f 0 g0 X0 / Y 0 / Z0

βX βY βZ X / 0 / 0 f Y g Z with (f 0, g0), the diagram

δB 1 HomC(X,B) / ExtE (Z,B)

1 HomC(βX ,B) ExtE (βZ ,B) 0 1 0 HomC(X ,B) 0 / ExtE (Z ,B) δB also commutes. δB is called the contravariant connecting morphism in B relative to (f, g). 192 CHAPTER 6. YONEDA-EXT-FUNCTORS

Lemma 6.15. Let (C, E) be an exact category and (f, g) ∈ E. For the diagram A γ f g X / Y / Z the following are equivalent:

i) [f, g]γ = 0 = [ωX , πA] ii) There is a morphism h: A → Y with γ = g ◦ h.

Proof. i) ⇒ ii) If we have a commutative diagram

ωX πA X / X q A / A

pY PB γ X / Y / Z f g it follows that γ = γ ◦ πAωA = g ◦ pY ◦ ωA hence γ = g ◦ h with h := pY ◦ ωA. ii) ⇒ i) Let k pA X / P / A

pY PB γ X / Y / Z f g be a commutative diagram. Then g◦h = γ◦idA, hence the universal property of the pullback yields a unique morphism λ: A → P with h = pY ◦ λ and idA = pA◦λ, hence pA is a retraction and we have [f, g]γ = [k, pA] = [ωX , πA] by 3.8.

By duality we get:

Lemma 6.16. Let (C, E) be an exact category and (f, g) ∈ E. For the diagram f g X / Y / Z

α C the following are equivalent:

i) α[f, g] = 0 = [ωX , πA] 6.1. YONEDA-EXT1 193

ii) There is a morphism h: Y → A with α = h ◦ f.

Theorem 6.17. Let (C, E) be an exact category and (f, g) ∈ E. Then for A ∈ Ob(C) the sequence

HomC(A,f) HomC(A,g) δA 1 0 / HomC(A, X) / HomC(A, Y ) / HomC(A, Z) / ExtE (A, X)

Ext1 (A,f) Ext1 (A,g) E 1 E 1 / ExtE (A, Y ) / ExtE (A, Z) is exact in (AB).

Proof. Since the functor HomC(A, −) is left exact, it only remains to be shown:

(1) HomC(A, g)(HomC(A, Y )) = ker(δA)

1 (2) δA(HomC(A, Z)) = ker(ExtE (A, f)) 1 1 1 (3) ExtE (ExtE (A, X)) = ker(ExtE (A, g))

By 6.15 we see that [f, g]g ◦ v = 0 for all v ∈ HomC(A, Y ), hence we have

HomC(A, g)(HomC(A, Y )) ⊆ ker(δA).

If [f, g]γ = 0 we have γ = g ◦ v for a v ∈ HomC(A, Y ) by 6.15, hence

HomC(A, g)(HomC(A, Y )) ⊇ ker(δA).

This shows (1). 1 For w ∈ HomC(A, Z) we have ExtE (A, f) ◦ δA(w) = f[f, g]w = 0, since 1 f[f, g] = 0, hence δA(HomC(A, Z)) ⊆ ker(ExtE (A, f)). 1 Let [r, s] ∈ ExtE (A, X) with f[r, s] = 0. Then we have the commutative diagram r s X / R / A

f PO β Y / Y q A / A ωY πA and it follows that πY ◦ β ◦ r = πY ◦ ωY ◦ f = f. Then the universal property of s gives rise to a unique morphism γ : A → Z with g ◦ πY ◦ β = γ ◦ s, hence the diagram r s X / R / A

α (1) γ X / Y / Z f g 194 CHAPTER 6. YONEDA-EXT-FUNCTORS with α := πY ◦ β is also commutative. By 3.13 the square (1) is a pullback 1 and thus [r, s] = [f, g]γ. This shows δA(HomC(A, Z)) ⊇ ker(ExtE (A, f)) and thus (2). Since 1 1 ExtE (A, g) ◦ ExtE (A, f)([r, s]) = g ◦ f[r, s] = 0[r, s] = 0 1 1 1 1 we have ExtE (ExtE (A, X)) ⊆ ker(ExtE (A, g)). Let [p, q] ∈ ExtE (A, Y ) with g[p, q] = 0, i.e. we have a commutative diagram

p q Y / T / A .

g PO β Z / Z q A / A ωZ πA

Let k : K → T be the kernel of πA ◦ β, then the universal property of k gives rise to a unique morphism r : X → K with k ◦ r = p ◦ f, hence the diagram

r q◦k X / K / A

f (1) k Y p / T q / A is commutative. Claim A. The morphism q ◦ k is a cokernel of r.

The morphism β is an admissible epimorphism by 3.15, hence πZ ◦ β is also an admissible epimorphism and thus (k, πZ ◦ β) ∈ E. Since the diagram f g X / Y / Z

r (1) p K / T / Z k πZ ◦β commutes, the square (1) is a pushout by 3.12. If then l : K → L is a morphism with l ◦ r = 0, the universal property of the pushout (1) gives rise to a unique morphism λ: T → L with l = λ ◦ k and λ ◦ p = 0. Then λ = µ ◦ q for a unique morphism µ: A → L by the universal property of q. Then l = λ◦k = µ◦q◦k. If µe: A → L is another morphism with l = µe◦q◦k, then λ ◦ k = µe ◦ q ◦ k and λ ◦ p = 0 = µe ◦ q ◦ p, hence it follows that λ = µe ◦ q and thus µ = µe which establishes the claim. Since k ◦ r = p ◦ f is an admissible monomorphism and r has a cokernel, it follows from 3.20 that r is an admissible monomorphism and thus (r, q ◦k) is an element of E. By 3.12 the square (1) is a pushout, hence f[r, q ◦k] = [p, q] 1 1 1 and thus ExtE (ExtE (A, X)) ⊇ ker(ExtE (A, g)) which shows (3) and thus the theorem. 6.2. YONEDA-EXTN 195

By duality we get:

Theorem 6.18. Let (C, E) be an exact category and (f, g) ∈ E. Then for B ∈ Ob(C) the sequence

HomC(g,B) HomC(f,B) δB 1 0 / HomC(Z,B) / HomC(Y,B) / HomC(X,B) / ExtE (Z,B)

Ext1 (g,B) Ext1 (f,B) E 1 E 1 / ExtE (Y,B) / ExtE (X,B) is exact in (AB).

6.2 Yoneda-Extn

Deﬁnition 6.19. Let (C, E) be an exact category and

fn f0 E : 0 / X / Yn−1 / ... / Y0 / Z / 0 an exact sequence. We then call n the length of E, X the left end of E and Z the right end of E. A morphism φ: E → E0 between exact sequences E and E0 of length n is a commutative diagram

fn f0 E : 0 / X / Yn−1 / ... / Y0 / Z / 0 .

φ α βn−1 β0 γ f 0 f 0 0 0 n 0 0 0 0 E : 0 / X / Y n−1 / ... / Y 0 / Z / 0

We say the morphism φ has fixed ends, if X = X0 and Z = Z0 as well as α = idX and γ = idZ . For X,Z ∈ Ob(C) we define M n(Z,X) to be the exact sequences of length n with right end Z and left end X. On M n(Z,X) we define the following equivalence relation:

0 0 E ∼ E :⇔ There is a sequence E = E0,E1,...,Ek−1,Ek = E of elements of M n(Z,X), so that for every 0 ≤ i ≤ k − 1

there is either a morphism Ei → Ei+1 with ﬁxed ends,

or a morphism Ei+1 → Ei with ﬁxed ends.

We deﬁne n n ExtE (Z,X) =: M (Z,X)/ ∼ n and denote by [E] the equivalence class of E in ExtE (Z,X). 196 CHAPTER 6. YONEDA-EXT-FUNCTORS

n Remark 6.20. If (C, E) is an exact category, the class ExtE (Z,X) need not be a set. However, if C is a small category or C has enough E-projective or n E-injective objects, then ExtE (Z,X) will be sets (see 6.44 at the end of this chapter).

Remark 6.21. If

fn f0 E : 0 / X / Yn−1 / ... / Y0 / Z / 0 is an exact sequence of length n with right end Z and

f 0 f 0 m 0 ... 0 0 0 F : 0 / Z / Ym−1 / / Y0 / Z / 0 is an exact sequence of length m with left end Z, then the composed sequence

f f f 0 ◦f f 0 f 0 n ... 1 m 0 0 m−1 ... 0 0 EF : 0 / X / / Y0 / Ym−1 / / Z / 0 ?? z= ?? zz ?? z 0 f0 ? zz fm ? zz Z

0 0 is an an exact sequence of length m + n, since (m1, f0), (fm, em−1) ∈ E and 0 fm ◦ f0 is an E-strict morphism. Especially: If

fn fn−1 f1 f0 E : 0 / X / Yn−1 / Yn−1 / ... / Y1 / Y0 / Z / 0 ; ? ww ~~ en−1 ww e1 ~~ wwmn−1 ~~ m1 ww ~~ In−1 I1 is the factorization of E with short exact sequences

fn en−1 En : 0 / X / Yn−1 / In−1 / 0

mi ei−1 Ei : 0 / Ii / Yi−1 / Ii−1 / 0 2 ≤ i ≤ n − 1

m1 f0 E1 : 0 / I1 / Y0 / Z / 0 we can write in the above notation:

E = EnEn−1 ...E2E1

Hence every exact sequence of length n is a composition of n sequences of length 1. 6.2. YONEDA-EXTN 197

Remark and Deﬁnition 6.22. Let (C, E) be an exact category, X,Z ∈ Ob(C) and

fn f0 E : 0 / X / Yn−1 / ... / Y0 / Z / 0 an exact sequence of length n. For a morphism γ : Z0 → Z we ﬁnd a commutative diagram

P / Z0 / 0 , > pZ0 ~~ ~~ ~~k ~~ p I1 Y0 PB γ ? @ ~~ @@ ~~ @@ ~~ e1 m1 @@ fn ~ f f0 / / Yn−1 / ... / Y / Y / / 0 E : 0 X 1 1 0 Z where f1 = m1 ◦e1 is the canonical factorization of the E-strict morphism f1, 0 0 (P, pY0 , pZ ) is the pullback of f0 and γ and (k, pZ ) ∈ E. Then the sequence

f k◦e p 0 n 1 Z 0 Eγ,P : 0 / X / Yn−1 / ... / Y1 / P / Z / 0 is also exact. If

f 0 f 0 0 0 n 0 ... 0 0 E : 0 / X / Yn−1 / / Y0 / Z / 0 is another exact sequence of length n with E ∼ E0, we obtain an exact sequence

0 0 0 f k ◦e q 0 0 n 0 ... 0 1 Z 0 Eγ,Q : 0 / X / Yn−1 / / Y1 / Q / Z / 0 by using a pullback construction as above (with (Q, q 0 , q 0 ) the pullback of Y0 Z 0 f0 and γ). 0 We want to show that Eγ,P ∼ Eγ,Q, and for that it suﬃces to show one can ﬁnd for all commutative diagrams of the form

b qZ0 B / Q / Z0 ? ? φ1 ?? }}} ?? }}}} 0 ? }}} b ?? }}} a pZ0 }}} A / P / Z0 γ @ @ @@ @@ @ pC @ qD γ 0 @ @@ a @@ @@ @@ @@ D / Z @@ > d | @ ~~ |||| @@ ~~ |||| @ ~ φ0 ||| @ ~~ |||| C c / Z 198 CHAPTER 6. YONEDA-EXT-FUNCTORS

0 0 with (a , c), (b , d) ∈ E, and where (P, pC , pZ0 ) is the pullback of c and γ and (Q, qD, qZ0 ) is the pullback of d and γ, a morphism λ: P → Q that complements the above diagram commutatively. Since in the above diagram we have γ ◦ pZ0 = c ◦ pC = d ◦ φ0 ◦ c the universal property of (Q, qD, qZ0 ) gives rise to a unique morphism λ: P → Q with 0 φ0 ◦ pC = qD ◦ λ and pZ0 = qZ0 ◦ λ . So it remains to be shown that λ ◦ a = b ◦ φ1 and for this it suﬃces that

a) qZ0 ◦ λ ◦ a = qZ0 ◦ b ◦ φ1

b) qD ◦ λ ◦ a = qD ◦ b ◦ φ1 because of the universal propert of the pullback. Since we have qZ0 ◦ λ ◦ a = pZ0 ◦ a = 0 and qZ0 ◦ b = 0, the assertion a) holds. Additionally we have

qD ◦ λ ◦ a = φ0 ◦ pC ◦ a 0 = φ0 ◦ a 0 = b ◦ φ1

= qD ◦ b ◦ φ1,

0 hence b) also holds. Thus it follows that Eγ,P ∼ Eγ,Q. The above shows, that the mapping

n n 0 ExtE (Z,X) → ExtE (Z ,X), [E] 7→ [Eγ,P ] =: [E]γ is well-deﬁned. By duality we ﬁnd a commutative diagram

fn fn−1 f0 E : 0 / X / Yn−1 / Yn−2 / ... / Y0 / Z / 0 DD z< DD zz e DD zzm n−1 DD zz n−1 s " z α PO Yn−1 In−1 y< yy yyc yy yy 0 / X0 / S sX0

0 for every morphism α: X → X , where fn−1 = mn−1 ◦ en−1 is the canonical 0 factorization of the E-strict morphism fn−1,(S, sYn−1 , sX ) is the pushput of fn and α and (sX0 , c) ∈ E. Then the sequence

s 0 m ◦c f 0 X n−1 0 ES,α : 0 / X / S / Yn−2 / ... / Y0 / Z / 0 is also exact and the mapping

n n 0 ExtE (Z,X) → ExtE (Z,X ), [E] 7→ [ES,α] =: α[E] is well-deﬁned. 6.2. YONEDA-EXTN 199

n Proposition 6.23. Let (C, E) be an exact category, [E] ∈ ExtE (Z,X) and α: X → X0, α0 : X0 → X00, γ : Z0 → Z, γ0 : Z00 → Z0 be morphisms. Then:

i) idX [E] = [E] = [E] idZ

ii) (α0 ◦ α)[E] = α0(α[E])

iii) [E](γ ◦ γ0) = ([E]γ)γ0

iv) (α[E])γ = α([E]γ)

Proof. i) is trivial, ii) follows from the commutative diagram

fn fn−1 f0 E : 0 / X / Yn−1 / Yn−2 / ... / Y0 / Z / 0 GG e O s GG n−1 α PO Yn−1 GG mn−1 GG sX0 c # X0 / S / In−1 ; ww α0 rS w PO ww0 ww c ww X00 / R rX00 and the transitivity of the pushout, and iii) is the dual statement of ii). iv) Clearly the assertion is true for n ≥ 3 since pullback and pushout then do no longer get ’into each other’s way’ and for n = 1 the assertion is true by 6.4.v), so it suﬃces to show the case n = 2. In this case we have the commutative diagrams

f2 f1 f0 E : 0 / X / Y1 / Y0 / Z / 0 @@ ~? @@ ~~ e1 @@ ~~m1 @ ~~ s α PO Y1 I1 ? ~~ ~~ ~~ c s 0 ~ f 0 X ~ 0 0 / X / S m ◦c / Y0 / Z / 0 @@ 1 ? O O @@ ~~ @@ ~~ c @@ ~~ m1 ~ p I1 Y0 PB γ @ @@ @@ k @@ @ H : 0 / X0 / S / P / Z0 / 0 sX0 k◦c pZ0 200 CHAPTER 6. YONEDA-EXT-FUNCTORS and f2 f1 f0 E : 0 / X / Y1 / Y0 / Z / 0 @@ ~? O O @@ ~~ e1 @@ ~~m1 @ ~~ p I1 Y0 PB γ @@ @@ @@ k @@ 0 0 / X / Y1 / P / Z / 0 f e ◦k pZ0 2 @@ 1 ~? @@ ~~ e @ ~~ 1 @@ ~~ k s ~ α PO Y1 I1 >> ~~ ~~ ~~ c ~~ H : 0 / X0 / S / P / Z / 0 sX0 k◦c pZ0 which show that (α[E])γ = [H] = α([E]γ). Notation 6.24. The above proposition allows us to write: α0 ◦ α[E] := (α0 ◦ α)[E] = α0(α[E]) [E]γ ◦ γ0 := [E](γ ◦ γ0) = ([E]γ)γ0 α[E]γ := (α[E])γ = α([E]γ) Remark 6.25. Let (C, E) be an exact category and A ∈ Ob(C). By 6.23 we have a covariant functor n n X 7→ ExtE (A, X) ExtE (A, −): C → (Class) , n , α 7→ ExtE (A, α) where n n n 0 ExtE (A, α): ExtE (A, X) → ExtE (A, X ), [E] 7→ α[E] and a contravariant functor n n Z 7→ ExtE (Z,A) ExtE (−,A): C → (Class) , n , γ 7→ ExtE (γ, A) where n n n 0 ExtE (γ, A): ExtE (Z,A) → ExtE (Z ,A), [E] 7→ [E]γ. Additionally 6.23.iv) shows that n op ExtE (−, −): C q C → (Class) is a bifunctor, i.e. for morphisms α: A → A0 and β : B0 → B the following diagram commutes:

Extn(B,α) n E n 0 ExtE (B,A) / ExtE (B,A )

n n 0 ExtE (β,A) ExtE (β,A ) n 0 n 0 0 ExtE (B ,A) n 0 / ExtE (B ,A ) ExtE (B ,α) 6.2. YONEDA-EXTN 201

Lemma 6.26. Let (C, E) be an exact category.

i) If E,E0 ∈ M n(Z,X then the sequence

f ⊕f 0 f ⊕f 0 0 n n 0 ... 0 0 0 E ⊕ E : 0 / X q X / Yn−1 q Yn−1 / / Y0 q Y0 / Z q Z / 0

is also exact.

ii) The mapping

n n n ⊕: ExtE (Z,X) q ExtE (Z,X) → ExtE (Z q Z,X q X) ([E], [E0]) 7→ [E ⊕ E0] =: [E] ⊕ [E0]

is well-deﬁned.

Proof. i) The sequence E ⊕ E0 has the following factorization

m ⊕m0 e ⊕e0 0 k+1 k+1 0 fk⊕fk 0 k−1 k−1 0 Ik+1 q Ik+1 / Yk q Yk / Yk−1 q Yk−1 / Ik−1 q Ik−1 KK p7 KKK ppp 0 KK pp 0 ek⊕ek K p mk⊕mk % 0 p Ik q Ik

0 0 0 0 for k = n−1,..., 0 (with In qIn = XqX, mn ⊕mn = fn ⊕fn, I0 qI0 = ZqZ 0 0 and e0 ⊕ e0 = f0 ⊕ f0). 0 0 0 0 By 3.11 we know that (fn ⊕ fn, en−1 ⊕ en−1), (m1 ⊕ m1, f0 ⊕ f0) and (mk ⊕ 0 0 mk, ek−1 ⊕ ek−1) for k = n − 1,..., 1 are elements of E, hence the sequence E ⊕ E0 is exact. ii) Let E,E0,E00,E000 ∈ M n(Z,X) be exact sequences with E ∼ E00 and E0 ∼ E000, i.e. there exist sequences

00 E = E0,E1,...,Ek−1,Ek = E 0 0 0 0 0 000 E = E0,E1,...,El−1,El = E of exact sequences of length n with morphisms with ﬁxed ends between them by deﬁnition 6.19. Then we have a sequence

0 0 0 0 0 00 0 E ⊕ E = E0 ⊕ E ,E1 ⊕ E ,...,Ek−1 ⊕ E ,Ek ⊕ E = E ⊕ E and a sequence

00 0 00 0 00 0 00 0 00 0 00 000 E ⊕ E = E ⊕ E0,E ⊕ E1,...,E ⊕ El−1,E ⊕ El = E ⊕ E with the induced morphisms between the steps of the sequence, hence it follows that E ⊕ E0 ∼ E00 ⊕ E000. 202 CHAPTER 6. YONEDA-EXT-FUNCTORS

Deﬁnition 6.27. Let (C, E) be an exact category and let X and Z be objects of C. From 6.26 and 6.22 it follows that the mapping

n n n +: ExtE (Z,X) q ExtE (Z,X) → ExtE (Z,X) 0 0 ([E], [E ]) 7→ ∆X ([E] ⊕ [E ])∇Z is well-deﬁned. [E] + [E0] is called the Baer-sum of [E] and [E0]. Remark and Notation 6.28. Let (C, E) be an exact category and let

hn h0 G: 0 / W / Un−1 / ... / U0 / X / 0

fn f0 E : 0 / X / Yn−1 / ... / Y0 / Z / 0

f 0 f 0 0 n 0 ... 0 0 E : 0 / X / Yn−1 / / Y0 / Z / 0

g g n 0 0 F : 0 / Z / Vn−1 / ... / V0 / Z / 0 be exact sequences with E ∼ E0. Since the morphisms of the relation ∼ have ﬁxed ends it follows that also EF ∼ E0F and GE ∼ GE0. If F 0 ∈ M k(Z,X) is another exact sequence of length k with F ∼ F 0, we have EF ∼ E0F ∼ E0F 0 hence [EF ] = [E0F 0]. We will write [E][F ] := [EF ].

n Lemma 6.29. Let (C, E) be an exact category and let [E], [E] ∈ ExtE (Z,X) with

[E] = [En][En−1] ... [Ei+1]([Ei]β)[Ei−1] ... [E2][E1] 0 [E ] = [En][En−1] ... [Ei+1][Ei](β[Ei−1]) ... [E2][E1] for an i ∈ {2, . . . , n}. Then [E] = [E0]. Proof. Because of 6.28 it is suﬃcient to show

([Ei]β)[Ei−1] = [Ei](β[Ei−1]). For this we consider the commutative diagrams

k pW 0 Ei,P,β : 0 / V / P / W 0 / 0

pYi PB β E : 0 / V / Y / W / 0 i f i g 6.2. YONEDA-EXTN 203 and

f 0 g0 Ei−1 : 0 / W 0 / Yi−1 / U / 0 . s β PO Yi−1 Ei−1,S,β : 0 / W / S / U / 0 sW c

0 Then we have pYi ◦ k = f, c ◦ sYi−1 = g and

0 0 0 sW ◦ g ◦ pYi = sW ◦ β ◦ pW = sYi−1 ◦ f ◦ pW hence the diagram

0 0 k f ◦pW 0 g Ei,P,βEi−1 : 0 / V / P / Yi−1 / U / 0 s pYi Yi−1 EiEi−1,S,β : 0 / V / Yi / S / U / 0 f sW ◦g c

is also commutative, which shows that Ei,P,βEi−1 ∼ EiEi−1,S,β. Then we have:

([Ei]β)[Ei−1] = [Ei,P,βEi−1]

= [EiEi−1,S,β]

= [Ei](β[Ei−1])

Remark 6.30. Let (C, E) be an exact category and let

fn f0 E : 0 / X / Yn−1 / ... / Y0 / Z / 0

φ φn−1 φ0

gn g0 F : 0 / X / Vn−1 / ... / V0 / Z / 0 be a morphism with ﬁxed ends. This commutative diagram induces the 204 CHAPTER 6. YONEDA-EXT-FUNCTORS following diagrams on the sequences of length 1:

fn en−1 X / Yn−1 / In−1 (1)

φn−1 λn−1 / V / J X gn n−1 0 n−1 en−1

mk ek−1 Ik / Yk / Ik−1 (2)

λk φk λk−1 Jk 0 / Vk 0 / Jk−1 mk ek−1

m1 f0 I1 / Y0 / Z (3)

λ1 φ0 J / V / 1 0 0 g0 Z m1

In the diagram (1) the morphism λn−1 is the induced morphism between the 0 cokernels of fn and gn with λn−1 ◦ en−1 = en−1 ◦ φn−1 and inductively we 0 ﬁnd morphisms λl between the cokernels of fl+1 and gl+1 with λl ◦el = el ◦φl for l = n − 2,..., 1. To show that the left hand squares of the diagrams (2) and (3) commute we consider the diagrams

fk Yk / Yk−1 @ = @@ zz @@ zz ek @@ zz mk @ zz Ik

φk λk φk−1 Jk e0 > DD m0 k ~~ D k ~~ DD ~~ DD ~~ D! V / Vk−1 k gk for k = n − 1,..., 2. Since we have

φk−1 ◦ mk ◦ ek = φk−1 ◦ fk

= gk ◦ φk 0 0 = mk ◦ ek ◦ φk 0 = mk ◦ λk ◦ ek 0 and ek is an epimorphism it follows that φk−1 ◦ mk = mk ◦ λk, which establishes the commutativity of (2) and (3). 6.2. YONEDA-EXTN 205

The commutative diagram (2) induces by 3.22 a commutative diagram

mk ek−1 Ik / Yk / Ik−1

0 λk (1) bk Jk / Dk / Ik−1 ik ck 00 b k (2) λk−1 Jk 0 / Vk 0 / Jk−1 mk ek−1 in C with exact rows, so that the diagrams (1) and (2) are pullback as well 00 0 as pushout squares and b k ◦ bk = φk. This shows that when we deﬁne

0 [En] := [fn, en−1] , [Fn] := [gn, en−1] 0 0 [Ek] := [mk, ek−1] , [Fk] := [mk, ek−1] for k = 2, . . . , n − 1 0 [E1] := [m1, f0] , [F1] := [m1, g0] we get the equations

[En] = [Fn]λn−1

λk = [Fk] for k = 2, . . . , n − 1

λ1[E1] = [F1] which allow us, with the help of 6.29 the following operations on [E]:

[E] = [En][En−1][En−2] ... [E2][E1]

= ([Fn]λn−1)[En−1][En−2] ... [E2][E1]

= [Fn](λn−1[En−1])[En−2] ... [E2][E1]

= [Fn]([Fn−1]λn−2)[En−2] ... [E2][E1] . .

= [Fn][Fn−1][Fn−2] ... ([F2]λ1)[E1]

= [Fn][Fn−1][Fn−2] ... [F2](λ1[E1])

= [Fn][Fn−1][Fn−2] ... [F2][F1] = [F ]

This shows that we can convert [E] into [F ] by a ﬁnite number of operations of the type of 6.29. n This generalizes to all [H], [G] ∈ ExtE (Z,X) with [H] = [G], so one could use it as an alternative deﬁnition of the equivalence relation ∼ on M n(Z,X).

Lemma 6.31. Let (C, E) be an exact category. The following equations hold whenever the terms are deﬁned: 206 CHAPTER 6. YONEDA-EXT-FUNCTORS

i) ([E] ⊕ [E0])([F ] ⊕ [F 0]) = [EF ] ⊕ [E0F 0]

ii) (α ⊕ α0)([E] ⊕ [E0]) = (α[E]) ⊕ (α0[E0])

iii) ([E] + [E0])[F ] = [EF ] + [E0F ]

iv) (α + α0)[E] = α[E] + α0[E]

v) α([E] + [E0]) = α[E] + α[E0]

vi) ([E] ⊕ [E0])(γ ⊕ γ0) = ([E]γ) ⊕ ([E0]γ0)

vii) [E]([F ] + [F 0]) = [EF ] + [EF 0] viii) [E](γ + γ0) = [E]γ + [E]γ0

ix) ([E] + [E0])γ = [E]γ + [E0]γ

Proof. Since vi) − ix) are the dual statements of ii) − v) it suﬃces to show i) − v). i) is a consequence of the diagram

... 0 (gm−1◦f0)⊕(gm−1◦f0) 0 ... / Y0 q Y0 / Vm−1 q Vm−1 / LLL oo7 LL ooo 0LL oo 0 f0⊕f0 LL ogm−1⊕gm−1 L% ooo W q W 0 8 P qq PPP qqq PPP qqq PPP qq PPP 0 qq P( 0 being commutative. ii) The diagram

0 f ⊕f 0 fn−1⊕f 0 n n 0 n−1 0 ... 0 / X q X / Yn−1 q Yn−1 / Yn−2 q Yn−2 / PP e ⊕e0 n7 PPn−1 n−1 nn PPP nnn PP mnn ⊕m0 PP nn n−1 n−1 s ⊕s0 ' 0 n α⊕α0 (1) Yn−1 0 Yn−1 In−1 q In−1 nn6 nnn nnn 0 nnn c⊕c 00 000 0n 0 / X q X 0 / S q S s 00 ⊕s X X000 is commutative and it follows from the universal property of the two pushouts and the universal property of the coproduct that the square (1) is a pushout, hence the assertion holds. 6.2. YONEDA-EXTN 207 iii) For exact sequences [f, g] of length 1 the commutative diagram

d g X / Y / Z

∆X ∆Y ∆Z X q X / Y q Y / Z q Z f⊕f g⊕g together with 3.22 shows that

∆X [f, g] = ([f, g] ⊕ [f, g])∆Z .

Write

[E] = [En] ... [E1] 0 0 0 [E ] = [En] ... [E1] [F ] = [Fm] ... [F1] then we see with the help of the above equation, 6.29 and ii) that:

0 0 ([E] + [E ])[F ] = (∇X [E ⊕ E ]∆Z )[F ] 0 0 = ∇X ([En] ⊕ [En]) ... (([E1] ⊕ [E1])∆Z )[Fm] ... [F1] 0 0 = ∇X ([En] ⊕ [En]) ... ([E1] ⊕ [E1])(∆Z )[Fm]) ... [F1] 0 0 = ∇X ([En] ⊕ [En]) ... ([E1] ⊕ [E1])(([Fm] ⊕ [Fm])∆Z ) ... [F1] . . 0 0 = ∇X ([En] ⊕ [En]) ... ([E1] ⊕ [E1])([Fm] ⊕ [Fm]) ... ([F1] ⊕ [F1])∆Z 0 0 = ∇X ([En] ... [E1][Fm] ... [F1] ⊕ [En] ... [E1][Fm] ... [F1])∆Z 0 = ∇X ([EF ] ⊕ [E F ])∆Z = [EF ] + [E0F ] iv) From 6.10.ii) and iii) it follows that

0 0 (α + α )[E] = (α + α )[En] ... [E1] 0 = (α[En] + α [En])[En−1] ... [E1] 0 = α[En][En−1] ... [E1] + α [En][En−1] ... [E1] = α[E] + α0[E]. v) By 6.8 we know that for α: X → X0 we have

α ◦ ∇X = ∇X0 ◦ (α ⊕ α), 208 CHAPTER 6. YONEDA-EXT-FUNCTORS hence it follows with the help of i):

0 0 α([E] + [E ]) = α(∇X ([E] ⊕ [E ])∆Z ) 0 = (α ◦ ∇X )(([E] ⊕ [E ])∆Z ) 0 = (∇X0 ◦ (α ⊕ α))(([E] ⊕ [E ])∆Z ) 0 = ∇X0 ((α ⊕ α)([E] ⊕ [E ]))∆Z ) 0 = ∇X0 (α[E] ⊕ α[E ])∆Z ) = α[E] + α[E0].

Theorem 6.32. Let (C, E) be an exact category and X,Z ∈ Ob(C). The n n Baer-sum makes ExtE (Z,X) a big abelian groupExtE (Z,X). Proof. We start with the associativity. By 6.31.ii)+vi) we have

([E] + [F ]) + [G] = ∇X ([E] ⊕ [F ])∆Z + [G]

= ∇X (∇X ([E] ⊕ [F ])∆Z ⊕ [G])∆Z

= ∇X (∇X ⊕ idX )(([E] ⊕ [F ]) ⊕ [G])(∆Z ⊕ idZ )∆Z and

[E] + ([F ] + [G]) = [E] + ∇X ([F ] ⊕ [G])∆Z

= ∇X ([E] ⊕ ∇X ([F ] ⊕ [G])∆Z )∆Z

= ∇X (idX ⊕∇X )([E] ⊕ ([F ] ⊕ [G]))(idZ ⊕∆Z )∆Z and since matrix multiplication shows that

∇X (∇X ⊕ idX ) = ∇X (idX ⊕∇X )

(∆Z ⊕ idZ )∆Z = (idZ ⊕∆Z )∆Z and additionally the direct sum is associative, it follows that

([E] + [F ]) + [G] = [E] + ([F ] + [G]).

Next we will show the commutativity. Deﬁne

0 idX 0 idZ τX := , τZ := . idX 0 idZ 0 0 In the proof of 6.12 we have shown that for exact sequences [Ek], [Ek] of length 1 we have

0 0 τX ([Ek] ⊕ [Ek]) = ([Ek] ⊕ [Ek])τZ 6.2. YONEDA-EXTN 209 and additionally matrix multiplication yields

∇X ◦ τX = ∇X , τZ ◦ ∆Z = ∆Z .

Together with 6.31.i) and 6.29 it follows then that

0 0 [E] + [E ] = ∇X ([E] ⊕ [E ])∆Z 0 0 0 = ∇X ◦ τX ([En] ⊕ [En])([En−1] ⊕ [En−1]) ... ([E1] ⊕ [E1])∆Z 0 0 0 = ∇X (([En] ⊕ [En])τZ )([En−1] ⊕ [En−1]) ... ([E1] ⊕ [E1])∆Z 0 0 0 = ∇X ([En] ⊕ [En])(τZ ([En−1] ⊕ [En−1])) ... ([E1] ⊕ [E1])∆Z . = . 0 0 0 = ∇X ([En] ⊕ [En])([En−1] ⊕ [En−1]) ... ([E1] ⊕ [E1])τZ ◦ ∆Z 0 = ∇X ([E ] ⊕ [E])∆Z = [E0] + [E].

We will now show that the exact sequence

idX idZ E0 : 0 / X / X / 0 / ... / 0 / Z / Z / 0 acts as a neutral element for the Baer-sum. This follows from the fact that for an exact sequence

fn fn−1 f0 E : 0 / X / Yn−1 / Yn−2 / ... / Y0 / Z / 0 of length n the diagrams

idX ⊕fn ( 0 fn−1 ) E0 ⊕ E : 0 / X q X / X q Yn−1 / Yn−2 / ... KK x; KK( 0 en−1 ) xx KKK x KK xxmn−1 K% xx ∇X (1) ( fn idYn−1 )In−1 ss9 sss sssen−1 ss 0 / X / Yn−1 fn and Y0 / Z / 0 ; f0 m1 xx xx xx xx x f0 I1 (id ) (2) ∆Z ? F Y0 e ~ FF 1 ~~ FF ~ 0 FF ~~ ( ) FF ~ m1 # idZ ⊕f0 ... / Y1 / Z q Y0 / Z q Z / 0 ( 0 ) f1 210 CHAPTER 6. YONEDA-EXT-FUNCTORS are commutative, as simple calculations show. Hence the square (1) is a pushout by 3.12 and the square (2) is a pullback by 3.13. It follows that

[E0] + [E] = ∇X ([E0 ⊕ E])∆Z = [E], and therefore [E0] is the neutral element of the Baer-sum. For the additive inverse we consider the commutative diagram

fn fn−1 E : 0 / X / Yn−1 / Yn−2 / ... K ; KK xx KK xx en−1 KK xmn−1 KK xx 0 % x 0 PO (e ) In−1 n−1 9 ss ss πs ss In−1 ss 0E : 0 / X / X q In−1 / Yn−2 / ... ωX en−1

πX E0 : 0 / X / X / 0 / ... idX that also commutes on the right end, and which shows that

[E0] = 0[E] = (idX +(− idX ))[E] = [E] + (− idX )[E], hence (− idX )[E] is an additive inverse for [E].

Remark 6.33. Let (C, E) be an exact category, X, W, Z ∈ Ob(C) and m, n ∈ N with m, n ≥ 1. Then the pairing n m m+n σ : ExtE (W, X) q ExtE (Z,W ) → ExtE (Z,X) ([E], [F ]) 7→ [E][F ] is bilinear in the sense that we have

σ([E] + [E0], [F ]) = ([E] + [E0])[F ] = σ([E], [F ]) + σ([E0], [F ]) σ(α[E], [F ]) = (α[E])[F ] = α[E][F ] = α σ([E], [F ]) σ([E], [F ] + [F 0]) = [E]([F ] + [F 0]) = σ([E], [F ]) + σ([E], [F 0]) σ([E], [F ]γ) = [E]([F ]γ) = [E][F ]γ = σ([E], [F ]) γ by 6.31. The bilinearity of the above directly implies:

n Corollary 6.34. Let (C, E) be an exact category and [E] ∈ ExtE (Z,X). If [E] can be written as [E] = [G][H] with either [G] = 0 or [H] = 0, then [E] = 0. 6.2. YONEDA-EXTN 211

Corollary 6.35. Let (C, E) be an exact category and A ∈ Ob(C).

n0 i) If ExtE (A, X) = 0 for all X ∈ Ob(C) and for an n0 ≥ 1, then we have n ExtE (A, X) = 0 for all X ∈ Ob(C) and all n ≥ n0.

n0 ii) If ExtE (Z,A) = 0 for all Z ∈ Ob(C) and for an n0 ≥ 1, then we have n ExtE (Z,A) = 0 for all Z ∈ Ob(C) and all n ≥ n0.

n Proof. i) For [E] ∈ ExtE (A, X) with [E] = [En] ... [En0+1][En0 ][En0−1] ... [E1] we have [En0 ][En0−1] ... [E1] = 0 by assumption. Hence [E] = 0 by 6.34. Analogously one shows ii).

Corollary 6.36. Let (C, E) be an exact category and let A ∈ Ob(C).

n i) The functor ExtE (A, −) is eﬀacable for n ≥ 1.

n ii) The functor ExtE (−,A) is coeﬀacable for n ≥ 1.

Proof. It suﬃces to show i), since ii) is the dual statement. If I is an E- injective objective object of C every kernel-cokernel pair I / V / V is n split exact by 5.33. For [E] ∈ ExtE (A, I) with [E] = [En][En−1] ... [E2][E1] it follows that we have always [En] = 0 and therefore [E] = 0 by 6.34. This n shows that ExtE (A, I) = 0.

Theorem 6.37 (Schanuel-Lemma). Let (C, E) be an exact category and r s [E] ∈ ExtE (W, X), [F ] ∈ ExtE (Z,W ). Then the following are equivalent:

i) [E][F ] = 0

0 r 0 ii) There is a morphism φ: W → W and a [G] ∈ ExtE (W ,X) with [E] = [G]φ and φ[F ] = 0.

iii) There is a morphism ψ : W 0 → W with [F ] = ψ[H] and [E]ψ = 0.

Proof. ii) ⇒ i) Since φ[F ] = 0 we have

[E][F ] = ([G]φ)[F ] = [G](φ[F ]) = 0 by 6.34. iii) ⇒ i) Since [E]ψ = 0 we have

[E][F ] = [E](ψ[H])([E]ψ)[H] = 0 by 6.34. 212 CHAPTER 6. YONEDA-EXT-FUNCTORS

As a next step we will show ii) ⇒ iii) in the case r = s = 1, hence we have [E] = [fE, gE], [F ] = [fF , gF ] and [G] = [fG, gG] with exact sequences

fE gE E : 0 / X / Y1 / W / 0 ,

fF gF F : 0 / W / Y0 / Z / 0

fG gG G: 0 / X / Ve / W / 0

0 0 as well as φ: W → W . Since φ[F ] = 0 we ﬁnd h: Y0 → W with φ = h ◦ fF by 6.15, hence we have

[E] = [G]φ = [G](h ◦ fF ).

Since the diagram

fF gF W / Y0 / Z

fF ωY Y Y q Z 0 id / 0 / Z Y0 O ( gF idZ ) ( −g ) F idY0 0 −gF idZ Y / Y q Z / Z 0 ωY 0 πZ commutes, we have fF [F ] = 0. By substituting [G]h for [G] we can thus assume that fF = φ. Since [E] = [G]fF the diagram

0 0

fE gE E : 0 / X / Y1 / W / 0

p (1) fF G: 0 / X / V g / Y0 / 0 fG G q gF Z Z

0 0 commutes and (1) is a pullback as well as a pushout square by 3.13, we know by 3.14 that p is an admissible monomorphism, hence (p, q) ∈ E. It follows that gE[H] = [F ], which was what we wanted. Since iii) ⇒ ii) is the dual argument of ii) ⇒ iii), we have also shown 6.2. YONEDA-EXTN 213 iii) ⇒ ii) in the case r = s = 1. As a next step we will show i) ⇒ ii) in the case r = s = 1. By 6.30 we can transform the sequence EF by a ﬁnite number of operations of the type described in 6.29 into the zero sequence. We will show i) ⇒ ii) in the case r = s = 1 by induction on the minimal number N of such transformations. If EF is already 0, then ii) already holds with G = E and φ = id0. In addition iii) holds with H = F and ψ = id0. If N > 0, the following cases are possible:

a) E = E0η, so that E0(ηF ) needs one less operation of the type discussed in 6.29 to be transformed into 0 than the sequence EF .

b) F = ηF 0, so that (E0η)F ) needs one less operation of the type discussed in 6.29 to be transformed into 0 than the sequence EF .

In the case of a) the inductive assumption gives E0 = Gφ0 with φ0(ηF ) = 0, hence we have E = E0η = G(φ0 ◦ η) and (φ0 ◦ η)F = 0 which shows ii). In the case of b) the inductive assumption gives F 0 = ψ0H with (E0η)ψ0 = 0. Then F = ηF 0 = (η ◦ ψ0)H and E0(η ◦ ψ0) = 0, which shows iii). Since ii) and iii) are equivalent, we have thus shown the equivalence of i), ii) and iii) in the case r = s = 1. Now we will show the general case by induction on r + s: We assume i) ⇔ ii) ⇔ iii) in the case r + s < n for n > 2 and deﬁne r + s = n. ii) ⇒ iii) We have [E] = [G]φ with φ[F ] = 0. For s > 1 we write F = F 1F s−1, where F 1 has length 1 and F s−1 has length s − 1. Then φ[F 1][F s−1] = 0, hence the inductive assumption yields a [Hs−1] and an η with [F s−1] = η[Hs−1] and φ[F 1]η = 0, hence we have [E][F 1]η = ([G]φ)[F 1]η = 0. Again the inductive assumption gives rise to a [H1] and a ψ with [F 1]η = ψ[H1] and [E]ψ = 0. It follows that

[F] = [F 1][F s−1] = [F 1]η[Hs−1] = ψ[H1][Hs−1] hence iii) holds with [H] := [H1][Hs−1] and ψ. In the case s = 1, we have r > 1 and thus can write [G]φ = [Gr−1][G1], where [Gr−1] has length r − 1 and [G1] has length 1. The case r = s = 1 then yields a [H1] and a ψ with [F ] = ψ[H1] and [G1]φ ◦ ψ. Then iii) also holds with these two. iii) ⇒ ii) is the dual argument of ii) ⇒ iii). i) ⇒ ii) + iii) We show this direction again by induction on the minimal number N of operations of the type discussed in 6.29 required to transform EF into 0. If EF is already 0, we have shown above that ii) and iii) hold. Let then EFg be sequence that requires one less operation than EF 214 CHAPTER 6. YONEDA-EXT-FUNCTORS to be transformed into 0. If this operation takes place solely in E or F , i.e. EFg = E0F or EFg = EF 0, then the inductive assumption already yields the result. Hence it is enough to distinguish the following cases:

a) EFg = E0(ηF ) with E = E0η.

b) EFg = (Eη)F 0 with F = ηF 0. In the ﬁrst case the inductive assumption yields [E0] = [G]φ0 with (φ0 ◦η)[F ]. If we deﬁne φ := φ0 ◦ η it follows that

[E] = [E0]η = [G](φ0 ◦ η) = [G]φ, hence ii) holds. In the second case we can, with the help of the inductive assumption, write [F 0] = ψ0[H] with [E](η ◦ ψ0) = 0. Hence, if we deﬁne ψ := η ◦ ψ0, we have

[F] = η[F 0] = (η ◦ ψ0)[H] = ψ[H], which shows iii). Since ii) and iii) are equivalent this ﬁnishes the proof of the theorem.

n Theorem 6.38. Let (C, E) be an exact category and [E] ∈ ExtE (Z,X). Then the following are equivalent:

i) [E] = 0

ii) There is an exact sequence F ∈ M n(Z,X) and morphisms

0 / FEo

with ﬁxed ends.

iii) There is an exact sequence G ∈ M n(Z,X) and morphisms

0 o G / E

with ﬁxed ends.

Proof. The implications ii) ⇒ i) and iii) ⇒ i) are trivial. i) ⇒ ii) Write [E] = [En][En−1] ... [E1] with

fn en−1 En : 0 / X / Yn−1 / In−1 / 0

mk ek−1 Ek : 0 / Ik / Yk−1 / Ik−1 / 0 for k = 2, . . . , n − 1

m1 f0 E1 : 0 / I1 / Y0 / Z / 0 . 6.2. YONEDA-EXTN 215

1 0 By 6.37 there is [Fn] ∈ ExtE (In−1,X) and a morphism

0 φn : In−1 → In−1 with [En] = [Fn]φn and φn[En−1] ... [E1] = 0. Again with 6.37 we ﬁnd 1 0 [Fn−1] ∈ ExtE (In−2, In − 1) and a morphism

0 φn−1 : In−1 → In−1 with φn[En−1] = [Fn−1]φn−1 and φn−1[En−2] ... [E1] = 0. Inductively we 1 0 ﬁnd [Fk] ∈ ExtE (Ik−1,Ik) and morphisms 0 φk : Ik−1 → Ik−1 with φk+1[Ek] = [Fk]φk and φk[Ek−1] ... [E1] = 0 for k = n − 2,..., 3. 1 0 Finally we ﬁnd [F2] ∈ ExtE (I2,I2) and a morphism

0 φ2 : I1 → I1 with φ3[E2] = [F2]φ2 and φ2[E1] = 0. Deﬁne then [F1] := φ2[E1] and [F ] := [Fn][Fn−1] ... [E2][E1]. Because of the equations

[En] = [Fn]φn

φk+1[Ek] = [Fk]φk for k = n − 1,..., 2

φ2[E1] = [F1] = 0 we ﬁnd commutative diagrams

fn en−1 En : 0 / X / Yn−1 / In−1 / 0

βn−1 φn Y 0 I0 Fn : 0 / X 0 / n−1 0 / n−1 / 0 fn en−1

mk ek−1 Ek : 0 / Ik / Yk−1 / Ik−1 / 0

φk+1 βk−1 φk 0 Y 0 I0 Fk : 0 / Ik 0 / k−1 0 / k−1 / 0 mk ek−1

m1 f0 E1 : 0 / I1 / Y0 / Z / 0

φ2 β0

0 0 F1 : 0 / I / I q Z / Z / 0 1 id 0 1 πZ ( I1 ) 0 216 CHAPTER 6. YONEDA-EXT-FUNCTORS with exact rows, which when composed give rise to a morphism

β : E → F := FnFn−1 ...F2F1 with ﬁxed ends. In addition the diagram

fn fn−1 f1 f0 E : 0 / X / Yn−1 / Yn−2 / ... / Y1 / Y0 / Z / 0

βn−1 βn−2 β1 β0 F : 0 / X / Y 0 / Y 0 / ... / Y 0 / I0 q Z / Z / 0 0 n−1 0 n−2 1 0 1 πZ fn f n−1 O e1 O O O ( 0 ) 0 fn ωZ

E0 : 0 / X / X / 0 / ... / 0 / Z / Z / 0 idX idZ commutes, which shows ii). i) ⇒ ii) is the dual argument of i) ⇒ ii).

Proposition 6.39. Let (C, E) be an exact category and let X and Z be two objects of C. For n ≥ 2 the following are equivalent: n i) ExtE (Z,X) = 0 ii) For every E ∈ M n(Z,X) there is a commutative diagram

fn fn−1 f1 f0 E : 0 / X / Yn−1 / Yn−2 / ... / Y1 / Y0 / Z / 0

φn−1 φn−2 φ1 φ0 / / Vn−1 / Vn−2 / ... / V / V / 0 F : 0 X gn gn−1 1 g1 0

n−1 with F ∈ M (V0,X). iii) For every E ∈ M n(Z,X) there is a commutative diagram

fn fn−1 f1 f0 E : 0 / X / Yn−1 / Yn−2 / ... / Y1 / Y0 / Z / 0

φn−1 φn−2 φ1 / / Vn−1 / Vn−2 / ... / V / Y / 0 F : 0 X gn gn−1 1 g1 0

n−1 with F ∈ M (Y0,X). n iv) For every E ∈ M (Z,X) there is a Vn−1 ∈ Ob(C) and a commutative diagram

gn−1 g1 g0 F : 0 / Vn−1 / Vn−2 / ... / V1 / V0 / Z / 0

φn−1 φn−2 φ1 φ0 E : 0 / X / Yn−1 / Yn−2 / ... / Y1 / Y0 / Z / 0 fn fn−1 f1 f0

n−1 with F ∈ M (Z,Vn−1). 6.2. YONEDA-EXTN 217

v) For every E ∈ M n(Z,X) there is a commutative diagram

gn g1 g0 F : 0 / Yn−1 / Vn−2 / ... / V1 / V0 / Z / 0

φn−2 φ1 φ0 E : 0 / X / Yn−1 / Yn−2 / ... / Y1 / Y0 / Z / 0 fn fn−1 f1 f0

n−1 with F ∈ M (Z,Yn−1). Proof. i) ⇒ ii) If E is an element of M n(Z,X), we have [E] = 0 by assumption, therefore by 6.38 there is a sequence

E / F o 0 of elements of M n(Z,X) and morphisms with ﬁxed ends, i.e. there is a commutative diagram

fn fn−1 f1 f0 E : 0 / X / Yn−1 / Yn−2 / ... / Y1 / Y0 / Z / 0

φn−1 φn−2 φ1 φ0 0 F : 0 / X / Wn−1 / Wn−2 / ... / W1 / W0 / Z / 0 gn O gn−1 O O g1 O g0 gn ψ0 0: 0 / X X / 0 / ... / 0 / Z Z / 0.

Therefore the morphism g0 is a retraction with right-inverse ψ0. Let then f f g g fk = mk ◦ ek and fk = mk ◦ ek be the canonical factorizations of the E-strict f morphisms fk and gk and let φek be the unique morphism with φk−1 ◦ mk = g f g mk ◦ φek and φek ◦ ek = ek ◦ φk+1 (see 6.30) for k = 1, . . . , n − 1. Then we have the commutative diagram

f f f m2 e1 m1 f0 0 / K2 / Y1 / K1 / Y0 / Z / 0

φe2 φ1 φe1 φ0 0 / K0 / W / K0 / W / / 0 2 g 1 g 1 g 0 g0 Z m2 e1 m1 induced by the upper right corner of the previous diagram, with f f f g g g (m2 , e1 ), (m1 , f0), (m2, e1), (m1, g0) ∈ E. Since g0 is a retraction, it follows g 0 that m1 is a coretraction by 3.8. Let then h: W0 → K1 be a morphism with g h ◦ m = id 0 and deﬁne µ := h ◦ φ . We have 1 K1 0 g f g f g g g m1 ◦ µ ◦ m1 = m1 ◦ h ◦ φ0 ◦ m1 = m1 ◦ h ◦ m1 ◦ φe1 = m1 ◦ φe1 f g and therefore µ ◦ m1 = φe1, since m1 is a monomorphism. Then the diagram

fn fn−1 f1 f0 E : 0 / X / Yn−1 / Yn−2 / ... / Y1 / Y0 / Z / 0

φn−1 φn−2 φ1 µ 0 0 / / Wn−1 / Wn−2 / ... / W / K / 0 F : 0 X gn gn−1 1 g 1 e1 218 CHAPTER 6. YONEDA-EXT-FUNCTORS is commutative, which shows ii). ii) ⇒ iii) Let

f f f m2 e1 m1 f0 0 / K2 / Y1 / K1 / Y0 / Z / 0 } ψe2 ψ1 }} ~}} ψ0 0 / K0 / V / V / 0 2 g 1 g1 0 m2

f f f g be the commutative diagram with (m2 , e1 ), (m1 , f0), (m2, g1) ∈ E induced by the right hand side of the diagram of the assumption and the canoncial factorizations of the E-strict morphisms of this diagram. Since g1 is an admissible epimorphism, the pullback (P, pY0 , pV1 ) of ψ0 and g1 does exist and by 3.13 we have a commutative diagram p 0 k Y0 K2 / P / Y0

pV1 ψ0 K0 / V / V 2 g 1 g1 0 m2 whose rows are elements of E. Since ψ0 ◦ f1 = g1 ◦ ψ1 the universal property of the pullback gives rise to a unique morphism λ: Y1 → P with f1 = pY0 ◦ λ and ψ1 = pV1 ◦ λ. Then we have also f f g pV1 ◦ λ ◦ m2 = ψ1 ◦ m2 = m2 ◦ ψe2 f f pY0 ◦ λ ◦ m2 = f1 ◦ m2 = 0 = pY0 ◦ k ◦ ψe2 f and therefore λ ◦ m2 = k ◦ ψe2 by the universal property of the pullback. This shows that the diagram

f m2 f1 K2 / Y1 / Y0

0 K2 / V1 / Y0 k pY0 is commutative and it ﬁts together with the rest of the diagram of the assumption to give the desired commutative diagram. n iii) ⇒ i) Let [E] be an element of ExtE (Z,X). If the diagram of the assumption is commutative, then the diagram

fn fn−1 f1 f0 E : 0 / X / Yn−1 / Yn−2 / ... / Y1 / Y0 / Z / 0 φ φn−1 φn−2 φ1 ( 0 ) 0 V V ... F : 0 / X g / n−1 g / n−2 / / V1 g / V0 q Z / Z / 0 n n−1 1 ( g0 1 ) O O O ( 0 ) O gn ωZ 0: 0 / X X / 0 / ... / 0 / Z Z / 0 6.2. YONEDA-EXTN 219 is also commutative, which shows that [E] = 0. By following the dual arguments of the proofs i) ⇒ ii) ⇒ iii) ⇒ i) one can analogously show i) ⇒ iv) ⇒ v) ⇒ i).

Corollary 6.40. Let (C, E) be an exact category and E,E0 ∈ M n(Z,X). Then the following are equivalent:

i) [E] = [E0]

n ii) There are exact sequences E1,E2 ∈ M (Z,X) and morphisms

0 E / E1 o E2 / E

with ﬁxed ends.

Proof. The implication ii) ⇒ i) is trivial. i) ⇒ ii) Since [E] = [E0] we have

0 [E] + (− idX )[E ] = 0.

By 6.38 there exist G, F ∈ M n(Z,X) and morphisms

0 0 o G / E + (− idX )E 0 0 0 / F o E + (− idX )E with ﬁxed ends. Then we also have morphisms

0 0 0 0 0 ⊕ E o G ⊕ E / (E + (− idX )E ) ⊕ E 0 0 E ⊕ 0 / E ⊕ F o E ⊕ (E + (− idX )E ) with ﬁxed ends. On these two equations we can operate with ∇X from the left and with ∆Z from the right to get

0 0 0 0 0 + E o G + E / (E + (− idX )E ) + E 0 0 E + 0 / E + F o E + (E + (− idX )E ) .

The proof of 6.32 shows

0 + E0 = E0 E + 0 = E 0 0 0 0 (E + (− idX )E ) + E = E + (E + (− idX )E ),

0 which proofs the assertion if we deﬁne E1 := E + F and E2 := G + E . 220 CHAPTER 6. YONEDA-EXT-FUNCTORS

Remark and Deﬁnition 6.41. Let (C, E) be an exact category and

f g E : X / Y / Z be an element of E. For A ∈ Ob(C) we have for n ≥ 1 the mapping

n n+1 δ := δ(n, E, A): ExtE (A, Z) → ExtE (A, X) [F ] 7→ [E][F ] and by 6.31 we have

0 0 0 0 δ([F ] + [F ]) = [E]([F ] + [F ]) = [E][F ] + [E][F ] = δ([F ]) + δA([F ]), hence δ is a group morphism. If α: A → B is a morphism in C, then the diagram

n δ(n,E,B) n+1 ExtE (B,Z) / ExtE (B,X)

Extn(α,Z) n+1 E ExtE (α,X) Extn(A, Z) / Extn(A, X) E δ(n,E,A) E commutes, i.e.

n n+1 δ := (δ(n, E, A))A∈Ob(C) : ExtE (−,Z) → ExtE (−,X) is a natural transformation. Additionally the commutative diagram

f g X / Y / Z

αX αY αZ X0 / Y 0 / Z0 f 0 g0 with (f 0, g0) ∈ E, induces a commutative diagram

n δ(n,E,A) n+1 ExtE (A, Z) / ExtE (A, X)

Extn(A,α ) n+1 E Z ExtE (A,αX ) Extn(A, Z0) / Extn+1(A, X0) E δ(n,E0,A) E by 3.22. δ(n, E, A) is called the covariant connecting morphism of degree n in A relative to [E]. By duality we get for B ∈ Ob(C) and n ≥ 1 a group homomorphism

0 0 n n+1 δ := δ (n, E, B): ExtE (X,B) → ExtE (Z,B) [H] 7→ [H][E] 6.2. YONEDA-EXTN 221 so that

0 0 n n+1 δ := (δ (n, E, B))B∈Ob(C) : ExtE (X, −) → ExtE (Z, −) is a natural transformation and for every commutative diagram

f 0 g0 X0 / Y 0 / Z0

βX βY βZ X / 0 / 0 f Y g Z with (f 0, g0) ∈ E, the diagram

0 n δ (n,E,B n+1 ExtE (X,B) / ExtE (Z,B)

Extn(β ,B) n+1 E X ExtE (βZ ,B) Extn(X0,B) / Extn+1(Z0,B) E δ0(n,E,B) E also commutes. δ0(n, E, B) is called the contravariant connecting morphism of degree n in B relative to [E].

Theorem 6.42 (Covariant Long Exact Sequence). Let (C, E) be an exact category and

f g E : 0 / X / Y / Z / 0 an element of E. Then for n ≥ 1 and A ∈ Ob(C) the sequence

Extn(A,f) Extn(A,g) n−1 δ n E n E n δ n+1 ExtE (A, Z) / ExtE (A, X) / ExtE (A, Y ) / ExtE (A, Z) / ExtE (A, X) is exact in (AB)..

Proof. Since f[E] = 0, g ◦ f = 0 and [E]g = 0 we have

n ExtE (A, f) ◦ δ(n − 1,E,A) = 0 n n ExtE (A, g) ◦ ExtE (A, f) = 0 n δ(n, E, A) ◦ ExtE (A, g) = 0. Thus it remains to be shown:

n n−1 (1) For [G] ∈ ExtE (A, X) with f[G] = 0, there exists [F ] ∈ ExtE (A, Z) with [G] = [E][F ].

n n (2) For [H] ∈ ExtE (A, Y ) with g[H] = 0, there exists [G] ∈ ExtE (A, X) with [H] = f[G]. 222 CHAPTER 6. YONEDA-EXT-FUNCTORS

n n (3) For [K] ∈ ExtE (A, Z) with [E][K] = 0, there exists [H] ∈ ExtE (A, Y ) with [K] = g[H].

n We will ﬁrst show the assertion (3): If we have [K] ∈ ExtE (A, Z) with 0 n 0 [E][K] = 0, then by 6.37 we ﬁnd [H ] ∈ ExtE (A ,Z) and a morphism ψ : Z → Z0 with [K] = ψ[H0] and [E]ψ = 0. Then it follows from 6.15 that we have an h: Z0 → Y with ψ = g ◦ h, hence

[K] = ψ[F ] = g(h[H0]),

0 n with h[H ] ∈ ExtE (A, Y ). The case n = 1 of the assertions (1) and (2) has already been proven in 6.17, so it suﬃces to show the case n ≥ 1 in both cases. n For [G] ∈ ExtE (A, X) with n ≥ 1 and f[G] = 0 we write [G] = [Gn][Gn−1] 1 with [Gn] ∈ ExtE (In−1,X). Then we have f[Gn][Gn−1] = 0 with

fn en−1 Gn : 0 / X / Yn−1 / In−1 / 0 . s f PO Yn−1 fGn : 0 / Y / S / In−1 / 0 sY c

If we apply (3) with regard to f[Gn] instead of [E], we ﬁnd [Fn−1] ∈ n−1 ExtE (A, S) with [Gn−1] = c[Fn−1], because of f[Gn][Gn−1] = 0. It follows that [G] = [Gn](c[Fn−1]) = ([Gn]c)[Fn−1]. 1 Since [Gn]c ∈ ExtE (A, Z) we have f[Gn]c = 0. Then we ﬁnd a morphism φ: S → Z with [Gn]c = [E]φ by the case n = 1 of (1). Thus we have

[G] = [E](φ[Fn−1])

n−1 with φ[Fn−1] ∈ ExtE (A, Z), which shows (1). n For [H] ExtE (A, Y ) with n ≥ 1 and g[H] = 0 we write [H] = [Hn][Hn−1] 1 0 with [Hn] ∈ ExtE (In−1,Y ). Then we have g[Hn][Hn−1] = 0 with

e0 hn n−1 0 Hn : 0 / Y / Hn−1 / In−1 / 0 .

g PO tYn−1

0 gHn : 0 / Z / T / In−1 / 0 tY c0

If we apply, as above, (3) with regard to g[Hn] instead of [E], we ﬁnd n−1 0 0 [Gn−1] ∈ ExtE (A, T ) with [Hn−1] = c [Gn−1]. Then [H] = ([Hn]c )[Gn−1] 0 1 0 with [Hn]c ∈ ExtE (S, Y ). Therefore we have g([Hn]c ) = 0 and if we apply 6.2. YONEDA-EXTN 223

1 0 the case n = 1 of (2), we ﬁnd [Gn] ∈ ExtE (S, X) with [Hn]c = f[Gn]. Then we have [H] = (f[Gn])[Gn−1] = f([Gn][Gn−1]) n with [Gn][Gn−1] =: [G] ∈ ExtE (A, X), which shows (2) and thus ﬁnishes the proof of the theorem.

By duality we obtain: Theorem 6.43 (Contravariant Long Exact Sequence). Let (C, E) be an exact category and

f g E : 0 / X / Y / Z / 0 an element of E. Then for n ≥ 1 and B ∈ Ob(C) the sequence

0 Extn(g,B) Extn(f,B) 0 n−1 δ n E n E n δ n+1 ExtE (X,B) / ExtE (Z,B) / ExtE (Y,B) / ExtE (X,B) / ExtE (Z,B) is exact in (AB). Remark 6.44. Let (C, E) be an exact category and

f g E : 0 / X / Y / Z / 0 an element of E. By combining 6.17 with 6.42 we get the long covariant exact sequence

1 0 → HomC(A, X) → HomC(A, Y ) → HomC(A, Z) → ExtE (A, X)

1 1 2 → ExtE (A, Y ) → ExtE (A, Z) → ExtE (A, X) → ... for every A ∈ Ob(C). Deﬁne

0 ExtE (A, −) := HomC(A, −), then it follows from the above and the naturality of the connecting morphisms 6.41 that n ExtE (A, −) := (ExtE (A, −))n≥0 is an exact cohomological δE -functor. If the category C has enough E- injective objects, it follows from 5.74 that ExtE (A, −) is even an universal n cohomological δE -functor, since ExtE (A, −) is eﬀacable for n > 0 by 6.36. Then it follows from 5.77 that there is an isomorphism of δE -functors ∼ ExtE (A, −) = R HomC(A, −).

Dually to the above we get by combining 6.18 with 6.43 the long contravariant exact sequence 224 CHAPTER 6. YONEDA-EXT-FUNCTORS

1 0 → HomC(Z,B) → HomC(Y,B) → HomC(X,B) → ExtE (Z,B)

1 1 2 → ExtE (Y,B) → ExtE (X,B) → ExtE (Z,B) → ... for every B ∈ Ob(C). Deﬁne

0 ExtE (−,A) := HomC(−,A), then it follows from the above and the naturality of the connecting morphisms 6.41 that n ExtE (−,A) := (ExtE (−,A))n≥0 is a contravariant exact cohomological δE -functor. If the category C has enough E-projective objects, it follows from 5.74 that ExtE (−,A) is even an n universal cohomological δE -functor, since ExtE (−,A) is coeﬀacable for n > 0 by 6.36. Then it follows from 5.77 that there is an isomorphism of δE -functors ∼ ExtE (A, −) = R HomC(−,A). Chapter 7

Appendix 1: Exactness in Categories of Projective Spectra

7.1 Categories of Projective Spectra

n Deﬁnition 7.1. Let C be a category. A projective spectrum X = (Xn,Xm) with values in C consists of a sequence (Xn)n∈N of objects of C and mor- n phisms Xm : Xm → Xn in C, that are deﬁned for n ≤ m, such that n i) Xn = idXn for all n ∈ N, k n k ii) Xn ◦ Xm = Xm for k ≤ n ≤ m. n n For two projective spectra X = (Xn,Xm) and Y = (Yn,Ym) a morphism f : X → Y of projective spectra is a sequence (fn)n∈N of morphisms in C n n fn : Xn → Yn such that fn ◦ Xm = Ym ◦ fm for n ≤ m, i.e. the diagram

fm Xm / Ym

n n Xm Ym Xn / Yn fn is commutative.

Remark and Deﬁnition 7.2. Let C be a category.

n n i) Given three projective spectra X = (Xn,Xm), Y = (Yn,Ym) and n Z = (Zn,Zm) and morphisms f : X → Y and g : Y → Z we can deﬁne the composite morphism

g ◦ f := (gn ◦ fn)n∈N : X → Z

225 226 CHAPTER 7. APPENDIX 1: PROJECTIVE SPECTRA

of projective spectra. This composition is associative. In addition n we have for every projective spectrum X = (Xn,Xm) the identity morphism idX := (idXn )n∈N. Therefore we have the category P (C) of projective spectra and morphisms of projective spectra.

ii) If the category C has a zero object 0, then the projective spectrum

(0, id0)n∈N is a zero object in P (C).

iii) If C is a preadditive category, we can deﬁne the sum of two morphisms f, g : X → Y as

f + g := (fn + gn)n∈N : X → Y

and this addition makes P (C) a preadditve category.

iv) If C has biproducts, then P (C) also possesses biproducts. In fact, for n n two projective spectra X = (Xn,Xm) and Y = (Yn,Ym) the projective spectrum n n X q Y := (Xn q Yn,Xm ⊕ Ym)

together with the morphisms of spectra πX := (πXn )n∈N, πY := (πYn )n∈N, ωX := (ωXn )n∈N and ωY := (ωYn )n∈N is a biproduct of X and Y in P (C).

v) It follows from ii) that if C is additive, then so is P (C).

vi) If C is a preadditive category that has kernels and f : X → Y is a morphism of projective spectra, then we have the projective spectrum n n ker f := (ker fn, kerm), where kerm : ker fm → ker fn for n ≤ m is the unique morphism making the diagram

kfm fm ker fm / Xm / Ym n n n kerm Xm Ym ker fn / Xn / Yn kfn fn

commutative (see 2.18). The morphism of spectra

kf := (kfn )n∈N : ker f → X

is then a kernel of f in P (C).

vii) If C is a preadditive category that has cokernels and f : X → Y is a morphism of projective spectra, then we have the projective spectrum 7.1. CATEGORIES OF PROJECTIVE SPECTRA 227

n n cok f := (cok fn, cokm), where cokm : cok fm → cok fn for n ≤ m is the unique morphism making the diagram

fm cfm Xm / Ym / cok fm n n n Xm Ym cokm Xn / Yn / cok fm fn cfn

commutative (see 2.18). The morphism of spectra

cf := (cfn )n∈N : Y → cok f

is then a cokernel of f in P (C). viii) It follows from v) and vi) that if C is a preabelian category, then so is P (C).

ix) If the category C has pullbacks, we can construct for two morphisms g : Y → Z and t: T → Z in P (C) the projective spectrum

n P = (Pn,Pm)n∈N,

such that (Pn, pYn , pTn ) is the pullback of gn and tn in C and where n Pm : Pm → Pn is the unique morphism making the diagram

pTn Pn / Tn P n = = m { zz { zzT n { pTm z m Pm / Tm tn

pYn tm pYm Y / Z n gn n {= z= {{ n zz n {{ Ym zz Zm Y / Z m gm m

commutative. Then pT := (pTn )n∈N and pY := (pYn )n∈N are morphisms of projective spectra and (P, pT , pY ) is a pullback of g and t in P (C).

x) If the category C has pushouts, we can construct for two morphisms f : X → Y and t: X → T in P (C) the projective spectrum

n S = (Sn,Sm)n∈N, 228 CHAPTER 7. APPENDIX 1: PROJECTIVE SPECTRA

such that (Sn, sYn , sTn ) is the pushout of fn and tn in C and where n Sm : Sm → Sn is the unique morphism making the diagram

f X n Y n n / n Xm y< {= yy {{ n yy fm {{ Ym s Xm / Ym Zn

tn sYm tm Tn / Sn < sTn = yy { y n { n yy Tm { Sm Tm / Sm sTm

commutative. Then sT := (sTn )n∈N and sY := (sYn )n∈N are morphisms of projective spectra and (S, sT , sY ) is a pullback of f and t in P (C).

Proposition 7.3. Let (C, E) be an exact category. Then the class

f g P (E) := { X / Y / Z | (fn, gn) ∈ E for all n ∈ N} is an exact structure on P (C).

Proof. It follows from 7.2.vi) and vii) that every (f, g) ∈ P (E) is a kernel- cokernel pair. Since a morphism in P (E) is an isomorphism if and only if it is an isomorphism at all the steps, it follows that P (E) is closed under isomorphisms, since E is so. In addition, the deﬁnition of the identities and the zero

idX idX object in P (C) show that the pairs X / X / 0 and 0 / X / X are f g f g elements of P (E) for all X ∈ Ob(P (C)). If X / Y / Z and X / Y / Z 0 are elements of P (E), then for each n ∈ N the morphism gn ◦ gn in C has a 0 0 kernel such that (ker gn ◦ gn, gn ◦ gn) ∈ E, because of E being an exact structure on C. It follows from 7.2.vi) that then (ker g0 ◦ g, g0 ◦ g) ∈ P (E), hence P (E) satisﬁes the axiom [E1op]. The dual argument of the above shows that P (E) also satisﬁes the axiom [E1]. f g Let X / Y / Z be an element of P (E). If t: T → Z is a morphism in

P (C), then we can for each n ∈ N construct the pullback (Pn, pTn , pYn ) of gn and tn in C and get a commutative diagram

kn pTn Xn / Pn / Tn

pYn tn Xn / Yn / Zn fn gn

with (kn, pTn ) ∈ E by 3.13. It follows from the construction of the pullback 7.1. CATEGORIES OF PROJECTIVE SPECTRA 229 and the kernel in P (C), that these diagrams ﬁt together to a diagram

k pT X / P / T

pY (1) t X / Y / Z f g in P (C), such that (1) is a pullback square. Then (k, pT ) ∈ P (E) by construction, which shows that P (E) satisﬁes the axiom [E2op]. The dual argument of the above shows that P (E) also satisﬁes [E2], hence P (E) is an exact structure on P (C).

Corollary 7.4. Let C be a preabelian category. The class

C f g P (Emax) := { X / Y / Z | (fn, gn) is a stable kernel-cokernel pair for all n ∈ N} is the maximal exact structure of the preabelian category P (C).

P (C) Proof. By 7.3 the maximal exact structure Emax of the preabelian category C P (C) is the class of all stable kernel-cokernel pairs in P (C). Since P (Emax) C P (C) is an exact structure on P (C) by 7.3, we have P (Emax) ⊆ Emax. It remains f g P (C) to show the other inclusion. Let X / Y / Z be an element of Emax. Let n0 ∈ N and let r : R → Zn0 be a morphism in C. Then we have a projective n spectrum T = (Tn,Tm)n∈N with Tn0 = R and Tn = 0 otherwise, as well as a morphism of projective spectra t: T → Z with tn0 = r and tn = 0 and tn = 0 otherwise. Construct the pullback (P, pT , pY ) of g and t in P (C), then we have a commutative diagram

k pT X / P / T

pY (1) t X / Y / Z f g in P (C) with (k, p ) ∈ EP (C) by 3.13. It follows that (k , p ) = (k , p ) T max n0 Tn0 n0 R is a kernel-cokernel pair, which shows that (fn, gn) is a stable kernel-cokernel pair in C.

Corollary 7.5. If C is a quasiabelian category, then so is P (C). Proof. Follows directly from 7.4 and the fact that a preabelian category is quasiabelian if and only if its maximal exact structure is the class of all kernel-cokernel pairs.

Proposition 7.6. If (C, E) is an exact category such that C is preabelian and has enough E-injective objects, then the category P (C) has enough P (E)-injective objects. 230 CHAPTER 7. APPENDIX 1: PROJECTIVE SPECTRA

n Proof. Let X = (Xn,Xm)n∈N be an object of P (C). Since C has enough E-injective objects we can choose for each n ∈ N an E-injective object In and an admissible monomorphism in : Xn → In. We have the projective n n n m n spectrum J = (Jn,Jm)n∈N, where Jn := qk=1Ik and Jm : qk=1 Ik → qk=1Ik for m ≥ n is the canonical morphism. We claim that J is a P (E)-injective n n object of P (C). Let Y = (Yn,Ym)n∈N and Z = (Zn,Zm)n∈N be two objects of P (C) and let f : Y → J and i: Y → Z be morphisms in P (C) such that i is an admissible monomorphism with regard to P (C). Consider the following commutative diagram (in which we ﬁrst ignore the dotted arrows):

f2 πI2 Y2 / I1 q I2 / I2 AA v: AA v i A v 2 A v h2 Z2

1 1 πI Y2 Z2 1 Z 1 H i1 }> h1 }} H }} H } H$ Y1 / I1 f1

Since I1, I2 are E-injective objects of C and because i1, i2 are admissible monomorphisms with regard to E we get morphisms h1 : Z1 → I1 and eh2 : Z2 → I2 with f1 = h1 ◦ i1 and πI2 ◦ f2 = eh2 ◦ i2. Deﬁne 1 h1 ◦ Z2 h2 := : Z2 → I1 q I2, eh2 1 then we have πI1 ◦ h2 = h1 ◦ Z2 , hence the universal property of the product also gives h2 ◦ i2 = f2 making the extended diagram above commutative. Continuing inductively in that way we construct a morphism h: Z → J of projective spectra with h ◦ i = f, which establishes the claim that J is P (E)-injective. n For the projective spectrum X = (Xn,Xm)n∈N we have the morphism j : X → J of projective spectra with

1 2 t n jn := (i1 ◦ Xn, i2 ◦ Xn, . . . , in) : Xn → qk=1Ik.

Then, for each n ∈ N, the composition πIn ◦ jn = in is an admissible monomorphism and since C is preabelian it follows from 3.20 that jn is also an admissible monomorphism. This shows that j is an admissible monomorphism in P (C) with regard to P (E) and therefore P (C) has enough P (E)- injective objects.

Corollary 7.7. If C is a quasiabelian category that has enough injective objects, then the quasiabelian category P (C) has enough injective objects. Proof. This is a direct consequence of 7.4 and 7.6. 7.2. THE PROJECTIVE LIMIT 231

7.2 The Projective Limit

n Deﬁnition 7.8. Let C be a category and let X = (Xn,Xm)n∈N be a projective spectrum with values in C.A projective limit of X is a pair (P, (pn))n∈N with

(P1) P is an object of C,

(P2) pn : P → Xn is a morphism in C for every n ∈ N, (P3) for n ≤ m the diagram

P pm { BB pn {{ BB }{{ B! Xm n / Xn Xm

is commutative, such that the following universal property is fulﬁlled:

If (Q, (qn))n∈N is another pair which has the properties (P 1), (P 2) and (P 3), there exists a unique morphism λ: Q → P such that the diagram

λ Q ______/ P AA } AA }} qn A ~}} pn Xn is commutative for all n ∈ N. We say that C possesses countable projective limits, if the projective limit of n every projective spectrum X = (Xn, σm)n∈N exists.

Notation 7.9. If (P, (pn))n∈N is a projective limit of a projective spectrum n X = (Xn,Xm)n∈N with values in a category C, it follows that (P, (pn))n∈N is uniquely determined up to a unique isomorphism, since it is deﬁned by a universal property (compare 2.9.ii)). Therefore from now on

Proj(X) := P

n will denote the projective limit of the projective spectrum X = (Xn,Xm)n∈N. Remark and Deﬁnition 7.10. Let C be a category that possesses countable projective limits and let f : X → Y be a morphism of projective spectra n n X = (Xn,Xm)n∈N and Y = (Yn,Ym)n∈N. Since the diagram

Proj(X) X fm◦p J mtt JJ tt XJJ ytt fn◦pn J$ Ym n / Yn Ym 232 CHAPTER 7. APPENDIX 1: PROJECTIVE SPECTRA is commutative for n ≤ m, the universal property of Proj(Y ) gives rise to a unique morphism Proj(f): Proj(X) → Proj(Y ), such that the diagram

Proj(f) Proj(X) / Proj(Y ) X Y pn pn Xn / Yn fn is commutative for all m ∈ N. It follows from the uniqueness of this induced morphism, that

• Proj(idX ) = idProj(X) for all X ∈ Ob(P (C)), • Proj(g ◦ f) = Proj(g) ◦ Proj(f) for all morphisms f : X → Y and g : Y → Z in P (C), i.e. we have a covariant functor X 7→ Proj(X) Proj: P (C) → C , . f 7→ Proj(f)

Example 7.11. In the category (LCS) of locally convex spaces and continous linear maps, the projective limit of a projective spectrum X = n (Xn,Xm)n∈N is the space

Y n Proj(X) := {(xn)n∈N ∈ Xn : Xm(xm) = xn for all m ≥ n}, n∈N equipped with the topology induced by the product topology, together with X the projections pn : Proj(X) → Xn onto the n-th components (cf. [31]). For a morphism of projective spectra f : X → Y , we have

Proj(f): Proj(X) → Proj(Y ), (xn)n∈N 7→ (fn(xn))n∈N. Consider the the composition of the functor Proj: P (LCS) → (LCS) and the forgetful functor (LCS) → (K − V ec), which we will also denote

Proj: P (LCS) → (K − V ec). By 7.5 the category P (LCS) is quasiabelian, hence the above functor Proj is an additive functor from an exact category to an abelian category. Since the category (LCS) has enough injective objects by 7.5.iii), it follows from 7.7 that P (LCS) has enough injective objects and therefore the right-derived functors k k Proj := R Proj: P (LCS) → (K − V ec) 7.2. THE PROJECTIVE LIMIT 233 are deﬁned for k ∈ Z by 5.48. Using the injective objects constructed in the proof of 7.6 one ﬁnds an ex- k n plicit description of Proj (X) for a given projective spectrum X = (Xn,Xm)n∈N, namely

• Projk(X) = 0 for k < 0,

• Proj0(X) = Proj(X),

• Projk(X) = 0 for k ≥ 2, and

• Proj1(X) ∼ (Q X )/ im(ψ), = n∈N n where ψ : Q X → Q X , (x ) 7→ (x − Xn (x )) (cf. n∈N n n∈N n n n∈N n n+1 n+1 n∈N [31]). Index

δE -functor, 171 of commutative rings with unity, cohomological 2 contravariant, 173 of complexes, 127 covariant, 171 of Hausdorff locally convex spaces, exact, 173 2 universal, 174 of locally convex spaces, 2 homological (DN), 79 contravariant, 173 (Ω), 79 covariant, 171 barelled, 79 exact, 173 bornological, 117 universal, 174 Fréchet, 79 metrizable, 79 abelian groups, see category Montel, 79 Baer-sum normed, 79 of Ext1, 186 nuclear, 79 of Extn, 202 Schwartz, 79 Banach spaces, see category of metric spaces, 2 bifunctor, 185 of sets, 2 Yoneda-Ext1, 185 of small categories, 13 Yoneda-Extn, 200 of topological spaces, 2 big abelian group, see quasicategory of topological vector spaces, 2 biproduct, 44 complete, 79 finite, 52 Hausdorff, 79 is self dual, 45 locally bounded, 79 of complexes, 127 pseudometrizable, 79 of vector spaces, 2 canonical factorization, 33 preabelian, 51 category, 1 preadditive, 25 abelian, 58 has biproducts, 44 additive, 50 with cokernels, 27 dual, see dual category with kernels, 27 exact, 67 quasiabelian, 115 homotopy, 140 semi-abelian, 58 large, 12 small, 12 of abelian groups, 2 subcategory, see subcategory of Banach spaces, 3 that has coproducts, 43

234 INDEX 235

that has products, 42 E-injective resolution, 158 that has pullbacks, 36 E-projective resolution, 159 that has pushouts, 37 additive, 26 closed under extensions, see three space coeffacable, 175 property completion, 12 codomain, 2 composite, 10 coimage, 29 constant, 10 coimage-object, 29 contravariant, 9 cokernel, 27 contravariant Hom-, 11 cokernel-object, 27 covariant, 9 semi-stable, 103 covariant Hom-, 11 commutative rings with unity, see cat- duality, 10 egory effacable, 175 complex, 126 essentially surjective, 14 exact, 134 exact, 121 composition law, 1 faithful, 14 construct, 2 forgetful, 11 coproduct, 42 full, 14 coretraction, 6 fully faithful, 14 homology, 131 domain, 2 identity, 10 dual category, 3–4 inclusion, 14 dual property, 4 injective, 123 duality principle for categories, 4 left derived, 159 Embedding Theorem for small exact additional, 165 category, 125 projective, 124 equivalence of categories, 17 projective limit, 232 exact category, see category right derived, 232 has enough E-pinjective objects, representable, 20 143 right derived, 159 has enough E-projective objects, additional, 164 142 semi-injective, 123 exact sequence, 90 semi-projective, 124 1 long, see long exact sequence Yoneda-Ext short, see kernel-cokernel pair contravariant, 185 exact structure, 66 covariant, 185 n maximal of a preabelian category, Yoneda-Ext 114 contravariant, 200 minimal, 69 covariant, 200 restricted, 77 Hausdorff locally convex spaces, see five lemma, 92 category short five lemma, 82 Hausdorff spaces, see category functor, 9 homology, 130 236 INDEX homotopic, see morphisms class of all, 1 homotopic to zero, see morphisms epimorphisms, 5, 6 homotopy equivalence, 138 monomorphisms, 5 horseshoe lemma, 153 of δE -functors, 172 of complexes, 126 image, 29 homotopic, 138 image-object, 29 homotopic to zero, 138 initial object, 8 of projective spectra, 225 isomorphic, 6 strict, 33 isomorphism, 6 E-strict, 88 ofδE -functors, 173 natural isomorphism, see natural trans- kernel, 27 formation kernel-object, 27 naturally isomorphic, 17 semi-stable, 103 natural transformation, 16 kernel-cokernel pair, 65 class of all, 16 split exact, 69 composition of, 16 stable, 109 natural isomorphism, 16 nine lemma, 84 locally convex spaces, see category noether isomorphism, 83 (DN), see category (Ω), see category objects, 1 barelled, see category E-injective, 142 bornological, see category E-projective, 142 Fréchet, see category injective, 143 metrizable, see category projective, 143 Montel, see category obscure axiom, 79 normed, see category nuclear, see category product, 42 Schwartz, see category projective limit, 231 long exact sequence projective spectrum, 225 contravariant pullback, 36 left derived, 170 pushout, 37 right derived, 170 Yoneda-Extn, 223 quasi-inverse, 17 covariant quasicategory, 13 left derived, 168 functor, 17 right derived, 166 of all categories, 13 Yoneda-Extn, 221 of big abelian groups, 188 homology, 134 of classes, 185 matrix-calculus, 54 reflects exactness, 121 metric spaces, see category resolution, 147 morphisms, 1 E-injective, 147 bimorphisms, 5 E-projective, 147 INDEX 237

comparison lemma, 149 left-, 147 right-, 147 retraction, 6

Schanuel-lemma, 211 sets, see category shelf lemma, 155 snake lemma, 93 extended, 101 split exact, see kernel-cokernel pair subcategory, 14 full preabelian, 51 full preadditive, 26 that reflects cokernels, 30 that reflects kernels, 30 that reflects coproducts, 44 that reflects products, 43 that reflects pullbacks, 49 that reflects pushouts, 50 three space property, 77 terminal object, 8 three space property, see subcategory toplogical vector spaces complete spaces, see category topological spaces, see category topological vector spaces, see category locally bounded, see category pseudometrizable, see category vector spaces, see category

Yoneda Embedding, 22 Yoneda-Ext1, 181 Yoneda-Extn, 195 zero morphism, 9 zero object, 8 238 INDEX Bibliography

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