On Universal Properties of Preadditive and Additive Categories

On universal properties of preadditive and additive categories

Karoubi envelope, additive envelope and tensor product

Bachelor’s Thesis

Mathias Ritter February 2016 II Contents

0 Introduction1 0.1 Envelope operations...... 1 0.1.1 The Karoubi envelope...... 1 0.1.2 The additive envelope of preadditive categories...... 2 0.2 The tensor product of categories...... 2 0.2.1 The tensor product of preadditive categories...... 2 0.2.2 The tensor product of additive categories...... 3 0.3 Counterexamples for compatibility relations...... 4 0.3.1 Karoubi envelope and additive envelope...... 4 0.3.2 Additive envelope and tensor product...... 4 0.3.3 Karoubi envelope and tensor product...... 4 0.4 Conventions...... 5

1 Preliminaries 11 1.1 Idempotents...... 11 1.2 A lemma on equivalences...... 12 1.3 The tensor product of modules and linear maps...... 12 1.3.1 The tensor product of modules...... 12 1.3.2 The tensor product of linear maps...... 19 1.4 Preadditive categories over a commutative ring...... 21

2 Envelope operations 27 2.1 The Karoubi envelope...... 27 2.1.1 Deﬁnition and duality...... 27 2.1.2 The Karoubi envelope respects additivity...... 30 2.1.3 The inclusion functor...... 33

III 2.1.4 Idempotent complete categories...... 34 2.1.5 The Karoubi envelope is idempotent complete...... 38 2.1.6 Functoriality...... 40 2.1.7 The image functor...... 46 2.1.8 Universal property...... 48 2.1.9 Karoubi envelope for preadditive categories over a commutative ring 55 2.2 The additive envelope of preadditive categories...... 59 2.2.1 Deﬁnition and additivity...... 59 2.2.2 The inclusion functor...... 65 2.2.3 Functoriality...... 67 2.2.4 The realisation functor...... 72 2.2.5 Universal property...... 74 2.2.6 Additive envelope for preadditive categories over a commutative ring 79

3 The tensor product 83 3.1 The tensor product of preadditive categories over a commutative ring... 83 3.1.1 Deﬁnition...... 83 3.1.2 Universal property...... 90 3.1.3 Functoriality...... 95 3.2 The tensor product of additive categories over a commutative ring..... 101 3.2.1 Deﬁnition...... 101 3.2.2 Universal property...... 101

4 Counterexamples for compatibility relations 107 4.1 Additive envelope and Karoubi envelope...... 107 4.2 Additive envelope and tensor product...... 109 4.3 Karoubi envelope and tensor product...... 111

IV Chapter 0

Introduction

0.1 Envelope operations

In the construction of our envelope operations we follow Nico Stein [7], in that we ﬁrst perform the envelope construction including its functorialities, and then use these functorialities to derive its universal properties.

0.1.1 The Karoubi envelope

We give an account of the construction of the Karoubi envelope of an additive category. This construction is due to Karoubi [3, II.1], see also [4, Theorem 6.10]. He calls it “enveloppe pseudo-ab´elienne“or “pseudo-abelian category“ associated with the given category. Suppose given an additive category A. It is not necessarily idempotent complete, that is, an idempotent morphism X ÝÑe X in A does not necessarily have an image; cf. Deﬁnition 45. We aim to endow A with images of all idempotents in a universal manner. More precisely, we construct an idempotent complete additive category Kar A and - an additive functor JA : A Kar A such that every additive functor F from A to an idempotent complete additive category B factors uniquely, up to isomorphism, over JA 1 as F “ F ˝ JA . JA A / Kar A

F 1 F & B Still more precisely, we have the equivalence of categories „ addrA, Bs addrKar A, Bs

β˚JA β- pU ˝ JAq ÝÝÝÑpV ˝ JAq pU V q, which is surjective on´ objects; cf. Theorem 78¯. The category Kar A is constructed as follows. Its objects are the pairs of the form pX, eq, where X P Ob A and X ÝÑe X is an idempotent. A morphism from pX, eq to pY, fq in

1 ϕ Kar A is given by a morphism X ÝÑ Y in A such that eϕf “ ϕ. Composition is then given by the composition in A; cf. Deﬁnition 33. More generally, we perform this construction for arbitrary, not necessarily additive categories.

0.1.2 The additive envelope of preadditive categories

We give an account of the construction of the additive envelope of a preadditive category. In the literature, it has been mentioned for example in [5, VII.2, ex. 6.(a)] or in [2, Def. 1.1.15]. Suppose given a preadditive category A, that is, a category whose morphism sets are abelian groups and whose composition is Z-bilinear. Note that A is not necessarily additive, that is, A does not necessarily have direct sums. We aim to endow A with direct sums in a universal manner. More precisely, we construct an additive category Add A - and an additive functor IA : A Add A such that every additive functor F from A to 1 an additive category B factors uniquely, up to isomorphism, over IA as F “ F ˝ IA .

IA A / Add A

F 1 F & B

Still more precisely, we have the equivalence of categories

„ addrA, Bs addrAdd A, Bs

β˚IA β- pU ˝ IAq ÝÝÝÑpV ˝ IAq pU V q, ´ ¯ which is surjective on objects; cf. Theorem 113. The category Add A is constructed as follows. Its objects are tuples of objects of A. Its morphisms are formal matrices having as entries morphisms of A. Composition is then given by the usual matrix multplication rule; cf. Deﬁnition 88. More generally, we perform this construction over an arbitrary commutative ground ring.

0.2 The tensor product of categories

0.2.1 The tensor product of preadditive categories

We give an account of the construction of the tensor product of preadditive categories as mentioned in [6, 16.7.4], where preadditive categories in our sense are called “additive“ by Schubert. Suppose given preadditive categories A and B. We construct a preadditive category AbB - and a Z-bilinear functor MA,B : A ˆ B A b B such that every Z-bilinear functor F

2 from A ˆ B to a preadditive category C factors uniquely over MA,B as F “ F ˝ MA,B .

MA,B A ˆ B / A b B

F F ' C More precisely, we have the isomorphism of categories

„ Z-bilrA ˆ B, Cs addrA b B, Cs

β˚MA,B β- pU ˝ MA,Bq ÝÝÝÝÝÑpV ˝ MA,Bq pU V q; ´ ¯ cf. Theorem 130. The category A b B is constructed as follows. Its objects are the pairs pA, Bq with A P Ob A and B P Ob B. We denote such a pair by A b B. For objects A1 b B1 and A2 b B2 of A b B, let

AbBpA1 b B1 ,A2 b B2q :“ ApA1 ,A2q b BpB1 ,B2q. Z Composition in A b B is then given on elementary tensors by

pa1 b b1qpa2 b b2q “ pa1a2q b pb1b2q;

cf. Deﬁnition 123. More generally, we perform this construction over an arbitrary commutative ground ring.

0.2.2 The tensor product of additive categories

We give an account of the construction of the tensor product of additive categories as mentioned in [2, Def. 1.1.15].

add Suppose given additive categories A and B. We construct an additive category A b B add add - and a Z-bilinear functor MA,B : A ˆ B A b B such that every Z-bilinear functor F add from A ˆ B to an additive category C factors uniquely, up to isomorphism, over MA,B as 1 add F “ F ˝ MA,B . add MA,B add A ˆ B / A b B

F 1 F & C More precisely, we have the equivalence of categories

add „ Z-bilrA ˆ B, Cs addrA b B, Cs

β˚M add add A,B add β- pU ˝ MA,Bq ÝÝÝÝÝÑpV ˝ MA,Bq pU V q, ˆ ˙ 3 which is surjective on objects; cf. Theorem 143. add The category A b B is constructed by ﬁrst taking the tensor product A b B of A and B as preadditive categories and then taking the additive envelope of A b B. In other words, add we have A b B “ AddpA b Bq; cf. §0.1.2 and §0.2.1. More generally, we perform this construction over an arbitrary commutative ground ring.

0.3 Counterexamples for compatibility relations

0.3.1 Karoubi envelope and additive envelope

Given a preadditive category A, in general we have

AddpKar Aq ﬁ KarpAdd Aq.

More precisely, consider the subring Λ :“ tpa, bq P Z ˆ Z : a ”5 bu of Z ˆ Z and the full preadditive subcategory A of Λ -free with Ob A :“ tΛ, 0u. Then AddpKar Aq ﬁ KarpAdd Aq; cf. Proposition 145.

0.3.2 Additive envelope and tensor product

Given a commutative ring R and preadditive categories pA, ϕAq and pB, ϕBq over R, in general we have pAdd Aq b pAdd Bq ﬁ AddpA b Bq. R R More precisely, consider R “ Q and the full Q-linear preadditive subcategory A of Q -mod with Ob A :“ tV P Ob Q -mod : dim V ‰ 1u; cf. Remark 32. Then pAdd Aq b pAdd Aq ﬁ AddpA b Aq; cf. Proposition 147. Q Q

0.3.3 Karoubi envelope and tensor product

Given a commutative ring R and preadditive categories pA, ϕAq and pB, ϕBq over R, in general we have pKar Aq b pKar Bq ﬁ KarpA b Bq. R R More precisely, consider R “ Q and the full Q-linear preadditive subcategory of Qpiq -mod with Ob A :“ tQpiq, 0u; cf. Remark 32. Then pKar Aq b pKar Aq ﬁ KarpA b Aq; cf. Proposition 149. Q Q

4 0.4 Conventions

We assume the reader to be familiar with elementary category theory. An introduction to this subject can be found in [5] or [6]. Some basic deﬁnitions and notations are given below. Concerning additive categories, we essentially follow Nico Stein [7]. Let A, B and C be categories.

1. All categories are supposed to be small (with respect to a suﬃciently big universe); cf. [6, §3.2 and §3.3].

2. We write Ob A for the set of objects and Mor A for the set of morphisms of A. Given A, B P Ob A, we denote the set of morphisms from A to B by ApA, Bq. The identity morphism of A P Ob A is written as 1A . If unambiguous, we often write 1 :“ 1A .

3. The composition of morphisms in A is written naturally:

f g fg f¨g A ÝÑ B ÝÑ C “ A ÝÑ C “ A ÝÝÑ C . ´ ¯ ´ ¯ ´ ¯ 4. The composition of functors is written traditionally:

A ÝÑF B ÝÑG C “ A ÝÝÑG˝F C . ´ ¯ ´ ¯ 5. Suppose given A, B P Ob A. If A and B are isomorphic in A, we write A – B. ϕ- If ϕ P ApA, Bq is an isomorphism, we often write A „ B. Given an isomorphism ´1 f P ApA, Bq, we write f P ApB,Aq for its inverse.

6. Given a functor F : A Ñ B and X,Y P Ob A, we write

FX,Y : ApX,Y q Ñ BpFX,FY q, ϕ ÞÑ F ϕ.

f 7. The opposite category (or dual category) of A is denoted by A˝. Given A ÝÑ B in f ˝ A, we write B ÝÑ A for the corresponding morphism in A˝.

8. We call A preadditive if it fullﬁlls the following conditions (1, 2).

(1) For A, B P Ob A, the set ApA, Bq carries the structure of an abelian group, written additively.

g1 f / h (2) For A / B / C / D in A, we have fpg1 ` g2qh “ fg1h ` fg2h. g2

Suppose A to be preadditive. Suppose given A, B P Ob A. We denote the zero morphism in ApA, Bq by 0A,B . If unambiguous, we often write 0 :“ 0A,B . Note that full subcategories of preadditive categories are preadditive.

9. We write Z for the set of integers. Given a P Z, we write Zěa :“ tb P Z : b ě au.

5 10. For a, b P Z, we write ra, bs :“ tc P Z : a ď c ď bu.

11. Let M be a ﬁnite set. The cardinality of M is denoted by |M|. 12. Suppose A to be preadditive. We use following variant of the Kronecker delta.

Let I be a set. Suppose given Ai P Ob A for i P I. Let

1Ai if i “ j, δAi,Aj :“ 0A ,A if i ‰ j, " i j

for i, j P I. If unambiguous, we often write δi,j :“ δAi,Aj .

13. Suppose A to be preadditive. Suppose given A1 ,A2 P Ob A. An object C P Ob A ι1 ι2 / o together with morphisms A1 o C / A2 is called a direct sum of A1 and A2 in π1 π2

A, if ι1π1 “ 1A1 , ι2π2 “ 1A2 and π1ι1 ` π2ι2 “ 1C . In the following way this is generalized to a ﬁnite number of objects.

Suppose given n P Zě0 . Suppose given Ai P Ob A for i P r1, ns. A direct sum of A1,...,An in A is a tuple pC, pπiqiPr1,ns , pιiqiPr1,nsq with C P Ob A, πi P ApC,Aiq and

ιi P ApAi ,Cq for i P r1, ns such that ιiπj “ δi,j for i, j P r1, ns and iPr1,ns πiιi “ 1C . Often the following matrix notation is used for morphisms betweenř direct sums.

Suppose given m, n P Zě0 . Suppose given A1,i and A2,j in Ob A for i P r1, ms and j P r1, ns. Suppose pC1 , pπ1,iqiPr1,ms , pι1,iqiPr1,msq to be a direct sum of A1,1,...,A1,m and pC2 , pπ2,jqjPr1,ns , pι2,jqjPr1,nsq to be a direct sum of A2,1,...,A2,n .

Suppose given f P ApC1 ,C2q.

Let fi,j :“ ι1,ifπ2,j P ApA1,i ,A2,jq for pi, jq P r1, ms ˆ r1, ns. Then

f “ π1,ifi,jι2,j . pi,jqPr1,msˆr1,ns ÿ We write f1,1 ... f1,n . . f “ pfi,jqi,j “ pfi,jqiPr1,ms,jPr1,ns “ . . . ˜fm,1 ... fm,n ¸ Omitted matrix entries are stipulated to be zero.

Suppose given n P Zě0 . Suppose given A1,i and A2,i in Ob A for i P r1, ns. Suppose fi given A1,i ÝÑ A2,i in A for i P r1, ns. We write

f 1 .. diagpfiqi “ diagpfiqiPr1,ns “ . . f ˆ n ˙

14. We call A P Ob A a zero object if | ApA, Bq| “ 1 “ | ApB,Aq| for B P Ob A. 15. The category A is called additive if the following conditions (1, 2, 3) hold.

(1) The category A is preadditive. (2) There exists a zero object in A.

6 (3) Every pair pA, Bq P pOb Aq ˆ pOb Aq has a direct sum in A. Note that in additive categories direct sums of arbitrary ﬁnite length exist.

16. Suppose A to be additive. We choose a zero object 0A P Ob A.

For n P Zě0 and A1,...,An P Ob A, we choose a direct sum

pAj qjPr1,ns pAj qjPr1,ns Ai , pπi qiPr1,ns , pιi qiPr1,ns ˜iPr1,ns ¸ à in A.

pAj qjPr1,ns pAj qjPr1,ns If unambiguous, we often write πi :“ πi and ιi :“ ιi for i P r1, ns. In particular, we choose

Ai , pπiqiPr1,1s , pιiqiPr1,1s “ A1 , p1A1 qiPr1,1s , p1A1 qiPr1,1s ˜iPr1,1s ¸ à ` ˘ and

Ai , pπiqiPr1,0s , pιiqiPr1,0s “ p0A , p q, p qq. ˜iPr1,0s ¸ à

We often write A1 ‘ ¨ ¨ ¨ ‘ An :“ iPr1,ns Ai . In matrix notation, we have À 0. . 0 pAj qjPr1,ns π “ pδi,kqkPr1,ns,lPr1,1s “ ¨1˛ i 0. . ˚ ‹ ˚0‹ ˝ ‚ and pA q j jPr1,ns 0 ... 0 1 0 ... 0 ιi “ pδi,lqkPr1,1s,lPr1,ns “ p q for i P r1, ns; cf. Stipulation 106 below.

17. Suppose A and B to be preadditive. A functor F : A Ñ B is called additive if ϕ / F pϕ ` ψq “ F ϕ ` F ψ for X / Y in A. ψ

α 18. Let F,G : A Ñ B be functors. Suppose given FX ÝÝÑX GX for X P Ob A. The tuple f pαX qXPOb A is called natural if αX ¨ Gf “ F f ¨ αY for X ÝÑ Y in A. A natural tuple F * is often called a transformation. We write α : F Ñ G, α : F ñ G or A α B. 4 G α A transformation α : F ñ G is called an isotransformation if FX ÝÝÑX GX is an α isomorphism in B for X P Ob A. We often write F „ +3 G. Suppose given functors F, G, H : A Ñ B. Suppose given transformations α : F ñ G and β : G ñ H.

7 We have the transformation αβ : F ñ H where pαβqX :“ αX βX for X P Ob A.

We have the transformation 1F : F ñ F where p1F qX :“ 1FX for X P Ob A.

F K ) ) Suppose given A α B γ C . We have the transformation 5 5 G L

pγ ˚ αq : pK ˝ F q ñ pL ˝ Gq

where pγ ˚ αqX :“ KαX ¨ γGX “ γFX ¨ LαX for X P Ob A.

γFX pK ˝ F qX / pL ˝ F qX

KαX LαX

γGX pK ˝ GqX / pL ˝ GqX

In particular, we have the transformations γ ˚ F :“ γ ˚ 1F and K ˚ α :“ 1K ˚ α.

19. A functor F : A Ñ B is called an isomorphism if there exists a functor G : B Ñ A ´1 with F ˝ G “ 1B and G ˝ F “ 1A . In this case, we write G “ F . A functor F : A Ñ B is called an equivalence if there exists a functor G : B Ñ A and α β isotransformations G ˝ F „ +3 1A and F ˝ G „ +3 1B . If there exists an equivalence F : A Ñ B, we call A and B equivalent and write A » B. A functor F : A Ñ B is called surjective on objects, if the map

Ob A Ñ Ob B,A ÞÑ FA

is surjective. A functor F : A Ñ B is called dense if given B P Ob B, there exists A P Ob A with FA – B.

A functor F : A Ñ B is called full if FX,Y is surjective for X,Y P Ob A.

A functor F : A Ñ B is called faithful if FX,Y is injective for X,Y P Ob A. Note that a functor F is an equivalence if and only if it is full, faithful and dense.

F G η ε 20. Let A ÝÑ B and B ÝÑ A be functors. Let 1A ÝÑpG ˝ F q and pF ˝ Gq ÝÑ 1B be transformations. We call pF, G, η, εq an adjunction, if the following diagrams commute. F η ηG F / F ˝ G ˝ F G / G ˝ F ˝ G

εF Gε F G In this case, F is called left adjoint to G and G is called right adjoint to F . We also write F % G. Furthermore, we call η the unit and ε the counit of the adjunction.

21. By rA, Bs we denote the functor category whose objects are the functors from A to B and whose morphisms are the transformations between such functors.

8 α / 22. Suppose B to be preadditive. For F / G in rA, Bs, we have the transformation β

α ` β : F Ñ G where pα ` βqX :“ αX ` βX for X P Ob A. The category rA, Bs is preadditive with respect to this addition.

23. Suppose A and B to be preadditive. By addrA, Bs we denote the full subcategory of rA, Bs with Ob addrA, Bs :“ tF P ObrA, Bs : F is additiveu. 24. Given X P Ob A, we call X ÝÑe X an idempotent in A if e2 “ e. A tuple pY, π, ιq e with Y P Ob A, π P ApX,Y q and ι P ApY,Xq is called an image of X ÝÑ X in A if the following diagram commutes.

e X / X > ι π π 1Y Y / Y The category A is called idempotent complete if every idempotent in A has an image in A; cf. Deﬁnitions1 and 45 below. 25. Suppose A to be idempotent complete. Given an idempotent X ÝÑe X in A, we choose an image pIm e, e,¯ e9q of e in A.

For X P Ob A, choose pIm 1X , 1X , p1X q q “ pX, 1X , 1X q; cf. Stipulation 52 below. ppp 26. Let R be a ring. By an R-module, we understand an R-left-module. The category of R-modules is denoted by R -Mod. By R -mod we denote the category of ﬁnitely generated R-modules. Let R -free be the full additive subcategory of R -mod with

n Ob R -free :“ tR : n P Zě0u. In particular, R -free is a skeleton of the category of ﬁnitely generated free R- modules.

27. Suppose I to be a set. Suppose given categories Ai for i P I. We denote the product

category by iPI Ai . For m P Zě0 , we also write A1 ˆ ¨ ¨ ¨ ˆ Am :“ iPr1,ms Ai and Aˆm :“ A. iPrś1,ms ś 28. Let R beś a commutative ring.

Given preadditive categories pD, ϕq and pE, ψq over R, we write R-linrD, Es for the full subcategory of rD, Es with

Ob R-linrD, Es :“ tF P ObrD, Es : F is R-linearu.

Given preadditive categories pD, ϕq, pE, ψq and pF, ηq over R, we write R-bilrD ˆ E, Fs for the full subcategory of rD ˆ E, Fs with

Ob R-bilrD ˆ E, Fs :“ tF P ObrD ˆ E, Fs : F is R-bilinearu. Cf. Deﬁnitions 23 and 27 below.

29. For m P Zě0 , the symmetric group on m symbols is denoted by Sm .

9 10 Chapter 1

Preliminaries

1.1 Idempotents

For this §1.1, let A be a category. Deﬁnition 1. Deﬁne Idem A :“ te P Mor A : e2 “ eu. The elements of Idem A are called idempotents.

f Remark 2. Suppose given idempotents X ÝÑe X and Y ÝÑ Y in A. Suppose given ϕ P ApX,Y q with eϕf “ ϕ. We have eϕ “ ϕ and ϕf “ ϕ.

Proof. We have eϕ “ eeϕf “ eϕf “ ϕ. Similarly, we have ϕf “ eϕff “ eϕf “ ϕ. ϕ Remark 3. Suppose given an isomorphism X ÝÑ Y in A. Suppose given an idempotent ϕ´1eϕ X ÝÑe X. We have the idempotent Y ÝÝÝÝÑ Y in A.

Proof. We have pϕ´1eϕq2 “ ϕ´1eϕϕ´1eϕ “ ϕ´1eeϕ “ ϕ´1eϕ. Remark 4. Suppose A to be preadditive. Suppose given an idempotent X ÝÑe X in A. ϕ e´ϕ Suppose given an idempotent X ÝÑ X with eϕe “ ϕ. We have the idempotent X ÝÝÑ X in A.

Proof. We have pe ´ ϕq2 “ e2 ´ eϕ ´ ϕe ` ϕ2 R“2 e ´ ϕ ´ ϕ ` ϕ “ e ´ ϕ.

Remark 5. Suppose given a category B and a functor A ÝÑF B. Suppose given idempotents e f X ÝÑ X and Y ÝÑ Y in A. Suppose given ϕ P ApX,Y q with eϕf “ ϕ. The following assertions (1, 2) hold:

(1) The morphism FX ÝÑF e FX is an idempotent in B. (2) We have F e ¨ F ϕ ¨ F f “ F ϕ.

Proof. Ad (1). We have F e ¨ F e “ F peeq “ F e. Ad (2). We have F e ¨ F ϕ ¨ F f “ F peϕfq “ F ϕ.

11 1.2 A lemma on equivalences

Lemma 6. Suppose given categories C and D. Suppose given a functor F : C Ñ D. Suppose F to be surjective on objects. Suppose given a map Ob D Ñ Ob C, X ÞÑ X1 such that FX1 “ X for X P Ob D. Suppose the following assertions (1, 2) to hold.

1 1 (1) For X,Y P Ob D, the map FX1,Y 1 : CpX ,Y q Ñ DpX,Y q, ϕ ÞÑ F ϕ is bijective. (2) For A, B P Ob C with FA “ FB, we have A – B.

Then F is an equivalence.

Proof. Since F is surjective on objects, F is dense. Therefore, it suﬃces to show that F is full and faithful.

Suppose given A, B P Ob C. We show that FA,B : CpA, Bq Ñ DpF A, F Bq, ϕ ÞÑ F ϕ is a bijection.

α- 1 β- 1 Since (2) holds, we have isomorphisms A „ pFAq and B „ pFBq . This gives a bijection

1 1 ´1 f : CpA, Bq Ñ CppFAq , pFBq q, ϕ ÞÑ α ϕβ. Furthermore, we obtain a bijection

´1 g : DpF A, F Bq Ñ DpF A, F Bq, ψ ÞÑ F α ¨ ψ ¨ F β .

Suppose given ϕ P CpA, Bq. We have

´1 ϕpf ¨ FpFAq1,pFBq1 ¨ gq “ pα ϕβqpFpFAq1,pFBq1 ¨ gq “ pF pα´1ϕβqqg “ F α ¨ F α´1 ¨ F ϕ ¨ F β ¨ F β´1 “ F ϕ

“ ϕFA,B .

Because (1) holds, FA,B “ f ¨ FpFAq1,pFBq1 ¨ g is a bijection.

1.3 The tensor product of modules and linear maps

For this §1.3, let R be a commutative ring.

1.3.1 The tensor product of modules

In this §1.3.1 we establish the tensor product of a finite number of modules. Instead of defining the tensor product of two modules and then defining the tensor product of a finite number of modules inductively, we follow the idea of Atiyah and Macdonald [1, Prop. 2.12˚] and do it at once. This is useful in §3.1.1.

12 For this §1.3.1, let n P Zě0 , Mi P Ob R -Mod for i P r1, ns and X P Ob R -Mod.

Deﬁnition 7. We say that a map f : iPr1,ns Mi Ñ X is R-multilinear, if the following property (ML) holds. ś

1 2 (ML) Suppose given i P r1, ns. Suppose given mi, mi P Mi. Suppose given mj P Mj for j P r1, nsztiu. Suppose given r1, r2 P R. Then we have

1 1 2 2 pm1, . . . , mi´1, r mi ` r mi , mi`1, . . . , mnqf 1 1 2 2 “ r pm1, . . . , mi´1, mi, mi`1, . . . , mnqf ` r pm1, . . . , mi´1, mi , mi`1, . . . , mnqf.

For n “ 2, we often write R-bilinear instead of R-multilinear.

Remark 8. Let f : iPr1,ns Mi Ñ X be a map. The following assertionsś (1, 2) are equivalent.

(1) The map f is R-multilinear.

(2) For i P r1, ns and pmjqjPr1,nsztiu P jPr1,nsztiu Mj , the map ś Mi Ñ X, a ÞÑ pm1, . . . , mi´1, a, mi`1, . . . , mnqf

is R-linear.

Deﬁnition 9. Deﬁne

R-mul ΠiPr1,nsMi ,X :“ tf :ΠiPr1,nsMi Ñ X : f is R-multilinearu. ` ˘ For f, g P R-mulpΠiPr1,nsMi ,Xq, we deﬁne

f ` g :ΠiPr1,nsMi Ñ X, m ÞÑ mf ` mg .

For f P R-mulpΠiPr1,nsMi ,Xq and r P R, we deﬁne

r ¨ f :ΠiPr1,nsMi Ñ X, m ÞÑ r ¨ pmfq .

Remark 10. The following assertions (1, 2, 3) hold.

(1) For f, g P R-mulpΠiPr1,nsMi ,Xq, we have f ` g P R-mulpΠiPr1,nsMi ,Xq.

(2) For f P R-mulpΠiPr1,nsMi ,Xq and r P R, we have rf P R-mulpΠiPr1,nsMi ,Xq.

(3) We have an R-module R-mulpΠiPr1,nsMi ,Xq with respect to the operations in (1, 2).

Example 11. We examine Deﬁnition 7 in the case n “ 0.

We have iPr1,ns Mi “ tp qu “: N, where p q is the empty tuple. Since the condition a map f : N Ñ X has to fulﬁll to be R-multilinear is empty, we ﬁnd that for m P M the map f : N Ñ M,ś p q ÞÑ m is R-multilinear.

13 pf q i iPr1,ns ˆn Lemma 12. Suppose given pXiqiPr1,ns ÝÝÝÝÝÑ pYiqiPr1,ns in pR -Modq . Suppose given h N P Ob R -Mod and g P R-mulpΠiPr1,nsYi ,Nq. Suppose given N ÝÑ P in R -Mod. Then

fi gh : Xi Ñ P ¨ ˛ iPr1,ns iPr1,ns ź ź is R-multilinear. ˝ ‚

1 2 1 2 Proof. Suppose given r , r P R. Suppose given i P r1, ns. Suppose given xi, xi P Xi . 2 1 Suppose given xj P Xj for j P r1, nsztiu. For j P r1, nsztiu, we deﬁne xj :“ xj :“ xj . We have

1 1 2 2 px1 , . . . ,xi 1 , r x ` r x , xi 1 , . . . , xnq fi gh ´ i i ` ¨¨ ˛ ˛ iPr1,ns ź 1 1 ˝2 ˝2 ‚ ‚ “ px1f1 , . . . , xi´1fi´1 , pr xi ` r xi qfi , xi`1fi`1, . . . , xnfnqgh 1 1 2 2 “ px1f1 , . . . , xi´1fi´1 , r pxifiq ` r pxi fiq, xi`1fi`1, . . . , xnfnqgh 1 1 2 2 “ pr ppxjfjqjPr1,nsgq ` r ppxj fjqjPr1,nsgqqh 1 1 2 2 “ r pxjfjqjPr1,nsgh ` r pxj fjqjPr1,nsgh

1 1 2 2 “ r px q fi gh ` r px q fi gh . ¨ j jPr1,ns ¨¨ ˛ ˛˛ ¨ j jPr1,ns ¨¨ ˛ ˛˛ iPr1,ns iPr1,ns ź ź ˝ ˝˝ ‚ ‚‚ ˝ ˝˝ ‚ ‚‚

Deﬁnition 13. Let C be the free R-module on the set iPr1,ns Mi . Let D be the R-submodule of C generated by the elements of the form ś 1 1 2 2 pm1, . . . , mi´1, r mi ` r mi , mi`1, . . . , mnq 1 1 ´r pm1, . . . , mi´1, mi , mi`1, . . . , mnq 2 2 ´r pm1, . . . , mi´1, mi , mi`1, . . . , mnq 1 2 1 2 for i P r1, ns, r , r P R, mi , mi P Mi , mj P Mj for j P r1, nsztiu.

Deﬁne the tensor product over R of M1,...,Mn as iPr1,ns Mi :“ C{D. R We also write M1 b ¨ ¨ ¨ b Mn :“ iPr1,ns Mi for theÂ tensor product. R R R Â If unambiguous, we often write M1 b ¨ ¨ ¨ b Mn :“ iPr1,ns Mi :“ iPr1,ns Mi . R Â Â For pm1, . . . , mnq P iPr1,ns Mi , we write

śm1 b ¨ ¨ ¨ b mn :“ biPr1,nsmi :“ pm1, . . . , mnq ` C. Let µ pM1,...,Mnq Mi ÝÝÝÝÝÝÝÑ iPr1,ns Mi , R iPr1,ns ź Â pm1, . . . , mnq ÞÑ biPr1,nsmi .

If unambiguous, we often write µ :“ µpM1,...,Mnq .

14 Lemma 14. The following assertions (1, 2) hold.

(1) The map µ is R-multilinear.

(2) The set Im µ “ biPr1,nsmi : pm1, . . . , mnq P iPr1,ns Mi is an R-linear generating

set for iPr1,ns M!i . ) R ś Â 1 2 Proof. Ad (1). Suppose given i P r1, ns. Suppose given mi , mi P Mi . Suppose given 1 2 2 1 r , r P R. Suppose given mj P Mj for j P r1, nsztiu. Let mj :“ mj :“ mj for j P r1, nsztiu. We have 1 1 2 2 pm1, . . . , mi´1, r mi ` r mi , mi`1, . . . , mnqµ 1 1 2 2 “ m1 b ¨ ¨ ¨ b mi´1 b pr mi ` r mi q b mi`1 b ¨ ¨ ¨ b mn 1 1 2 2 “ r bjPr1,nsmj ` r bjPr1,nsmj r1 m1 µ r2 m2 µ. “ `p jqjPr1,ns˘ ` ` p j qjPr1,n˘s Therefore, µ is R-multilinear.` ˘ ` ˘

Ad (2). Since iPr1,ns Mi is an R-linear basis of C, we obtain the R-linear generating set ś Im µ “ m b ¨ ¨ ¨ b m : pm , . . . , m q P M $ 1 n 1 n i, iPr1,ns & ź . % - “ pm , . . . , m q ` D : pm , . . . , m q P M $ 1 n 1 n i, iPr1,ns & ź . for iPr1,ns Mi “ C{D. % - ExampleÂ 15. We examine Deﬁnition 13 in the case n “ 0.

Then we have C “ Rxp qy – R and D “ 0. Therefore, we obtain iPr1,0s Mi – R. R Â Lemma 16. Let Y P Ob R -Mod. Suppose given f P R-mulp iPr1,ns Mi,Y q. ¯ ¯ There exists a unique R-linear map f : iPr1,ns Mi Ñ Y suchś that µf “ f. R Â Proof. Since iPr1,ns Mi is an R-linear basis of C, the map f extends to an R-linear map ˆ f : C Ñ Y, pm1, . . . , mnq ÞÑ pm1, . . . , mnqf. Since f is R-multilinear, D is contained in ˆ śˆ ¯ Ker f. Thus, f induces a unique R-linear map f : iPr1,ns Mi Ñ X with ¯ Âˆ pm1 b ¨ ¨ ¨ b mnqf “ pm1, . . . , mnqf “ pm1, . . . , mnqf ¯ for pmiqiPr1,ns P iPr1,ns Mi . Therefore, we have µf “ f.

Suppose given anś R-linear map g : iPr1,ns Mi Ñ X such that µg “ f. Then we have µg “ f “ µf¯. Since Im µ is an R-linear generating set for M by Lemma 14.(2), Â iPr1,ns i we obtain g “ f. Â

15 Lemma 17. We have the R-linear isomorphism

R-mulp iPr1,ns Mi ,Xq ÐÝ R -Modp iPr1,ns Mi ,Mq R ś µg ÞÝÑ g. Â

Proof. By Lemma 14.(1) and Lemma 12, this map is welldeﬁned. By Lemma 16, this map is bijective. We show that it is R-linear.

1 2 Suppose given r , r P R. Suppose given f, g P R -Modp iPr1,ns Mi ,Mq. Suppose given R mi P Mi for i P r1, ns. We have Â

1 2 1 2 pm1, . . . , mnqµpr f ` r gq “ pm1 b ¨ ¨ ¨ b mnqpr f ` r gq 1 2 “ pm1 b ¨ ¨ ¨ b mnqpr fq ` pm1 b ¨ ¨ ¨ b mnqpr gq 1 2 “ r pm1 b ¨ ¨ ¨ b mnqf ` r pm1 b ¨ ¨ ¨ b mnqg 1 2 “ r pm1, . . . , mnqµf ` r pm1, . . . , mnqµg 1 2 “ pm1, . . . , mnqpr µfq ` pm1, . . . , mnqpr µgq 1 2 “ pm1, . . . , mnqpr µf ` r µgq.

Remark 18. Suppose that n ě 1. Suppose given x P iPr1,ns Mi. There exist s P Zě0 R and mi,j P Mi for pi, jq P r1, ns ˆ r1, ss with Â

x “ biPr1,nsmi,j . jPr1,ss ÿ

Proof. Since Im µ is an R-linear generating set for iPr1,ns Mi , there exist s P Zě0 , rj P R for j P r1, ss and m P M for pi, jq P r1, ns ˆ r1, ss with i,j i Â

x “ rj ¨ pm1,j b ¨ ¨ ¨ b mn,jq “ prjm1,jq b ¨ ¨ ¨ b mn,j . jPr1,ss jPr1,ss ÿ ÿ

Lemma 19. Suppose given j P r2, ns. There exists a unique R-linear isomorphism

- ψ : Mi b Mi „ Mi iPr1, j ´ 1s R iPrj, ns iPr1, ns ˆR ˙ ˆR ˙ R â â â with

pm1 b ¨ ¨ ¨ b mj´1q b pmj b ¨ ¨ ¨ b mnq ψ “ m1 b ¨ ¨ ¨ b mn

for pm1 , . . . , mn`q P iPr1,ns Mi . ˘ ś Proof. Deﬁne

T1 :“ Mi , T2 :“ Mi , T :“ Mi . iPr1,j´1s iPrj,ns iPr1,ns â â â 16 Let µ :“ µM1,...,Mn : iPr1,ns Mi Ñ T, µ :“ µ : M Ñ T , 1 M1,...,Mj´1 ś iPr1,j´1s i 1 µ :“ µ : M Ñ T , 2 Mj ,...,Mn śiPrj,ns i 2 µ :“ µ : T ˆ T Ñ T b T , 3 T1,T2 1 ś 2 1 2 µ˜ :“ pµ1 ˆ µ2qµ3 : iPr1,ns Mi Ñ T1 b T2 ; cf. Deﬁnition 13. ś We show thatµ ˜ is R-multilinear.

1 2 1 2 Suppose given r , r P R. Suppose given k P r1, ns. Suppose given mk, mk P Mk. Suppose 1 2 given mi P Mi for i P r1, nsztku. For i P r1, nsztku we deﬁne mi :“ mi :“ mi. Case k P r1, j ´ 1s. We have

1 1 2 2 pm1, . . . ,mk´1, r mk ` r mk, mk`1, . . . , mnqµ˜ 1 1 2 2 “ ppm1, . . . , mk´1, r mk ` r mk, mk´1, . . . , mj´1qµ1, pmj, . . . , mnqµ2q µ3 1 1 2 2 “ pm1 b ¨ ¨ ¨ b mk´1 b pr mk ` r mkq b mk`1 ¨ ¨ ¨ b mj´1q b pbiPrj,nsmiq 1 1 2 2 “ r ¨ pbiPr1,j´1smiq ` r ¨ pbiPr1,j´1smi q b pbiPrj,nsmiq 1 1 2 2 “ `r pbiPr1,j´1smiq b pbiPrj,nsmiq ` r pb˘ iPr1,j´1smi q b pbiPrj,nsmiq 1 1 2 2 “ r `¨ pmiqiPr1,ns µ˜ ` r ¨ pmi qiPr˘ 1,ns µ`˜ . ˘ Case k P rj, ns. This` follows analoguosly˘ ` to the ﬁrst˘ case. Therefore,µ ˜ is R-multilinear.

By Lemma 16, there exists a unique R-linear map ϕ : T Ñ T1 b T2 with

pbiPr1,nsmiqϕ “ pm1, . . . , mnqµ˜ “ pbiPr1,j´1smiq b pbiPrj,nsmiq

for pm1, . . . , mnq P iPr1,ns Mi.

For m “ pm1, . . . , mśj´1q P iPr1,j´1s Mi , we deﬁne ś mf : Mi Ñ T, paj, . . . , anq ÞÑ pm1, . . . , mj´1, aj, . . . , anqµ . iPrj,ns ź First we show that mf is R-multilinear. Suppose given r1, r2 P R. Suppose given k P rj, ns. 1 2 1 2 Suppose given ak, ak P Mk. Suppose given ai P Mi for i P rj, nsztku. Deﬁne ai :“ ai :“ ai 1 1 1 2 2 2 for i P rj, nsztku. Let a “ paj, . . . , anq and a :“ paj , . . . , anq. We have

1 2 2 2 paj, . . . , ak´1,r ak ` r ak, ak`1, . . . , anqpmfq 1 1 2 2 “ pm1, . . . , mj´1, aj, . . . , ak´1, r ak ` r ak, ak`1, . . . , anqµ “ r1pm, a1qµ ` r2pm, a2qµ “ r1pa1pmfqq ` r2pa2pmfqq . Thus, mf is R-multilinear. Therefore, we have the map

f : Mi Ñ R-mul Π Mi,T , m ÞÑ mf; iPrj,ns iPr1,j´1s ź ˆ ˙ 17 cf. Deﬁnition 9. Now we show that f is R-multilinear.

1 2 1 2 Suppose given r , r P R. Suppose given k P r1, j ´ 1s. Suppose given mk, mk P Mk. 1 2 Suppose given mi P Mi for i P r1, j ´ 1sztku. For i P r1, j ´ 1sztku deﬁne mi :“ mi :“ mi. 1 1 1 2 2 2 Furthermore, we deﬁne m “ pm1, . . . , mj´1q and m :“ pm1, . . . , mj´1q.

Suppose given a P iPrj,ns Mi. We have

ś 1 1 2 2 a pm1, . . . , mk´1,r mk ` r mk, mk`1, . . . , mj´1qf m , . . . , m , r1m1 r2m2, m , . . . , m , a µ ` “ p 1 k´1 k ` k ˘ k`1 j´1 q “ r1pm1, aqµ ` r2pm2, aqµ “ r1papm1fqq ` r2papm2fqq “ apr1pm1fq ` r2pm2fqq .

Thus, we have

1 1 2 2 1 1 2 2 pm1, . . . , mk´1, r mk ` r mk, mk`1, . . . , mkqf “ r pm fq ` r pm fq. Therefore, f is R-multilinear. ˆ By Lemma 16, there exists a unique R-linear map f : T1 Ñ R-mulpΠiPrj,nsMi,T q with ˆ µ1f “ f. So, ˆ pbiPr1,j´1smiqf : pmiqiPrj,ns ÞÑ biPr1,nsmi. ˆ Suppose given x P T1. Since xf P R-mulpΠiPrj,nsMi ,T q, there exists a unique R-linear map ˆ ˆ ˆ xf : T2 Ñ T with µ2xf “ xf; cf. Lemma 16.(1). Deﬁne ˆ g : T1 ˆ T2 Ñ T, px, yq ÞÑ y xf . We show that g is R-bilinear. ´ ¯

1 2 1 2 1 2 Suppose given r , r P R. Suppose given x, x , x P T1. Suppose given y, y , y P T2. Since fˆ is R-linear, we have

pr1x1 ` r2x2, yqg “ y pr1x1 ` r2x2qfˆ ´ ¯ “ y r1px1fˆq ` r2px2fˆq L“17 y´pr1 ¨ x1fˆ` r2 ¨ x2fˆq¯ D“9 r1 ¨ y x1fˆ ` r2 ¨ y x2fˆ “ r1 ¨ px´1, yq¯g ` r2 ¨ px´2, yqg.¯

Since xfˆ is R-linear, we have

px, r1y1 ` r2y2qg “ pr1y1 ` r2y2qxfˆ “ r1 ¨ y1 xfˆ ` r2 ¨ y2 xfˆ “ r1 ¨ px, y1qg ` r2 ¨ px, y2qg. ´ ¯ ´ ¯ Thus, g is R-bilinear.

Therefore, there exists a unique R-linear map ψ : T1bT2 Ñ T with µ3ψ “ g; cf. Lemma 16.

18 Suppose given mi P Mi for i P r1, ns. We have

ppbiPr1,j´1smiq b pbiPrj,nsmiqqψ “ ppbiPr1,j´1smiq, pbiPrj,nsmiqqg ˆ “ pbiPrj,nsmiqpbiPr1,j´1smiqf ˆ “ pmj, . . . , mnqppbiPr1,j´1smiqfq

“ pmj, . . . , mnqppm1, . . . , mj´1qfq

“ pm1, . . . , mj´1, mj, . . . , mnqµ

“ biPr1,nsmi . Thus, we have

pbiPr1,nsmiqpϕψq “ ppbiPr1,j´1smiq b pbiPrj,nsmiqqψ “ biPr1,nsmi and

ppbiPr1,j´1smiq b pbiPrj,nsmiqqpψϕq “ pbiPr1,nsmiqϕ “ pbiPr1,j´1smiq b pbiPrj,nsmiq.

Since Im µ is an R-linear generating set of T and Im µ3 is an R-linear generating set of

T1 b T2 ; cf. Lemma 14.(2); we have ϕψ “ 1T and ψϕ “ 1T1bT2 . Thus, ϕ and ψ are mutually inverse R-linear isomorphisms.

1.3.2 The tensor product of linear maps

For this §1.3.2, let n P Zě0.

pf q i iPr1,ns ˆn Deﬁnition 20 (and Lemma). Suppose given pMiqiPr1,ns ÝÝÝÝÝÑpNiqiPr1,ns in pR -Modq . We have the R-linear map

biPr1,nsfi iPr1,ns Mi ÝÝÝÝÝÑ iPr1,ns Ni , R R Â Â biPr1,nsmi,j ÞÝÑ biPr1,nspmi,jfiq . jPr1,ms jPr1,ms ÿ ÿ We often write f1 b ... b fn :“ bifi :“ biPr1,nsfi .

Proof. By Lemma 12,

p iPr1,ns fiqµN1,...,Nn Mi ÝÝÝÝÝÝÝÝÝÝÝÝÝÑ Ni , ś iPr1,ns iPr1,ns ź â pmiqiPr1,ns ÞÝÑ biPr1,nspmifiq,

is R-multilinear. For brevity, we write f :“ iPr1,ns fi µN1,...,Nn . By Lemma 16, there exists a unique R-linear´ś map ¯ ¯ f : Mi Ñ Ni iPr1,ns iPr1,ns â â 19 ¯ with pbiPr1,nsmiqf “ pmiqiPr1,nsf “ biPr1,nspmifiq for pmiqiPr1,ns P iPr1,ns Mi .

Suppose given jPr1,mspbiPr1,nsmi,jq P iPr1,ns Mi . We have ś ř Â ¯ ¯ b mi,j f “ pb mi,jqf “ b pmi,jfiq. ¨ iPr1,ns ˛ iPr1,ns iPr1,ns jPr1,ms jPr1,ms jPr1,ms ÿ ÿ ÿ ˝ ‚ ¯ Thus, the assertion follows with biPr1,nsfi :“ f. Deﬁnition 21 (and Lemma). We have the functor

iPr1,ns ˆn R pR -Modq ÝÝÝÝÑÂ R -Mod,

pfiqiPr1,ns biPr1,nsfi pMiqiPr1,ns ÝÝÝÝÝÑpNiqiPr1,ns ÞÑ iPr1,ns Mi ÝÝÝÝÝÑ iPr1,ns Ni . ˆ ˙ ˆ R R ˙ Â Â If unambiguous, we often write :“ iPr1,ns :“ iPr1,ns. R Â Â Â Proof. Suppose given

pfiqiPr1,ns pgiqiPr1,ns pMiqiPr1,ns ÝÝÝÝÝÑpNiqiPr1,ns ÝÝÝÝÝÑpPiqiPr1,ns in pR -Modqˆn. We have b 1pMiqiPr1,ns “ p1Mi qiPr1,ns ÞÝÑ biPr1,ns1Mi .

Suppose given biPr1,nsmi P iPr1,ns Mi . We have m 1 Â m 1 m m 1 . pbiPr1,ns iqpbiPr1,ns Mi q “ biPr1,nsp i Mi q “ biPr1,ns i “ pbiPr1,ns iq iPr1,ns Mi Â By Lemma 14.(2), we have

1 b 1 1 . pMiqiPr1,ns ÞÝÑ biPr1,ns Mi “ iPr1,ns Mi Â Suppose given biPr1,nsmi P iPr1,ns Mi. We have Â pbiPr1,nsmiqppbiPr1,nsfiqpbiPr1,nsgiqq “ ppbiPr1,nsmiqpbiPr1,nsfiqqpbiPr1,nsgiq

“ pbiPr1,nspmifiqqpbiPr1,nsgiq

“ biPr1,nsppmifiqgiq

“ biPr1,nsppmiqpfigiqq

“ pbiPr1,nsmiqpbiPr1,nspfigiqq. By Lemma 14.(2), we have

b ppfiqiPr1,ns ¨ pgiqiPr1,nsq ÞÝÑ pbiPr1,nsfiqpbiPr1,nsgiq b “ biPr1,nspfigiq ÞÝÑ pfigiqiPr1,ns . Thus, we have a functor indeed.

Â 20 1.4 Preadditive categories over a commutative ring

In this §1.4 we establish the notion of a preadditive category over a commutative ring. For this type of categories we shall deﬁne the tensor product below; cf. §3.1.

For this §1.4, let R be a commutative ring.

Remark 22. Let A be a preadditive category. Then End 1A “ rA,Asp1A, 1Aq is a commutative ring.

Proof. Recall that End 1A is a ring. We have to prove commutativity.

Suppose given α, β P End 1A. Suppose given X P Ob A. We have βX P ApX,Xq. Since α is natural, we have the following commutative diagram.

1AβX 1AX / 1AX

αX αX

1AβX 1AX / 1AX

Thus, we have αX βX “ βX αX . Since X P Ob A was arbitrary, we have αβ “ βα. Therefore, End 1A is commutative.

Deﬁnition 23.

(1) Let A be a preadditive category. Let ϕ : R Ñ End 1A be a ring morphism. We call pA, ϕq a preadditive category over R or an R-linear preadditive category. We often refer to just A as a preadditive category over R. f g Suppose given X ÝÑ Y ÝÑ Z in A. Suppose given r P R. We often write

r ¨ f :“ rf :“ prϕqX f.

We have

rpfgq “ prϕqX pfgq “ pprϕqX fqg “ prfqg

“ fprϕqY g “ fpprϕY qgq “ fprgq.

(2) Let pA, ϕq be a preadditive category over R. Let A be additive. We call pA, ϕq an additive category over R or an R-linear additive category.

(3) Let pC, ϕq and pD, ψq be preadditive categories over R. Let F P addrC, Ds. We call F an R-linear functor if

F prfq “ F pprϕqX ¨ fq “ prψqFX ¨ F f “ rpF fq

f for X ÝÑ Y in C and r P R.

21 Remark 24. Let pA, ϕq be an additive category over R. Suppose given

pfi,j qi,j A1,i ÝÝÝÝÑ A2,j iPr1,ms jPr1,ns à à in A. Suppose given r P R. We have

r ¨ pfi,jqi,j “ pr ¨ fi,jqi,j .

Proof. We have

pr ¨ fi,jqi,j “ pr ¨ ι1,ifπ2,jqi,j

“ pprϕqA1,i ι1,ifπ2,jqi,j

“ pι1,iprϕqA1,1‘¨¨¨‘A1,m fπ2,jqi,j

“ pι1,ipr ¨ fqπ2,jqi,j

“ ppr ¨ fqi,jqi,j

“ r ¨ pfi,jqi,j

Remark 25. Let pC, ϕq and pD, ψq be preadditive categories over R. Let F P ObrC, Ds. The following assertions (1, 2) are equivalent.

(1) The functor F is R-linear.

f1 1 / 2 (2) For C / C in C and r1, r2 P R, we have F pr1f1 ` r2f2q “ r1F f1 ` r2F f2 . f2

Remark 26. Let pA, ϕq be a preadditive category over R. Suppose given X,Y P Ob A. We have the ring morphism

ε : End 1A Ñ EndZ p ApX,Y qq

f α f fα α ÞÑ pX ÝÑ Y q ÞÑ pX ÝÝÑX Y q “ pX ÝÝÑY Y q . ´ ¯ Therefore, we have the R-module p ApX,Y q, ϕεq, i.e. for r P R and f P ApX,Y q, we have a module operation r ¨ f :“ prϕqX f “ fprϕqY .

Proof. Suppose given α P End 1A. Since α is natural, we have αX f “ fαY for f P ApX,Y q.

The map f ÞÑ αX f is a Z-linear endomorphism of ApX,Y q since A is preadditive. Thus, ε is a welldeﬁned map. We show that ε is a ring morphism. We have fp1 εq “ 1 f “ f for f P pX,Y q. Thus, we have 1 ε “ 1 . End 1A X A End 1A EndZp ApX,Y qq

Suppose given α, β P End 1A . We have

fppα ` βqεq “ pα ` βqX f “ pαX ` βX qf “ αX f ` βX f “ fpαεq ` fpβεq “ fpαε ` βεq

22 and fppαβqεq “ pαβqX f “ αX βX f “ αX fβY “ pfpαεqqβY “ fppαεqpβεqq for f P ApX,Y q. Thus, ε is a ring morphism.

Therefore, we have a ring morphism ϕε : R Ñ EndZ p ApX,Y qq.

Thus, we have an R-module p ApX,Y q, ϕεq. Deﬁnition 27. Let pA, ϕq, pB, ψq and pC, τq be preadditive categories over R. Let F : A ˆ B ÝÑ C be a functor. The functor F is called R-bilinear if

F pr1f1 ` r2f2, gq “ r1F pf1, gq ` r2F pf2, gq,

F pf, r1g1 ` r2g2q “ r1F pf, g1q ` r2F pf, g2q

f1 g1 1 / 2 3 f 4 1 / 2 3 g 4 for r1, r2 P R, A / A and A ÝÑ A in A and for B / B and B ÝÑ B in B. f2 g2

Remark 28. Let pA, ϕq, pB, ψq and pC, τq be preadditive categories over R. Let F : A ˆ B ÝÑ C be a functor. The following assertions (1, 2) are equivalent.

(1) The functor F is R-bilinear.

(2) Suppose given A P Ob A and B P Ob B. Then

1 g 2 1 F pA,gq “ F p1A,gq 2 FA : B Ñ C, B ÝÑ B ÞÑ F pA, B q ÝÝÝÝÝÝÝÝÝÝÑ F pA, B q ,

1 f 2 1 F pf,Bq “ F pf,1B q 2 FB : A Ñ C,` A ÝÑ A ˘ ÞÑ` F pA ,Bq ÝÝÝÝÝÝÝÝÝÝÑ F pA ,Bq˘ are R-linear functors. ` ˘ ` ˘

Proof. Ad (1) ñ (2). Because of the symmetry of the situation, we only have to show that FB is an R-linear functor.

1 f1 2 f2 3 First of all, we show that FB is a functor. Suppose given A ÝÑ A ÝÑ A in A. We have

FBpf1f2q “ F pf1f2, 1Bq “ F ppf1, 1Bq ¨ pf2, 1Bqq “ F pf1, 1Bq ¨ F pf2, 1Bq “ FBf1 ¨ FBf2 . Furthermore, we have

1 1 1 1 FB1A “ F p1A , 1Bq “ 1F pA ,Bq “ 1FB A .

Thus, FB is a functor.

f1 1 / 2 Now we show that FB is R-linear. Suppose given A / A in A and r1, r2 P R. We f2 have

FBpr1f1 ` r2f2q “ F pr1f1 ` r2f2, 1Bq “ r1F pf1, 1Bq ` r2F pf2, 1Bq “ r1FBf1 ` r2FBf2 .

Thus, FB is R-linear by Remark 25. f1 g1 1 / 2 3 f 4 1 / 2 Ad (2) ñ (1). Suppose given r1, r2 P R, A / A and A ÝÑ A in A, B / B g g2 and B3 ÝÑ B4 in B. f2

23 We have

F pr1f1 ` r2f2, gq “ F ppr1f1 ` r2f2, 1B3 q ¨ p1A2 , gqq

“ F pr1f1 ` r2f2, 1B3 q ¨ F p1A2 , gq

“ FB3 pr1f1 ` r2f2q ¨ F p1A2 , gq

“ pr1FB3 f1 ` r2FB3 f2q ¨ F p1A2 , gq

“ pr1F pf1, 1B3 q ` r2F pf2, 1B3 qq ¨ F p1A2 , gq

“ r1F pf1, 1B3 q ¨ F p1A2 , gq ` r2F pf2, 1B3 q ¨ F p1A2 , gq

“ r1F pf1, gq ` r2F pf2, gq . Similarly, we obtain

F pf, r1g1 ` r2g2q “ F pp1A3 , r1g1 ` r2g2q ¨ pf, 1B2 qq

“ F p1A3 , r1g1 ` r2g2q ¨ F pf, 1B2 q

“ FA3 pr1g1 ` r2g2q ¨ F pf, 1B2 q

“ pr1FA3 g1 ` r2FA3 g2q ¨ F pf, 1B2 q

“ pr1F p1A3 , g1q ` r2F p1A3 , g2qq ¨ F pf, 1B2 q

“ r1F p1A3 , g1q ¨ F pf, 1B2 q ` r2F p1A3 , g2q ¨ F pf, 1B2 q

“ r1F pf, g1q ` r2F pf, g2q . Thus, F is R-bilinear. Remark 29. Let pA, ϕq, pB, ψq, pC, τq, pD, δq, pE, ηq and pF, ζq be preadditive categories over R. Suppose given R-linear functors A ÝÑF C and B ÝÑG D. Suppose given an R-bilinear functor C ˆ D ÝÑH E. Suppose given an R-linear functor E ÝÑK F. Then K˝H˝pF ˆGq A ˆ B ÝÝÝÝÝÝÝÝÑ F is R-bilinear.

f1 g1 1 / 2 3 f 4 1 / 2 3 g 4 Proof. Suppose given r1, r2 P R, A / A and A ÝÑ A in A, B / B and B ÝÑ B in B. f2 g2 We have pK ˝ H ˝ pF ˆ Gqqpf, r1g1 ` r2g2q “ pK ˝ HqpF f, Gpr1g1 ` r2g2qq “ pK ˝ HqpF f, r1Gg1 ` r2Gg2q “ Kpr1HpF f, Gg1q ` r2HpF f, Gg2qq “ r1pK ˝ HqpF f, Gg1q ` r2pK ˝ HqpF f, Gg2q “ r1pK ˝ H ˝ pF ˆ Gqqpf, g1q ` r2pK ˝ H ˝ pF ˆ Gqqpf, g2q and pK ˝ H ˝ pF ˆ Gqqpr1f1 ` r2f2, gq “ pK ˝ HqpF pr1f1 ` r2f2q, Ggq “ pK ˝ Hqpr1F f1 ` r2F f2, Ggq “ Kpr1HpF f1, Ggq ` r2HpF f2, Ggqq “ r1pK ˝ HqpF f1, Ggq ` r2pK ˝ HqpF f2, Ggq “ r1pK ˝ H ˝ pF ˆ Gqqpf1, gq ` r2pK ˝ H ˝ pF ˆ Gqqpf2, gq.

24 Lemma 30. Suppose given preadditive categories pA, ϕAq and pB, ϕBq over R. Suppose given an R-linear functor F : A Ñ B. Suppose given an adjunction pF, G, η, εq. Then G is R-linear.

Proof. We show that ψB ApGB, GBq ÝÝÑ BpF GB, Bq,

u ÞÑ F u ¨ εB is an R-linear map for B P Ob B. 1 1 Suppose given B P Ob B. Suppose given u, u P ApGB, GBq. Suppose given r, r P R. Since F is R-linear, we have

1 1 1 1 pru ` r u qψB “ F pru ` r u q ¨ εB 1 1 “ prF u ` r F u qεB 1 1 “ rF u ¨ εB ` r F u ¨ εB 1 1 “ rpuψBq ` r pu ψBq.

Thus, ψB is R-linear. For B P Ob B, let ˜ ψB ApGB, GBq ÐÝÝ BpF GB, Bq,

ηGB ¨ Gv ÞÑ v . We show that we have mutually inverse R-linear isomorphisms

ψB „ / ApGB, GBq o BpF GB, Bq ˜ ψB for B P Ob B.

Suppose given B P Ob B. Suppose given u P ApGB, GBq and v P BpF GB, Bq. We have ˜ ˜ uψBψB “ pF u ¨ εBqψB

“ ηGB ¨ GpF u ¨ εBq

“ ηGB ¨ GF u ¨ GεB

“ u ¨ ηGB ¨ GεB “ u and ˜ vψBψB “ pηGB ¨ GvqψB

“ F pηGB ¨ Gvq ¨ εB

“ F ηGB ¨ F Gv ¨ εB

“ F ηGB ¨ εF GB ¨ v “ v.

25 ˜ ˜ Thus, we have mutually inverse bijections ψB and ψB. Since ψB is R-linear, so is ψB. ´ ˜ For B P Ob B, we now write ψB “ ψB. b / ˜ 1 Suppose given B / B in B. Suppose given r, r P R. b1 We have to show that Gprb ` r1b1q “! rGb ` r1Gb1; cf. Remark 25. We have

1 1 1 1 Gprb ` r b q “ ηGB ¨ GεB ¨ Gprb ` r b q 1 1 ´ “ pεB ¨ prb ` r b qqψB 1 1 ´ “ prεBb ` r εBb qψB ´ 1 1 ´ “ rpεBbqψ ` r pεBb qψB 1 1 “ r ¨ ηGB ¨ GεB ¨ Gb ` r ¨ ηGB ¨ GεB ¨ Gb “ rGb ` r1Gb1.

Corollary 31. Suppose given preadditive categories pA, ϕAq and pB, ϕBq over R. Suppose given an R-linear functor F : A Ñ B. Suppose given a functor G : B Ñ A. Suppose given α β isotransformations pF ˝ Gq „ +3 1B and pG ˝ F q „ +3 1A . Then G is R-linear.

Proof. By [6, Remark 16.5.9], we have F % G. Thus, the claim follows from Lemma 30.

Remark 32. Suppose pA, ϕAq to be a preadditive category over R. Suppose B to be a full subcategory of A. Then B is a preadditive category by restriction. We have the ring morphism

ψ End 1A ÝÑ End 1B ,

pαX qXPOb A ÞÑ pαX qXPOb B .

Thus, pB, ϕA ¨ ψq is a preadditive category over R.

26 Chapter 2

Envelope operations

2.1 The Karoubi envelope

2.1.1 Deﬁnition and duality

For this §2.1.1, let A be a category.

Definition 33 (and Lemma). We shall define a category Kar A as follows. Let 2 Ob Kar A :“ tpX, eq : X P Ob A, e P ApX,Xq with e “ eu. For pX, eq and pY, fq in Ob Kar A, we define

Kar AppX, eq, pY, fqq :“ t ep ϕ qf : ϕ P ApX,Y q and ϕ “ eϕfu.

If unambiguous, we often write ϕ :“ ep ϕ qf for ep ϕ qf P Kar AppX, eq, pY, fqq. For pX, eq, pY, fq and pZ, gq in Ob Kar A,

ep ϕ qf P Kar AppX, eq, pY, fqq and fp ψ qg P Kar AppY, fq, pZ, gqq, we deﬁne composition by ep ϕ qf fp ψ qg :“ ep ϕψ qg . For pX, eq P Ob Kar A, we deﬁne

1pX,eq :“ ep e qe .

We call Kar A the Karoubi envelope of A. This deﬁnes a category Kar A.

ep ϕ q p ψ qg p ρ q Proof. Suppose given pX, eq ÝÝÝÝÑpf Y, fq ÝÝÝÝÑpf Z, gq ÝÝÝÑpg h W, hq in Kar A. ϕ ψ ρ Thus, we have X ÝÑ Y ÝÑ Z ÝÑ W in A with

eϕf “ ϕ, fψg “ ψ and gρh “ ρ.

27 Using these equalities, we obtain eϕψg R“2 ϕψ. Therefore, we have

ep ϕψ qg P Kar AppX, eq, pZ, gqq.

Moreover, we obtain

p ep ϕ qf fp ψ qgq gp ρ qh “ ep ϕψ qg gp ρ qh “ ep pϕψqρ qh

“ ep ϕpψρq qh “ ep ϕ qf fp ψρ qh “ ep ϕ qf p fp ψ qg gp ρ qhq.

We have 1pY,fq “ fp f qf P Kar AppY, fq, pY, fqq, because f “ fff. Furthermore, we have

R2 1pY,fq fp ψ qg “ fp f qf fp ψ qg “ fp fψ qg “ fp ψ qg .

Similarily, we obtain

R2 ep ϕ qf 1pY,fq “ ep ϕ qf fp f qf “ ep ϕf qf “ ep ϕ qf .

Lemma 34. We have the isomorphism of categories

pKarpAqq˝ - KarpA˝q

˝ ep ϕ q ˝p ϕ˝ q ˝ pX, eq ÐÝÝÝÝpf Y, fq - pX, e˝q ÐÝÝÝÝÝpf e Y, f ˝q ˆ ˙ ˆ ˙ ˝ ep ϕ q ˝p ϕ˝ q ˝ pX, eq ÐÝÝÝÝf pY, fq pX, e˝q ÐÝÝÝÝÝf e pY, f ˝q . ˆ ˙ ˆ ˙

e f Proof. Suppose given idempotents X ÝÑ X and Y ÝÑ Y in A. Suppose given ϕ P ApX,Y q. Assume eϕf “ ϕ. Then we have f ˝ϕ˝e˝ “ peϕfq˝ “ ϕ˝. Assume f ˝ϕ˝e˝ “ ϕ˝. This implies

eϕf “ ppeϕfq˝q˝ “ pf ˝ϕ˝e˝q˝ “ pϕ˝q˝ “ ϕ.

Thus, we have eϕf “ ϕ if and only if f ˝ϕ˝e˝ “ ϕ˝. Therefore, we obtain welldeﬁned maps on objects and morphisms in both directions.

p ψ q ep ϕ q Suppose given pW, dq ÝÝÝÑpd e X, eq ÝÝÝÝÑpf Y, fq in Kar A. We have

˝ ˝ ˝ D33 ˝ ep ϕ qf dp ψ qe “ p dp ψ qe ep ϕ qf q “ dp ψϕ qf ˝ ˝ ˝ D33 ˝ ˝ ÞÑ f ˝p pψϕq qd˝ “ f ˝p ϕ ψ qd˝ “ f ˝p ϕ qe˝ e˝p ψ qd˝

28 and

˝ ˝ ˝ ˝ D33 ˝ f ˝p ϕ qe˝ e˝p ψ qd˝ “ f ˝p ϕ ψ qd˝ “ f ˝p pψϕq qd˝ ˝ ˝ D33 ˝ ˝ ÞÑ dp ψϕ qf “ p dp ψ qe ep ϕ qf q “ ep ϕ qf dp ψ qe .

˝ D33 ˝ In pKar Aq we have 1pX,eq “ ep e qe .

˝ D33 ˝ In KarpA q we have 1pX,e˝q “ e˝p e qe˝ . We obtain ˝ ˝ 1pX,eq “ ep e qe ÞÑ e˝p e qe˝ “ 1pX,e˝q and ˝ ˝ 1pX,e˝q “ e˝p e qe˝ ÞÑ ep e qe “ 1pX,eq . Therefore, we have functors indeed. They are mutual inverses by deﬁnition. Thus, we have isomorphisms of categories.

ϕ- Remark 35. Suppose given an isomorphism X „ Y in A. Suppose given idempotents e P ApX,Xq and f P ApY,Y q with eϕ “ ϕf. We have mutually inverse isomorphisms

ep eϕf qf pX, eq „ / pY, fq o ´1 fp fϕ e qe

in Kar A.

Proof. We have eeϕff “ eϕf and ffϕ´1ee “ fϕ´1e. ´1 Therefore, we have ep eϕf qf P Kar AppX, eq, pY, fqq and fp fϕ e qe P Kar AppY, fq, pX, eqq. Furthermore, we have

´1 D33 ´1 ´1 ep eϕf qf fp fϕ e qe “ ep eϕffϕ e qe “ ep eϕfϕ e qe

´1 D33 “ ep eeϕϕ e qe “ ep e qe “ 1pX,eq . Similarily, we obtain

´1 D33 ´1 ´1 fp fϕ e qe ep eϕf qf “ fp fϕ eeϕf qf “ fp fϕ eϕf qf

´1 D33 “ fp fϕ ϕff qf “ fp f qf “ 1pY,fq . Therefore, we have mutually inverse isomorphisms.

ϕ- Remark 36. Suppose given an isomorphism X „ Y in A. Suppose given an idempotent ´1 e P ApX,Xq. Deﬁne f :“ ϕ eϕ. Then we have mutually inverse isomorphisms

ep eϕ qf pX, eq „ / pY, fq o ´1 fp ϕ e qe

in Kar A.

29 Proof. By Remark 3, f is an idempotent. By applying ϕ from the left to f “ ϕ´1eϕ, we obtain ϕf “ eϕ. Thus, the claim follows from Remark 35. Note that

eϕf “ eϕϕ´´1eϕ “ eϕ and fϕ´1e “ ϕ´1eϕϕ´1e “ ϕ´´1e.

2.1.2 The Karoubi envelope respects additivity

The main purpose of the Karoubi envelope construction as presented in §0.1.1 is to complete an additive category with respect to its idempotents. Thus, the Karoubi envelope of an additive category should be additive. In this §2.1.2, we prove that this is the case.

For this §2.1.2, let A be a category.

Lemma 37. Let A have a zero object A. Then pA, 1Aq is a zero object in Kar A.

Proof. Suppose given pB, fq P ObpKar Aq. We have to show that

| Kar AppB, fq, pA, 1Aqq| “ 1 “ | Kar AppA, 1Aq, pB, fqq| .

By assumption, there exists exactly one morphism ϕ in ApB,Aq. We have fϕ1A P ApB,Aq.

Thus, we have fϕ1A “ ϕ, i.e. fp ϕ q1A P Kar AppB, fq, pA, 1Aqq. In consequence, we have | Kar AppB, fq, pA, 1Aqq| ě 1.

Suppose given ρ P Kar AppB, fq, pA, 1Aqq. By Deﬁnition 33, there exists ψ P ApB,Aq with

ρ “ fp ψ q1A . Since ApB,Aq “ tϕu, we have ϕ “ ψ. Therefore, we obtain ρ “ fp ψ q1A “

fp ϕ q1A . In consequence, we have | Kar AppB, fq, pA, 1Aq| ď 1.

Thus, we have | Kar AppB, fq, pA, 1Aqq| “ 1.

Dually, we obtain | Kar AppA, 1Aq, pB, fqq| “ 1.

Deﬁnition 38 (and Lemma). Suppose A to be preadditive. Suppose given pX, eq and pY, fq in Ob Kar A.

For ep ϕ qf and ep ψ qf in Kar AppX, eq, pY, fqq, we deﬁne

ep ϕ qf ` ep ψ qf :“ ep ϕ ` ψ qf .

The Karoubi envelope Kar A is preadditive with respect to this addition. In particular, we have ´ ep ϕ qf “ ep ´ϕ qf ,

for ep ϕ qf P Mor Kar A.

Proof. First we show that Kar AppX, eq, pY, fqq is an abelian group carrying the addition deﬁned above.

It suﬃces to show that U :“ tϕ P ApX,Y q : eϕf “ ϕu is a subgroup of ApX,Y q.

30 We have e0X,Y f “ 0X,Y , i.e. 0X,Y P U. Suppose given ϕ, ψ P U. We have

ϕ ´ ψ “ eϕf ´ eψf “ epϕ ´ ψqf.

Thus, we have ϕ ´ ψ P U. In consequence, Kar AppX, eq, pY, fqq is an abelian group.

p ϕ q fp ψ1 qg ρ e f / gp qh Suppose given pX, eq / pY, fq / pZ, gq / pW, hq in Kar A. We have fp ψ2 qg

ep ϕ qf p fp ψ1 qg ` fp ψ2 qgq gp ρ qh “ ep ϕ qf fp ψ1 ` ψ2 qg gp ρ qh

“ ep ϕpψ1 ` ψ2qρ qh

“ ep ϕψ1ρ ` ϕψ2ρ qh

“ ep ϕψ1ρ qh ` ep ϕψ2ρ qh

“ ep ϕ qf fp ψ1 qg gp ρ qh ` ep ϕ qf fp ψ2 qg gp ρ qh .

Lemma 39. Suppose A to be preadditive. Suppose given A1 ,A2 P Ob A. Let A1 and A2 have a direct sum ι1 ι2 / o A1 o C / A2 π1 π2 in A.

e1 e2 Suppose given idempotents A1 ÝÑ A1 and A2 ÝÑ A2 in A.

Deﬁne e :“ π1e1ι1 ` π2e2ι2 P ApC,Cq. Then

e1p e1ι1 qe e2p e2ι2 qe / o pA1 , e1q o pC, eq / pA2 , e2q ep π1e1 qe1 ep π2e2 qe2

is a direct sum of pA1 , e1q and pA2 , e2q in Kar A.

Proof. We have

ι1π1 “ 1A1

ι2π2 “ 1A2

ι1π2 “ 0A2

ι2π1 “ 0A1 . We will use these equalities throughout this proof without further notice. We calculate

2 e “ π1e1ι1π1e1ι1 ` π1e1ι1π2e2ι2 ` π2e2ι2π1e2ι1 ` π2e2ι2π2e2ι2

“ π1e1e1ι1 ` 0 ` 0 ` π2f2f2ι2 “ e.

Thus, we have pC, eq P Ob Kar A. We show that

eip eiιi qe P Kar AppAi , eiq, pC, eqq and ep πiei qei P Kar AppC, eq, pAi , eiqq

31 for i P r1, 2s. Pars pro toto, consider i “ 1. We have

e1 ¨ e1ι1 ¨ e “ e1e1ι1π1e1ι1 ` e1e1ι1π2e2ι2 “ e1e1ι1 ` 0 “ e1ι1

and e ¨ π1e1 ¨ e1 “ π1e1ι1π1e1e1 ` π2e2ι2π1e1e1 “ π1e1e1 ` 0 “ π1e1 . Furthermore, we have

ep π1e1 qe1 e1p e1ι1 qe ` ep π2e2 qe2 e2p e2ι2 qe “ ep π1e1e1ι1 qe ` ep π2e2e2ι2 qe

“ ep π1e1ι1 ` π2e2ι2 qe

“ ep e qe

“ 1pC,eq .

Finally, we have

e1p e1ι1 qe ep π1e1 qe1 “ e1p e1ι1π1e1 qe1 “ e1p e1 qe1 “ 1pA1 ,e1q .

and

e2p e2ι2 qe ep π2e2 qe2 “ e2p e2ι2π2e2 qe2 “ e2p e2 qe2 “ 1pA2 ,e2q .

e1 Corollary 40. Suppose A to be additive. Suppose given idempotents A1 ÝÑ A1 and e2 A2 ÝÑ A2 in A.

e1 0 Deﬁne e :“ 0 e2 P ApA1 ‘ A2 ,A1 ‘ A2q. Then ` ˘ e1p pe1 0q qe e2p p0 e2 q qe / o pA1 , e1q o pA1 ‘ A2, eq / pA2 , e2q e1 0 ep qe ep qe 0 1 e2 2 ´ ¯ ´ ¯

is a direct sum of pA1 , e1q and pA2 , e2q in Kar A.

Proof. Since p1 0q p0 1q / o A1 o A1 ‘ A2 / A2 1 0 0 1 ´ ¯ ´ ¯ is a direct sum of A1 and A2 in A, this follows from Lemma 39. Proposition 41. Suppose A to be an additive. Then Kar A is additive.

Proof. Since A is additive, it has a zero object. Thus, Kar A has a zero object; cf. Lemma 37. Since A is additive, it is preadditive. Thus, Kar A is preadditive; cf. Deﬁnition 38. Suppose given pX, eq and pY, fq in Ob Kar A. By Corollary 40, pX, eq and pY, fq have a direct sum in Kar A. Thus, Kar A is additive.

32 2.1.3 The inclusion functor

The inclusion functor will play the role of the universal functor of the Karoubi envelope construction; cf. §2.1.8.

For this §2.1.3, let A be a category. Deﬁnition 42 (and Lemma). We have the functor J A ÝÑA Kar A

ϕ 1Xp ϕ q1X pX ÝÑ Y q ÞÑ pX, 1X q ÝÝÝÝÝÝÑpY, 1Y q . ˆ ˙ We call JA the inclusion functor of A in Kar A.

If unambiguous, we often write J :“ JA.

ϕ ψ Proof. Suppose given X ÝÑ Y ÝÑ Z in A. 2 Since 1X “ 1X , we obtain a welldeﬁned map on objects.

Since 1X ϕ1Y “ ϕ, we obtain a welldeﬁned map on morphisms. Furthermore, we have D33 Jpϕψq “ 1Xp ϕψ q1Z “ 1Xp ϕ q1Y 1Yp ψ q1Z “ Jϕ ¨ Jψ. Finally, we have D33 J p1X q “ 1Xp 1X q1X “ 1pX,1X q “ 1JX .

Lemma 43. The following assertions (1, 2) hold.

(1) The inclusion functor JA is full.

(2) The inclusion functor JA is faithful.

Proof. Suppose given A, B P Ob A.

Ad (1). Suppose given ρ P Kar AppA, 1Aq, pB, 1Bqq. By Deﬁnition 33, there exists ϕ P D42 ApA, Bq with ρ “ 1Ap ϕ q1B . In particular, we have Jϕ “ 1Ap ϕ q1B “ ρ. Therefore, J is full.

Ad (2). Suppose given ϕ, ψ P ApA, Bq with Jϕ “ 1Ap ϕ q1B “ 1Ap ψ q1B “ Jψ. By Deﬁnition 33, we have ϕ “ ψ. Therefore, J is faithful.

Lemma 44. Suppose A to be preadditive. Then JA is additive.

Proof. Since A is preadditive, so is Kar A; cf. Deﬁnition 38. ϕ / Suppose given A / B in A. We have ψ

D42 D38 D42 Jpϕ ` ψq “ 1Ap ϕ ` ψ q1B “ 1Ap ϕ q1B ` 1Ap ψ q1B “ Jϕ ` Jψ,

33 2.1.4 Idempotent complete categories

In §0.1.1 we stated that the Karoubi envelope completes a category by endowing it with images of its idempotents. In this §2.1.4, we specify this by deﬁning images of idempotents and introducing idempotent complete categories.

For this §2.1.4, let A be a category. Deﬁnition 45. Let X ÝÑe X be an idempotent in A. A tuple pY, π, ιq with Y P Ob A, π P ApX,Y q and ι P ApY,Xq, is called an image of e if the following diagram commutes.

e X / X > π ι π

1Y Y / Y If every idempotent in A has an image, we call A idempotent complete. Remark 46. Suppose given an idempotent X ÝÑe X in A with image pY, π, ιq. Suppose ´1 ϕ- ϕ eϕ given Z P Ob A and an isomorphism X „ Z. Then Z ÝÝÝÝÑ Z is an idempotent with image pY, ϕ´1π, ιϕq.

Proof. By Remark 3, ϕ´1eϕ is an idempotent. Since pY, π, ιq is an image of e, we have the following commutative diagram.

e X / X > π ι π

1Y Y / Y Since ϕ is an isomorphism, we obtain the following commutative diagram.

ϕ´1eϕ Z / Z ? ιϕ ϕ´1π ϕ´1π 1Y Y / Y Thus, pY, ϕ´1π, ιϕq is an image of ϕ´1eϕ.

Remark 47. Suppose given categories B and C. Suppose given a functor F : B Ñ C. Suppose given an idempotent X ÝÑe X in B with image pY, π, ιq. Then FX ÝÑF e FX is an idempotent in C with image pF Y, F π, F ιq.

Proof. By Remark 5.(1), F e is an idempotent in C. Since pY, π, ιq is an image of e, we have the following commutative diagram in B.

e X / X > π ι π

1Y Y / Y

34 Since F is a functor, we have the following commutative diagram in C.

F e FX / FX ; F π F ι F π

# 1FY # FY / FY

Thus, pF Y, F π, F ιq is an image of F e.

Remark 48. Suppose given categories B and C. Suppose given an equivalence F : B Ñ C. Then B is idempotent complete if and only if C is idempotent complete.

Proof. Because of the symmetry of the situation, it suﬃces to show that if B is idempotent complete, so is C. Assume B to be idempotent complete. Suppose given an idempotent X ÝÑe X in C. ϕ- Since F is dense, there exists A P Ob B with FA – X. Let FA „ X be an isomorphism. ϕeϕ´1 By Remark 3, we have an idempotent FA ÝÝÝÝÑ FA in C. ´1 Since F is full, there exists f P BpA, Aq with F f “ ϕeϕ . We have F pf 2q “ pF fq2 “ ϕeϕ´1 ¨ ϕeϕ´1 “ ϕeϕ´1 “ F f. Since F is faithful, we obtain f 2 “ f. f Therefore, we have an idempotent A ÝÑ A in B. Since B is idempotent complete, there exists an image pB, π, ιq of f in B. By Remark 47, pF B, F π, F ιq is an image of F f “ ϕeϕ´1 in C. By Remark 46, pF B, ϕ´1 ¨ F π, F ι ¨ ϕq is an image of e “ ϕ´1 ¨ F f ¨ ϕ in C.

Lemma 49. Let X ÝÑe X be an idempotent in A with image pY, π, ιq. The following assertions (1, 2) hold.

(1) The morphism X ÝÑπ Y is a retraction, in particular π is epic.

(2) The morphism Y ÝÑι X is a coretraction, in particular ι is monic.

Proof. Recall that ιπ “ 1Y ; cf. Deﬁnition 45. Therefore, π is a retraction of ι. From the same equation we obtain that ι is a coretraction of π.

35 1 Lemma 50. Suppose given idempotents X ÝÑe X and X1 ÝÑe X1 in A. Suppose given an 1 1 1 1 1 image pY, π, ιq of e. Suppose given an image pY , π , ι q of e . Suppose given f P ApX,X q with efe1 “ f. Deﬁne g :“ ιfπ1. The following assertions (1, 2, 3) hold:

(1) The following diagram commutes.

π ι X / Y / X f g f π1 ι1 X1 / Y 1 / X1

1 (2) Suppose given g˜ P ApY,Y q such that the following diagram commutes.

π ι X / Y / X f g˜ f π1 ι1 X1 / Y 1 / X1 We have g˜ “ g. (3) Suppose f to be an isomorphism. Then g is an isomorphism with g´1 “ ι1f ´1π. Furthermore, the following diagram commutes.

π ι X / Y / X O O O f ´1 g´1 f ´1 π1 ι1 X1 / Y 1 / X1

Proof. Ad (1). We have

1 1 1 1 1 1 1 1 1 πg “ πιfπ “ efπ “ fe π “ fπ ι π “ fπ 1Y 1 “ fπ . Therefore, the left quadrangle commutes. Furthermore, we have

1 1 1 1 gι “ ιfπ ι “ ιfe “ ιef “ ιπιf “ 1Y ιf “ ιf. Therefore, the right quadrangle commutes. Ad (2). We have πgι1 “ πιf “ πgι˜ 1. Since π is epic and ι1 is monic; cf. Lemma 49; we obtain g “ g˜. Ad (3). We have

1 ´1 1 1 ´1 1 ´1 ´1 gι f π “ ιfπ ι f π “ ιfe f π “ ιeff π “ ιπι1X π “ 1Y ιπ “ 1Y and

1 ´1 1 ´1 1 1 ´1 1 1 ´1 1 1 1 1 1 1 ι f πg “ ι f πιfπ “ ι f efπ “ ι f fe π “ ι 1X1 π ι π “ 1Y 1 1Y 1 “ 1Y 1 . Therefore, g is an isomorphism with inverse g´1 “ ι1f ´1π. The commutativity of the stated diagram follows from (1).

36 Corollary 51. Let X ÝÑe X be an idempotent in A. Suppose given images pY, π, ιq and pY 1, π1, ι1q of e. The following assertions (1, 2) hold.

ιπ1 „ / 1 (1) We have mutually inverse isomorphisms Y o Y . ι1π (2) The following diagram commutes. Y : O π ι

$ X ι1π o ιπ1 X : π1 ι1 $ Y 1

Proof. Ad (1). By Lemma 50, we obtain the following commutative diagrams with mutually inverse isomorphisms ιπ1 and ι1π.

π ι π ι X / Y / X X / Y / X O O O 1 1 1X ιπ 1X 1X ι π 1X π1 ι1 π1 ι1 X / Y 1 / X1 X / Y 1 / X Ad (2). Combining these commutative diagrams, we obtain the commutative diagram stated above. Stipulation 52. Suppose given an idempotent complete category B. Suppose given an idempotent X ÝÑe X in B. Recall that we have chosen an image pIm e, e,¯ e9q of e in B. Thus, the following diagram commutes.

e X / X < e¯ e9 e¯

" 1Im e " Im e / Im e Recall that we have chosen

pIm 1X , 1X , 19X q “ pX, 1X , 1X q for X P Ob B; cf. Convention no. 25. If unambiguous, we also refer to just Im e as the image of e . Lemma 53. Let A be additive and idempotent complete. Suppose given an idempotent X ÝÑe X. The following assertions (1, 2, 3) hold.

(1) We have e9p1 ´ eq “ 0 and p1 ´ eq e¯ “ 0. ppp (2) We have mutually inverse isomorphisms

e9 p1´eq Im e Im 1 e ´ p ¯ / X . p q ‘ p ´ q o pe¯ p1´eqq

37 (3) The following quadrangle is commutative.

1 0 0 0 Impeq ‘ Imp1 ´ eq ´ ¯ / Impeq ‘ Imp1 ´ eq O O

e e o 9 o 9 pe¯ p1éqq p1éq pe¯ p1éqq p1éq ´ p ¯ ´ p ¯ e X / X

Proof. Ad (1). We have 0 “ ep1 ´ eq “ e¯e9p1 ´ eqp1 ´ eq . By Lemma 49,e ¯ is epic and p1 ´ eq is monic. Therefore, 0 “ e9p1 ´ eq. ppp ppp The other equality follows from the symmetry of the situation. Ad (2). Since (1) holds, we have

e9 e9e¯ e9p1éq 1 0 e¯ p1éq . p1éq p q “ p1éq e¯ p1éq p1éq “ 0 1 p ´ p p ¯ Moreover, we obtain ` ˘ ` ˘

e9 pe¯ p1´eqq p1´eq “ e¯e9 ` p1 ´ eqp1 ´ eq “ e ` p1 ´ eq “ 1. p p Ad (3). Using (1), we compute` ˘p1 ´ eq e “ p1 ´ eq e¯e9 “ 0. ppp ppp Therefore, we have

e9 ee9 e9 1 0 e9 p1éq e “ p1éq e “ 0 “ 0 0 p1éq . p p p Thus, the claim follows` from˘ (2). ` ˘ ` ˘ ` ˘ ` ˘

2.1.5 The Karoubi envelope is idempotent complete

After introducing idempotent complete categories in § 2.1.4, we can now show that the Karoubi envelope is idempotent complete. Furthermore, in Proposition 56 we give a necessary and suﬃcient condition for a category C to be idempotent complete. This condition justiﬁes the notion Karoubi envelope for Kar C.

For this §2.1.5, let A be a category.

p ϕ q Lemma 54. Suppose given an idempotent pA, eq ÝÝÝÑpe e A, eq in Kar A.

Then ppA, ϕq, ep ϕ qϕ , ϕp ϕ qeq is an image of ep ϕ qe . In particular, Kar A is idempotent complete.

2 D33 2 2 Proof. Since ep ϕ qe “ ep ϕ qe “ ep ϕ qe , we have ϕ “ ϕ; cf. Deﬁnition 33. Therefore, we have pA, ϕq P Ob Kar A. R2 Moreover, we have ϕϕe “ ϕe “ ϕ. Thus, ϕp ϕ qe P Kar AppA, ϕq, pA, eqq.

38 R2 We also have eϕϕ “ eϕ “ ϕ. Thus, ep ϕ qϕ P Kar AppA, eq, pA, ϕqq. Therefore, we can consider the following diagram in Kar A.

ep ϕ qe pA, eq / pA, eq : ep ϕ qϕ ϕp ϕ qe ep ϕ qϕ

$ ϕp ϕ qϕ“1pA,ϕq $ pA, ϕq / pA, ϕq

We have D33 ep ϕ qϕ ϕp ϕ qe “ ep ϕϕ qe “ ep ϕ qe .

Furthermore, we have

D33 D33 ϕp ϕ qe ep ϕ qϕ “ ϕp ϕϕ qϕ “ ϕp ϕ qϕ “ 1pA,ϕq .

Thus, the diagram stated above is commutative.

Therefore, ppA, ϕq, ep ϕ qϕ , ϕp ϕ qeq is an image of ep ϕ qe; cf. Deﬁnition 45.

Remark 55. Suppose given an idempotent X ÝÑe X in A. Suppose e to have an image pY, π, ιq in A. We have mutually inverse isomorphisms

ep π q1Y X, e / Y, 1 p q o p Y q 1Y p ι qe

in Kar A.

D42 Proof. Since pY, π, ιq is an image of e in A, pJY, Jπ, Jιq “ ppY, 1Y q, 1Xp π q1Y , 1Yp ι q1X q D42 is an image of Je “ 1Xp e q1X in Kar A; cf. Remark 47.

By Lemma 54, ppX, eq, 1Xp e qe, ep e q1X q is an image of 1Xp e q1X in Kar A. By Corollary 51, we have mutually inverse isomorphisms

ep e q1X 1Xp π q1Y X, e / Y, 1 . p q o p Y q 1Y p ι q1X 1Xp e qe

Since we have

D33 D33 ep e q1X 1Xp π q1Y “ ep eπ q1Y “ ep π q1Y and 1Yp ι q1X 1Xp e qe “ 1Yp ιe qe “ 1Yp ι qe , the claim follows.

39 Proposition 56. Suppose given a category B. The following assertions (1, 2) are equivalent.

(1) The category B is idempotent complete.

(2) The inclusion functor JB : B Ñ Kar B is an equivalence.

Proof. Ad (1)ñ(2). By Lemma 43, it suﬃces to show that J is dense. Suppose given pX, eq P Ob Kar B. Since B is idempotent complete, there exists an image Im e of e in B. D42 R55 We have JpIm eq “ pIm e, 1Im eq – pX, eq. Thus, J is dense. Ad (2)ñ(1). By Lemma 54, Kar B is idempotent complete. Thus, B is idempotent complete; cf. Remark 48.

2.1.6 Functoriality

In this §2.1.6 we deﬁne the Karoubi envelope construction Kar for functors and transformations. Furthermore, we establish functoriality properties of Kar, which could be expressed by saying that it is turned into a 2-functor.

For this §2.1.6, let A, B and C be categories. Deﬁnition 57 (and Lemma). Suppose given a functor F : A Ñ B. Let Kar A ÝÝÝÑKar F Kar B

p ϕ q p F ϕ q pX, eq ÝÝÝÝÑpe f Y, fq ÞÑ pF X, F eq ÝÝÝÝÝÝÑpF e F f F Y, F fq . ˆ ˙ ˆ ˙ This deﬁnes a functor Kar F : Kar A Ñ Kar B. So, we have pKar F qpX, eq “ pF X, F eq for pX, eq P Ob Kar A. Furthermore, we have pKar F q ep ϕ qf “ F ep F ϕ qF f for ep ϕ qf P Mor Kar A.

Proof. By Remark 5, we obtain welldeﬁned maps on objects and morphisms; cf. Deﬁnition 33.

p ϕ q p ψ q Suppose given pX, eq ÝÝÝÝÑpe f Y, fq ÝÝÝÝÑpf g Z, gq in Kar A. We have D33 D33 pKar F q1pX,eq “ pKar F q ep e qe “ F ep F e qF e “ 1pF X,F eq “ 1pKar F qpX,eq . Furthermore, we have D33 pKar F q p ep ϕ qf fp ψ qgq “ pKar F q ep ϕψ qg “ F ep F pϕψq qF g “ F ep F ϕ ¨ F ψ qF g D33 “ F ep F ϕ qF f F fp F ψ qF g “ pKar F q ep ϕ qf ¨ pKar F q fp ψ qg . Therefore, Kar F is a functor indeed.

40 Lemma 58. Suppose given a functor F : A Ñ B. The following assertions (1, 2, 3) hold.

(1) Suppose F to be faithful. Then Kar F is faithful. (2) Suppose F to be full. Then Kar F is full. (3) Suppose F to be full, faithful and dense. Then Kar F is full, faithful and dense. I.e. if F is an equivalence, so is Kar F .

Proof. Suppose given pX, eq, pY, fq P Ob Kar A.

Ad (1). Suppose given ep ϕ qf and ep ψ qf in Kar AppX, eq, pY, fqq with

F ep F ϕ qF f “ pKar F q ep ϕ qf “ pKar F q ep ψ qf “ F ep F ψ qF f . By Deﬁnition 33, we obtain F ϕ “ F ψ. Since F is faithful, we obtain ϕ “ ψ. Therefore, we have ep ϕ qf “ ep ψ qf . Thus, Kar F is faithful.

Ad (2). Suppose given F ep ρ qF f P Kar BppF X, F eq, pF Y, F fqq.

Then we have ρ P BpFX,FY q. Since F is full, there exists ϕ P ApX,Y q with F ϕ “ ρ. Deﬁneϕ ˜ :“ eϕf. Then we have eϕf˜ “ eeϕff “ eϕf “ ϕ.˜

1 Therefore, we have ep ϕ qf P Kar AppX, eq, pY, fqq.

Since F ep ρ qF f P Kar BppF X, F eq, pF Y, F fqq, we have F ϕ˜ “ F peϕfq “ F e ¨ F ϕ ¨ F f “ F e ¨ ρ ¨ F f “ ρ. In consequence, we have

D57 pKar F q ep ϕ˜ qf “ F ep F ϕ˜ qF f “ F ep ρ qF f . Thus, Kar F is full. Ad (3). Since (1) and (2) hold, it suﬃces to show that Kar F is dense. Suppose given pA, fq P Kar B. Then we have A P Ob B.

ρ- Since F is dense, there exists X P Ob A with FX – A. Let FX „ A be an isomorphism. Deﬁne f 1 :“ ρfρ´1. By Remark 3, we have pF X, f 1q P Ob Kar B. By Remark 36, we have 1 1p f ρf q 1 f f an isomorphism pF X, f q „ / pA, fq . 1 Since F is full, there exists e P ApX,Xq with F e “ f . We have F pe2q “ F e ¨ F e “ F e. Since F is faithful, we obtain e2 “ e. Therefore, we have pX, eq P Ob Kar A. In consequence, we have

pKar F qpX, eq D“57 pF X, F eq “ pF X, f 1q – pA, fq . Thus, Kar F is dense.

41 Lemma 59. The following assertions (1, 2) hold.

(1) We have Kar 1A “ 1Kar A .

(2) Suppose given functors A ÝÑF B ÝÑG C. We have pKar Gq ˝ pKar F q “ KarpG ˝ F q.

p ϕ q Proof. Suppose given pX, eq ÝÝÝÝÑpe f Y, fq in Kar A. Ad (1). We have

1 ϕ ep ϕ qf D57 1Apeqp A q1Apfq pKar 1Aq pX, eq ÝÝÝÝÑpY, fq “ p1AX, 1Aeq ÝÝÝÝÝÝÝÝÝÑp1AY, 1Afq ˆ ˙ ˆ ˙ p ϕ q “ pX, eq ÝÝÝÝÑpe f Y, fq ˆ ˙ ep ϕ qf “ 1Kar A pX, eq ÝÝÝÝÑpY, fq . ˆ ˙

Ad (2). We have

p ϕ q ppKar Gq ˝ pKar F qq pX, eq ÝÝÝÝÑpe f Y, fq ˆ ˙ p F ϕ q D“57 pKar Gq pF X, F eq ÝÝÝÝÝÝÑpF e F f F Y, F fq ˆ ˙ p GpF ϕq q D“57 pGpFXq,GpF eqq ÝÝÝÝÝÝÝÝÝÝÝÝÑpGpF eq GpF fq GpFY q,GpF fq ˆ ˙ p pG˝F qϕ q “ ppG ˝ F qX, pG ˝ F qeq ÝÝÝÝÝÝÝÝÝÝÝÝÝÝÑpppG˝F qe pG˝F qf G ˝ F qY, pG ˝ F qfq ˆ ˙ p ϕ q D“57 pKarpG ˝ F qq pX, eq ÝÝÝÝÑpe f Y, fq . ˆ ˙

42 Deﬁnition 60 (and Lemma). Suppose given F ÝÑα F˜ in rA, Bs. For pX, eq P Ob Kar A, we deﬁne ˜ α¯pX,eq :“ F e ¨ αX ¨ F e.

For pX, eq P Ob Kar A, we have

˜ ˜ F ep α¯pX,eq qF˜ e P Kar B pF X, F eq, pFX, F eq . ´ ¯

Suppose given pY, fq and pZ, gq in Ob Kar A and fp ϕ qg P Kar AppY, fq, pZ, gqq. The following diagram is commutative.

F fp F ϕ qF g pF Y, F fq / pF Z, F gq

F fp α¯pY,fq qF˜ f F gp α¯pZ,gq qF˜ g

˜ F˜ fp F ϕ qF˜ g pFY,˜ F˜ fq / pFZ,˜ F˜ gq

Deﬁne Kar α : α¯ “ F ep pX,eq qF˜ e pX,eqPOb Kar A ˜ “ ` F ep F e ¨ αX ¨˘F e qF˜ e . pX,eqPOb Kar A ´ Kar α ¯ We have the transformation Kar F ùùùñ Kar F˜.

Proof. Suppose given pX, eq P Ob Kar A. We have ˜ ˜ ˜ ˜ F e ¨ α¯pX,eq ¨ F e “ F e ¨ F e ¨ αX ¨ F e ¨ F e “ F e ¨ αX ¨ F e “ α¯pX,eq .

˜ ˜ Therefore, we have F ep α¯pX,eq qF˜ e P Kar B pF X, F eq, pFX, F eq . Furthermore, we have ´ ¯

D33 ˜ F fp F ϕ qF g F gp α¯pZ,gq qF˜ g “ F fp F ϕ ¨ F g ¨ αZ ¨ F g qF˜ g ˜ “ F fp F f ¨ F ϕ ¨ αZ ¨ F g qF˜ g ˜ ˜ “ F fp F f ¨ αY ¨ F ϕ ¨ F g qF˜ g ˜ ˜ “ F fp F f ¨ αY ¨ F f ¨ F ϕ qF˜ g D33 ˜ “ F fp α¯pY,fq qF˜ f ¨ F˜ fp F ϕ qF˜ g . Therefore, the diagram stated above is commutative. Thus, Kar α is a transformation.

α0 α1 Lemma 61. Suppose given F P ObrA, Bs. Suppose given F0 ÝÑ F1 ÝÑ F2 in rA, Bs. The following assertions (1, 2) hold.

(1) We have Kar 1F “ 1Kar F .

(2) We have Karpα0α1q “ pKar α0qpKar α1q.

43 Proof. Suppose given pX, eq P Ob Kar A. Ad (1). We have D60 D33 pKar 1F qpX,eq “ F ep F e ¨ p1F qX ¨ F e qF e “ F ep F e qF e “ 1pF X,F eq .

Ad (2). We have D60 α0α1pX,eq “ F0e ¨ pα0qX ¨ pα1qX ¨ F2e “ F0e ¨ F0e ¨ pα0qX ¨ pα1qX ¨ F2e ¨ F2e “ F e ¨ pα0qX ¨ F1e ¨ F1e ¨ pα1qX ¨ F2e D60 “ α0pX,eqα1pX,eq . Thus, we obtain D60 Karpα0α1qpX,eq “ F0ep α0α1pX,eq qF2e “ F0ep α0pX,eqα1pX,eq qF2e D33 D60 “ F0ep α0pX,eq qF1e F1ep α1pX,eq qF2e “ pKar α0qpX,eqpKar α1qpX,eq .

Remark 62. We have the functor rA, Bs ÝÝÑKar rKar A, Kar Bs

F ÝÑα G ÞÑ Kar F ÝÝÝÑKar α Kar G ; cf. Lemma 61. ´ ¯ ´ ¯ Lemma 63. Suppose A and B to be preadditive. The following assertions (1, 2) hold.

(1) Suppose given F P Ob addrA, Bs. Then Kar F P Ob addrKar A, Kar Bs.

α / ˜ (2) Suppose F / F in addrA, Bs. We have Karpα ` α˜q “ Kar α ` Karα ˜. α˜

Proof. Since A and B are preadditive, Kar A and Kar B are preadditive by Deﬁnition 38.

Ad (1). Suppose given pX, eq, pY, fq P Ob Kar A. Suppose given ep ϕ qf and ep ψ qf in Kar AppX, eq, pY, fqq. We have D38 pKar F q p ep ϕ qf ` ep ψ qf q “ pKar F q ep ϕ ` ψ qf D57 “ F ep F pϕ ` ψq qF f “ F ep F ϕ ` F ψ qF f D38 “ F ep F ϕ qF f ` F ep F ψ qF f D57 “ pKar F q ep ϕ qf ` pKar F q ep ψ qf . Ad (2). Suppose given pX, eq P Ob Kar A. We have D60 ˜ pKarpα ` α˜qqpX,eq “ F ep F e ¨ pα ` α˜qX ¨ F e qF˜ e ˜ “ F ep F e ¨ pαX ` α˜X q ¨ F e qF˜ e ˜ ˜ “ F ep F e ¨ αX ¨ F e ` F e ¨ α˜X ¨ F e qF˜ e D38 ˜ ˜ “ F ep F e ¨ αX ¨ F e qF˜ e ` F ep F e ¨ α˜X ¨ F e qF˜ e D60 “ pKar αqpX,eq ` pKarα ˜qpX,eq

“ pKar α ` Karα ˜qpX,eq .

44 Remark 64. Suppose A and B to be preadditive. We have the additive functor

Kar addrA, Bs ÝÝÑ addrKar A, Kar Bs

F ÝÑα G ÞÑ Kar F ÝÝÝÑKar α Kar G ; ´ ¯ ´ ¯ cf. Remark 62 and Lemma 63.

α- 1 Remark 65. Suppose given an isotransformation F „ F in rA, Bs.

Kar α 1 Then Kar F „ / Kar F is an isotransformation in rKar A, Kar Bs.

´1 L61 ´1 L59 Proof. We have pKar αqpKar α q “ Karpαα q “ Kar 1F “ 1Kar F .

Remark 66. We can now give a new proof for Lemma 58.(3). Suppose given an equivalence F : A Ñ B. Then Kar F : Kar A Ñ Kar B is an equivalence.

Proof. Since F is an equivalence, there exists a functor G : B Ñ A and isotransformations α β pG ˝ F q „ +3 1A and pF ˝ Gq „ +3 1B . By Remark 65, we obtain the isotransformation

L59 Kar α pKar G ˝ Kar F q “ KarpG ˝ F q „ +3 1Kar A .

Similarily, we have the isotransformation

L59 Kar β pKar F ˝ Kar Gq “ KarpF ˝ Gq „ +3 1Kar B ;

cf. Remark 65. Thus, Kar F is an equivalence.

45 Lemma 67. Suppose given functors F, F˜ : A Ñ B and G, G˜ : B Ñ C. Suppose given transformations α : F ñ F˜ and β : G ñ G˜. We have Karpβ ˚ αq “ pKar βq ˚ pKar αq.

Proof. Suppose given pX, eq P Ob Kar A. We have

ppKar βq ˚ pKar αqqpX,eq

“ pKar GqpKar αqpX,eq ¨ pKar βqpKar F˜qpX,eq D60 ¯ “ pKar Gq F ep α¯pX,eq qF˜ e ¨ G F˜ ep β FX,˜ F˜ e q G˜ F˜ e D57 p ˝ q p q p ˝ q ˜ ˜ ˜ ˜ “ pKar Gq F ep F e ¨ αX ¨ F e qF˜ e ¨ pG˝F˜qep pG ˝ F qe ¨ βFX˜ ¨ pG ˝ F qe qpG˜˝F˜qe D57 ˜ ˜ ˜ ˜ “ pG˝F qep pG ˝ F qe ¨ GαX ¨ pG ˝ F qe qpG˝F˜qe ¨ pG˝F˜qep pG ˝ F qe ¨ βFX˜ ¨ pG ˝ F qe qpG˜˝F˜qe ˜ ˜ ˜ ˜ “ pG˝F qep pG ˝ F qe ¨ GαX ¨ pG ˝ F qe ¨ pG ˝ F qe ¨ βFX˜ ¨ pG ˝ F qe qpG˜˝F˜qe ˜ ˜ ˜ “ pG˝F qep pG ˝ F qe ¨ GαX ¨ pG ˝ F qe ¨ βFX˜ ¨ pG ˝ F qe qpG˜˝F˜qe ˜ ˜ “ pG˝F qep pG ˝ F qe ¨ pG ˝ F qe ¨ GαX ¨ βFX˜ ¨ pG ˝ F qe qpG˜˝F˜qe ˜ ˜ “ pG˝F qep pG ˝ F qe ¨ GαX ¨ βFX˜ ¨ pG ˝ F qe qpG˜˝F˜qe ˜ ˜ “ pG˝F qep pG ˝ F qe ¨ pβ ˚ αqX ¨ pG ˝ F qe qpG˜˝F˜qe D60 “ pG˝F qep pβ ˚ αqpX,eq qpG˜˝F˜qe D60 “ pKarpβ ˚ αqqpX,eq .

2.1.7 The image functor

For this §2.1.7, let A be an idempotent complete category. Deﬁnition 68 (and Lemma). Recall that in Stipulation 52 we assigned to every idempotent X ÝÑe X in A an image pIm e, e,¯ e9q. We have the functor

I Kar A ÝÑA A

ep ϕ qf eϕ9 f¯ ppX, eq ÝÝÝÝÑpY, fqq ÞÑ pIm e ÝÝÑ Im fq.

We call IA the image functor of A.

If unambiguous, we often write I :“ IA .

Proof. Since A is idempotent complete, we obtain a welldefined map on objects. By Definition 33 and Stipulation 52, we obtain a welldefined map on morphisms.

p ϕ q p ψ q Suppose given pX, eq ÝÝÝÝÑpe f Y, fq ÝÝÝÝÑpf g Z, gq in Kar A.

We have I1pX,eq “ I ep e qe “ ee9 e¯ “ e9e¯e9e¯ “ 1Im e1Im e “ 1Im e “ 1IpX,eq. Furthermore, we have ¯ Ip ep ϕ qf fp ψ qgq “ I ep ϕψ qg “ eϕψ9 g¯ “ eϕfψ9 g¯ “ eϕ9 ffψ9 g¯ “ I ep ϕ qf ¨ I fp ψ qg .

46 Remark 69. We have IA ˝ JA “ 1A ; cf. Deﬁnitions 42 and 68 and Stipulation 52. Lemma 70. The following assertions (1, 2, 3) hold.

(1) The image functor IA is faithful.

(2) The image functor IA is full.

(3) The image functor IA is surjective on objects. In particular, IA is dense.

Thus, IA is an equivalence.

Proof. Suppose given pX, eq and pY, fq in Ob Kar A.

Ad (1). Suppose given ep ϕ qf and ep ψ qf in Kar AppX, eq, pY, fqq with ¯ ¯ eϕ9 f “ I ep ϕ qf “ I ep ψ qf “ eψ9 f.

Applyinge ¯ from the left and f9 from the right to this equation, we obtain

ϕ “ eϕf “ e¯eϕ9 f¯f9 “ e¯eψ9 f¯f9 “ eψf “ ψ.

Thus, I is faithful. D68 Ad (2). Suppose given ϕ P ApIpX, eq,IpY, fqq “ ApIm e, Im fq.

Considerρ ˜ :“ eρ¯ f9 P ApX,Y q. We have ¯ eρf˜ “ eeρ¯ ff9 “ e¯e9eρ¯ f9ff9 “ e¯ ¨ 1Im e ¨ ρ ¨ 1Im f ¨ f9 “ eρ¯ f9 “ ρ.˜

Therefore, we have ep ρ˜ qf P Kar AppX, eq, pY, fqq. Furthermore, we have

D68 ¯ ¯ I ep ρ˜ qf “ e9ρ˜f “ e9eρ¯ f9f “ 1Im e ¨ ρ ¨ 1Im f “ ρ.

Thus, I is full.

Ad (3). Suppose given Z P Ob A. We have IpZ, 1Z q “ Im 1Z “ Z; cf. Stipluation Stipu- lation 52. Thus, I is surjective on objects.

Lemma 71. Suppose A to be preadditive. Then IA is additive.

Proof. Since A is preadditive, so is Kar A; cf. Deﬁnition 38.

ep ϕ qf Suppose given X, e / Y, f in Kar A. p q / p q ep ψ qf We have D38 I p ep ϕ qf ` ep ψ qf q “ I ep ϕ ` ψ qf D“68 e9pϕ ` ψqf¯ “ eϕ9 f¯` eψ9 f¯ D68 “ I ep ϕ qf ` I ep ψ qf .

47 2.1.8 Universal property

For this §2.1.8, let A and B be categories. Let B be idempotent complete. Deﬁnition 72 (and Lemma). We have the functor

rA, Bs Ñ rKar A, Bs

α IB˚Kar α pF ÝÑ Gq ÞÑ pIB ˝ Kar F q ÝÝÝÝÝÑpIB ˝ Kar Gq . ´ ¯ Cf. Deﬁnitions 57, 60 and 68.

α 1 1 For F ÝÑ G in rA, Bs, we write F :“ IB ˝ Kar F and α :“ IB ˚ Kar α. Suppose given F ÝÑα G in rA, Bs. We have

1 Kar A ÝÑF B

p ϕ q F e F ϕ F f X, e e f Y, f Im F e p q ¨ ¨ Im F f . p q ÝÝÝÝÑp q ÞÑ ÝÝÝÝÝÝÝÝÝÑp ˆ ˙ ˆ ˙ Suppose given pX, eq P Ob Kar A. We have

1 αpX,eq “ pF eq ¨ αX ¨ Ge. p

β Proof. Suppose given F ÝÑα G ÝÑ H in rA, Bs.

p ϕ q Suppose given pX, eq ÝÝÝÝÑpe f Y, fq in Kar A. We have

ϕ ϕ 1 ep qf ep qf F pX, eq ÝÝÝÝÑpY, fq “ pIB ˝ Kar F q pX, eq ÝÝÝÝÑpY, fq ˆ ˙ ˆ ˙ D57 F ep F ϕ qF f “ IB pF X, F eq ÝÝÝÝÝÝÑpF Y, F fq ˆ ˙ F e F ϕ F f D68 Im F e p q ¨ ¨ Im F f . “ ÝÝÝÝÝÝÝÝÝÑp ˆ ˙ Furthermore, we have

1 αpX,eq “ pIB ˚ Kar αqpX,eq “ IB pKar αqpX,eq D60 “ IB `p F ep F e ¨ αX ˘¨ Ge qGeq D68 “ pF eq ¨ F e ¨ αX ¨ Ge ¨ Ge ppp “ pF eq ¨ αX ¨ Ge. ppp Thus, we have

1 p1F qpX,eq “ pF eq ¨ p1F qX ¨ F e ppp “ pF eq ¨ F e ppp “ 1ImpF eq

“ 1F 1X .

48 1 Therefore, we have p1F q “ 1F 1 . Moreover, we have

1 pαβqpX,eq “ pF eq ¨ pαβqX ¨ He ppp “ pF eq ¨ αX ¨ βX ¨ He ppp “ pF eq ¨ F e ¨ αX ¨ βX ¨ He ppp “ pF eq ¨ αX ¨ Ge ¨ βX ¨ He ppp “ pF eq ¨ αX ¨ Ge ¨ pGeq ¨ βX ¨ He 1 ppp 1 ppp “ αpX,eq ¨ βpX,eq . In consequence, we have pαβq1 “ α1β1. Thus, we have a functor indeed.

Lemma 73. Suppose given F ÝÑα G in rA, Bs. The following assertions (1, 2) hold.

1 (1) We have F ˝ JA “ F .

1 (2) We have α ˚ JA “ α.

ϕ Proof. Ad (1). Suppose given X ÝÑ Y in A. We have

p ϕ q 1 ϕ D42 1 1X 1Y pF ˝ JAq X ÝÑ Y “ F pX, 1X q ÝÝÝÝÝÝÑpY, 1Y q ˆ ˙ ´ ¯ F 1 F ϕ F 1 D72 Im F 1 p X q ¨ ¨ Y Im F 1 “ p X q ÝÝÝÝÝÝÝÝÝÝÝÑp p Y q ˆ F 1 ¨ F ϕ ¨ F 1 ˙ S“52 FX ÝÝÝÝÝÝÝÝÝÝÑX Y FY ϕ “ F´ X ÝÑ Y . ¯ ´ ¯ Ad (2). Suppose given X P Ob A. We have

1 D42 1 D72 S52 pα ˚ JAqX “ αpX,1 q “ pF 1X q ¨ αX ¨ G1X “ F 1X ¨ αX ¨ G1X “ αX . X ppp

Lemma 74. Suppose given F P ObrA, Bs. The following assertions (1, 2, 3) hold.

(1) If F is faithful, so is F 1.

(2) If F is full, so is F 1.

(3) If F is dense, so is F 1.

Proof. Ad (1). Since F is faithful, so is Kar F ; cf. Lemma 58.(1). By Lemma 70.(1), IB is 1 faithful. Thus, F “ IB ˝ Kar F is faithful.

Ad (2). Since F is full, so is Kar F ; cf. Lemma 58.(2). By Lemma 70.(2), IB is full. Thus, 1 F “ IB ˝ Kar F is full.

49 Ad (3). Suppose given B P Ob B. Since F is dense, there exists A P Ob A with FA – B. We have 1 1 L73.p1q F pA, 1Aq “ pF ˝ JAqA “ FA – B. Thus, F 1 is dense.

Proposition 75. Recall that A and B are categories and that B is idempotent complete; cf. Deﬁnition 45. Recall that Kar A is idempotent complete; cf. Lemma 54. The following assertions (1, 2, 3) hold.

(1) We have JA P ObrA, Kar As. 1 α 1 α 1 1 Suppose given F ÝÑ G in rA, Bs. We have F ÝÑ G in rKar A, Bs with F ˝ JA “ F , 1 1 G ˝ JA “ G and α ˚ JA “ α; cf. Deﬁnition 72 and Lemma 73. 1 1 1 Suppose given β P rKar A,BspF ,G q with β ˚ JA “ α. Then we have β “ α .

JA A / Kar A

G α AI α1 F 1 +3 G1 F Ù * B

(2) Suppose given U, V P ObrKar A, Bs with U ˝ JA “ V ˝ JA. Then U – V . (3) We have the equivalence of categories

Λ rA, Bs ÐÝÝÝA,B rKar A, Bs

β˚JA β pU ˝ JAq ÝÝÝÑpV ˝ JAq ÞÑ pU ÝÑ V q, ´ ¯ that is surjective on objects.

If unambiguous, we often write Λ:“ ΛA,B.

1 Proof. Ad (1). Suppose given pX, eq P Ob Kar A. We show that αpX,eq “ βpX,eq. By assumption, we have

β β J α α1 J α1 . pX,1X q “ p ˚ AqX “ X “ p ˚ AqX “ pX,1X q Since α1 is natural, the following diagram commutes.

F 1 p e q F 1 p e q 1 1X e 1 e 1X 1 F pX, 1X q / F pX, eq / F pX, 1X q

1 1 1 β X,1 “ α α β X,1 “ α p X q pX,1X q pX,eq p X q pX,1X q

1 1 G 1 p e qe G ep e q1 1 X 1 X 1 G pX, 1X q / G pX, eq / G pX, 1X q

50 Since β is natural, the following diagram commutes.

F 1 p e q F 1 p e q 1 1X e 1 e 1X 1 F pX, 1X q / F pX, eq / F pX, 1X q

1 1 β X,1 “ α β β X,1 “ α p X q pX,1X q pX,eq p X q pX,1X q

1 1 G 1 p e qe G ep e q1 1 X 1 X 1 G pX, 1X q / G pX, eq / G pX, 1X q

1Xp e q1X By Lemma 54, ppX, eq, 1Xp e qe, ep e q1X q is an image of pX, 1X q ÝÝÝÝÝÑpX, 1X q in Kar A. 1 1 1 Thus, pF pX, eq,F 1Xp e qe ,F ep e q1X q is an image of

F 1 e 1 1Xp q1X 1 F pX, 1X q ÝÝÝÝÝÝÝÑ F pX, 1X q ˆ ˙ in B; cf. Remark 47.

1 1 1 Similarily, pG pX, eq,G 1Xp e qe ,G ep e q1X q is an image of

G1 e 1 1Xp q1X 1 G pX, 1X q ÝÝÝÝÝÝÝÑ G pX, 1X q ˆ ˙ in B; cf. Remark 47.

1 In consequence, we obtain αpX,eq “ βpX,eq ; cf. Lemma 50.(2). 1 Ad (2). Suppose we have shown that pW ˝ JAq – W for W P ObrKar A, Bs. Then we have 1 1 U – pU ˝ JAq “ pV ˝ JAq – V.

1 Therefore, it suﬃces to show that pW ˝ JAq – W for W P ObrKar A, Bs. Suppose given W P ObrKar A, Bs. Suppose given pX, eq P Ob Kar A.

By Lemma 54 and Remark 47, pW pX, eq,W 1Xp e qe ,W ep e q1X q is an image of W 1Xp e q1X “ pW ˝ JAqe in B. By Corollary 51, we have mutually inverse isomorphisms

ppW ˝JAqeq ¨ W 1Xp e qe 1 D72 / pW ˝ JAq pX, eq “ ImppW ˝ JAqeq o ppp W pX, eq W ep e q1X ¨ pW ˝JAqe

in B. Deﬁne

δ :“ W ep e q1X ¨ pW ˝ JAqe . pX,eqPOb Kar A ´ ¯ p ϕ q We show that δ is natural. Suppose given pX, eq ÝÝÝÝÑpe f Y, fq in Kar A.

51 We have 1 δpX,eq ¨ pW ˝ JAq ep ϕ qf D72 “ W ep e q1X ¨ pW ˝ JAqe ¨ ppW ˝ JAqeq ¨ pW ˝ JAqϕ ¨ pW ˝ JAqf ppp “ W ep e q1X ¨ pW ˝ JAqe ¨ pW ˝ JAqϕ ¨ pW ˝ JAqf

“ W ep e q1X ¨ W 1Xp e q1X ¨ W 1Xp ϕ q1Y ¨ pW ˝ JAqf

“ W p ep e q1X ¨ 1Xp e q1X ¨ 1Xp ϕ q1Y q ¨ pW ˝ JAqf

“ W ep eeϕ q1Y ¨ pW ˝ JAqf

“ W ep ϕf q1Y ¨ pW ˝ JAqf

“ W p ep ϕ qf ¨ fp f q1Y q ¨ pW ˝ JAqf

“ W ep ϕ qf ¨ W fp f q1Y ¨ pW ˝ JAqf

“ W ep ϕ qf ¨ δpY,fq .

δ 1 Thus, we have W „ +3 pW ˝ JAq . Ad (3). Suppose given F P ObrA, Bs. Because (1) holds, we have F 1 P rKar A, Bs and 1 1 ΛF “ F ˝ JA “ F . Therefore, Λ is surjective on objects. Suppose given F,G P ObrA, Bs. Since (1) holds, we have a bijection 1 1 rKar A,BspF ,G q Ñ rA,BspF,Gq, β ÞÑ Λβ. Suppose given U, V P ObrKar A, Bs with ΛU “ ΛV . Since (2) holds, we have U – V . Therefore, Λ is an equivalence by Lemma 6.

Lemma 76. Suppose A and B to be preadditive. Suppose given F P Ob addrA, Bs. We 1 have F P Ob addrKar A, Bs.

Proof. Since A is preadditive, so is Kar A; cf. Deﬁnition 38.

Since F is additive, so is Kar F ; cf. Lemma 63. By Lemma 71, IB is additive. Thus, 1 D72 F “ IB ˝ Kar F is additive. Proposition 77. Suppose A and B to be preadditive categories. Recall that B is idempotent complete; cf. Deﬁnition 45. Recall that Kar A is preadditive; cf. Deﬁnition 38; and idempotent complete; cf. Lemma 54. The following assertions (1, 2, 3) hold.

(1) We have JA P addrA, Kar As; cf. Lemma 44. 1 α 1 α 1 Suppose given F ÝÑ G in addrA, Bs. We have F ÝÑ G in addrKar A, Bs with 1 1 1 F ˝ JA “ F , G ˝ JA “ G and α ˚ JA “ α; cf. Deﬁnition 72, Lemmas 73 and 76. 1 1 1 Suppose given β P addrKar A, BspF ,G q with β ˚ JA “ α. Then we have β “ α .

JA A / Kar A

G α AI α1 F 1 +3 G1 F Ù * B

52 (2) Suppose given U, V P Ob addrKar A, Bs with U ˝ JA “ V ˝ JA. Then U – V . (3) We have the equivalence of categories

ΛA,B addrA, Bs ÐÝÝÝ addrKar A, Bs

β˚JA β pU ˝ JAq ÝÝÝÑpV ˝ JAq ÞÑ pU ÝÑ V q , ´ ¯ that is surjective on objects.

If unambiguous, we often write Λ:“ ΛA,B.

Proof. Ad (1). This follows from Proposition 75.(1). Ad (2). This follows from Proposition 75.(2).

1 Ad (3). Suppose given F P Ob addrA, Bs. Because (1) holds, we have F P addrKar A, Bs 1 1 and ΛF “ F ˝ JA “ F . Therefore, Λ is surjective on objects.

Suppose given F,G P Ob addrA, Bs. Since (1) holds, we have a bijection

1 1 addrKar A,BspF ,G q Ñ addrA,BspF,Gq, β ÞÑ Λβ.

Suppose given U, V P Ob addrKar A, Bs with ΛU “ ΛV . Since (2) holds, we have U – V . Therefore, Λ is an equivalence by Lemma 6. Theorem 78. Suppose A and B to be additive categories. Recall that B is idempotent complete; cf. Deﬁnition 45. Recall that Kar A is additive; cf. Proposition 41; and idempotent complete; cf. Lemma 54. The following assertions (1, 2, 3) hold.

(1) We have JA P addrA, Kar As; cf. Lemma 44. 1 α 1 α 1 Suppose given F ÝÑ G in addrA, Bs. We have F ÝÑ G in addrKar A, Bs with 1 1 1 F ˝ JA “ F , G ˝ JA “ G and α ˚ JA “ α; cf. Deﬁnition 72, Lemmas 73 and 76. 1 1 1 Suppose given β P rKar A,BspF ,G q with β ˚ JA “ α. Then we have β “ α .

JA A / Kar A

G α AI α1 F 1 +3 G1 F Ù * B

(2) Suppose given U, V P Ob addrKar A, Bs with U ˝ JA “ V ˝ JA. Then U – V . (3) We have the equivalence of categories

ΛA,B addrA, Bs ÐÝÝÝ addrKar A, Bs

β˚JA β pU ˝ JAq ÝÝÝÑpV ˝ JAq ÞÑ pU ÝÑ V q , ´ ¯ 53 that is surjective on objects.

If unambiguous, we often write Λ:“ ΛA,B.

Proof. Since additive categories are preadditive, this follows from Proposition 77.

Proposition 79. Suppose given an idempotent complete additive category C. Suppose given a full subcategory D of C. Let D1 be the full subcategory of C with

Ob D1 :“ tA P Ob C : DB P Ob C DD P Ob D : A ‘ B – Du.

The following assertions (1, 2, 3) hold.

(1) The category D is a full subcategory of D1.

(2) The category D1 is idempotent complete.

(3) Consider the additive functor

D ÝÑF D1

f f A ÝÑ B ÞÑ A ÝÑ B . ´ ¯ ´ ¯ Then 1 Kar D ÝÑF D1

ep ϕ qf eϕ9 f¯ pA, eq ÝÝÝÝÑpB, fq ÞÑ Im e ÝÝÑ Im f , ˆ ˙ ´ ¯ is an additive equivalence; cf. Deﬁnition 72.

Proof. Ad (1). Since D is a full subcategory of C, it suffices to show that Ob D Ď Ob D1. Suppose given D P Ob D. Since C is additive, there exists a zero object N P Ob C. We have D ‘ N – D. Thus, we have D P Ob D1. Ad (2). Suppose given an idempotent A ÝÑe A in D1. Since C is idempotent complete, we have an image pIm e, e,¯ e9q of e in C. Since D1 is a full subcategory of C, it suffices to show that Im e P Ob D1. By Lemma 53.(2), we have Im e ‘ Imp1 ´ eq – A in C. Since A P Ob D1, there exist B P Ob C and D P Ob D with A ‘ B – D. Thus, we have Im e ‘ pImp1 ´ eq ‘ Bq – D. Therefore, we have Im e P Ob D1. Ad (3). Since (1) holds, F is additive, full and faithful. Since (2) holds, D1 is idempotent complete. Thus, F 1 exists; cf. Definition 72. By Lemma 76, F 1 is additive. By Lemma 74.(1, 2), F 1 is full and faithful. Thus, it suffices to show that F 1 is dense.

1 Suppose given A1 P Ob D .

54 ϕ There exist A2 P Ob C, D P Ob D and an isomorphism A1 ‘ A2 „ / D in C. So, we 1 have A1 ,A2 P Ob D . ´1 1 0 Consider e :“ ϕ 0 0 ϕ P DpD,Dq. We have

2 ´1 1 0 ´1 1 0 ´1 1 0 1 0 ´1 1 0 e “ ϕ` ˘0 0 ϕϕ 0 0 ϕ “ ϕ 0 0 0 0 ϕ “ ϕ 0 0 ϕ “ e. Thus, e is an idempotent` ˘ in D.` Since˘ (1) holds,` ˘e `is˘ an idempotent` ˘ in D1. Since D1 is idempotent complete; cf. (2); e has an image pIm e, e,¯ e9q in D1. We have the following commutative diagram in C.

e D / D > ´1 1 p1 0qϕ ϕ 0 ´ ¯ ´1 1 ϕ 0 1A ´ ¯ 1 A1 / A1

1 1 1 Since D is a full subcategory of C, this diagram is in D . Thus, A1 is an image of e in D . 1 By Corollary 51, we have A1 – Im e in D . By Deﬁnition 72, we have

1 F pD, eq “ ImpF eq “ Im e – A1 . Thus, F 1 is dense.

2.1.9 Karoubi envelope for preadditive categories over a commutative ring

For this §2.1.9, let R be a commutative ring. Deﬁnition 80 (and Lemma). Let A be a preadditive category. We have the ring morphism ψA End 1A ÝÝÑ End 1Kar A α ÞÑ Kar α.

Proof. Since A is preadditive, so is Kar A; cf. Deﬁnition 38.

1 Suppose given α, α P End 1A. For brevity, we write ψ :“ ψA . We have L63.p2q pα ` α1qψ “ Karpα ` α1q “ Kar α ` Kar α1 “ αψ ` α1ψ and L61.p2q pαα1qψ “ Karpαα1q “ pKar αqpKar α1q “ pαψqpα1ψq. Furthermore, we have

L61.p1q L59.p1q p1End 1A qψ “ p11A qψ “ Kar 11A “ 1Kar 1A “ 11Kar A “ 1End 1Kar A . Thus, ψ is a ring morphism.

55 Lemma 81. The following assertions (1, 2) hold.

(1) Let pA, ϕAq be a preadditive category over R. Then pKar A, ϕAψAq is a preadditive category over R; cf. Deﬁnition 80.

(2) Let pA, ϕAq be an additive category over R. Then pKar A, ϕAψAq is an additive category over R; cf. Deﬁnition 80.

Proof. Ad (1). By Definition 80, ψA : End 1A Ñ End 1Kar A is a ring morphism. Thus, ϕAψA : R Ñ End 1Kar A is a ring morphism. Therefore, pKar A, ϕAψAq is a preadditive category over R; cf. Definition 23.(1). Ad (2). By Proposition 41, Kar A is additive. Thus, the assertions follows from (1) and Definition 23.(2).

ep α qf Remark 82. Suppose given a preadditive category pA, ϕAq over R. Suppose given pX, eq ÝÝÝÑ pY, fq in Kar A. Suppose given r P R. We have r ¨ ep α qf “ ep r ¨ α qf .

Proof. For brevity, we write ϕ :“ ϕA and ψ :“ ψA. We have

D23.p1q r ¨ ep α qf “ pprϕqψqpX,eq ¨ ep α qf D80 “ pKarprϕqqpX,eq ¨ ep α qf D60 “ ep e ¨ prϕqX ¨ e qe ¨ ep α qf “ ep prϕqX ¨ ee ¨ α qf “ ep prϕqX ¨ α qf D23.p1q “ ep r ¨ α qf .

Lemma 83. Suppose given a preadditive category pA, ϕAq over R. The inclusion functor JA : A Ñ Kar A is R-linear.

Proof. By Lemma 81.(1), pKar A, ϕAψAq is a preadditive category over R. By Lemma 44, J is additive. Suppose given r P R. Suppose given X ÝÑα Y in A. We have

D42 R82 D42 Jpr ¨ αq “ 1Xp r ¨ α q1Y “ r ¨ 1Xp α q1Y “ r ¨ Jα.

Thus, J is R-linear; cf. Deﬁnition 23.(3).

Lemma 84. Suppose given preadditive categories pA, ϕAq and pB, ϕBq over R. Suppose given F P Ob R-linrA, Bs. Then Kar F P Ob R-linrKar A, Kar Bs.

Proof. By Lemma 81.(1), pKar A, ϕAψAq and pKar B, ϕBψBq are preadditive categories over R. By Lemma 63.(1), Kar F is additive.

56 p α q Suppose given r P R. Suppose given pX, eq ÝÝÝÑpe f Y, fq in Kar A. We have

R82 pKar F qpr ¨ ep α qf q “ pKar F q ep r ¨ α qf D57 “ F ep F pr ¨ αq qF f

“ F ep r ¨ F α qF f R82 “ r ¨ F ep F α qF f D57 “ r ¨ pKar F q ep α qf .

Thus, Kar F is R-linear; cf. Deﬁnition 23.(3).

Lemma 85. Suppose given a preadditive category pB, ϕBq over R. Suppose B to be idempotent complete. The image functor IB : Kar B Ñ B is R-linear; cf. Deﬁnition 68.

Proof. By Lemma 81.(1), pKar B, ϕBψBq is a preadditive category over R. By Lemma 71, I is additive.

p α q Suppose given r P R. Suppose given pX, eq ÝÝÝÑpe f Y, fq in Kar B. We have

R82 Ipr ¨ ep α qf q “ I ep r ¨ α qf D“68 e9 ¨ pr ¨ αq ¨ f¯ D23.p1q ¯ “ e9 ¨ prϕqX ¨ α ¨ f ¯ “ prϕqX ¨ e9 ¨ α ¨ f R82 “ r ¨ I ep α qf .

Thus, I is R-linear; cf. Deﬁnition 23.(3).

Lemma 86. Suppose given preadditive categories pA, ϕAq and pB, ϕBq over R. Suppose B to be idempotent complete. Suppose given F P Ob R-linrA, Bs. 1 We have F P Ob R-linrKar A, Bs; cf. Deﬁnition 72.

Proof. By Lemma 81.(1), pKar A, ϕAψAq and pKar B, ϕBψBq are preadditive categories over R.

1 Recall that F “ IB ˝ Kar F ; cf. Deﬁnition 72.

Since F is R-linear, so is Kar F ; cf. Lemma 84. Furthermore, IB is R-linear; cf. Lemma 85. 1 Thus, F “ IB ˝ Kar F is R-linear.

Proposition 87. Suppose given preadditive categories pA, ϕAq and pB, ϕBq over R. Sup- pose B to be idempotent complete; cf. Deﬁnition 45.

Recall that pKar A, ϕAψAq is a preadditive category over R; cf. Lemma 81.(1). Recall that Kar A is idempotent complete; cf. Lemma 54. The following assertions (1, 2, 3) hold.

57 α (1) We have JA P Ob R-linrA, Kar As. Suppose given F ÝÑ G in R-linrA, Bs. We have 1 1 α 1 1 1 1 F ÝÑ G in R-linrKar A, Bs with F ˝ JA “ F , G ˝ JA “ G and α ˚ JA “ α; cf. Deﬁnition 72, Lemmas 73 and 86. 1 1 1 Suppose given β P R-linrKar A, BspF ,G q with β ˚ JA “ α. Then we have β “ α .

JA A / Kar A

G α AI α1 F 1 +3 G1 F Ù * B

(2) Suppose given U, V P Ob R-linrKar A, Bs with U ˝ JA “ V ˝ JA. Then U – V . (3) We have the equivalence of categories

ΛA,B R-linrA, Bs ÐÝÝÝ R-linrKar A, Bs

β˚JA β pU ˝ JAq ÝÝÝÑpV ˝ JAq ÞÑ pU ÝÑ V q , ´ ¯ that is surjective on objects.

If unambiguous, we often write Λ:“ ΛA,B.

Proof. Ad (1). This follows from Lemma 83 and Proposition 75.(1). Ad (2). This follows from Proposition 75.(2).

1 Ad (3). Suppose given F P Ob R-linrA, Bs. Because (1) holds, we have F P R-linrKar A, Bs 1 1 and ΛF “ F ˝ JA “ F . Therefore, Λ is surjective on objects.

Suppose given F,G P Ob R-linrA, Bs. Since (1) holds, we have a bijection

1 1 R-linrKar A,BspF ,G q Ñ R-linrA,BspF,Gq, β ÞÑ Λβ.

Suppose given U, V P Ob R-linrKar A, Bs with ΛU “ ΛV . Since (2) holds, we have U – V . Therefore, Λ is an equivalence by Lemma 6.

58 2.2 The additive envelope of preadditive categories

2.2.1 Deﬁnition and additivity

For this §2.2.1, let A be a preadditive category.

Deﬁnition 88 (and Lemma). We shall write tuples of objects and tuples of morphisms of A in square brackets. We shall deﬁne a category Add A as follows. Let

Ob Add A :“ trA1,...,Ams : m P Zě0 , Ai P Ob A for i P r1, msu.

We often write A1 ‘ ¨ ¨ ¨ ‘ Am :“ Ai :“ rA1,...,Ams P Ob Add A. iPr1,ms Ð We write NAdd A :“ r s for the empty tuple of objects; cf. Lemma 89 below.

For A1,1 ‘ ¨ ¨ ¨ ‘ A1,m and A2,1 ‘ ¨ ¨ ¨ ‘ A2,n in Ob Add A, let

Add ApA1,1 ‘ ¨ ¨ ¨ ‘ A1,m ,A2,1 ‘ ¨ ¨ ¨ ‘ A2,nq

:“ rfi,jsiPr1,ms,jPr1,ns : fi,j P ApA1,i ,A2,jq for pi, jq P r1, ms ˆ r1, ns . ( For rfi,jsiPr1,ms,jPr1,ns P Mor Add A, we often write

f1,1 ... f1,n . . f :“ . . :“ rfi,jsi,j :“ rfi,jsiPr1,ms,jPr1,ns . «fm,1 ... fm,nﬀ

Omitted entries are stipulated to be zero.

For A1,1 ‘¨ ¨ ¨‘A1,m and A2,1 ‘¨ ¨ ¨‘A2,m in Ob Add A and fi P ApA1,i ,A2,iq for i P r1, ms, we often write f1 .. diagrfisi :“ diagrfisiPr1,ms :“ . . f „ m ϕ For A1 ÝÑ A2 in A, we often write

rϕs :“ rϕsiPr1,1s,jPr1,1s P Add AprA1s, rA2sq.

For A1,1 ‘ ¨ ¨ ¨ ‘ A1,m , A2,1 ‘ ¨ ¨ ¨ ‘ A2,n and A3,1 ‘ ¨ ¨ ¨ ‘ A3,p in Ob Add A,

f “ rfi,jsi,j P Add ApA1,1 ‘ ¨ ¨ ¨ ‘ A1,m ,A2,1 ‘ ¨ ¨ ¨ ‘ A2,nq and

g “ rgj,ksj,k P Add ApA2,1 ‘ ¨ ¨ ¨ ‘ A2,n ,A3,1 ‘ ¨ ¨ ¨ ‘ A3,pq, we deﬁne composition by

fg “ f ¨ g “ rfi,jsi,j ¨ rgj,ksj,k :“ fi,jgj,k . » ﬁ jPr1,ns ÿ iPr1,ms,kPr1,ps – ﬂ 59 For A1 ‘ ¨ ¨ ¨ ‘ Am P Ob Add A, let

1 .. 1A ‘¨¨¨‘Am :“ rδi,jsi,j “ diagr1A siPr1,ns “ . . 1 i 1 „ 

For A1,1‘¨ ¨ ¨‘A1,m and A2,1‘¨ ¨ ¨‘A2,n in Ob Add A, rfi,jsi,j and rgi,jsi,j in Add ApA1,1 ‘ ¨ ¨ ¨ ‘ A1,m ,A2,1 ‘ ¨ ¨ ¨ ‘ A2,nq, we deﬁne

rfi,jsi,j ` rgi,jsi,j :“ rfi,j ` gi,jsi,j .

We call Add A the additive envelope of A. This deﬁnes a preadditive category Add A.

Proof. First, we show that Add A is a category. Consider f g h ‘ A1,i ÝÑ ‘ A2,j ÝÑ ‘ A3,k ÝÑ ‘ A4,l iPr1,ms jPr1,ns kPr1,ps lPr1,qs in Add A. We calculate

1A A ¨ g “ rδi,jsi,j ¨ rgj,ksj,k “ δi,jgj,k “ rgi,ksi,k “ g. 2,1‘¨¨¨‘ 2,n » ﬁ jPr1,ms ÿ i,k – ﬂ Similarily, we have

f ¨ 1A A “ rfi,jsi,j ¨ rδj,ksj,k “ fi,jδj,k “ rfi,ksi,k “ f. 2,1‘¨¨¨‘ 2,n » ﬁ jPr1,ns ÿ i,k – ﬂ Furthermore, we obtain

fpghq “ rfi,jsi,j ¨ prgj,ksj,k ¨ rhk,lsk,lq

“ rfi,jsi,j ¨ gj,khk,l kPr1,ps « ﬀj,l ř

“ fi,j gj,khk,l jPr1,ns kPr1,ps « ˜ ¸ﬀi,l ř ř

“ fi,jpgj,khk,lq pj,kqPr1,nsˆr1,ps « ﬀi,l ř

“ pfi,jgj,kqhk,l pj,kqPr1,nsˆr1,ps « ﬀi,l ř 60 “ fi,jgj,k hk,l kPr1,ps jPr1,ns « ˜ ¸ ﬀi,l ř ř

“ fi,jgj,k ¨ rhk,lsk,l jPr1,ns « ﬀi,k ř “ prfi,jsi,j ¨ rgj,ksj,kq ¨ rhk,lsk,l “ pfgqh. Thus, Add A is a category. Now we show that Add A is preadditive.

Suppose given ‘ A1,i and ‘ A2,j in Ob Add A. Since A is preadditive, ApA1,i ,A2,jq iPr1,ms jPr1,ns is an abelian group for pi, jq P r1, ms ˆ r1, ns. By deﬁnition of the addition in Add A, we have

Add Ap ‘ A1,i , ‘ A2,jq “ ApA1,i,A2,jq i 1,m j 1,n Pr s Pr s pi,jqPr1,msˆr1,ns à as abelian groups, so that Add Ap ‘ A1,i , ‘ A2,jq is an abelian group. iPr1,ms jPr1,ns Suppose given

g f / h ‘ A1,i / ‘ A2,j / ‘ A3,k / ‘ A4,l iPr1,ms jPr1,ns g˜ kPr1,ps lPr1,qs in Add A. We calculate

fpg ` g˜qh “ rfi,jsi,j ¨ prgj,ksj,k ` rg˜j,ksj,kq ¨ rhk,lsk,l

“ rfi,jsi,j ¨ rgj,k ` g˜j,ksj,k ¨ rhk,lsk,l

“ fi,jpgj,k ` g˜j,kqhk,l » ﬁ pj,kqPr1,nsˆr1,ps ÿ i,l – ﬂ

“ pfi,jgj,khk,l ` fi,jg˜j,khk,lq » ﬁ pj,kqPr1,nsˆr1,ps ÿ i,l – ﬂ

“ fi,jgj,khk,l ` fi,jg˜j,khk,l » fi » fi pj,kqPr1,nsˆr1,ps pj,kqPr1,nsˆr1,ps ÿ i,l ÿ i,l – fl – fl “ rfi,jsi,j ¨ rgj,ksj,k ¨ rhk,lsk,l ` rfi,jsi,j ¨ rg˜j,ksj,k ¨ rhk,lsk,l “ fgh ` fgh.˜

Thus, Add A is preadditive.

61 Lemma 89. We have a zero object NAdd A in Add A; cf. Deﬁnition 88.

Proof. Suppose given A1 ‘ ¨ ¨ ¨ ‘ An P Ob Add A. We have to show that

| Add ApA1 ‘ ¨ ¨ ¨ ‘ An,NAdd Aq| “ 1 “ | Add ApNAdd A,A1 ‘ ¨ ¨ ¨ ‘ Anq| . We have Add ApA1 ‘ ¨ ¨ ¨ ‘ An,NAdd Aq “ tr su; cf. Deﬁnition 88; and Add ApNAdd A,A1 ‘ ¨ ¨ ¨ ‘ Anq “ tr su; cf. Deﬁnition 88.

Remark 90. Let F and G be ﬁnite totally ordered sets. For m :“ |F|, n :“ |G|, we have σ τ unique monotone bijections r1, ms ÝÑ F, r1, ns ÝÑ G. Given A1,ζ P Ob A for ζ P F and A2,η P Ob A for η P G, we write ‘ A1,ζ :“ ‘ Aiσ and ‘ A2,η :“ ‘ Ajτ . ζPF iPr1,ms ηPG jPr1,ns

Given fζ,η P ApA1,ζ ,A2,ηq for pζ, ηq P F ˆ G, we write rfζ,ηsζ,η :“ rfiσ,jτ si,j.

Lemma 91. Suppose given m, n P Zě0 . Deﬁne

I1 :“ tp1, iq : i P r1, msu ,

I2 :“ tp2, jq : j P r1, nsu and

I :“ I1 Y I2 , ordered lexicographically.

Suppose given A1,1 ‘ ¨ ¨ ¨ ‘ A1,m and A2,1 ‘ ¨ ¨ ¨ ‘ A2,n in Ob Add A. Then

A1,1 ‘ ¨ ¨ ¨ ‘ A1,m O 1 .. . 1 0 ... 0 » 1fi .. . . rδζ,θsζPI,θPI1 “ 0 ... 0 . . . “rδθ,ζ sθPI1,ζPI —. .ffi « 1 0 ... 0ff —. .ffi — ffi —0 ... 0ffi – fl A1,1 ‘ ¨ ¨ ¨ ‘ A1,m ‘ A2,1 ‘ ¨ ¨ ¨ ‘ A2,n O 0 ... 0 . . . . 0 ... 0 1 »0 ... 0fi . . .. rδζ,θsζPI,θPI2 “ 1 . . . “rδθ,ζ sθPI2,ζPI — .. ffi «0 ... 0 1 ff — . ffi — ffi — 1ffi – fl A2,1 ‘ ¨ ¨ ¨ ‘ A2,n

is a direct sum of A1,1 ‘ ¨ ¨ ¨ ‘ A1,m and A2,1 ‘ ¨ ¨ ¨ ‘ A2,n in Add A.

Proof. We calculate

rδθ,ζ sθPI1,ζPI ¨ rδζ,ηsζPI,ηPI1 “ δθ,ζ δζ,η “ rδθ,ηsθPI1,ηPI1 “ 1A1,1‘¨¨¨‘A1,m . «ζPI ﬀ ÿ θPI1,ηPI1 62 Similarily, we obtain

rδθ,ζ sθPI2,ζPI ¨ rδζ,ηsζPI,ηPI2 “ δθ,ζ δζ,η “ rδθ,ηsθPI2,ηPI2 “ 1A2,1‘¨¨¨‘A2,n . «ζPI ﬀ ÿ θPI2,ηPI2 Furthermore, we have

rδζ,ηsζPI,ηPI1 ¨ rδη,ξsηPI1,ξPI ` rδζ,ηsζPI,ηPI2 ¨ rδη,ξsηPI2,ξPI

“ δζ,ηδη,ξ ` δζ,ηδη,ξ ηPI ηPI « 1 ﬀζPI,ξPI « 2 ﬀζPI,ξPI ř ř

“ δζ,ηδη,ξ ηPI « ﬀζPI,ξPI ř “ rδζ,ξsζPI,ξPI

“ 1A1,1‘¨¨¨‘A1,m‘A2,1‘¨¨¨‘A2,n .

Proposition 92. The additive envelope Add A of A is additive.

Proof. By Deﬁnition 88, Add A is preadditive. By Lemma 89, Add A has a zero object. By Lemma 91, all objects A, B P Ob Add A have a direct sum in Add A. Thus, Add A is additive.

Remark 93. Suppose given A1,...,Am P Ob A. Let

pAj qjPr1,ms ιi :“rδi,lskPr1,1s,lPr1,ms“r0 ... 0 1 0 ... 0s / rAis o A1 ‘ ¨ ¨ ¨ ‘ Am 0. . »0fi pAj qj 1,m π Pr s :“rδ s “ 1 i k,i kPr1,ms,lPr1,1s — ffi —0.ffi —.ffi —.ffi —0ffi — ffi – fl for i P r1, ms. If unambiguous, we often write

pAj qjPr1,ms pAj qjPr1,ms ιi :“ ιi and πi :“ πi for i P r1, ms.

Then pA1 ‘ ¨ ¨ ¨ ‘ Am , pπiqiPr1,ms , pιiqiPr1,msq is a direct sum of rA1s,..., rAms in Add A.

Proof. We have

πi ¨ ιi “ rδk,iskPr1,ms,lPr1,1s ¨ rδi,jslPr1,1s,jPr1,ms iPr1,ms iPr1,ms ÿ ÿ “ rδk,i ¨ δi,jskPr1,ms,jPr1,ms iPr1,ms ÿ

“ δk,iδi,j » ﬁ iPr1,ms ÿ kPr1,ms,jPr1,ms – ﬂ “ rδk,jskPr1,ms,jPr1,ms

“ 1A1‘¨¨¨‘Am .

63 Suppose given i, j P r1, ms. We have

ιi ¨ πj “ rδi,ls ¨ rδl,js “ δi,lδl,j “ rδi,js. kPr1,1s,lPr1,ms lPr1,ms,nPr1,1s » ﬁ lPr1,ms ÿ – ﬂ Thus, we have

ιi ¨ πi “ r1Ai s “ 1rAis and

ιi ¨ πj “ r0Ai,Aj s “ 0rAis,rAj s for j P r1, msztiu.

Remark 94. Suppose given m P Zě0. Suppose given Ai P Ob A for i P r1, ms. The following assertions (1, 2, 3) hold.

rδi,jσsi,j / (1) Suppose given σ P Sm. Then Ai o Ajσ are mutually inverse isomorphisms in Add A. iPr1,ms rδkσ,lsk,l jPr1,ms Ð Ð rδi,j si,j / (2) Suppose Am to be a zero object in A. Then Ai o Aj are mutually inverse isomorphisms in Add A. iPr1,ms rδk,lsk,l jPr1,m´1s Ð Ð

(3) Suppose A1 ,...,Am to have a direct sum pC, pπiqiPr1,ms, pιiqiPr1,msq in A.

rπ1 ... πms / Then rCs o Ai are mutually inverse isomorphisms in Add A. ι1 iPr1,ms . » . ﬁ Ð ιm – ﬂ Remark 95. Suppose given a ring R. Suppose B to be a full preadditive subcategory of R -free with R P Ob B. Then

R -free ÝÑF Add B

pf q rf s Rm ÝÝÝÝÑi,j i,j Rn ÞÑ R ÝÝÝÝÑi,j i,j R ˆ ˙ ˜iPr1,ms jPr1,ns ¸ Ð Ð is an equivalence.

Proof. For x P R we also write x for the map R Ñ R, y ÞÑ yx. First we show that F is a functor. pf q pg q Suppose given Rm ÝÝÝÝÑi,j i,j Rn ÝÝÝÝÑj,k j,k Rp in R -free. We have

F 1 m F diag 1 diag 1 1 1 m . p R q “ p qi “ r si “ iPr1,ms R “ F pR q Ð 64 Furthermore, we have

F ppfi,jqi,j ¨ pgj,kqj,kq “ F fi,jgj,k ¨ ˛ jPr1,ns ÿ j,k ˝ ‚

“ fi,jgj,k » ﬁ jPr1,ns ÿ j,k – ﬂ “ rfi,jsi,j ¨ rgj,ksj,k

“ F pfi,jqi,j ¨ F pgj,kqj,k . Thus, F is a functor. We show that F is full. m n m n Suppose given R ,R P Ob R -free and f “ rfi,jsi,j P Add BpF pR q,F pR qq. By Def- inition 88, we have fi,j P BpR,Rq “ R for pi, jq P r1, ms ˆ r1, ns. Thus, we have f “ rfi,jsi,j “ F pfi,jqi,j . Therefore, F is full. We show that F is faithful. m n m n Suppose give R and R in Ob R -free, pfi,jqi,j and pgi,jqi,j in R -freepR ,R q with

rfi,jsi,j “ F pfi,jqi,j “ F pgi,jqi,j “ rgi,jsi,j .

By Deﬁnition 88, we obtain fi,j “ gi,j for pi, jq P r1, ms ˆ r1, ns. Thus, we have pfi,jqi,j “ pgi,jqi,j. Therefore, F is faithful. We show that F is dense.

Suppose given B1 ‘ ¨ ¨ ¨ ‘ Bm P Ob Add B. Since B is a subcategory of R -free, there exist ki ki P Zě0 with Bi “ R for i P r1, ms. Deﬁne n :“ iPr1,ms ki . We have

k1 km n B1 ‘ ¨ ¨ ¨ ‘ Bm “ R ‘ ¨ ¨ ¨ ‘ R ř– rR s – R. R94.p3q R94.p3q iPr1,ns ð n Thus, we have B1 ‘ ¨ ¨ ¨ ‘ Bm – R “ F pR q. Therefore, F is dense. iPr1,ns Ð 2.2.2 The inclusion functor

The inclusion functor will play the role of the universal functor of the additive envelope construction; cf. §2.2.5. In Proposition 99 we give a necessary and suﬃcient condition for a preadditive category C to be additive. This condition justiﬁes the notion additive envelope for Add C.

For this §2.2.2, let A be a preadditive category. Deﬁnition 96 (and Lemma). We have the additive functor

I A ÝÑA Add A

ϕ rϕs pA ÝÑ Bq ÞÑ rAs ÝÑrBs . ´ ¯ 65 We call IA the inclusion functor of A in Add A.

If unambiguous, we often write I :“ IA . Cf. also Remark 94.(3).

ϕ ψ Proof. Suppose given A ÝÑ B ÝÑ C in A. We have

Ipϕψq “ rϕψs “ rϕsrψs “ Iϕ ¨ Iψ.

Furthermore, we have Ip1Aq “ r1As “ 1IA . Thus, I is a functor. ϕ / Suppose given A / B in A. We have ψ

Ipϕ ` ψq “ rϕ ` ψs “ rϕs ` rψs “ Iϕ ` Iψ .

Thus, I is additive. Lemma 97. The following assertions (1, 2) hold.

(1) The inclusion functor IA is full.

(2) The inclusion functor IA is faithful.

Proof. Suppose given A, B P Ob A.

Ad (1). Suppose given f P Add AprAs, rBsq. By Deﬁnition 88, there exists ϕ P ApA, Bq with f “ rϕs “ Iϕ. Thus, I is full.

Ad (2). Suppose given ϕ, ψ P ApA, Bq with Iϕ “ Iψ. We have

rϕs “ Iϕ “ Iψ “ rψs.

Therefore, we obtain ϕ “ ψ. Thus, I is faithful.

Lemma 98. Suppose A to be additive. Then IA : A Ñ Add A is dense.

Proof. Suppose given A1 ‘ ¨ ¨ ¨ ‘ An P Ob Add A. Since A is additive, we have a direct sum

Ai , pπiqiPr1,ns , pιiqiPr1,ns ˜iPr1,ns ¸ à of A1 ,...,An in A.

Since IA is an additive functor, we have a direct sum

I Ai , pIπiqiPr1,ns , pIιiqiPr1,ns ˜ ˜iPr1,ns ¸ ¸ à 66 of IA1 “ rA1s,...,IA1 “ rAns in Add A.

By Remark 93, A1 ‘ ¨ ¨ ¨ ‘ An is also a direct sum of rA1s,..., rAns in Add A.

Therefore, IpA1 ‘ ¨ ¨ ¨ ‘ Anq – A1 ‘ ¨ ¨ ¨ ‘ An . Thus, IA is dense.

Proposition 99. Suppose given a preadditive category C. The following assertions (1, 2) are equivalent.

(1) The preadditive category C is additive.

I (2) The inclusion functor C ÝÑC Add C is an equivalence.

Proof. Ad (1)ñ(2). By Lemma 97, IC is full and faithful. Since C is additive, IC is dense; cf. Lemma 98. Thus, IC is an equivalence. Ad (2)ñ(1). By Proposition 92, Add C is additive. Thus, C » Add C is additive.

2.2.3 Functoriality

In this §2.2.3 we deﬁne the additive envelope construction Add for additive functors and transformations between them. Furthermore, we establish functoriality properties of Add, which could be expressed by saying that it is turned into a 2-functor.

For this §2.2.3, let A, B and C be preadditive categories.

Deﬁnition 100 (and Lemma). Let F P Ob addrA, Bs. We have the additive functor

Add A ÝÝÝÑAdd F Add B

rfi,j si,j rF fi,j si,j A1,i ÝÝÝÝÑ A2,j ÞÑ FA1,i ÝÝÝÝÝÑ FA2,j . ˜iPr1,ms jPr1,ns ¸ ˜iPr1,ms jPr1,ns ¸ Ð Ð Ð Ð So, we have

pAdd F q Ai “ FAi ˜iPr1,ms ¸ iPr1,ms ð ð for Ai P Ob Add A. iPr1,ms Furthermore,Ð we have

pAdd F qrfi,jsi,j “ rF fi,jsi,j for rfi,jsi,j P Mor Add A.

67 f g Proof. Suppose given ‘ A1,i ÝÑ ‘ A2,j ÝÑ ‘ A3,k in Add A. We have iPr1,ms jPr1,ns kPr1,ps

pAdd F qpfgq “ pAdd F q fi,jgj,k ¨» ﬁ ˛ jPr1,ns ÿ i,k ˝– ﬂ ‚

“ F fi,jgj,k » ¨ ˛ﬁ jPr1,ns ÿ i,k – ˝ ‚ﬂ

“ F pfi,jgj,kq » ﬁ jPr1,ns ÿ i,k – ﬂ

“ F fi,j ¨ F gj,k » ﬁ jPr1,ns ÿ i,k – ﬂ “ rF fi,jsi,j ¨ rF gj,ksj,k “ pAdd F qf ¨ pAdd F qg.

Furthermore, we have

pAdd F q1A1,1‘¨¨¨‘A1,m “ pAdd F qrδi,jsi,j

“ rF δi,jsi,j

“ rδi,jsi,j

“ 1FA1,1‘¨¨¨‘FA1,m

“ 1pAdd F qpA1,1‘¨¨¨‘A1,mq .

Thus, Add F is a functor.

f / Consider ‘ A1,i / ‘ A2,j in Add A. We calculate iPr1,ms g jPr1,ns

pAdd F qpf ` gq “ pAdd F q prfi,j ` gi,jsi,jq

“ rF pfi,j ` gi,jqsi,j

“ rF fi,j ` F gi,jsi,j

“ rF fi,jsi,j ` rF gi,jsi,j “ pAdd F qf ` pAdd F qg.

Thus, Add F is additive.

Lemma 101. Let F P Ob addrA, Bs. The following assertions (1, 2, 3) hold.

(1) Suppose F to be full. Then Add F is full.

(2) Suppose F to be faithful. Then Add F is faithful.

(3) Suppose F to be dense. Then Add F is dense.

68 Proof. Suppose given ‘ A1,i , ‘ A2,j P Ob Add A. iPr1,ms jPr1,ns Ad (1). Suppose given

rϕi,jsi,j P Add B pAdd F q ‘ A1,i , pAdd F q ‘ A1,j i 1,m j 1,n ˆ ˆ Pr s ˙ ˆ Pr s ˙˙

“ Add B ‘ FA1,i , ‘ FA1,j . i 1,m j 1,n ˆ Pr s Pr s ˙

Since F is full, there exists fi,j P ApA1,i,A2,jq with ϕi,j “ F fi,j for pi, jq P r1, ms ˆ r1, ns. With f “ rfi,jsi,j P Add Ap ‘ A1,i , ‘ A2,jq, we obtain iPr1,ms jPr1,ns

pAdd F qf “ pAdd F qrfi,jsi,j “ rF fi,jsi,j “ rϕi,jsi,j .

Thus, Add F is full.

Ad (2). Suppose given f, g P Add A ‘ A1,i , ‘ A2,j with pAdd F qf “ pAdd F qg. i 1,m j 1,n ˆ Pr s Pr s ˙ We have F fi,j “ F gi,j for pi, jq P r1, ms ˆ r1, ns. Since F is faithful, we obtain fi,j “ gi,j for pi, jq P r1, ms ˆ r1, ns. Therefore, we have f “ g. Thus, Add F is faithful.

Ad (3). Suppose given B1 ‘¨ ¨ ¨‘Bn P Ob Add B. Since F is dense, there exists Ai P Ob A ϕ-i with FAi – Bi for i P r1, ns. Let FAi „ Bi be an isomorphism for i P r1, ns. We have mutually inverse isomorphisms

diagrϕisi FA1 ‘ ¨ ¨ ¨ ‘ FAn / B1 ‘ ¨ ¨ ¨ ‘ Bn . o ´1 diagrpϕiq si

Therefore, we have

pAdd F qpA1 ‘ ¨ ¨ ¨ ‘ Anq “ FA1 ‘ ¨ ¨ ¨ ‘ FAn – B1 ‘ ¨ ¨ ¨ ‘ Bn .

Thus, Add F is dense.

F G Lemma 102. Suppose given additive functors A / B / C. The following assertions (1, 2) hold.

(1) We have Add 1A “ 1Add A .

(2) We have AddpG ˝ F q “ pAdd Gq ˝ pAdd F q.

rfi,j si,j Proof. Suppose given iPr1,ms A1,i ÝÝÝÝÑ jPr1,ns A2,j in Add A.

Ð Ð69 Ad (1). We have

rfi,j si,j r1Afi,j si,j pAdd 1Aq A1,i ÝÝÝÝÑ A2,j “ 1AA1,i ÝÝÝÝÝÝÑ 1AA2,j ˜iPr1,ms jPr1,ns ¸ ˜iPr1,ms jPr1,ns ¸ ð ð ð ð rfi,j si,j “ A1,i ÝÝÝÝÑ A2,j ˜iPr1,ms jPr1,ns ¸ ð ð rfi,j si,j “ 1Add A A1,i ÝÝÝÝÑ A2,j . ˜iPr1,ms jPr1,ns ¸ ð ð Ad (2). We have

rfi,j si,j pAddpG ˝ F qq A1,i ÝÝÝÝÑ A2,j ˜iPr1,ms jPr1,ns ¸ Ð Ð rpG˝F qfi,j si,j “ pG ˝ F qA1,i ÝÝÝÝÝÝÝÝÑ pG ˝ F qA2,j ˜iPr1,ms jPr1,ns ¸ Ð Ð rF fi,j si,j “ pAdd Gq FA1,i ÝÝÝÝÝÑ FA2,j ˜iPr1,ms jPr1,ns ¸ Ð Ð rfi,j si,j “ ppAdd Gq ˝ pAdd F qq A1,i ÝÝÝÝÑ A2,j . ˜iPr1,ms jPr1,ns ¸ Ð Ð

α ˜ Deﬁnition 103 (and Lemma). Suppose given F ÝÑ F in addrA, Bs. Let

pAdd αqA1‘¨¨¨‘An :“ diagrαAi siPr1,ns

for A1 ‘ ¨ ¨ ¨ ‘ An P Ob Add A, being a morphism from

D100 ˜ D100 ˜ FAi “ pAdd F q Ai to FAi “ pAdd F q Ai . iPr1,ns ˜iPr1,ns ¸ iPr1,ns ˜iPr1,ns ¸ Ð Ð Ð Ð Let Add α : Add α . “ pp qA1‘¨¨¨‘An qA1‘¨¨¨‘AnPOb Add A This deﬁnes a transformation Add α : Add F ñ Add F˜.

f Proof. Suppose given ‘ A1,i ÝÑ ‘ A2,j in Add A. iPr1,ms jPr1,ns We have ˜ ˜ pAdd αqA1,1‘¨¨¨‘A1,m ¨ pAdd F qf “ diagrαA1,i siPr1,ms ¨ rF fi,jsi,j ˜ “ rαA1,i ¨ F fi,jsi,j

“ rF fi,j ¨ αA2,j si,j

“ rF fi,jsi,j ¨ diagrαA2,j sjPr1,ns

“ pAdd F qf ¨ pAdd αqA2,1‘¨¨¨‘A2,n . Thus, Add α is natural.

70 Lemma 104. Suppose given additive functors F , F0, F1, F2 : A Ñ B. The following assertions (1, 2, 3) hold.

(1) We have Add 1F “ 1Add F .

(2) Suppose given transformations α, α˜ : F0 ñ F1. We have

Addpα ` α˜q “ Add α ` Addα. ˜

(3) Suppose given transformations α0 : F0 ñ F1 and α1 : F1 ñ F2. We have

Addpα0α1q “ pAdd α0qpAdd α1q.

Proof. Suppose given A1 ‘ ¨ ¨ ¨ ‘ Am P Ob Add A. Ad (1). We compute

pAdd 1F qA1‘¨¨¨‘Am “ diagrp1F qAi siPr1.ms

“ diagr1FAi siPr1,ms

“ 1FA1‘¨¨¨‘FAm

“ 1pAdd F qpA1‘¨¨¨‘FAmq 1 . “ p Add F qA1‘¨¨¨‘Am

Ad (2). We have

pAddpα ` α˜qqA1‘¨¨¨‘Am “ diagrpα ` α˜qAi siPr1,ms

“ diagrαAi ` α˜Ai siPr1,ms

“ diagrαAi siPr1,ms ` diagrα˜Ai siPr1,ms

“ pAdd αqA1‘¨¨¨‘Am ` pAddα ˜qA1‘¨¨¨‘Am .

Ad (3). We have

pAddpα0α1qqA1‘¨¨¨‘Am “ diagrpα0α1qAi siPr1,ms

“ diagrpα0qAi pα1qAi siPr1,ms

“ diagrpα0qAi siPr1,ms ¨ diagrpα1qAi siPr1,ms

“ pAdd α0qA1‘¨¨¨‘Am ¨ pAdd α1qA1‘¨¨¨‘Am

“ ppAdd α0qpAdd α1qqA1‘¨¨¨‘Am .

Lemma 105. Suppose given additive functors F, F˜ : A Ñ B and G, G˜ : B Ñ C. Suppose given transformations α : F ñ F˜ and β : G ñ G˜. Then we have Addpβ ˚ αq “ pAdd βq ˚ pAdd αq.

Proof. Suppose given A1 ‘ ¨ ¨ ¨ ‘ Am P Ob Add A.

71 We calculate

pAddpβ ˚ αqqA1‘¨¨¨‘Am “ diagrpβ ˚ αqAi siPr1,ms diag Gα β “ r Ai ¨ FA˜ i siPr1,ms diag Gα diag β “ r Ai siPr1,ms ¨ r FA˜ i siPr1,ms Add G Add α Add β “ p qp qA1‘¨¨¨‘Am ¨ p qpAdd F˜qpA1‘¨¨¨‘Amq

“ ppAdd βq ˚ pAdd αqqA1‘¨¨¨‘Am .

2.2.4 The realisation functor

In this §2.2.4 we introduce the realisation functor, mapping formal direct sums as constructed in §2.2.1 to direct sums, provided the latter exist. In Deﬁnition 110, the realisation functor is used for the construction of the induced functors and transformations.

For this §2.2.4, let A be an additive category.

Stipulation 106. We recall no. 16. Recall that we have chosen a zero object 0A in A.

Suppose given m P Zě0 . Suppose given Ai P Ob A for i P r1, ms. Recall that we have chosen a direct sum

pAj qjPr1,ms pAj qjPr1,ms Ai , pπi qiPr1,ms , pιi qiPr1,ms ˜iPr1,ms ¸ à of A1 ,...,Am in A. Recall that we have chosen

Ai , pπiqiPr1,1s , pιiqiPr1,1s “ A1 , p1A1 qiPr1,1s , p1A1 qiPr1,1s ˜iPr1,1s ¸ à ` ˘ and

Ai , pπiqiPr1,0s , pιiqiPr1,0s “ p0A , p q, p qq . ˜iPr1,0s ¸ à Deﬁnition 107 (and Lemma). We have the additive functor

R Add A ÝÝÑA A

rfi,j si,j pfi,j qi,j ‘ A1,i ÝÝÝÝÑ ‘ A2,j ÞÑ A1,i ÝÝÝÝÑ A2,j . i 1,m j 1,n ˆ Pr s Pr s ˙ ˜iPr1,ms jPr1,ns ¸ À À We call RA the realisation functor from Add A to A.

If unambiguous, we often write R :“ RA .

72 Proof. Suppose given f g ‘ A1,i ÝÑ ‘ A2,j ÝÑ ‘ A3,k iPr1,ms jPr1,ns kPr1,ps in Add A. We have

Rpfgq “ R prfi,jsi,j ¨ rgj,ksj,kq

“ R fi,jgj,k ¨» ﬁ ˛ jPr1,ns ÿ i,k ˝– ﬂ ‚

“ fi,jgj,k ¨ ˛ jPr1,ns ÿ i,k ˝ ‚ “ pfi,jqi,j ¨ pgj,kqj,k

“ R prfi,jsi,jq ¨ R prgj,ksj,kq “ Rf ¨ Rg.

Furthermore, we obtain

Rp1A1‘¨¨¨‘Am q “ R diagr1Ai siPr1,ms “ diagp1Ai qiPr1,ms “ 1A1‘¨¨¨‘Am “ 1RpA1‘¨¨¨‘Amq .

Thus, R is a functor. ` ˘

f / Suppose given ‘ A1,i / ‘ A2,j in Add A. iPr1,ms f˜ jPr1,ns We have

˜ ˜ ˜ ˜ ˜ Rpf ` fq “ R rfi,j ` fi,jsi,j “ pfi,j ` fi,jqi,j “ pfi,jqi,j ` pfi,jqi,j “ Rf ` Rf. ´ ¯ Thus, R is additive.

Lemma 108. The following assertions (1, 2, 3) hold.

(1) The realisation functor RA is full.

(2) The realisation functor RA is faithful.

(3) The realisation functor RA is surjective on objects. In particular, RA is dense.

Thus, RA is an equivalence.

Proof. Suppose given A1,1 ‘ ¨ ¨ ¨ ‘ A1,m and A2,1 ‘ ¨ ¨ ¨ ‘ A2,n in Ob Add A.

Ad (1). Suppose given pfi,jqi,j P ApA1,1 ‘ ¨ ¨ ¨ ‘ A1,m ,A2,1 ‘ ¨ ¨ ¨ ‘ A2,nq. We have

R prfi,jsi,jq “ pfi,jqi,j .

Thus, R is full.

73 Ad (2). Suppose given f, g P Add ApA1,1 ‘ ¨ ¨ ¨ ‘ A1,m ,A2,1 ‘ ¨ ¨ ¨ ‘ A2,nq with Rf “ Rg. We have pfi,jqi,j “ Rf “ Rg “ pgi,jqi,j . We obtain fi,j “ gi,j for pi, jq P r1, ms ˆ r1, ns. Therefore, we have f “ rfi,jsi,j “ rgi,jsi,j “ g. Thus, R is faithful. Ad (3). Suppose given A P Ob A. We have rAs P Ob Add A; cf. Deﬁnition 88. We obtain RrAs “ A; cf. Stipulation 106. Thus, R is surjective on objects.

Remark 109. We have RA ˝ IA “ 1A .

ϕ Proof. Suppose given A ÝÑ B in A. We have

ϕ D96 rϕs D107 ϕ ϕ pR ˝ IqpA ÝÑ Bq “ R rAs ÝÑrBs “ pA ÝÑ Bq “ 1ApA ÝÑ Bq. S106 ´ ¯

2.2.5 Universal property

For this §2.2.5, let A be a preadditive category. Let B be an additive category.

Deﬁnition 110 (and Lemma). We have the functor

addrA, Bs Ñ addrAdd A, Bs

α RB˚Add α pF ÝÑ Gq ÞÑ pRB ˝ Add F q ÝÝÝÝÝÝÑpRB ˝ Add Gq . ´ ¯ α 1 1 For F ÝÑ G in addrA, Bs, we write F :“ RB ˝ Add F and α :“ RB ˚ Add α. α Suppose given F ÝÑ G in addrA, Bs. We have

1 Add A ÝÑF B

rfi,j si,j pF fi,j qi,j ‘ A1,i ÝÝÝÝÑ ‘ A2,j ÞÑ FA1,i ÝÝÝÝÝÑ FA2,j . i 1,m j 1,n ˆ Pr s Pr s ˙ ˜iPr1,ms jPr1,ns ¸ À À Suppose given A1 ‘ ¨ ¨ ¨ ‘ Am P Ob Add A. We have

1 αA1‘¨¨¨‘Am “ diagpαAi qiPr1,ms .

1 Proof. Since RB and Add F are additive, cf. Deﬁnitions 107 and 100, F “ RB ˝ Add F is additive. α β Suppose given F ÝÑ G ÝÑ H in addrA, Bs. We have

1 αA1‘¨¨¨‘Am “ pRB ˚ Add αqA1‘¨¨¨‘Am

“ RBpAdd αqA1‘¨¨¨‘Am

“ RBpdiagrαAi siq

“ diagpαAi qi .

74 Thus, we have

1 1 p1F qA1‘¨¨¨‘Am “ diagpp1F qAi qi “ diagp1FAi qi “ 1FA1‘¨¨¨‘FAm “ 1F pA1‘¨¨¨‘Amq .

Furthermore, we have

1 pαβqA1‘¨¨¨‘Am “ diagppαβqAi qi

“ diagpαAi ¨ βAi qi

“ diagpαAi qi ¨ diagpβAi qi 1 1 “ αA1‘¨¨¨‘Am ¨ βA1‘¨¨¨‘Am .

Thus, we have a functor indeed. Furthermore, we have

1 rfi,j si,j rfi,j si,j F ‘ A1,i ÝÝÝÝÑ ‘ A2,j “ pRB ˝ Add F q ‘ A1,i ÝÝÝÝÑ ‘ A2,j i 1,m j 1,n i 1,m j 1,n ˆ Pr s Pr s ˙ ˆ Pr s Pr s ˙ rF fi,j si,j “ RB ‘ FA1,i ÝÝÝÝÝÑ ‘ FA2,j i 1,m j 1,n ˆ Pr s Pr s ˙

pF fi,j qi,j “ FA1,i ÝÝÝÝÝÑ FA2,j . ˜iPr1,ms jPr1,ns ¸ à à

Lemma 111. Suppose given F P Ob addrA, Bs. The following assertions (1, 2, 3) hold.

(1) If F is full, so is F 1.

(2) If F is faithful, so is F 1.

(3) If F is dense, so is F 1.

Proof. Ad (1). Since F is full, so is Add F ; cf. Lemma 101.(1). By Lemma 108.(1), RB is 1 full. Thus, F “ RB ˝ Add F is full.

Ad (2). Since F is faithful, so is Add F ; cf. Lemma 101.(2). By Lemma 108.(2), RB is 1 faithful. Thus, F “ RB ˝ Add F is faithful.

Ad (3). Since F is dense, so is Add F ; cf. Lemma 101.(3). By Lemma 108.(3), RB is 1 dense. Thus, F “ RB ˝ Add F is dense.

α Lemma 112. Suppose given F ÝÑ G in addrA, Bs. The following assertions (1, 2) hold.

1 (1) We have F ˝ IA “ F .

1 (2) We have α ˚ IA “ α.

Cf. Deﬁnition 96.

75 ϕ Proof. Ad (1). Suppose given X ÝÑ Y in A. We have

1 ϕ 1 rϕs pF ˝ IAqpX ÝÑ Y q “ F prXs ÝÑrY sq

pF ϕq D“110 FX ÝÝÝÝÑi,j FY ˜iPr1,1s jPr1,1s ¸ F ϕ S“106 FXÀ ÝÝÑ FY À ϕ “ ´F pX ÝÑ Y q. ¯

Ad (2). Suppose given X P Ob A. We have

α1 I α1 p ˚ AqX “ IAX 1 “ αrXs D110 “ diagpαX qiPr1,1s S106 “ αX .

Theorem 113. Recall that A is a preadditive category and that B is an additive category. Recall that Add A is additive; cf. Proposition 92. The following assertions (1, 2, 3) hold.

(1) We have IA P addrA, Add As; cf. Deﬁnition 96.

1 α 1 α 1 Suppose given F ÝÑ G in addrA, Bs. We have F ÝÑ G in addrAdd A, Bs with 1 1 1 F ˝ IA “ F , G ˝ IA “ G and α ˚ IA “ α; cf. Deﬁnition 110 and Lemma 112. 1 1 1 Suppose given β P addrAdd A,BspF ,G q with β ˚ IA “ α. Then we have β “ α .

IA A / Add A

G α AI α1 F 1 +3 G1 F Ù * B

(2) Suppose given U, V P Ob addrAdd A, Bs with U ˝ IA “ V ˝ IA. Then U – V .

(3) We have the equivalence of categories

ΦA,B addrA, Bs ÐÝÝÝ addrAdd A, Bs

β˚IA β pU ˝ IA ÝÝÝÑ V ˝ IAq ÞÑ pU ÝÑ V q,

that is surjective on objects.

If unambiguous, we often write Φ:“ ΦA,B.

76 Proof. Ad (1). Suppose given X1 ‘ ¨ ¨ ¨ ‘ Xm P Ob Add A. 1 We show that αX1‘¨¨¨‘Xm “ βX1‘¨¨¨‘Xm . 1 1 Let βi,j :“ pβX1‘¨¨¨‘Xm qi,j and αi,j :“ pαX1‘¨¨¨‘Xm qi,j for i, j P r1, ms. 1 Suppose given e, f P r1, ms. We show that βe,f “ αe,f . We have 1 1 αe,f “ pαX1‘¨¨¨‘Xm qe,f “ pdiagpαXi qiPr1,msqe,f “ αXe δe,f .

1 1 We have F pX1 ‘¨ ¨ ¨‘Xmq “ FX1 ‘¨ ¨ ¨‘FXm and G pX1 ‘¨ ¨ ¨‘Xmq “ GX1 ‘¨ ¨ ¨‘GXm . Since β is natural, the following diagram commutes.

α1 “α “β 1 e,e Xe rXes 1 F rXes “ FXe / G rXes “ GXe

1 1 pδe,j qiPr1,1s,jPr1,ms “F prδe,j siPr1,1s,jPr1,msq G prδe,j siPr1,1s,jPr1,msq “pδe,j qiPr1,1s,jPr1,ms

βX1‘¨¨¨‘Xm “pβj,kqjPr1,ms,kPr1,ms FX1 ‘ ¨ ¨ ¨ ‘ FXm / GX1 ‘ ¨ ¨ ¨ ‘ Xm

pδk,f qkPr1,ms,lPr1,1s

1 G rXf s “ GXf

Therefore, we obtain

βe,f “ pδe,jqiPr1,1s,jPr1,ms ¨ pβj,kqjPr1,ms,kPr1,ms ¨ pδk,f qkPr1,ms,lPr1,1s 1 “ αe,e ¨ pδe,jqiPr1,1s,jPr1,ms ¨ pδj,f qjPr1,ms,lPr1,1s

1 “ α ¨ δe,j ¨ δj,f e,e ¨ ˛ jPr1,ms ÿ iPr1,1s,lPr1,1s 1 ˝ ‚ “ αe,e ¨ δe,f

“ αXe ¨ δe,f 1 “ αe,f . Thus, β “ α1. 1 Ad (2). Suppose we have shown that pW ˝ IAq – W for W P Ob addrAdd A, Bs. Then we have 1 1 U – pU ˝ IAq “ pV ˝ IAq – V. 1 Therefore, it suﬃces to show that pW ˝ IAq – W for W P Ob addrAdd A, Bs.

Suppose given W P Ob addrAdd A, Bs.

Suppose given A1 ‘ ¨ ¨ ¨ ‘ Am P Ob Add A. By Deﬁnition 110, we have

1 pW ˝ IAq pA1 ‘ ¨ ¨ ¨ ‘ Amq “ ppW ˝ IAqA1q ‘ ¨ ¨ ¨ ‘ ppW ˝ IAqAmq “ W rA1s ‘ ¨ ¨ ¨ ‘ W rAms.

77 By Remark 93 and additivity of W ,

pW pA1 ‘ ¨ ¨ ¨ ‘ Amq , pW πiqiPr1,ms , pW ιiqiPr1,msq is a direct sum of W rA1s,...,W rAms in B. With the notation of Remark 93, we have mutually inverse isomorphisms

pW πiqi“pW π1 ... W πmq / W pA1 ‘ ¨ ¨ ¨ ‘ Amq o W rA1s ‘ ¨ ¨ ¨ ‘ W rAms. W. ι1 pW ιiqi“ . ˜W ιm¸ Deﬁne

pAj qjPr1,ms pδA1‘¨¨¨‘Am qA ‘¨¨¨‘A POb Add A :“ W ιi ; 1 m i 1,m ,k 1,1 Pr s Pr s A ‘¨¨¨‘A POb Add A ˆ´ ¯ ˙ 1 m cf. Remark 93.

rfi,j si,j Suppose given A1,1 ‘ ¨ ¨ ¨ ‘ A1,m ÝÝÝÝÑ A2,1 ‘ ¨ ¨ ¨ ‘ A2,n in Add A. We have A δ W ιp 1,xqxPr1,ms W δ A1,1‘¨¨¨‘A1,m “ p k qkPr1,ms,lPr1,1s “ r k,jsiPr1,1s,jPr1,ms kPr1,ms,lPr1,1s and ` ˘ A δ W ιp 2,xqxPr1,ns W δ . A2,1‘¨¨¨‘A2,n “ p k qkPr1,ns,lPr1,1s “ r k,jsiPr1,1s,jPr1,ns kPr1,ns,lPr1,1s Furthermore, we have ` ˘ 1 D110 pW ˝ IAq rfi,jsi,j “ ppW ˝ IAqfi,jqi,j “ pW rfi,jsqi,j . We show that the following quadrangle is commutative.

pW rfi,j sqiPr1,ms,jPr1,ns W rA1,1s ‘ ¨ ¨ ¨ ‘ W rA1,ms / W rA2,1s ‘ ¨ ¨ ¨ ‘ W rA2,ns

pA1,xqxPr1,ms pA2,yqyPr1,ns pW ιk qkPr1,ms,lPr1,1s pW ιk qkPr1,ns,lPr1,1s

W prfi,j siPr1,ms,jPr1,nsq W pA1,1 ‘ ¨ ¨ ¨ ‘ A1,mq / W pA2,1 ‘ ¨ ¨ ¨ ‘ A2,nq We have pA1,xqxPr1,ms pW ιk qkPr1,ms,lPr1,1s ¨ W rfi,jsiPr1,ms,jPr1,ns

pA1,xqxPr1,ms ` ˘ “ W ιk ¨ W rfi,jsiPr1,ms,jPr1,ns kPr1,ms,lPr1,1s ´ ¯ pA1,xqxPr1,ms ` ˘ “ W ιk ¨ rfi,jsiPr1,ms,jPr1,ns kPr1,ms,lPr1,1s ´ ´ ¯¯ W δ f “ r k,ispPr1,1s,iPr1,ms ¨ r i,jsiPr1,ms,jPr1,ns kPr1,ms,lPr1,1s ` ` ˘˘

“ W δk,i ¨ fi,j ¨ ¨» ﬁ ˛˛ iPr1,ms p 1,1 ,j 1,n ˚ ˚ ÿ Pr s Pr s‹‹kPr1,ms,lPr1,1s ˝ ˝– ﬂ ‚‚ W f “ r k,jspPr1,1s,jPr1,ns kPr1,ms,lPr1,1s ` ˘ 78 and pA2,yqyPr1,ns pW rfi,jsqiPr1,ms,jPr1,ns ¨ pW ιj qjPr1,ns,lPr1,1s

pA2,yqyPr1,ns “ W rfi,js ¨ ι ¨ ¨ j ˛˛ jPr1,ns ÿ iPr1,ms,lPr1,1s ˝ ˝ ‚‚

“ W rfi,js ¨ rδj,ks ¨ ¨ pPr1,1s,kPr1,ns˛˛ jPr1,ns ÿ iPr1,ms,lPr1,1s ˝ ˝ ‚‚

“ W rfi,j ¨ δj,ks ¨ ¨ pPr1,1s,kPr1,ns˛˛ jPr1,ns ÿ iPr1,ms,lPr1,1s ˝ ˝ ‚‚

“ W fi,j ¨ δj,k ¨ » ﬁ ˛ jPr1,ns p 1,1 ,k 1,n ˚ ÿ Pr s Pr s‹iPr1,ms,lPr1,1s ˝ – ﬂ ‚ “ pW rfi,kspPr1,1s,kPr1,nsqiPr1,ms,lPr1,1s .

1 δ Thus, δ is natural. Therefore, pW ˝ IAq „ +3 W is an isotransformation. 1 In consequence, we have W – pW ˝ IAq . 1 Ad (3). Suppose given F P Ob addrA, Bs. Because (1) holds, we have F P addrAdd A, Bs 1 1 with ΦF “ F ˝ IA “ F . Therefore, Φ is surjective on objects.

Suppose given F,G P Ob addrA, Bs. Since (1) holds, we have a bijection

1 1 addrAdd A,BspF ,G q Ñ addrA,BspF,Gq, β ÞÑ Φβ.

Suppose given U, V P Ob addrAdd A, Bs with ΦU “ ΦV . Since (2) holds, we have U – V . Therefore, Φ is an equivalence by Lemma 6.

2.2.6 Additive envelope for preadditive categories over a commutative ring

For this §2.2.6, let R be a commutative ring.

Deﬁnition 114 (and Lemma). Suppose given a preadditive category A. We have the ring morphism ψA End 1A ÝÝÑ End 1Add A α ÞÑ Add α.

Proof. Suppose given α, α˜ P End 1A. For brevity, we write ψ :“ ψA.

79 We have

L104.p2q pα ` α˜qψ “ Addpα ` α˜q “ Add α ` Addα ˜ “ αψ ` αψ˜ and

L104.p3q pαα˜qψ “ Addpαα˜q “ pAdd αqpAddα ˜q “ pαψqpαψ˜ q.

Furthermore, we have

L104.p1q L102.p1q p1End 1A qψ “ p11A qψ “ Add 11A “ 1Add 1A “ 11Add A “ 1End 1Add A .

Thus, ψ is a ring morphism.

Lemma 115. Suppose given a preadditive category pA, ϕAq over R. Then pAdd A, ϕAψAq is an additive category over R; cf. Deﬁnition 114.

Proof. By Proposition 92, Add A is additive. By Deﬁnition 114, we have a ring morphism ϕAψA : R Ñ End 1Add A.

Thus, pAdd A, ϕAψAq is an additive category over R.

Remark 116. Suppose given a preadditive category pA, ϕAq over R. Suppose given

rfi,j si,j A1,1 ‘ ¨ ¨ ¨ ‘ A1,m ÝÝÝÝÑ A2,1 ‘ ¨ ¨ ¨ ‘ A2,n

in Add A. Suppose given r P R. We have r ¨ rfi,jsi,j “ rr ¨ fi,jsi,j.

Proof. For brevity, we write ϕ :“ ϕA and ψ :“ ψA. We have

r ¨ rfi,jsi,j “ prpϕψqqA1,1‘¨¨¨‘A1,m ¨ rfi,jsi,j

“ pAddprϕqqA1,1‘¨¨¨‘A1,m ¨ rfi,jsi,j

“ diagrprϕqA1,i si ¨ rfi,jsi,j

“ rprϕqA1,i ¨ fi,jsi,j

“ rr ¨ fi,jsi,j .

80 Lemma 117. Suppose given a preadditive category pA, ϕAq over R. The inclusion functor IA : A Ñ Add A is R-linear.

Proof. By Deﬁnition 96, I is additive. f Suppose given X ÝÑ Y in A. Suppose given r P R. We have

Ipr ¨ fq “ rr ¨ fs R“116 r ¨ rfs “ r ¨ If.

Thus, I is R-linear; cf. Deﬁnition 23.(3).

Lemma 118. Suppose given preadditive categories pA, ϕAq and pB, ϕBq over R. Suppose given F P Ob R-linrA, Bs. Then Add F P Ob R-linrAdd A, Add Bs.

Proof. By Deﬁnition 100, Add F is additive.

rfi,j si,j Suppose given A1,1 ‘ ¨ ¨ ¨ ‘ A1,m ÝÝÝÝÑ A2,1 ‘ ¨ ¨ ¨ ‘ A2,n in Add A. Suppose given r P R. We have R116 pAdd F qpr ¨ rfi,jsi,jq “ pAdd F qrr ¨ fi,jsi,j “ rF pr ¨ fi,jqsi,j “ rr ¨ F fi,jsi,j R116 “ r ¨ rF fi,jsi,j “ r ¨ pAdd F qrfi,jsi,j . Thus, Add F is R-linear; cf. Deﬁnition 23.(3).

Lemma 119. Suppose given an additive category pB, ϕBq over R. The realisation functor RB : Add B Ñ B is R-linear.

Proof. By Deﬁnition 107, RB is additive.

rfi,j si,j Suppose given B1,1 ‘ ¨ ¨ ¨ ‘ B1,m ÝÝÝÝÑ B2,1 ‘ ¨ ¨ ¨ ‘ B2,n in Add B. Suppose given r P R. We have

R116 R24 Rpr ¨ rfi,jsi,jq “ Rrr ¨ fi,jsi,j “ pr ¨ fi,jqi,j “ r ¨ pfi,jqi,j “ r ¨ Rrfi,jsi,j .

Thus, R is R-linear; cf. Deﬁnition 23.(3).

Lemma 120. Suppose given a preadditive category pA, ϕAq over R. Suppose given an additive category pB, ϕBq over R. Suppose given an R-linear functor F : A Ñ B. Then F 1 : Add A Ñ B is R-linear.

Proof. By Lemma 115, pAdd A, ϕAψAq and pAdd B, ϕBψBq are additive categories over R. 1 Recall that F “ RB ˝ Add F ; cf. Deﬁnition 110.

Since F is R-linear, so is Add F ; cf. Lemma 118. Furthermore, RB is R-linear; cf. 1 Lemma 119. Thus, F “ RB ˝ Add F is R-linear.

81 Proposition 121. Suppose given a preadditive category pA, ϕAq over R. Suppose given an additive category pB, ϕBq over R. Recall that pAdd A, ϕAψAq is an additive category over R; cf. Lemma 115. The following assertions (1, 2, 3) hold.

(1) We have IA P Ob R-linrA, Add As; cf. Lemma 117. 1 α 1 α 1 Suppose given F ÝÑ G in R-linrA, Bs. We have F ÝÑ G in R-linrAdd A, Bs with 1 1 1 F ˝ IA “ F , G ˝ IA “ G and α ˚ IA “ α; cf. Deﬁnition 110 and Lemmas 112 and 120. 1 1 1 Suppose given β P R-linrAdd A,BspF ,G q with β ˚ IA “ α. Then we have β “ α .

IA A / Add A

G α AI α1 F 1 +3 G1 F Ù * B

(2) Suppose given U, V P Ob R-linrAdd A, Bs with U ˝ IA “ V ˝ IA. Then U – V . (3) We have the equivalence of categories

ΦA,B R-linrA, Bs ÐÝÝÝ R-linrAdd A, Bs

β˚IA β pU ˝ IA ÝÝÝÑ V ˝ IAq ÞÑ pU ÝÑ V q ,

that is surjective on objects.

If unambiguous, we often write Φ:“ ΦA,B.

Proof. Ad (1). This follows from Theorem 113.(1). Ad (2). This follows from Theorem 113.(2).

1 Ad (3). Suppose given F P Ob R-linrA, Bs. Because (1) holds, we have F P R-linrAdd A, Bs 1 1 and ΦF “ F ˝ JA “ F . Therefore, Φ is surjective on objects.

Suppose given F,G P Ob R-linrA, Bs. Since (1) holds, we have the bijection

1 1 R-linrAdd A,BspF ,G q Ñ R-linrA,BspF,Gq, β ÞÑ Φβ.

Suppose given U, V P Ob R-linrKar A, Bs with ΦU “ ΦV . Since (2) holds, we have U – V . Therefore, Φ is an equivalence by Lemma 6.

82 Chapter 3

The tensor product

3.1 The tensor product of preadditive categories over a commutative ring

For this §3.1, let R be a commutative ring. We often write b instead of b; cf. §1.3. R

3.1.1 Deﬁnition

For this §3.1.1, let pA, ϕq and pB, ψq be preadditive categories over R.

Deﬁnition 122 (and Lemma). Let Ai P Ob A and Bi P Ob B for i P r1, 3s. There exists a unique R-bilinear map

κA1,A2,A3,B1,B2,B3 ApA1,A2q b BpB1,B2q ˆ ApA2,A3q b BpB2,B3q ÝÝÝÝÝÝÝÝÝÝÝÑ ApA1,A3qb BpB1,B3q R R R ˆ ˙ ˆ ˙ with 1 1 2 2 1 2 1 2 pa b b , a b b qκA1,A2,A3,B1,B2,B3 “ pa a q b pb b q

1 2 1 2 for a P ApA1,A2q, a P ApA2,A3q, b P BpB1,B2q, b P BpB2,B3q.

Proof. Deﬁne

λ ApA1,A2q ˆ BpB1,B2q ˆ ApA2,A3q ˆ BpB2,B3q ÝÑ ApA1,A3q b BpB1,B3q pa1, b1, a2, b2q ÞÑ pa1a2q b pb1b2q.

We show that λ is R-multilinear.

1 1 Pars pro toto, consider the ﬁrst component. Suppose given a1, a2 P ApA1,A2q. Suppose 1 2 2 given r1, r2 P R. Suppose given b P BpB1,B2q, a P ApA2,A3q and b P BpB2,B3q.

83 We have

1 1 1 2 2 1 1 2 1 2 pr1a1 ` r2a2, b , a , b qλ “ ppr1a1 ` r2a2qa q b pb b q 1 2 1 2 1 2 “ pr1a1a ` r2a2a q b pb b q 1 2 1 2 1 2 1 2 “ r1ppa1a q b pb b qq ` r2ppa2a q b pb b qq 1 2 1 2 1 2 1 2 “ r1ppa1, a , b , b qλq ` r2ppa2, a , b , b qλq.

Deﬁne µ1 :“ µApA1,A2q, BpB1,B2q, ApA2,A3q, BpB2,B3q; cf. Deﬁnition 13. By Lemma 16, we obtain a unique R-linear map

¯ λ : ApA1,A2q b BpB1,B2q b ApA2,A3q b BpB2,B3q Ñ ApA1,A3q b BpB1,B3q

¯ with µ1λ “ λ.

Deﬁne µ2 :“ µApA1,A2qb BpB1,B2q, ApA2,A3qb BpB2,B3q; cf. Deﬁnition 13. By Lemma 19, there exists a unique R-linear isomorphism

ξ p ApA1,A2q b BpB1,B2qqbp ApA2,A3q b BpB2,B3qq ÝÑ ApA1,A2qb BpB1,B2qb ApA2,A3qb BpB2,B3q

with ppa1 b b1q b pa2 b b2qqξ “ a1 b b1 b a2 b b2,

1 1 2 2 for a P ApA1,A2q, b P BpB1,B2q, a P ApA2,A3q, b P BpB2,B3q. ¯ Deﬁne κ :“ κA1,A2,A3,B1,B2,B3 :“ µ2ξλ. We have

1 1 2 2 1 1 2 2 ¯ pa b b , a b b qκ “ pa b b , a b b qµ2ξλ “ ppa1 b b1q b pa2 b b2qqξλ¯ “ pa1 b b1 b a2 b b2qλ¯ “ pa1a2q b pb1b2q,

1 1 2 2 for a P ApA1,A2q, b P BpB1,B2q, a P ApA2,A3q, b P BpB2,B3q. ¯ Since µ2 is R-bilinear and ξ, λ are R-linear, κ is R-bilinear; cf. Lemma 12. Suppose given an R-bilinear map

1 κ : p ApA1,A2q b BpB1,B2qq ˆ p ApA2,A3q b BpB2,B3qq Ñ ApA1,A3q b BpB1,B3q

with pa1 b b1, a2 b b2qκ1 “ pa1a2q b pb1b2q “ pa1 b b1, a2 b b2qκ

1 2 1 2 for a P ApA1,A2q, a P ApA2,A3q, b P BpB1,B2q, b P BpB2,B3q. Suppose given

1 1 2 2 a b b , a b b P p ApA1,A2q b BpB1,B2qq ˆ p ApA2,A3q b BpB2,B3qq . ¨ i i j j ˛ iPr1,ms jPr1,ns ÿ ÿ ˝ ‚ 84 We have

a1 b b1 , a2 b b2 κ1 “ pa1 b b1 , a2 b b2qκ1 ¨ i i j j ˛ i i j j iPr1,ms jPr1,ns pi,jqPr1,msˆr1,ns ÿ ÿ ÿ ˝ ‚ 1 1 2 2 “ pai b bi , aj b bj qκ pi,jqPr1,msˆr1,ns ÿ

“ a1 b b1 , a2 b b2 κ . ¨ i i j j ˛ iPr1,ms jPr1,ns ÿ ÿ ˝ ‚ Thus, we have κ1 “ κ.

Deﬁnition 123 (and Lemma). We shall deﬁne a category A b B as follows. R Let Ob A b B :“ Ob A ˆ Ob B. R We write A b B :“ pA, Bq for pA, Bq P Ob A b B. R

For A1 b B1 and A2 b B2 in Ob A b B, we deﬁne R

AbBpA1 b B1 ,A2 b B2q :“ ApA1 ,A2q b BpB1 ,B2q. R R Since this is an R-module, it is in particular an abelian group.

For A1 b B1 , A2 b B2 and A3 b B3 in Ob A b B, R

ζ P AbBpA1 b B1,A2 b B2q and η P AbBpA2 b B2,A3 b B3q, R R

we deﬁne composition by

ζη :“ pζ, ηqκA1,A2,A3,B1,B2,B3 P AbBpA1 b B1 ,A3 b B3q; R

cf. Deﬁnition 122. Recall that in particular

pa1 b b1qpa2 b b2q “ pa1a2q b pb1b2q

1 1 2 2 for a P ApA1 ,A2q, b P BpB1 ,B2q, a P ApA2 ,A3q, b P BpB2 ,B3q.

For A b B P Ob A b B, we deﬁne 1AbB :“ 1A b 1B . R We call A b B the tensor product over R of A and B. R If unambiguous, we often write A b B :“ A b B. R This deﬁnes a preadditive category A b B. R

Proof. First we show that A b B is a category.

85 Suppose given

1 1 2 2 3 3 ζ “ aibbi η “ aj bbj θ “ ak bbk iPr1,m1s jPr1,m2s kPr1,m3s A1 b B1 ÝÝÝÝÝÝÝÝÝÑř A2 b B2 ÝÝÝÝÝÝÝÝÝÝÑř A3 b B3 ÝÝÝÝÝÝÝÝÝÝÝÑř A4 b B4

in A b B.

For brevity, we write κi,j,k :“ κAi,Aj ,Ak,Bi,Bj ,Bk ; cf. Deﬁnition 122.

We have 1A2bB2 “ 1A2 b 1B2 P ApA2,A2q b BpB2,B2q “ AbBpA2 b B2,A2 b B2q. Furthermore, we have

ζη “ pζ, ηqκ1,2,3 P ApA1,A3q b BpB1,B3q “ AbBpA1 b B1,A3 b B3q .

We calculate

1 1 ζ ¨ 1A B “ a b b , 1A b 1B κ1,2,2 2b 2 ¨ i i 2 2 ˛ iPr1,m1s ÿ ˝ 1 1 ‚ “ pai b bi , 1A2 b 1B2 q κ1,2,2 iPr1,m1s ÿ 1 1 “ pai1A2 q b pbi1B2 q iPr1,m1s ÿ 1 1 “ ai b bi iPr1,m1s ÿ “ ζ and

2 2 1A B ¨ η “ 1A b 1B , a b b κ2,2,3 2b 2 ¨ 2 2 j j ˛ jPr1,m2s ÿ ˝ 2 2 ‚ “ 1A2 b 1B2 , aj b bj κ2,2,3 jPr1,m2s ÿ ` ˘ 2 2 “ p1A2 aj q b p1B2 bj q jPr1,m2s ÿ 2 2 “ aj b bj jPr1,m2s ÿ “ η .

86 Moreover, we have

pζηqθ “ pζ, ηqκ1,2,3 , θ κ1,3,4 ` ˘ 1 1 2 2 “ a b b , a b b κ1,2,3 , θ κ1,3,4 ¨¨ i i j j ˛ ˛ iPr1,m1s jPr1,m2s ÿ ÿ ˝˝ ‚ ‚ 1 1 2 2 “ pa b b , a b b qκ1,2,3 , θ κ1,3,4 ¨ i i j j ˛ pi,jqPr1,m1sˆr1,m2s ÿ ˝ ‚ 1 2 1 2 3 3 “ pa a q b pb b q , a b b κ1,3,4 ¨ i j i j k k ˛ pi,jqPr1,m1sˆr1,m2s kPr1,m3s ÿ ÿ ˝ 1 2 1 2 3 3 ‚ “ paiaj q b pbibj q , ak b bk κ1,3,4 pi,j,kqPr1,m1sˆr1,m2sˆr1,m3s ÿ ` ˘

1 2 3 1 2 3 “ ppaiaj qak q b ppbibj qbk q pi,j,kqPr1,m1sˆr1,m2sˆr1,m3s ÿ 1 2 3 1 2 3 “ paipaj ak qq b pbipbj bk qq pi,j,kqPr1,m1sˆr1,m2sˆr1,m3s ÿ 1 1 2 3 2 3 “ ai b bi , paj ak q b pbj bk q κ1,2,4 pi,j,kqPr1,m1sˆr1,m2sˆr1,m3s ÿ ` ˘

1 1 2 3 2 3 “ a b b , pa a q b pb b q κ1,2,4 ¨ i i j k j k ˛ iPr1,m1s pj,kqPr1,m2sˆr1,m3s ÿ ÿ ˝ ‚ 2 2 3 3 “ ζ, pa b b , a b b qκ2,3,4 κ1,2,4 ¨ j j k k ˛ pj,kqPr1,m2sˆr1,m3s ÿ ˝ ‚ 2 2 3 3 “ ζ, a b b , a b b κ2,3,4 κ1,2,4 ¨ ¨ j j k k ˛ ˛ jPr1,m2s kPr1,m3s ÿ ÿ ˝ ˝ ‚ ‚ “ pζ, pη, θqκ2,3,4qκ1,2,4 “ ζpηθq .

Now we show that A b B is preadditive. Suppose given

η1 ζ / θ A1 b B1 / A2 b B2 / A3 b B3 / A4 b B4 η2

in A b B.

87 We have

ζpη1 ` η2qθ “ ppζ, η1 ` η2qκ1,2,3 , θqκ1,3,4

“ ppζ, η1qκ1,2,3 ` pζ, η2qκ1,2,3, θqκ1,3,4

“ ppζ, η1qκ1,2,3 , θqκ1,3,4 ` ppζ, η2qκ1,2,3 , θqκ1,3,4

“ ζη1θ ` ζη2θ .

Lemma 124. We have the ring morphism

ϕbψ R ÝÝÝÑ End 1AbB R

r ÞÑ prpϕ b ψqAbBqAbBPOb AbB :“ pprϕqA b 1BqAbBPOb AbB R R

“ p1A b prψqBqAbBPOb AbB . R

Thus, pA b B, ϕ b ψq is a preadditive category over R. R

Proof. Suppose given r P R. Suppose given A b B P Ob A b B. We have

prϕqA b 1B “ pprϕAq1Aq b 1B

“ pr ¨ 1Aq b 1B

“ r ¨ p1A b 1Bq

“ 1A b pr ¨ 1Bq

“ 1A b pprψqB1Bq

“ 1A b prψqB

P ApA, Aq b BpB,Bq “ AbBpA b B,A b Bq.

We show that rpϕ b ψq is natural.

88 ζ“ aibbi iPr1,ms Suppose given A1 b B1 ÝÝÝÝÝÝÝÝÑř A2 b B2 in A b B. We have

prpϕ b ψqqA B ¨ ζ “ pprϕqA b 1B q ¨ ai b bi 1b 1 1 1 ¨ ˛ iPr1,ms ÿ ˝ ‚ “ pprϕqA1 b 1B1 qpai b biq iPr1,ms ÿ

“ pprϕqA1 aiq b p1B1 biq iPr1,ms ÿ

“ paiprϕqA2 q b pbi1B2 q iPr1,ms ÿ

“ pai b biqpprϕqA2 b 1B2 q iPr1,ms ÿ

“ ai b bi ¨ pprϕqA b 1B q ¨ ˛ 2 2 iPr1,ms ÿ ˝ ‚ “ ζ ¨ prpϕ b ψqqA2bB2 . Thus, rpϕ b ψq is natural. Therefore, ϕ b ψ is a welldeﬁned map. Now we show that ϕ b ψ is a ring morphism. We have

p1Rpϕ b ψqqAbB “ p1RϕqA b 1B

“ 1A b 1B

“ 1AbB ,

for A b B P Ob A b B. Thus, we have 1Rpϕ b ψq “ 1End 1AbB . We have

ppr ` sqpϕ b ψqqAbB “ ppr ` sqϕqA b 1B

“ pprϕqA ` psϕqAq b 1B

“ prϕqA b 1B ` psϕqA b 1B

“ prpϕ b ψqqAbB ` pspϕ b ψqqAbB , for A b B P Ob A b B. Thus, pr ` sqpϕ b ψq “ rpϕ b ψq ` spϕ b ψq. We have

pprsqpϕ b ψqqAbB “ pprsqϕqA b 1B

“ pprϕqApsϕqAq b p1B1Bq

“ prϕqA b 1B ¨ psϕqA b 1B

“ prpϕ b ψqqAbB ¨ pspϕ b ψqqAbB , for A b B P Ob A b B. Thus, prsqpϕ b ψq “ rpϕ b ψq ¨ spϕ b ψq. Therefore, ϕ b ψ is a ring morphism.

89 3.1.2 Universal property

For this §3.1.2, let pA, ϕq and pB, ψq be preadditive categories over R. Deﬁnition 125 (and Lemma). We have the R-bilinear functor

M A ˆ B ÝÝÝÑA,B A b B R

pa,bq abb ppA1,B1q ÝÝÑpA2,B2qq ÞÑ pA1 b B1 ÝÝÑ A2 b B2q.

If unambiguous, we often write M :“ MA,B .

pa1,b1q pa2,b2q Proof. Suppose given pA1,B1q ÝÝÝÑpA2,B2q ÝÝÝÝÑpA3,B3q in A ˆ B. We have

Mppa1, b1qpa2, b2qq “ Mpa1a2, b1b2q “ pa1a2q b pb1b2q “ pa1 b b1qpa2 b b2q “ Mpa1, b1q ¨ Mpa2, b2q .

Furthermore, we have

M1pA,Bq “ Mp1A, 1Bq “ 1A b 1B “ 1AbB “ 1MpA,Bq .

Thus, M is a functor.

a1 / b 1 2 Suppose given A1 / A2 in A. Suppose given B1 ÝÑ B2 in B. Suppose given r , r P R. a2 We have

Mpr1a1 ` r2a2, bq “ pr1a1 ` r2a2q b b “ r1pa1 b bq ` r2pa2 b bq “ r1Mpa1, bq ` r2Mpa2, bq .

b1 a / 1 2 Suppose given A1 ÝÑ A2 in A. Suppose given B1 / B2 in A. Suppose given r , r P R. b2

Mpa, r1b1 ` r2b2q “ a b pr1b1 ` r2b2q “ r1pa b b1q ` r2pa b b2q “ r1Mpa, b1q ` r2Mpa, b2q .

Thus, M is R-bilinear; cf. Deﬁnition 27.

Remark 126. The functor MA,B : A ˆ B Ñ A b B is bijective on objects. In particular, R MA,B is dense. Deﬁnition 127 (and Lemma). Let pC, ωq be an R-linear preadditive category. Let F : A ˆ B Ñ C be an R-bilinear functor. The following assertions (1, 2, 3) hold.

90 (1) We have the R-linear functor

A b B ÝÑF C R

aibbi F pai ,biq iPr1,ms iPr1,ms A1 b B1 ÝÝÝÝÝÝÝÑř A2 b B2 ÞÑ F pA1,B1q ÝÝÝÝÝÝÝÝÑř F pA2,B2q . ˜ ¸ ˜ ¸

(2) We have F ˝ MA,B “ F .

F˜ ˜ ˜ (3) Suppose given an R-linear functor A b B ÝÑ C with F ˝ MA,B “ F . Then F “ F . R

Proof. Ad (1). Suppose given A1 b B1 and A2 b B2 in Ob A b B. Since F is an R-bilinear functor, we have the R-bilinear map

F pA1,B1q,pA2,B2q AˆBppA1,B1q, pA2,B2qq “ ApA1,A2q ˆ BpB1,B2q ÝÝÝÝÝÝÝÝÝÑ CpF pA1,B1q,F pA2,B2qq pa, bq ÞÑ F pa, bq.

By Lemma 16, we obtain a unique R-linear map

FpA1,B1q,pA2,B2q : ApA1,A2q b BpB1,B2q Ñ CpF pA1,B1q,F pA2,B2qq

with µ ApA1,A2q, BpB1,B2qFpA1,B1q,pA2,B2q “ FpA1,B1q,pA2,B2q, i.e.

pa b bqFpA1,B1q,pA2,B2q “ pa, bqFpA1,B1q,pA2,B2q “ F pa, bq

for a P ApA1,A2q and b P BpB1,B2q.

For ζ “ ai b bi P AbBpA1 b B1 ,A2 b B2q we have iPr1,ms ř

ζFpA1,B1q,pA2,B2q “ F pai , biq . iPr1,ms ÿ

Thus, F is welldeﬁned on morphisms. Now we show that F is a functor. Suppose given

1 1 2 2 ζ “ aibbi η “ aj bbj iPr1,m1s jPr1,m2s A1 b B1 ÝÝÝÝÝÝÝÝÝÑř A2 b B2 ÝÝÝÝÝÝÝÝÝÝÑř A3 b B3

91 in A b B. We calculate

F pζηq “ F pa1 a2q b pb1 b2q ¨ i j i j ˛ pi,jqPr1,m1sˆr1,m2s ÿ ˝ 1 2 1 2 ‚ “ F paiaj , bibj q pi,jqPr1,m1sˆr1,m2s ÿ 1 1 2 2 “ F pai , biq ¨ F paj , bj q pi,jqPr1,m1sˆr1,m2s ÿ

“ F pa1 , b1 q ¨ F pa2 , b2q ¨ i i ˛ ¨ j j ˛ iPr1,m1s jPr1,m2s ÿ ÿ “ ˝F ζ ¨ F η . ‚ ˝ ‚

Furthermore, we have

F 1 F 1 1 F 1 , 1 F 1 1 1 . A1bB1 “ p A1 b B1 q “ p A1 B1 q “ pA1 ,B1q “ F pA1 ,B1q “ F pA1bB1q

Thus, F is a functor. Now we show that F is R-linear.

ζ1 / Suppose given A1 b B1 / A2 b B2 in A b B. Suppose given r1, r2 P R. We have ζ2

F pr1ζ1 ` r2ζ2q “ pr1ζ1 ` r2ζ2qFpA1,B1q,pA2,B2q

“ r1 pζ1qFpA1,B1q,pA2,B2q ` r2 pζ2qFpA1,B1q,pA2,B2q “ r1 `¨ F ζ1 ` r2 ¨ F ζ2 . ˘ ` ˘

Thus, F is R-linear by Remark 25.

pa,bq Ad (2). Suppose given pA1 ,B1q ÝÝÑpA2 ,B2q in A ˆ B. We have

pa,bq abb pF ˝ Mq pA1 ,B1q ÝÝÑpA2 ,B2q “ F A1 b B1 ÝÝÑ A2 b B2

´ ¯ ´ F pa,bq ¯ “ F pA1 ,B1q ÝÝÝÑ F pA2 B2q

´ pa,bq ¯ “ F pA1 ,B1q ÝÝÑpA2 ,B2q . ´ ¯ Ad (3). Suppose given A b B P Ob A b B. We have

F˜pA b Bq “ pF˜ ˝ MqpA, Bq “ F pA, Bq “ F pA b Bq .

92 ζ“ aibbi iPr1,ms Suppose given A1 b B1 ÝÝÝÝÝÝÝÝÑř A2 b B2 in A b B. We obtain ˜ ˜ F ζ “ F pai b biq iPr1,ms ÿ “ F pai, biq iPr1,ms ÿ “ F pai b biq iPr1,ms ÿ “ F ζ .

Thus, we have F˜ “ F . Lemma 128. Suppose given an R-linear preadditive category pC, ωq. Suppose given an R-bilinear functor F : A ˆ B Ñ C. The following assertions (1, 2) hold.

(1) Suppose F to be full. Then F is full.

(2) Suppose F to be dense. Then F is dense.

Proof. Ad (1). Suppose given A1 b B1 and A2 b B2 in Ob A b B.

D127.p2q Suppose given ρ P CpF pA1 b B1q, F pA2 b B2qq “ CpF pA1 ,B1q,F pA2 ,B2qq.

Since F is full, there exists pa, bq P AˆBppA1 ,B1q, pA2 ,B2qq with F pa, bq “ ρ. D127.p2q We have a b b P AbBpA1 b B1 ,A2 b B2q and F pa b bq “ F pa, bq “ ρ. Thus, F is full. Ad (2). Suppose given C P Ob C. Since F is dense, there exists pA, Bq P Ob A ˆ B with D{127.p2q F pA, Bq – C. We have F pA b Bq “ F pA, Bq – C. Thus, F is dense. Deﬁnition 129 (and Lemma). Let pC, ωq be an R-linear preadditive category. α Suppose given F ÝÑ G in R-bilrA ˆ B, Cs. Deﬁne α :“ pαpA,BqqAbBPOb AbB . R The following assertions (1, 2, 3) hold.

α (1) We have F ÝÑ G in R-linrA b B, Cs. R

(2) We have α ˚ MA,B “ α.

α˜ (3) Suppose given F ÝÑ G in R-linrA b B, Cs with α˜ ˚ MA,B “ α. Then α˜ “ α. R

Proof. Ad (1). We have

α α F pA, Bq ÝÝÝÝÑpA,Bq GpA, Bq “ F pA b Bq ÝÝÝÑAbB GpA b Bq ´ ¯ ´ ¯ for A b B P Ob A b B, It remains to show that α is natural.

93 ζ“ aibbi iPr1,ms Suppose given A1 b B1 ÝÝÝÝÝÝÝÝÑř A2 b B2 in A b B. We have

αA1bB1 ¨ Gζ “ αpA1 ,B1q ¨ Gpai b biq iPr1,ms ÿ

“ αpA1 ,B1qGpai , biq iPr1,ms ÿ

“ F pai , biqαpA2 ,B2q iPr1,ms ÿ

“ F pai b biq ¨ α ¨ ˛ pA2 ,B2q iPr1,ms ÿ ˝ ‚ “ F ζ ¨ αA2bB2 .

Ad (2). Suppose given pA, Bq P Ob A ˆ B. We have

pα ˚ MqpA,Bq “ αMpA,Bq “ αAbB “ αpA,Bq .

Ad (3). Suppose given A b B P Ob A b B. We have

α˜AbB “ α˜MpA,Bq “ pα˜ ˚ MqpA,Bq “ αpA,Bq “ αAbB .

Theorem 130. Recall that pA, ϕq and pB, ψq are preadditive categories over R. Recall that pA b B, ϕ b ψq is a preadditive category over R; cf. Deﬁnition 123 and Lemma 124. R Suppose given an R-linear preadditive category pC, ωq. The following assertions (1, 2) hold.

(1) We have MA,B P Ob R-bilrA ˆ B, A b Bs; cf. Deﬁnition 125. R α Suppose given F ÝÑ G in R-bilrA ˆ B, Cs.

There exist unique R-linear functors F, G : A b B Ñ C with F ˝ MA,B “ F and R G ˝ MA,B “ G; cf. Deﬁnition 127. α There exists a unique transformation F ùñ G with α ˚ MA,B “ α; cf. Deﬁnition 129.

MA,B A ˆ B / A b B R G α BJ α F +3 G F Ù * C

94 (2) We have the isomorphism of categories

R ΨA,B,C R-bilrA ˆ B, Cs ÐÝÝÝÝ R-linrA b B, Cs R

β˚MA,B β U ˝ MA,B ÝÝÝÝÝÑ V ˝ MA,B ÞÑ U ÝÑ V ´ ¯ ´ ¯ with inverse R-bilrA ˆ B, Cs ÝÑ R-linrA b B, Cs R

pF ÝÑα Gq ÞÑ pF ÝÑα Gq

R If unambiguous, we often write Ψ:“ ΨA,B,C :“ ΨA,B,C .

Proof. Ad (1). This follows from Deﬁnition 127 and Deﬁnition 129. Ad (2). This follows from (1).

3.1.3 Functoriality

In this §3.1.3, we deﬁne the tensor product for R-linear functors between preadditive categories over R and for transformations between them. Furthermore, we establish functoriality properties of the tensor product, which could be expressed by saying that it is turned into a 2-bifunctor.

For this §3.1.3, let pA, ϕAq, pB, ϕBq, pC, ϕCq, pD, ϕDq, pE, ϕE q and pF, ϕF q be preadditive categories over R.

Deﬁnition 131 (and Lemma). Let F : A Ñ C and G : B Ñ D be R-linear functors. We have the R-linear functor

A b B ÝÝÝÑF bG C b D R R

iPr1,ns aibbi iPr1,nspF aiqbpGbiq A1 b B1 ÝÝÝÝÝÝÝÝÑ A2 b B2 ÞÑ FA1 b GB1 ÝÝÝÝÝÝÝÝÝÝÝÝÑ FA2 b GB2 . ř ř ˆ ˙ ˆ ˙ So, we have pF b GqpA b Bq “ FA b FG for A b B P ObpA b Bq. R Furthermore, we have

pF b Gq ai b bi “ pF aiq b pGbiq ¨ ˛ iPr1,ns iPr1,ns ÿ ÿ ˝ ‚ for iPr1,ns ai b bi P MorpA b Bq. R ř 95 Proof. Suppose given A1 b B1 and A2 b B2 in Ob A b B. Since F and G are R-linear functors, we have

pFA1,A2 ,GB1,B2 q p ApA1 ,A2q , BpB1 ,B2qq ÝÝÝÝÝÝÝÝÝÝÑ CpFA1 ,FA2q ˆ DpGB1 GB2q

in pR -Modqˆ2. By Deﬁnitions 20 and 123, we have

FA1,A2 b GB1,B2 AbBpA1 b B1 ,A2 b B2q ÝÝÝÝÝÝÝÝÝÝÑ CbDpFA1 b GB1 ,FA2 b GB2q

ai b bi ÞÑ pF aiq b pGbiq iPr1,ns iPr1,ns ÿ ÿ in R -Mod. Thus, F b G is welldeﬁned on morphisms. Now we show that F b G is a functor. 1 1 2 2 ζ“ iPr1,ms aibbi η“ jPr1,ks aj bbj Suppose given A1 b B1 ÝÝÝÝÝÝÝÝÝÝÑ A2 b B2 ÝÝÝÝÝÝÝÝÝÝÑ A3 b B3 in A b B. We have ř ř

pF b Gqpζηq “ pF b Gq pa1 a2q b pb1 b2q ¨ i j i j ˛ pi,jqPr1,msˆr1,ks ÿ ˝ 1 2 1 2 ‚ “ F paiaj q b Gpbibj q pi,jqPr1,msˆr1,ks ÿ 1 2 1 2 “ pF aiqpF aj q b pGbiqpGbj q pi,jqPr1,msˆr1,ks ÿ 1 1 2 2 “ pF aiq b pGbiq ¨ pF ai q b pGbj q iPr1,ms jPr1,ks ÿ ÿ “ pF b Gqζ ¨ pF b Gqη.

Suppose given A b B P Ob A b B. We have

pF b Gq1AbB “ pF b Gqp1A b 1Bq

“ F 1A b G1B

“ 1FA b 1GB

“ 1FAbGB

“ 1pF bGqpAbBq .

Thus, F b G is indeed a functor.

Since pF b GqA1bB1,A2bB2 “ FA1,A2 b GB1,B2 is R-linear for A1 b B1 and A2 b B2 in Ob A b B, we conclude that F b G is R-linear.

96 Lemma 132. Suppose given R-linear functors A ÝÑF C ÝÑH E and B ÝÑG D ÝÑI F. The following assertions (1, 2) hold.

(1) We have 1A b 1B “ 1AbB . (2) We have pH b Iq ˝ pF b Gq “ pH ˝ F q b pI ˝ Gq.

Proof. Ad (1). Suppose given A b B P ObpA b Bq. We have

p1A b 1BqpA b Bq “ 1AA b 1BB “ A b B “ 1AbBpA b Bq.

Suppose given A1 b B1 and A2 b B2 in ObpA b Bq. We have

p1A b 1BqA1bB1 ,A2bB2 “ p1AqA1 ,A2 b p1BqB1 ,B2

“ 1 ApA1 ,A2q b 1 BpB1 ,B2q D21 “ 1 ApA1,A2qb BpB1,B2q

“ 1 AbBpA1bB1 ,A2bB2q

“ p1AbBqA1bB1 ,A2bB2 .

Ad (2). Suppose given A b B P ObpA b Bq. We have

ppH b Iq ˝ pF b GqqpA b Bq “ pH b IqpFA b GBq “ ppH ˝ F qAq b ppI ˝ GqBq “ ppH ˝ F q b pI ˝ GqqpA b Bq.

Suppose given A1 b B1 and A2 b B2 in ObpA b Bq. We have

ppH b Iq ˝ pF b GqqA1bB1 ,A2bB2

“ pF b GqA1bB1 ,A2bB2 ¨ pH b IqpFA1qbpGB1q , pFA2qbpGB2q

“ pFA1 ,A2 b GB1 ,B2 qpHFA1 ,FA2 b IGB1 , GB2 q

D21 “ pFA1 ,A2 ¨ HFA1 ,FA2 q b pGB1 ,B2 ¨ IGB1 , GB2 q

“ pH ˝ F qA1 ,A2 b pI ˝ GqB1 ,B2

“ ppH ˝ F q b pI ˝ GqqA1bB1 ,A2bB2 .

Deﬁnition 133 (and Lemma). Let F,F 1 : A Ñ C and G, G1 : B Ñ D be R-linear α β functors. Let F ùñ F 1 and G ùñ G1 be transformations. We have the transformation αbβ F b G ùùñ F 1 b G1 with 1 1 pα b βqAbB “ αA b βB P CbDppF b GqpA b Bq, pF b G qpA b Bqq R for A b B P Ob A b B. R

97 Proof. Suppose given A b B P ObpA b Bq. We have

α β FA ÝÝÑA F 1A and GB ÝÑB G1B.

Since we have pF bGqpAbBq “ FAbGB and pF 1 bG1qpAbBq “ F 1AbG1B, we obtain

α bβ pF b GqpA b Bq ÝÝÝÝÑpA B F 1 b G1qpA b Bq.

We have to show that α b β is natural.

ζ“ iPr1,ns aibbi Suppose given A1 b B1 ÝÝÝÝÝÝÝÝÝÝÑ A2 b B2 in A b B. We have ř

pF b Gqζ ¨ pα b βqA B “ pF aiq b pGbiq pαA b βB q 2b 2 ¨ ˛ 2 2 iPr1,ns ÿ ˝ ‚ “ pF ai ¨ αA2 q b pGbi ¨ βB2 q iPr1,ns ÿ 1 1 “ pαA1 ¨ F aiq b pβB1 ¨ G biq iPr1,ns ÿ

1 1 “ pαA b βB q pF aiq b pG biq 1 1 ¨ ˛ iPr1,ns ÿ ˝ 1 1 ‚ “ pα b βqA1bB1 ¨ pF b G qζ.

Lemma 134. Let F,F 1,F 2 : A Ñ C and G, G1,G2 : B Ñ D be R-linear functors. Suppose α α1 β β1 given transformations F ùñ F 1 ùñ F 2 and G ùñ G1 ùñ G2. The following assertions (1, 2) hold.

(1) We have 1F b 1G “ 1F bG . (2) We have pα b βqpα1 b β1q “ pαα1q b pββ1q.

Proof. Ad (1). Suppose given A b B P ObpA b Bq. We have

p1F b 1GqAbB “ p1F qA b p1GqB “ 1FA b 1GB “ 1FAbGB “ 1pF bGqpAbBq “ p1F bGqAbB .

Ad (2). Suppose given A b B P ObpA b Bq. We have

1 1 1 1 ppα b βqpα b β qqAbB “ pα b βqAbB ¨ pα b β qAbB 1 1 “ pαA b βBqpαA b βBq 1 1 “ pαAαAq b pβBβBq 1 1 “ pαα qA b pββ qB 1 1 “ ppαα q b pββ qqAbB .

98 Lemma 135. Suppose given R-linear functors F,F 1 : A Ñ C and G, G1 : B Ñ D. Suppose α 1 β 1 αbβ 1 1 given isotransformations F „ +3 F and G „ +3 G . Then pF b Gq „ +3 pF b G q is an ´1 ´1 1 1 α bβ isotransformation with inverse pF b G q „ +3 pF b Gq .

Proof. We have

´1 ´1 L134.p2q ´1 ´1 L134.p1q pα b βqpα b β q “ pαα q b pββ q “ 1F b 1G “ 1F bG

and ´1 ´1 L134.p2q ´1 ´1 L134.p1q pα b β qpα b βq “ pα αq b pβ βq “ 1F 1 b 1G1 “ 1F 1bG1 .

Lemma 136. Suppose given R-linear equivalences F : A Ñ C and G : B Ñ D. Then F b G : A b B Ñ C b D is an R-linear equivalence. R R

Proof. By Deﬁnition 131, it suﬃces to show that F b G is an equivalence. Since F : A Ñ C is an equivalence, there exists F 1 : C Ñ A and isotransformations

1 α 1 γ pF ˝ F q „ +3 1A and pF ˝ F q „ +3 1C .

Since G : B Ñ D is an equivalence, there exists G1 : D Ñ B and isotransformations

1 β 1 δ pG ˝ Gq „ +3 1B and pG ˝ G q „ +3 1D .

By Corollary 31, F 1 and G1 are R-linear. By Lemma 135, we have isotransformations

1 1 L132.p2q 1 1 αbβ L132.p1q pF b G q ˝ pF b Gq “ pF ˝ F q b pG ˝ Gq „ +3 1A b 1B “ 1AbB and

1 1 L132.p2q 1 1 γbδ L132.p1q pF b Gq ˝ pF b G q “ pF ˝ F q b pG ˝ G q „ +3 1C b 1D “ 1CbD .

Thus, F b G is an equivalence.

Lemma 137. Suppose given

F H G I ) ) ) ) A α C γ E and B β D δ F 5 5 5 5 F 1 H1 G1 I1

with R-linear functors F,F 1, G, G1,H,H1,I and I1. We have pγ ˚ αq b pδ ˚ βq “ pγ b δq ˚ pα b βq.

99 Proof. Suppose given A b B P Ob A b B. R We have

D133 ppγ ˚ αq b pδ ˚ βqqAbB “ pγ ˚ αqA b pδ ˚ βqB 1 1 “ pγFA ¨ H αAq b pδGB ¨ I βBq

1 1 “ pγFA b δGBq ¨ pH αA b I βBq

D131 1 1 “ pγ b δqFAbGB ¨ pH b I qpαA b βBq D133

D131 1 1 “ pγ b δqpF bGqpAbBq ¨ pH b I qpα b βqAbB D133

“ ppγ b δq ˚ pα b βqqAbB .

100 3.2 The tensor product of additive categories over a commutative ring

For this § 3.2, let R be a commutative ring. Furthermore, let pA, ϕAq and pB, ϕBq be additive categories over R.

3.2.1 Deﬁnition

Deﬁnition 138 (and Lemma). We call

add A b B :“ Add A b B R R ˆ ˙ the additive tensor product over R of A and B. add add If unambiguous, we often write A b B :“ A b B. R

add This deﬁnes an additive category A b B, pϕA b ϕBqψAbB over R; cf. Lemmas 124 R R and 115. ˆ ˙

Proof. By Lemma 124, pA b B, ϕA b ϕBq is a preadditive category over R. By Propo- sition 92, AddpA b Bq is additive. By Lemma 115, pAddpA b Bq, pϕA b ϕBqψAbBq is a preadditive category over R.

3.2.2 Universal property

Deﬁnition 139 (and Lemma). Deﬁne

add MA,B :“ IAbB ˝ MA,B ; R

cf. Deﬁnitions 96 and 125. add add If unambiguous, we often write M :“ MA,B. add add We have the R-bilinear functor MA,B : A ˆ B Ñ A b B. R

Proof. By Deﬁnition 125, M “ MA,B is R-bilinear. By Lemma 117, I “ IAbB is R-linear. Thus, M add “ I ˝ M is R-bilinear; cf. Remark 29.

Deﬁnition 140 (and Lemma). We have the functor

R ΓA,B,C add R-bilrA ˆ B, Cs ÝÝÝÝÑ R-linrA b B, Cs R

1 1 1 pF ÝÑα Gq ÞÑ F ÝÑα G ; ´ ¯ 101 cf. Deﬁnitions 110, 127 and 129. R If unambiguous, we often write Γ :“ ΓA,B,C :“ ΓA,B,C. α Suppose given F ÝÑ G in R-bilrA ˆ B, Cs. Suppose given

fk,i,j bgk,i,j » ﬁ kPr1,mi,j s ř i,j A1,i b B1,i ÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÑ– ﬂ A2,j b B2,j iPr1,ms jPr1,ns ð ð add in A b B. We have R

fk,i,j bgk,i,j » ﬁ kPr1,mi,j s 1 ř i,j F ¨ A1,i b B1,i ÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÑ– ﬂ A2,j b B2,j˛ ˚iPr1,ms jPr1,ns ‹ ˚ ð ð ‹ ˚ ‹ ˝ ‚ F pfk,i,j ,gk,i,j q ¨ ˛ kPr1,mi,j s ř i,j “ ¨ F pA1,i ,B1,iq ÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÑ˝ ‚ F pA2,j ,B2,jq˛. ˚iPr1,ms jPr1,ns ‹ ˚ à à ‹ ˚ ‹ ˝ ‚ add Suppose given Ai b Bi P ObpA b Bq. We have iPr1,ms R Ð 1 α pA1bB1q‘¨¨¨‘pAmbBmq “ diagpαpAi ,BiqqiPr1,ms .

Proof. By Definition 127.(1), F is R-linear. By Lemma 120, F 1 is R-linear. Therefore, we obtain a welldefined map on objects. ´1 Consider the functor ΨA,B,C ; cf. Theorem 130. Consider the restriction of the functor from Definition 110; cf. Lemma 120. Then Γ is the composition of these functors. We have

fk,i,j bgk,i,j » ﬁ kPr1,mi,j s 1 ř i,j F ¨ A1,i b B1,i ÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÑ– ﬂ A2,j b B2,j˛ iPr1,ms jPr1,ns ˚ ‹ ˚ Ð Ð ‹ ˚ ‹ ˝ ‚ F fk,i,j bgk,i,j ¨ ¨ ˛˛ kPr1,mi,j s D110 ř i,j “ ¨ F pA1,i b B1,iq ÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÑ˝ ˝ ‚‚ F pA2,j b B2,jq˛ iPr1,ms jPr1,ns ˚ ‹ ˚ À À ‹ ˚ ‹ ˝ ‚ F pfk,i,j ,gk,i,j q ¨ ˛ D127. 1 kPr1,mi,j s p q ř i,j “ ¨ F pA1,i ,B1,iq ÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÑ˝ ‚ F pA2,j ,B2,jq˛. iPr1,ms jPr1,ns ˚ ‹ ˚ À À ‹ ˚ ‹ ˝ ‚ 102 Furthermore, we have

1 D110 α pA1bB1q‘¨¨¨‘pAmbBmq “ diagpαAibBi qi D129.p1q “ diagpαpAi ,Biqqi .

Lemma 141. Suppose given F P Ob R-bilrA ˆ B, Cs. The following assertions (1, 2) hold.

(1) If F is full, so is F 1.

(2) If F is dense, so is F 1.

Proof. Ad (1). This follows from Lemma 128.(1) and Lemma 111.(1). Ad (2). This follows from Lemma 128.(2) and Lemma 111.(3).

α Lemma 142. Suppose given F ÝÑ G in R-bilrA ˆ B, Cs. The following assertions (1, 2) hold.

1 add (1) We have F ˝ MA,B “ F .

1 add (2) We have α ˚ MA,B “ α.

add Proof. Recall that M “ IAbB ˝ MA,B ; cf. Deﬁnition 139. Ad (1). We have

1 add 1 L112.p1q D127.p2q F ˝ M “ pF ˝ IAbBq ˝ MA,B “ F ˝ MA,B “ F.

Ad (2). We have

1 add 1 L112.p2q D129.p2q α ˚ M “ pα ˚ IAbBq ˚ MA,B “ α ˚ MA,B “ α.

Theorem 143. Recall that pA, ϕAq and pB, ϕBq are additive categories over R. Recall add that pA b B, pϕA b ϕBqψAbBq is an additive category over R; cf. Deﬁnition 138. R R

Suppose given an additive category pC, ϕCq over R. The following assertions (1, 2, 3) hold.

add (1) We have MA,B P Ob R-bilrA ˆ B, Cs; cf. Deﬁnition 139.

α 1 α1 1 add Suppose given F ÝÑ G in R-bilrA ˆ B, Cs. We have F ÝÑ G in R-linrA b B, Cs with R 1 add 1 add 1 add F ˝MA,B “ F , G ˝MA,B “ G and α ˚MA,B “ α; cf. Deﬁnition 140 and Lemma 142.

103 1 1 add Suppose given β P add pF , G q with β ˚ MA,B “ α. Then we have β “ α. rA b B,Cs R

add MA,B add A ˆ B / A b B R

G α AI α1 F 1 +3 G1 F Ù ) C

add add add (2) Suppose given U, V P Ob R-linrA b B, Cs with U ˝ MA,B “ V ˝ MA,B. Then U – V . R

(3) We have the equivalence of categories

R ΩA,B,C add R-bilrA ˆ B, Cs ÐÝÝÝÝ R-linrA b B, Cs R

β˚M add add A,B add β ppU ˝ MA,Bq ÝÝÝÝÝÑpV ˝ MA,Bqq ÞÑ pU ÝÑ V q ,

that is surjective on objects.

R If unambiguous, we often write Ω:“ ΩA,B,C :“ ΩA,B,C.

Proof. Ad (1). This follows from (3). Ad (2). This follows from (3). Ad (3). We have equivalences of categories

add ΦAbB,C ΨA,B,C R-linrA b B, Cs ÝÝÝÝÑ R-linrA b B, Cs ÝÝÝÝÑ R-bilrA ˆ B, Cs; cf. Proposition 121.(3) and Theorem 130.(2).

β add Suppose given U ÝÑ V in R-linrA b B, Cs. To visualize the situation, we give the following diagram.

add MA,B

MA,B IAbB ) add A ˆ B / A b B / A b B

U˝IAbB U U˝pIAbB˝MA,Bq “U˝M add A,B &. C

104 We have

β β˚IAbB pΨA,B,C ˝ ΦAbB,Cq U ÝÑ V “ ΨA,B,C pU ˝ IAbBq ÝÝÝÝÑpV ˝ IAbBq

´ ¯ ´ pβ˚IAbBq˚MA,B ¯ “ ppU ˝ IAbBq ˝ MA,Bq ÝÝÝÝÝÝÝÝÝÝÑppV ˝ IAbBq ˝ MA,Bq ˆ ˙ β˚pIAbB˝MA,Bq “ pU ˝ pIAbB ˝ MA,Bqq ÝÝÝÝÝÝÝÝÝÝÑpV ˝ pIAbB ˝ MA,Bqq ˆ ˙ β M add add ˚ A,B add “ pU ˝ MA,Bq ÝÝÝÝÝÑpV ˝ MA,Bq ˆ ˙ β “ ΩA,B,C U ÝÑ V . ´ ¯ Thus, we have the equivalence of categories Ω “ ΩA,B,C “ ΨA,B,C ˝ ΦAbB,C .

Now Φ “ ΦAbB,C is surjective on objects; cf. Proposition 121.(3). Moreover, Ψ “ ΨA,B,C is bijective on objects; cf. Theorem 130.(2). Thus, Ω “ Ψ ˝ Φ is surjective on objects.

105 106 Chapter 4

Counterexamples for compatibility relations

4.1 Additive envelope and Karoubi envelope

Lemma 144. There exists an idempotent complete preadditive category A such that Add A is not idempotent complete.

More precisely, consider the subring Λ:“ tpa, bq P Z ˆ Z : a ”5 bu of Z ˆ Z and the full preadditive subcategory A of Λ -free with Ob A :“ tΛ, 0u. Then A is preadditive and idempotent complete, but Add A is not idempotent complete.

Proof. For x P Λ we also write x for the Λ-linear map Λ Ñ Λ, y ÞÑ yx.

Suppose given an idempotent pa, bq P ApΛ, Λq. We have

pa2, b2q “ pa, bq2 “ pa, bq.

Since a, b P Z, we have a, b P t0, 1u. Thus, we have pa, bq P tp1, 1q, p0, 0qu because a ”5 b. Therefore, we have p1,1q p0,0q 1 Idem A “ tΛ ÝÝÑ Λ, Λ ÝÝÑ Λ, 0 ÝÑ0 0u. Furthermore, p1,1q pΛ, p1, 1q, p1, 1qq is an image of Λ ÝÝÑ Λ, p0,0q p0, 0Λ,0, 00,Λq is an image of Λ ÝÝÑ Λ, 10 p0, 10, 10q is an image of 0 ÝÑ 0. Thus, A is idempotent complete. We want to show that Add A is not idempotent complete. By Remarks 48 and 95, it suﬃces to show that Λ -free is not idempotent complete.

p1,26q p0,10q e:“ p0,´65q p0,´25q Consider Λ2 ÝÝÝÝÝÝÝÝÝÝÝÝÝÑˆ ˙ Λ2 in Λ -free. We have

2 p1,26q p0,10q p1,26q p0,10q p1,26q p0,10q e “ p0,´65q p0,´25q p0,´65q p0,´25q “ p0,´65q p0,´25q “ e. ´ ¯ ´ ¯ ´ ¯ 107 Thus, e is an idempotent. Assume pX, π, ιq to be an image of e in Λ -free. Consider Λ -free ÝÑG Z -free

pa ,b q pa q Λm ÝÝÝÝÝÝÑij ij i,j Λn ÞÑ Zm ÝÝÝÝÑij i,j Zn . ˆ ˙ ˆ ˙ We show that G is a functor. ppa ,b qq ppc ,d qq Suppose given Λm ÝÝÝÝÝÝÝÑij ij i,j Λn ÝÝÝÝÝÝÝÑjk jk j,k Λp in Λ -free. We have

Gp1Λm q “ Gpppδi,j, δi,jqqi,jq “ pδi,jqi,j “ 1Zm “ 1GpΛmq . Furthermore, we have

G pppaij, bijqqi,j ¨ ppcjk, djkqqj,kq “ G paij, bijqpcjk, djkq ¨ ˛ jPr1,ns ÿ i,k ˝ ‚

“ G aijcjk , bijdjk ¨¨ ˛˛ jPr1,ns jPr1,ns ÿ ÿ i,k ˝˝ ‚‚

“ aijcjk ¨ ˛ jPr1,ns ÿ i,k ˝ ‚ “ paijqi,j ¨ pcjkqjk

“ Gpppaij, bijqqi,jq ¨ Gpppcjk, djkqqj,kq. Thus, G is a functor. Furthermore, G is bijective on objects. By Remark 47, pGX, Gπ, Gιq is an image of Ge in Z -free. We have

p1,26q p0,10q 1 0 Ge “ G p0,´65q p0,´25q “ 0 0 . ´ ¯ Therefore, we have GX “ Z. Thus, we have X “ Λ. ` ˘ In consequence, we have

pu1,v1q π and ι px1,y1q px2,y2q “ pu2,v2q “ p q ´ ¯ for some pu1, v1q, pu2, v2q, px1, y1q, px2, y2q P Λ. Since pΛ, π, ιq is an image of e, the following diagram commutes.

p1,26q p0,10q p0,´65q p0,´25q Λ2 ˆ ˙ / Λ2 ? pu1,v1q ppx1,y1q px2,y2qq pu2,v2q ˆ ˙ pu1,v1q pu2,v2q ˆ ˙ p1,1q Λ / Λ

108 Therefore, we have pu1x1,v1y1q pu1x2,v1y2q p1,26q p0,10q pu2x1,v2y1q pu2x2,v2y2q “ p0,´65q p0,´25q and ´ ¯ ´ ¯ px1u1 ` x2u2, y1v1 ` y2v2q “ p1, 1q. Thus, u1x1 “ 1. Since we may multiply π and ι by p´1q, we may assume that u1 “ x1 “ 1. 1 1 1 1 We have y ”5 x “ 1 and v ”5 u “ 1. Since we have v1y1 “ 26, we obtain pv1, y1q “ p1, 26q or pv1, y1q “ p26, 1q. Since we have v1y2 “ 10, we obtain v1 “ 1 and therefore y1 “ 26. But we have v2y1 “ ´65, in contradiction to y1 “ 26. Thus, e has no image in Λ -free. Therefore, Λ -free is not idempotent complete.

Proposition 145. There exists a preadditive category A such that

AddpKar Aq ﬁ KarpAdd Aq.

More precisely, consider the subring Λ:“ tpa, bq P Z ˆ Z : a ”5 bu of Z ˆ Z and the full preadditive subcategory A of Λ -free with Ob A :“ tΛ, 0u. Then AddpKar Aq ﬁ KarpAdd Aq.

Proof. Assume that AddpKar Aq » KarpAdd Aq. J By Lemma 144, A is idempotent complete. Thus, A ÝÑA Kar A is an equivalence; cf. Proposition 56.

AddpJAq Since JA is an additive equivalence, cf. Lemma 44, so is Add A ÝÝÝÝÝÑ AddpKar Aq; cf. Deﬁnition 100 and Lemma 101. Thus, we have Add A » AddpKar Aq » KarpAdd Aq. By Lemma 54, KarpAdd Aq is idempotent complete. Therefore, Add A is idempotent complete, cf. Remark 48, in contradiction to Lemma 144. Thus, AddpKar Aq ﬁ KarpAdd Aq.

4.2 Additive envelope and tensor product

Lemma 146. There exists a commutative ring R and additive categories pA, ϕAq and pB, ϕBq over R such that pA b B, ϕA b ψBq is not additive. R More precisely, consider R “ Q and the full subcategory A of Q -mod with Ob A :“ tV P Ob Q -mod : dim V ‰ 1u. Then A is Q-linear; cf. Remark 32. Furthermore, A is an additive category over Q, but A b A is not additive. Q

109 Proof. Suppose given A, B P Ob A. Then dim A ‰ 1 ‰ dim B. Since Q -mod is additive, there exists a direct sum C of A and B in Q -mod. Since dim A ‰ 1 ‰ dim B, we have dim C “ dim A ` dim B ‰ 1. Thus, we have C P Ob A. Therefore, A is additive. We have the Q-linear functor

A ÝÑF Q -mod

f f pA ÝÑ Bq ÞÑ pA ÝÑ Bq. Furthermore, we have the Q-bilinear functor

pQ -modq ˆ pQ -modq ÝÑG Q -mod

pf,gq fbg pA, Bq ÝÝÑpC,Dq ÞÑ A b B ÝÝÑ C b D . Q Q ´ ¯ ˆ ˙ By Remark 29, we have the Q-bilinear functor

G˝pF ˆF q A ˆ A ÝÝÝÝÝÑ Q -mod

pf,gq fbg pA, Bq ÝÝÑpC,Dq ÞÑ A b B ÝÝÑ C b D . Q Q ´ ¯ ˆ ˙ By Deﬁnition 127, we have the R-linear functor

A b A ÝÑH Q -mod Q

fibgi fibgi pA b Bq ÝÝÝÝÝÝÝÝÑpiPr1,ns C b Dq ÞÑ A b B ÝÝÝÝÝÝÝÝÑiPr1,ns C b D . ř Q ř Q ˆ ˙ ˆ ˙ Assume Q2 b Q2 and Q3 b Q3 to have a direct sum X b Y in A b A. Q Since H is additive, HpX b Y q “ X b Y is a direct sum of Q

HpQ2 b Q2q “ Q2 b Q2 and HpQ3 b Q3q “ Q3 b Q3 Q Q in Q -mod. We have

pdim Xqpdim Y q “ dimpX b Y q Q “ dimpQ2bQ2q ` dimpQ3bQ3q Q Q “ 4 ` 9 “ 13. We obtain dim X “ 1 or dim Y “ 1, in contradiction to X,Y P Ob A. Thus, Q2 b Q2 and Q3 b Q3 have no direct sum in A b A. Q Therefore, A b A is not additive. Q

110 Proposition 147. There exists a commutative ring R and preadditive categories pA, ϕAq and pB, ϕBq over R such that pAdd Aq b pAdd Bq ﬁ AddpA b Bq. R R More precisely, consider R “ Q and the full subcategory A of Q -mod with Ob A :“ tV P Ob Q -mod : dim V ‰ 1u; Then A is Q-linear; cf. Remark 32. Furthermore, pAdd Aq b pAdd Aq ﬁ AddpA b Aq. Q Q

Proof. Assume that pAdd Aq b pAdd Aq » AddpA b Aq. Q Q

I By Lemma 146, A is additive. Thus, A ÝÑA Add A is an equivalence; cf. Proposition 99.

By Lemma 117, IA is Q-linear. Therefore, we have the equivalence I I A b A ÝÝÝÝÑpAb A Add Aq b pAdd Aq; Q Q cf. Lemma 136. Thus, we have A b A » pAdd Aq b pAdd Aq » AddpA b Aq. Q Q Q By Proposition 92, AddpA b Aq is additive. Therefore, A b A is additive, in contradiction to Lemma 146. Q Q Thus, pAdd Aq b pAdd Aq ﬁ AddpA b Aq. Q Q

4.3 Karoubi envelope and tensor product

Lemma 148. There exists a commutative ring R and idempotent complete preadditive categories pA, ϕAq and pB, ϕBq over R such that pA b B, ϕA b ψBq is not idempotent complete. R More precisely, consider R “ Q and the full subcategory of Qpiq -mod with Ob A :“ tQpiq, 0u. Then A is Q-linear; cf. Remark 32. Furthermore, A is idempotent complete, but A b A is not idempotent complete. Q

Proof. For x P Qpiq we also write x for the map Qpiq Ñ Qpiq, y ÞÑ yx. Since Qpiq is a ﬁeld, we have

1 Idem A “ tQpiq ÝÑ1 Qpiq, Qpiq ÝÑ0 Qpiq, 0 ÝÑ0 0u. Furthermore, pQpiq, 1, 1q is an image of Qpiq ÝÑ1 Qpiq, 0 p0, 0Qpiq,0, 00,Qpiqq is an image of Qpiq ÝÑ Qpiq, 10 p0, 10, 10q is an image of 0 ÝÑ 0.

111 Thus, A is idempotent complete. We have Ob A b A “ tQpiq b Qpiq, Qpiq b 0, 0 b Qpiq, 0 b 0u. Q ˆ ˙ Since 0 is a zero object in A, we have zero objects

Qpiq b 0 – 0 b Qpiq – 0 b 0 in AbA; cf. Definition 123. Q e:“ 1 p1b1ì b iq Consider Qpiq b Qpiq ÝÝÝÝÝÝÝÝÝÑ2 Qpiq b Qpiq in AbA. We have Q 1 1 e2 “ p1 b 1 ` i b i ` i b i `p´1q b p´1qq “ p2p1 b 1q ` 2pi b iqq “ e. 4 4 Thus, e is an idempotent. Since Q is a field and p1, iq is a Q-linear basis of Qpiq, we have 1 b 1 ‰ e ‰ 0. In particular, Qpiq b 0, 0 b Qpiq and 0 b 0 cannot be images of e. Assume X :“ Qpiq b Qpiq to be an image of e. There exist X ÝÑπ X and X ÝÑι X such that the following diagram commutes.

e X / X > π ι π

1X X / X

Since AbApX,Xq “ Qpiq b Qpiq is commutative, we conclude Q Q

e “ πι “ ιπ “ 1X “ 1 b 1, a contradiction. Thus, e has no image in A b A. Therefore, A b A is not idempotent complete. Q Q

Proposition 149. There exists a commutative ring R and preadditive categories pA, ϕAq and pB, ϕBq over R such that pKar Aq b pKar Bq ﬁ KarpA b Bq. R R More precisely, consider R “ Q and the full subcategory of Qpiq -mod with Ob A :“ tQpiq, 0u. Then A is Q-linear; cf. Remark 32. Furthermore, pKar Aq b pKar Aq ﬁ KarpA b Aq. Q Q

Proof. Assume that pKar Aq b pKar Aq » KarpA b Aq. Q Q J By Lemma 148, A is idempotent complete. Thus, A ÝÑA Kar A is an equivalence; cf. Proposition 56.

By Lemma 84, JA is Q-linear.

112 By Lemma 136, we have the Q-linear equivalence

J J A b A ÝÝÝÝÑpAb A Kar Aq b pKar Aq. Q Q

Thus, A b A » pKar Aq b pKar Aq » KarpA b Aq. Q Q Q Since KarpA b Aq is idempotent complete, cf. Lemma 54, so is A b A; cf. Remark 48. Q Q This is a contradiction to Lemma 148. Thus, pKar Aq b pKar Aq ﬁ KarpA b Aq. Q Q

113 114 Bibliography

[1] Atiyah, M.F., Macdonald, I.G., Introduction to Commutative Algebra, Addison-Wesley, 1969. [2] Bakalov, B., Kirillov, A.A., Lectures on Tensor Categories and Modular Functors, AMS, 2001. [3] Karoubi, M., Fondements de la K-théorie(Notes de séminaire),Facultédes Sciences d’Alger, 1966. [4] Karoubi, M., K-Theory, Springer, 1978. [5] Mac Lane, S., Categories for the Working Mathematician, Springer, 1971. [6] Schubert, H., Categories, Springer, 1972. [7] Stein, N., Adelman’s abelianisation of an additive category (Bachelor’s Thesis), Stuttgart, 2012.

115 Zusammenfassung

Karoubih¨ulle

Wir beschreiben die Konstruktion der Karoubihülleeiner additiven Kategorie, wie von Max Karoubi in [3, II.1] eingeführt;siehe auch [4, Theorem 6.10]. Sei A eine additive Kategorie. Wir konstruieren eine idempotentvollständigeadditive Kategorie Kar A und einen additiven Funktor JA : A Ñ Kar A so, dass jeder additive Funktor F von A in eine idempotentvollständigeadditive Kategorie B eindeutig, bis auf 1 Isomorphie, über JA als F ˝ JA “ F faktorisiert.

JA A / Kar A

F 1 F & B

Genauer erhalten wir folgende Aquivalenz¨ von Kategorien.

„ addrA, Bs addrKar A, Bs

β˚JA β- pU ˝ JA ÝÝÝÑ V ˝ JAq pU V q.

Diese Aquivalenz¨ ist surjektiv auf Objekten; cf. Theorem 78.

Additive H¨ulle

Wir beschreiben die Konstruktion der additiven Hülleeiner präadditiven Kategorie. Diese Konstruktion wird in [5, VII.2, ex. 6.(a)] und in [2, Def. 1.1.15] erwähnt. Sei A eine präadditive, nicht notwendigerweise additive Kategorie. Wir konstruieren eine additive Kategorie Add A und einen additiven Funktor IA : A Ñ Add A so, dass jeder additive Funktor F von A in eine additive Kategorie B eindeutig, bis auf Isomorphie, 1 über IA als F ˝ IA “ F faktorisiert.

IA A / Add A

F 1 F & B

Genauer erhalten wir folgende Aquivalenz¨ von Kategorien.

„ addrA, Bs addrAdd A, Bs

β˚IA β- pU ˝ IA ÝÝÝÑ V ˝ IAq pU V q.

Diese Aquivalenz¨ ist surjektiv auf Objekten; cf. Theorem 113.

116 Tensorprodukt pr¨aadditiver Kategorien

Wir beschreiben die Konstruktion des Tensorproduktes präadditiver Kategorien, wie in [6, 16.7.4] erwähnt. Seien A und B präadditive Kategorien. Wir konstruieren eine präadditive Kategorie AbB und einen Z-bilinearen Funktor MA,B : A ˆ B Ñ A b B so, dass jeder Z-bilineare Funktor F von A ˆ B in eine präadditive Kategorie C eindeutig über MA,B als F ˝ MA,B “ F faktorisiert. MA,B A ˆ B / A b B

F F ' C Genauer erhalten wir folgenden Isomorphismus von Kategorien.

„ Z-bilrA ˆ B, Cs addrA b B, Cs

β˚MA,B β- pU ˝ MA,B ÝÝÝÝÝÑ V ˝ MA,Bq pU V q;

cf. Theorem 130.

Tensorprodukt additiver Kategorien

Wir beschreiben die Konstruktion des Tensorproduktes additiver Kategorien, wie in [2, Def. 1.1.15] erw¨ahnt.

add Seien A und B additive Kategorien. Wir konstuieren eine additive Kategorie A b B und add add einen Z-bilinearen Funktor MA,B : A ˆ B Ñ A b B so, dass jeder Z-bilineare Funktor add F von A ˆ B in eine additive Kategorie C eindeutig, bis auf Isomorphie, ¨uber MA,B als 1 add F ˝ MA,B “ F faktorisiert. add MA,B add A ˆ B / A b B

F 1 F & C Genauer erhalten wir folgende Aquivalenz¨ von Kategorien.

add „ Z-bilrA ˆ B, Cs addrA b B, Cs

β˚M add add A,B add β- pU ˝ MA,B ÝÝÝÝÝÑ V ˝ MA,Bq pU V q.

Diese Aquivalenz¨ ist surjektiv auf Objekten; cf. Theorem 143.

117 Gegenbeispiele f¨urKompatibilit¨atsrelationen

F¨urpr¨aadditive Kategorien A gilt im Allgemeinen

KarpAdd Aq ﬁ AddpKar Aq.

Genauer, betrachte den Teilring Λ :“ tpa, bq P Z ˆ Z : a ”5 bu von Z ˆ Z und die volle pr¨aadditive Teilkategorie A von Λ -free mit Ob A :“ tΛ, 0u. Dann gilt AddpKar Aq ﬁ KarpAdd Aq; cf. Proposition 145.

Fürkommutative Ringe R und präadditive Kategorien pA, ϕAq und pB, ϕBq über R gilt im Allgemeinen AddpA b Bq fi pAdd Aq b pAdd Bq. R R Genauer, betrachte R “ Q und die volle Teilkategorie A von Q -mod mit

Ob A :“ tV P Ob Q -mod : dim V ‰ 1u.

Dann ist A Q-linear; cf. Remark 32. Außerdem gilt pAdd Aq b pAdd Aq ﬁ AddpA b Aq; Q Q cf. Proposition 147.

Fürkommutative Ringe R und präadditive Kategorien pA, ϕAq und pB, ϕBq über R gilt im Allgemeinen KarpA b Bq fi pKar Aq b pKar Bq. R R Genauer, betrachte R “ Q und die volle Teilkategorie A von Qpiq -mod mit Ob A :“ tQpiq, 0u. Dann ist A Q-linear; cf. Remark 32. Außerdem gilt pKar Aq b pKar Aq fi KarpA b Aq; cf. Proposition 149. Q Q

118 Hiermit versichere ich,

(1) dass ich meine Arbeit selbstst¨andigverfasst habe,

(2) dass ich keine anderen als die angegebenen Quellen benutzt habe und alle wörtlich oder sinngemäßaus anderen Werken übernommenen Aussagen als solche gekenn- zeichnet habe,

(3) dass die eingereichte Arbeit weder vollst¨andignoch in wesentlichen Teilen Gegen- stand eines anderen Pr¨ufungsverfahrens gewesen ist und

(4) dass das elektronische Exemplar mit den anderen Exemplaren ¨ubereinstimmt.

Stuttgart, Februar 2016 Mathias Ritter

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