bardes, stefan - factorizable enriched categories and applications.pdf

Journal of Algebra 348 (2011) 416445

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actorizable enriched categories and applications

ura Brdes, Dragos Stefan

iversity of Bucharest, Faculty of Mathematics and Informatics, Bucharest, 14 Academiei Street, Ro-010014, Romania

r t i c l e i n f o a b s t r a c t

ticle history:ceived 14 May 2011ailable online 13 October 2011mmunicated by Nicols Andruskiewitsch

SC:imary 18D20condary 18D10, 16Sxx

ywords:riched categoryisting systemisted tensor productatched paircrossed product

We dene the twisted tensor product of two enriched categories,which generalizes various sorts of products of algebraic structures,including the bicrossed product of groups, the twisted tensorproduct of (co)algebras and the double cross product of bialgebras.The key ingredient in the denition is the notion of simple twistingsystems between two enriched categories. To give examples ofsimple twisted tensor products we introduce matched pairs ofenriched categories. Several other examples related to ordinarycategories, posets and groupoids are also discussed.

2011 Elsevier Inc. All rights reserved.

ntents

troduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4171. Preliminaries and notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4192. Factorizable M-categories and twisting systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4233. Matched pairs of enriched categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4324. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436knowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444ferences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445

Corresponding author.E-mail addresses: [email protected] (A. Brdes), [email protected] (D. Stefan).

21-8693/$ see front matter 2011 Elsevier Inc. All rights reserved.

i:10.1016/j.jalgebra.2011.08.034

A. Brdes, D. Stefan / Journal of Algebra 348 (2011) 416445 417

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The most convenient way to explain what we mean by the factorization problem of an algebraicructure is to consider a concrete example. Chronologically speaking, the rst problem of this typeas studied for groups, see for instance [Mai,Ore,Za,Sz,Tak]. Let G be a group. Let H and K denoteo subgroups of G . One says that G factorizes through H and K if G = HK and H K = 1. Therefore,e factorization problem for groups means to nd necessary and sucient conditions which ensureat G factorizes through the given subgroups H and K . Note that, if G factorizes through H andthen the multiplication induces a canonical bijective map : H K G, which can be used to

ansport the group structure of G on the Cartesian product of H and K . We shall call the resultingoup structure the bicrossed product of H and K , and we shall denote it by H K . The identityement of H K is (1,1), and its group law is uniquely determined by the twisting map

R : K H H K , R(k,h) := 1(kh).

bviously, R is induced by a couple of functions : K H H and : K H K such that R(k,h) = h,k h). Using this notation the multiplication on H K can be written as

(h,k) (h,k)= (h(k h), (k h)k).e group axioms easily imply that (H, K ,,) is a matched pair of groups, in the sense of [Tak].nversely, any bicrossed product H K factorizes through H and K . In conclusion, a group G fac-rizes through H and K if and only if it is isomorphic to the bicrossed product H K associated tocertain matched pair (H, K ,,).Similar products are known in the literature for many other algebraic structures. In [Be], for a dis-

ibutive law : G F F G between two monads, Jon Beck dened a monad structure on F G,hich can be regarded as a sort of bicrossed product of F and G with respect to the twisting naturalansformation .The twisted tensor product of two K-algebras A and B with respect to a K-linear twisting map

: B K A AK B was investigated for instance in [Ma1,Tam,CSV,CIMZ,LPoV,JLPvO]. It is the analo-us in the category of associative and unital algebras of the bicrossed product of groups. The classicalnsor product of two algebras, the graded tensor product of two graded algebras, skew algebras,ash products, Ore extensions, generalized quaternion algebras, quantum ane spaces and quantumri are all examples of twisted tensor products.Another class of examples, including the Drinfeld double and the double crossed product ofmatched pair of bialgebras, comes from the theory of Hopf algebras, see [Ma2]. Some of thesenstructions have been generalized for bialgebras in monoidal categories [BD] and bimonads [BV].Enriched categories have been playing an increasingly important role not only in Algebra, but alsoAlgebraic Topology and Mathematical Physics, for instance. They generalize usual categories, lineartegories, Hopf module categories and Hopf comodule categories. Monoids, algebras, coalgebras andalgebras may be regarded as enriched categories with one object.Our aim in this paper is to categorify the factorization problem, i.e. to answer the question whenenriched category factorizes through a couple of enriched subcategories. Finding a solution at this

vel of generality would allow us to approach in an unifying way all factorization problems thate have already mentioned. Moreover, it would also provide a general method for producing newn-trivial examples of enriched categories.In order to dene factorizable enriched categories, we need some notation. Let C be a small en-

ched category over a monoidal category (M,,1). Let S denote the set of objects in C . For them-objects in C we use the notation xC y . The composition of morphisms and the identity mor-isms in C are dened by the maps xc

yz : xC y yCz xCz and 1x : 1 xCx , respectively. Fortails, the reader is referred to the next section. We assume that A and B are two enriched sub-

418 A. Brdes, D. Stefan / Journal of Algebra 348 (2011) 416445

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extegories of C . The inclusion functor : A C is given by a family { xy}x,yS of morphisms in M ,here xy : x A y xC y . If is the inclusion of B in C , then for x, y and u in S we dene

xuy : x Au u B y xC y, x yz = xcuy ( xu uy).

ssuming that all S-indexed families of objects in M have a coproduct it follows that the mapsuy }uS yield a unique morphism

xy :uS

x Au u B y xC y .

e say that C factorizes through A and B if and only if all xy are invertible. An enriched categoryis called factorizable if it factorizes through A and B , for some A and B .In Theorem 2.3, our rst main result, under the additional assumption that the tensor product onis distributive over the direct sum, we show that to every M-category C that factorizes throughand B corresponds a twisting system between B and A, that is a family R := { xR yz }x,y,zS oforphisms

xRyz : xB y y Az

uS

x Au u Bz

hich are compatible with the composition and identity maps in A and B in a certain sense.Trying to associate to a twisting system R := { xR yz }x,y,zS an M-category we encountered some

culties due to the fact that, in general, the image of xRyz is too big. Consequently, in this paper we

cus on the particular class of twisting systems for which there is a function | | : S S S Sch that the image of xR

yz is included into x A|xyz| |xyz|Bz, for every x, y, z S . These twisting

stems are characterized in Proposition 2.5. A more precise description of them is given in Corol-ry 2.7, provided that M satises an additional condition (), see 2.6. A similar result is obtained inrollary 2.9 for a linear monoidal category.In this way we are led in 2.10 to the denition of simple twisting systems. For such a twisting

stem R between B and A, in Theorem 2.14 we construct an M-category A R B which factorizesrough A and B . Since it generalizes the twisted tensor product of algebras, A R B will be callede twisted tensor product of A and B .In the third section we consider the case when M is the monoidal category of coalgebras in

braided category M . In this setting, we prove that there is an one-to-one correspondence betweenmple twisting systems and matched pair of enriched categories, see 3.6 for the denition of thetter notion. We shall refer to the twisted tensor product of a matched pair as the bicrossed product.construction, the bicrossed product is a category enriched over M , but we prove that it is enricheder M as well.More examples of twisted tensor products of enriched categories are given in the last part of theper. By denition, usual categories are enriched over Set, the category of sets. Actually, they are en-ched over the monoidal category of coalgebras in Set. Hence, simple twisting systems and matchedirs are equivalent notions for usual categories. Moreover, if A and B are thin categories (that iseir hom-sets contain at most one morphism), then we show that any twisting system between Bd A is simple, so it corresponds to a uniquely determined matched pair of categories. We use thissult to investigate the twisting systems between two posets.Our results may be applied to algebras in a monoidal category M , which are enriched categories

ith one object. Therefore, we are also able to recover all bicrossed product constructions that wescussed at the beginning of this introduction.Finally, we prove that the bicrossed product of two groupoids is also a groupoid, and we give an

ample of factorizable groupoid with two objects.


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(1(2Preliminaries and notation

Mainly for xing the notation and the terminology, in this section we recall the denition ofriched categories, and then we give some example that are useful for our work.

. Monoidal categories. Throughout this paper (M,,1,a, l, r) will denote a monoidal category withsociativity constraints aX,Y ,Z : (X Y ) Z X (Y Z) and unit constraints lX : 1 X X and: X 1 X . The class of objects of M will be denoted by M0. Mac Lanes Coherence Theorem

ates that given two parenthesized tensor products of some objects X1, . . . , Xn in M (with possiblebitrary insertions of the unit object 1) there is a unique morphism between them that can beritten as a composition of associativity and unit constraints, and their inverses. Consequently, allese parenthesized tensor products can be identied coherently, and the parenthesis, associativitynstraints and unit constraints may be omitted in computations. Henceforth, we shall always ignoreem. The identity morphism of an object X in M will be denoted by the same symbol X .By denition, the tensor product is a functor. In particular, for any morphisms f : X Y and

: X Y in M the following equations hold(f Y ) (X f )= f f = (Y f ) ( f X ). (1)

If the coproduct of a family {Xi}iI of objects in M exists, then it will be denoted as a pairiI Xi, {i}iI ), where the maps i : Xi

iI Xi are the canonical inclusions.

2. The opposite monoidal category. If (M,,1,a, l, r) is a monoidal category, then one con-ructs the monoidal category (Mo,o,1o,ao, lo, ro) as follows. By denition, Mo and M share theme objects and identity morphisms. On the other hand, for two objects X, Y in M , one takesomMo (X, Y ) := HomM (Y , X). The composition of morphisms in Mo

: HomMo (Y , Z) HomMo (X, Y ) HomMo (X, Z)dened by the formula f g := g f , for any f : Z Y and g : Y X in M . The monoidalructure is dened by X o Y = X Y and 1o = 1. The associativity and unit constraints in Mo areven by aoX,Y ,Z = a1X,Y ,Z , lo = l1X and ro = r1X . If, in addition M is braided monoidal, with braidingX,Y : X Y Y X then Mo is also braided, with respect to the braiding o dened by oX,Y :=X,Y )

1.

enition 1.3. Let S be a set. We say that a monoidal category M is S-distributive if every S-indexedmily of objects in M has a coproduct, and the tensor product is distributive to the left and toe right over any such coproduct. More precisely, M is S-distributive if for any family {Xi}iS theproduct (

iS Xi, {i}iS) exists and, for an arbitrary object X ,(

X (

iSXi

), {X i}iS

)and

((iS

Xi

) X, {i X}iS

)

e the coproducts of {X Xi}iS and {Xi X}iS , respectively. Note that all monoidal categories aredistributive, provided that S is a singleton (i.e. the cardinal of S is 1).

4. Enriched categories. An enriched category C over (M,,1), or an M-category for short, consists:

) A class of objects, that we denote by C0. If C0 is a set we say that C is small.) A hom-object xC y in M , for each x and y in C0. It plays the same role as HomC (y, x), the set of

morphisms from y to x in an ordinary category C .


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) A morphism xcyz : xC y yCz xCz , for all x, y and z in C0.

) A morphism 1x : 1 xCx , for all x in C0.

denition one assumes that the diagrams in Fig. 1 are commutative, for all x, y, z and t in C0. Themmutativity of the square means that the composition of morphisms in C , dened by { xcyz }z,y,zC0 ,associative. We shall say that 1x is the identity morphism of x C0.An M-functor : C C is a couple (0, { xy}x,yC0 ), where 0 : C0 C 0 is a function andy : xC y xC y is a morphism in M for any x, y C0, where for simplicity we denoted 0(u)u , for any u C0. By denition, 0 and xy must satisfy the following conditions

xx 1Cx = 1Dx and xdy

z ( xy yz) = xz xc yz .

5. To work easier with tensor products of hom-objects in M-categories we introduce some newtation. Let S be a set and for every i = 1, . . . ,n + 1 we pick up a family { x Xiy}x,yS of objects in. If x1, . . . , xn+1 S then the tensor product x0 X1x1 x1 X2x2 xn1 Xnxn xn Xn+1xn+1 will be denotedx0 X

1x1 X

2x2 xn Xn+1xn+1 . Assuming that M is S-distributive and xing x0 and xn+1, one can construct

ductively the iterated coproduct

x0 X1x1

xn1 Xnxn Xn+1xn+1 :=x1S

xnS

x0 X1x1 xn1 Xnxn Xn+1xn+1 . (2)

is not dicult to see that this object is a coproduct of { x0 X1x1 xn Xn+1xn+1}(x1,...,xn)Sn . Moreover, asconsequence of the fact that the tensor product is distributive over the direct sum, we have

x0 X1x1

xn1 Xnxn Xn+1xn+1 =x1S

xnS

x0 X1x(1) X

2x(2) x(n) Xn+1xn+1 (3)

r any permutation of the set {1,2, . . . ,n}. The inclusion of x0 X1x1 xn Xn+1xn+1 into the coproductned in (2) is also inductively constructed as the composition of the following two arrows

x0 X1x1 x1 X1x2 xn Xn+1xn+1 x0 X1x1 x1 X1x2 xn Xn+1xn+1

x1S

x 0 X1x1 x1 X1x2 xn Xn+1xn+1 ,

here the rst morphism is the tensor product between the identity of x0 X1x1 and the inclusion of

X1x2 xn Xn+1xn+1 into x1 X1x2 xn Xn+1xn+1 . Clearly, for every xn+1 S ,

x0 X1x1

xn1 Xnxn Xn+1xn+1 :=x0S

x0 X1x1

xn1 Xnxn Xn+1xn+1

the coproduct of { x0 X1x1 xn Xn+1xn+1}(x0,x1,...,xn)Sn . The objects x0 X1x1 xn1 Xnxn Xn+1xn+1 and

1 n n+1Xx1 xn1 Xxn Xxn+1 are analogously dened.


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foThisA similar notation will be used for morphisms. Let us suppose that xiy is a morphism in M with

urce x Xiy and target xYiy , where x, y S and i {1, . . . ,n + 1}. We set

x11x2

2x3 xnn+1xn+1 := x01x1 xnn+1xn+1 .

the universal property of coproducts, { x01x1 xnn+1xn+1 }(x1xn)Sn induces a unique map x01x1 1

n1xn

nxn+1 that commutes with the inclusions. In a similar way one constructs

x01x1

xn1nxnn+1xn+1 , x01x1 xn1nxnn+1xn+1 and x01x1 xn1nxnn+1xn+1 .

make the above notation clearer, let us have a look at some examples. Let A and B be two M-tegories such that A0 = B0 = S . Recall that the hom-objects in A and B are denoted by x A y andy . Hence, x A y =

xS x A y . We also have x A y Bz At = x A y y Bz z At and

x Ay Bz At =yS

zS

x Ay Bz At =zS

yS

x Ay Bz At =y,zS

x Ay Bz At .

nce we have agreed to use the same notation for an object and its identity map, we can writeyz Atu instead of Idx B y yz Id z At tu , for any morphisms yz and tu in M . The mapsyz : x A y Az x Az and xayz : x A y Az x Az are induced by the composition in A, that is bye set { xayz }zS . For example, the former map is uniquely dened such that its restriction toy Az and x,z xayz coincide for all x S , where x,z is the inclusion of x Az into x Az . Similarly,yz : x A y Az x Az is the unique map whose restriction to x A y Az is xayz , for all y S .For more details on enriched categories the reader is referred to [Ke]. We end this section givingme examples of enriched categories.

6. The category Set. The category of sets is monoidal with respect to the Cartesian product. Theit object is a xed singleton set, say {}. The coproduct in Set is the disjoint union. Since thesjoint union and the Cartesian product commute, Set is S-distributive for any set S . Clearly, a Set-tegory is an ordinary category. If C is such a category, then an element f xC y will be thought ofa morphism from y to x, and it will be denoted by f : y x, as usual. In this case we shall sayat y (respectively x) is the domain or the source (respectively the codomain or the target) of f . Theme notation and terminology will be used for arbitrary M-categories, whose objects are sets.

7. The category KKK-Mod. Let K be a commutative ring. The category of K-modules is monoidalith respect to the tensor product of K-modules. The unit object is K, regarded as a K-module. Thisonoidal category is S-distributive for any S . By denition, a K-linear category is an enriched categoryer K-Mod.

8. The category -Mod-. Let be a K-algebra and let -Mod- denote the category of leftr right) modules over K o , where o is the opposite algebra of . Thus, M is an object in-Mod- if, and only if, it is a left and a right -module and these structures are compatible in thense that

a m =m a and (x m) y = x (m y)

r all a K, x, y and m M . A morphism in -Mod- is a map of left and right -modules.e category of -bimodules is monoidal with respect to () (). The unit object in -Mod-

, regarded as a -bimodule. This monoidal category also is S-distributive for any S .


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1.a9. The category H -Mod. Let H be a bialgebra over a commutative ring K. The category of left H-odules is monoidal with respect to () K (). If M and N are H-modules, then the H-action on N is given by

h m n =

h(1) m h(2) n.

the above equation we used the -notation (h) =h(1) h(2) . The unit object is K, which isH-module with the trivial action, induced by the counit of H . This category is S-distributive, for

y S . An enriched category over H-Mod is called H-module category.

10. The category Comod-H . Dually, the category of right H-comodules is monoidal with respect to) K (). The coaction on and M K N is dened by

(m n) =

m0 n0 m1n1,

here (m) = m0 n0 , and a similar -notation was used for (n). This category is S-stributive, for any set S . By denition, an H-comodule category is an enriched category overmod-H .

11. The category [A, A]. Let A be a small category, and let [A, A] denote the category of all endo-nctors of A. Therefore, the objects in [A, A] are functors F : A A, while the set F [A, A]G containsl natural transformations : G F . The composition in this category is the composition of natu-l transformations. The category [A, A] is monoidal with respect to the composition of functors. If: F G and : F G are natural transformations, then the natural transformations F and are given by

F : F F G G , (F )x := F (x),G : G F G G , (G)x := G(x).

e can now dene the tensor product of and by

:= G F = G F.

en if A is S-distributive, [A, A] may not have this property. In spite of the fact that, by as-mption, any S-indexed family in [A, A] has a coproduct, in general this does not commute withe composition of functors. Nevertheless, as we have already noticed, [A, A] is S-distributive if| = 1.This remark will allow us to apply our main results to an [A, A]-category C with one object x.

ence F := xCx is an endofunctor of A, and the composition and the identity morphisms in C areiquely dened by natural transformations

: F F F and : IdA F .

e commutativity of the diagrams in Fig. 1 is equivalent in this case with the fact that (F ,, ) is aonad, see [Be] for the denition of monads. In conclusion, monads are in one-to-one correspondence[A, A]-categories with one object.

12. The category Opmon(M). Let (M,,1) be a monoidal category. An opmonoidal functor is

triple (F , , ) that consists of


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evthsyFig. 2. The denition of opmonoidal functors.

) A functor F : M M .) A natural transformation := {x,y}(x,y)M0M0 , with x,y : F (x y) F (x) F (y).) A map : F (1) 1 in M .

addition, the transformations and are assumed to render commutative the diagrams in Fig. 2.opmonoidal transformation : (F , , ) (F , , ) is a natural map : F F such that, for

bitrary objects x and y in M ,

(x y) x,y = x,y xy and 1 = .

bviously the composition of two opmonoidal transformations is opmonoidal, and the identity ofopmonoidal functor is an opmonoidal transformation. The resulting category will be denoted by

pmon(M). For two opmonoidal functors (F , , ) and (F , , ) one denes

(F , , ) (F , , ) := (F F , F ,F F (), F ()),here F ,F = {F (x),F (y)}x,yM0 . On the other hand, if : F G and G are opmonoidalansformations, then := G F is opmonoidal too. One can see easily that denesmonoidal structure on Opmon(M) with unit object (IdM , {Idxy}x,yM0 , Id1).

3. The categories Alg(M) and Coalg(M). Let (M,,1,) be a braided monoidal category withaiding := {x,y}(x,y)M0M0 , where x,y : x y y x. The category Alg(M) of all algebras inis monoidal too. Recall that an algebra in M is an M-category with one object. As in 1.11, suchcategory is uniquely determined by an object X in M and two morphisms m : X X X (theultiplication) and u : 1 X (the unit). The commutativity of the diagrams in Fig. 1 means that thegebra is associative and unital. If (X,m,u) and (X ,m,u) are algebras in M , then X X is angebra in M with multiplication

(m m) (X X ,X X ) : (X X ) (X X ) X X

d unit u u : 1 X X .The monoidal category Coalg(M) of coalgebras in M can be dened in a similar way. Alternatively,e may take Coalg(M) := Alg(Mo)o . Note that the monoidal category of coalgebras in M and theonoidal category of algebras in Mo are opposite each other.It is not hard to see that Coalg(M) is S-distributive, provided that M is so.

Factorizable M-categories and twisting systems

In this section we dene factorizable M-categories and twisting systems. We shall prove that toery factorizable system corresponds a certain twisting system. Under a mild extra assumption one monoidal category M , we shall also produce enriched categories using a special class of twisting

stems that we call simple.


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Throughout this section S denotes a xed set. We assume that all M-categories that we work withe small, and that their set of objects is S .

1. Factorizable M-categories. Let C be a small enriched category over (M,,1). We assume thatis S-distributive. Suppose that A and B are M-subcategories of C . Note that, by assumption, A0 =

0 = C0 = S . For x, y and u in S we dene

xuy : x A u B y xC y, xuy := xcuy xuy, (4)

here : A C and : B C denote the corresponding inclusion M-functors. By the universaloperty of coproducts, for every x and y in S , there is xy : x Au B y xC y such that

xy x uy = xuy , (5)

here x uy is the canonical inclusion of x Au B y into x Au B y . Note that by the universal property of

products xy = xcuy xuy , as we have xuy x uy = x uy xuy and xcuy x uy = xcuy , whereuy denotes the inclusion of xCuC y into xCuC y .We shall say that C factorizes through A and B if xy is an isomorphism, for all x and y in S . Bynition, an M-category C is factorizable if it factorizes through A and B , where A and B are certain-subcategories of C .

2. The twisting system associated to a factorizable M-category. Let C be an enriched category overmonoidal category (M,,1). We assume that M is S-distributive. The family R := { xR yz }x,y,zSmorphisms xR

yz : xB y Az x Au B y is called a twisting system if the four diagrams in Fig. 3 are

mmutative for all x, y, z and t in S .Let us briey explain the notation that we used in these diagrams. As a general rule, we omit

l subscripts and superscripts denoting elements in S , and which are attached to a morphism. Thembol is also omitted. For example, a and 1A (respectively b and 1B ) stand for the suitable com-sition maps and identity morphisms in A (respectively B). The identity morphism of an objectM is denoted by I . Thus, by Ia : xB y Az At xB y At we mean xB y yazt . On the other hand,: x Av Au Bt x Au Bt is a shorthand notation for xavu Bt , which in turn is the unique map induced{ x ut xavu Bt}u,vS . We shall keep the foregoing notation in all diagrams that we shall work with.We claim that to every factorizable M-category C corresponds a certain twisting system. By de-

tion, the map xy constructed in (5) is invertible for all x and y in S . Let xy denote the inverse ofy . For x, y and z in S , we can now dene

xRyz : xB y Az x Au Bz, xR yz := xz xc yz xyz. (6)

eorem 2.3. If C is a factorizable enriched category over an S-distributive monoidal category M , then the

aps in (6) dene a twisting system.


Prw

Si

foxCin(B

{ xx

Si

so{ x

H

Thdiasoof. Let us rst prove that the rst diagram in Fig. 3 is commutative. We x x, y, z and t in S , ande consider the following diagram.

nce the tensor product in a monoidal category is a functor, that is in view of (1), we have

xC yCut xyu Bt = xyut, (7)r any u in S . Hence by the universal property of the coproduct and the construction of the mapsyCut , xyu Bt and xyut we deduce that the relation which is obtained by replacing u with u(7) holds true. This means that the square (A) is commutative. Proceeding similarly one shows that) is commutative as well. Furthermore, xcvuCt , xvut and xu Bt are induced by { xcvu uCt}uS ,vu ut}uS and { xu u Bt}uS , respectively. Hence their composite := xcvuCt xvut u Bt is induced by {u}uS , where

u =(xc

vu uCt

) ( xvu ut) ( xu u Bt) = ( xcvu xvu xu) ut= ( xu xu) ut .

nce xu is the inverse of xu it follows that u = xCut , for every u S . In conclusion

xcvuCt xvut xu Bt = xCut,

(C) is a commutative square. Since is an M-functor it follows that { xCvcut x vut}u,vS and vt x Avbut }u,vS are equal. Therefore these families induce the same morphism, that is

xCvcut x vut = x vt x Avbut .

ence (D) is commutative too. Since the composition of morphisms in C is associative, we have

xcvt xCvcut = xcut xcvuCt and xc yt xC ycut = xcut xc yuCt .

ese equations imply that (E) and (F) are commutative. Summarizing, we have just proved that allagrams (A)(F) are commutative. By diagram chasing it results that the outer square is commutativewell, that is

u y y

xt x Avbt xRu Bt = xct xyt .


Le

winti

In

thm

A

Sicoco

2.evato

isinS3

fo

Prbeonft composing and right composing both sides of this equation by xt and xB y Rzt , respectively, yield

x Avbut xR yubt xB y Rzt = xt xc yt xyt xB y Rzt

= xt xc yt xyt xB yt xB yczt xB yzt= xt xc yt xC yczt xyzt,

here for the second and third relations we used the denition of y Rzt and that yt and yt areverses each other. On the other hand, the denition of xRzt , the fact that is a functor and associa-vity of the composition in C imply the following sequence of identities

xRzt xbyz At = xt xczt xzt xbyz At

= xt xczt xc yz Ct xyzt= xt xc yt xC yczt xyzat .

conclusion, the rst diagram in Fig. 3 is commutative. Taking into account the denition of xRxy ,

e identity xx 1Bx = 1x and the compatibility relation between the composition and the identityorphisms in an enriched category, we get the following sequence of equations

xy xRxy 1Bx Ay = xy xy xcxy xxy 1Bx Ay = xcxy 1xy = xy .

nalogously, using the denition of xy and the properties of identity morphisms, we get

xy x yy x Ay1By = x yy x Ay1By = xc yy xyy x Ay1By = xc yy xy1y = xy .

nce xy is an isomorphisms, in view of the above computations, it follows that the third diagram ismmutative as well. One can prove in a similar way that the remaining two diagrams in Fig. 3 aremmutative. 4. We have noticed in the introduction that to every twisting system of groups (or, equivalently,ery matched pair of groups) one associates a factorizable group. Trying to prove a similar result fortwisting system R between the M-categories B and A we have encountered some diculties duethe fact that, in general, the image of the map

xRyz : xB y Az

uS

x Au Bz

not included into a summand x Au Bz , for some u S that depends on x, y and z. For this reason,this paper we shall investigate only those twisting systems for which there are a function | | : S and the maps x R yz : xB y Az x A|xyz|Bz such that

xRyz = x |xyz|z x R yz , (8)

r all x, y, z, S . For them we shall use the notation (R, | |).

oposition 2.5. Let M be a monoidal category which is S-distributive. Let | | : S3 S and { x R yz }x,y,zSa function and a set of maps as above. The family { xR yz }x,y,zS dened by (8) is a twisting system if and

ly if, for any x, y, z, t S, the following relations hold:


Prdi

Thv Btivhomthpr

m

2.saXan

{Xasob

it

CoAfox|xy|yzt||t x A|xy|yzt||b|yzt|t x R y|yzt|Bt xB y Rzt = x |xzt|t x R zt xbyz At, (9)

x|xyz|zt||t xa|xyz|||xyz|zt|Bt x A|xyz| R zt x R yz At = x |xyt|t x R yt xB yazt , (10)

x|xxy|y x Rxy

(1Bx x Ay

)= x yy ( x Ay 1By), (11)x

|xyy|y x R yy

(xB y 1Ay

)= x xy (1Ax xB y). (12)oof. We claim that { x R yz }x,y,zS satisfy (9) if and only if { xR yz }x,y,zS render commutative the rstagram in Fig. 3. Indeed, let us consider the following diagram.

e squares (B) and (C) are commutative by the denition of xRyu : xB y Au x Av Bu and vbut :

u Bt v Bt . Hence the hexagon (A) is commutative if and only if the outer square is commuta-e. This proves our claim as (A) and the outer square in Fig. 3 are commutative if and only if (9)lds and the rst diagram in Fig. 3 is commutative, respectively. Similarly one shows that the com-utativity of the second diagram from Fig. 3 is equivalent to (10). On the other hand, obviously,e third and fourth diagrams in Fig. 3 are commutative if and only if (11) and (12) hold, so theoposition is proved. The inclusion maps make dicult to handle Eqs. (9)(12). In some cases we can remove these

orphisms by imposing more conditions on the map | | or on the monoidal category M .

6. The assumption (). Let M be a monoidal category which is S-distributive. We shall say that Mtises the hypothesis () if for any coproduct (

iS Xi, {i}iS) in M and any morphisms f : X

i and f : X Xi such that i f = i f , then either X is an initial object in M , or f = f d i = i .The prototype for the class of monoidal categories that satisfy the condition () is Set. Indeed, leti}iS be a family of sets, and let i denote the inclusion of Xi into the disjoint union iS Xi . Wesume that f : X Xi and f : X Xi are functions such that X in not the empty set, the initialject of Set, and i f = i f . Then in view of the computation(

i, f (x))= (i f )(x) = (i f )(x) = (i, f (x))

follows that f = f and i = i .

rollary 2.7. Let M be an S-distributive monoidal category. Let A and B be two M-categories such that0 = B0 = S. Given a function | | : S3 S and the maps { x R yz }x,y,zS as in 2.4, let us consider the

llowing four conditions:


(

(i

(i

(i

hy

PrththSi

norei

2.thH

obfin

anarf

pr

Core

(

(i

(i(ii) If xB y Bz At is not an initial object, then |xy|yzt|| = |xzt| and

x A|xzt|b|yzt|t x R y|yzt|Bt xB y Rzt = x R zt xbyz At; (13)

i) If xB y Az At is not an initial object, then ||xyz|zt| = |xyt| and

xa|xyz||xyt| Bt x A|xyz| R zt x R yz At = x R yt xB yazt ; (14)

ii) If x A y is not an initial object, then |xxy| = y and

x Rxy (1Bx x Ay

)= x Ay 1By ; (15)v) If xB y is not an initial object, then |xyy| = x and

x Ryy (xB y 1Ay

)= 1Ax xB y . (16)The above conditions imply the relations (9)(12). Under the additional assumption that M satises thepothesis (), the reversed implication holds as well.

oof. Let us prove that the condition (i) implies the relation (9). In the case when xB y Bz At = is is clear, as both sides of (9) are morphisms from an initial object to x Au Bt . Let us supposeat xB y Bz At = . By composing both sides of (13) with x |xy|yzt||t = x |xzt|t we get the equation (9).milarly, the conditions (ii), (iii) and (iv) imply the relations (10), (11) and (12), respectively.Let us assume that M satises the hypothesis (). We claim that (9) implies (i). If xB y Bz At ist an initial object we take f and f to be the left-hand side and the right-hand side of (13),spectively. We also set i := |xy|yzt|| and i := |xzt|. In view of (), it follows that f = f and= i , so our claim has been proved. We conclude the proof in the same way. 8. KKK-linear monoidal categories. Recall that M is K-linear if its hom-sets are K-modules, and bothe composition and the tensor product of morphisms are K-bilinear maps. For instance, K-Mod,-Mod, Comod-H and -Mod- are S-distributive linear monoidal categories, for any set S .Note that the () condition fail in a K-linear monoidal category M . Indeed let us pick up anject X , which is not an initial object, and a coproduct (

iS Xi, {i}iS) in M . If f : X Xi and : X Xi are the zero morphisms, then of course i f = i f , but neither i = i nor f = f ,

general.Nevertheless, the relations (9)(12) can also be simplied if M is a linear monoidal category. Fory coproduct (

iS Xi, {i}iS) in M and every i S , there is a map i :

iS Xi Xi such that

i i = Xi and i j = 0, provided that j = i. Hence, supposing that f : X Xi and f : X Xie morphisms such that i f = i f , we must have either i = i and f = f , or i = i and = 0 = f .Using the above property of linear monoidal categories, and proceeding as in the proof of theevious corollary, we get the following result.

rollary 2.9. Let M be an S-distributive K-linear monoidal category. If A and B are M-categories, then thelations (9)(12) are equivalent to the following conditions:

i) If |xy|yzt|| = |xzt| then the relation (13) holds; otherwise, each side of this identity has to be the zeromap;

i) If ||xyz|zt| = |xyt|, then the relation (14) holds; otherwise, each side of this identity has to be the zeromap;

ii) If |xxy| = y, then the relation (15) holds; otherwise, each side of this identity has to be the zero map;

v) If |xyy| = x, then the relation (16) holds; otherwise, each side of this identity has to be the zero map.


2.of

| aso

Atois

la

2.

Le

asprx A

fo

2.anteBy

K

ofdo

Lebe

(1(210. Simple twisting systems. The proper context for constructing an enriched category A R B outa special type of twisting system R is provided by Corollary 2.7.By denition, the couple (R, | |) is a simple twisting system between B and A if the function

| : S3 S and the maps { x R yz }x,y,zS as in 2.4 satisfy the conditions (i)(iv) in Corollary 2.7. Aspart of the denition, we also assume that x A|xyz|Bz is not an initial object whenever xB y Az is not.The latter technical assumption will be used to prove the associativity of the composition in

R B , our categorical version of the twisted tensor product of two algebras, which we are goingdene in the next subsection. Note that for Set this condition is superuous (if the source of x R

yz

not empty, then its target cannot be the empty set).For a simple twisting system (R, | |) we dene the maps xR yz using the relation (8). By Corol-

ry 2.7 and Proposition 2.5 it follows that R := { xR yz }x,y,zS is a twisting system.

11. The category A R B . For a simple twisting system (R, | |) we set

(A R B)0 := S and x(A R B)y :=uS

x Au u B y = x Au B y .

t us x three elements x, y and z in S . By denition x Au B y Av Bz :=

u,vS x Au B y Av Bz , and

x Au B y Av Bz = x Au B y y Av Bz

M is S-distributive. Via this identication, the canonical inclusion of x Au B y Av Bz into the co-oduct x Au B y Av Bz corresponds to x uy

vz = x uy y vz . Thus, there is a unique morphism xcyz :

u B y Av Bz x Au Bz such that

xcyz x uy vz = x |uyv|z xa|uyv|bz x Au R yv Bz,

r all u, v S . Finally, we set 1x := x xx (1Ax 1Bx ), and we dene

xy := x yy (x Ay 1By

)and xy := x xy

(1Ax xB y

).

12. Domains. To show that the above data dene an enriched monoidal category A R B we needextra hypothesis on M . By denition, a monoidal category M is a domain in the case when the

nsor product of two objects is an initial object if and only if at least one of them is an initial object.convention, a monoidal category that has no initial objects is a domain as well.Obviously Set is a domain. If K is a eld, then K-Mod is a domain. Keeping the assumption on

, the categories H-Mod and Comod-H are domains, as their tensor product is induced by that oneK-Mod. On the other hand, if K is not a eld, then K-Mod and -Mod- are not necessarilymains. For instance, Z-Mod = Z-Mod-Z is not a domain.

mma 2.13. Let M be an S-distributive monoidal domain. Let (R, | |) denote a simple twisting systemtween B and A.

) If x Au B y Av Bz Aw Bt = then |uyq| = |pvq| = |pzw|, where p = |uyv| and q = |vzw|.

) In the following diagram all squares are well dened and commutative.


PrInobu Bth

ThbeabHof

BythAsqco

Thda

Prth

If{ xco

isdeoof. Since M is a domain it follows that any subfactor of x Au B y Av Bz Aw Bt is not an initial object.particular v Bz Aw = . Thus, by the denition of simple twisting systems, v AqBw is not an initialject. In conclusion, v Aq and qBw are not initial objects in M . Since u B y Av = it follows thaty Av Aq = . In view of the denition of simple twisting systems (the second condition) we deduceat |pvq| = |uyq|. The other relation can be proved in a similar way.Let f and g denote the following two morphisms

f := x Auap|pvq|BqBt x Au Ap Rvq Bt x Au R yv AqBt and g := x Au R yq Bt x Au B yavq Bt .

e target of f is x Au A|pvq|BqBt , while the codomain of g is x Au A|uyq|BqBt . These two objects maydifferent for some elements x,u, t, p and q in S . Thus, in general, it does not make sense to speakout the square (R). On the other hand, we have seen that |uyq| = |pvq|, if p = |uyv| and q = |vzw|.ence (R) is well dened for these values of p and q. Furthermore, since u B y Av Aq = , by denitionsimple twisting systems we have

uap|pvq|Bq u Ap Rvq u R yv Aq = u R yq u B yavq . (17)

tensoring both sides of the above relation with x Au on the left and with qBt on the right we getat f = g , i.e. (R) is commutative. Analogously, one shows that (L) is well dened and commutative.ll other squares are well dened by construction, their arrows targeting to the right objects. Theuares (F) are commutative since the tensor product is a functor. The remaining squares (A) aremmutative by associativity. eorem 2.14. Let M be an S-distributive monoidal domain. If (R, | |) is a simple twisting system, then theta in 2.11 dene an M-category A R B that factorizes through A and B .

oof. Let us assume that x Au B y Av Bz Aw Bt = . In view of the previous lemma, the outer square ine diagram from Lemma 2.13(2) is commutative. It follows that

xcyt x Au B yczt x uy vz wt = xczt xc yz Aw Bt x uy vz wt .

x Au B y Av Bz Aw Bt = this identity obviously holds. Since x Au B y Av Bz Aw Bt is the coproduct ofAuB y Av Bz Aw Bt}u,v,wS , with the canonical inclusions { x uy vz wt }u,v,wS , we deduce that themposition in A R B is associative.We apply the same strategy to show that 1x := x xx (1Ax 1Bx ) is a left identity map of x, thatwe have xcxy (1x x Au B y) = x Au B y for any y. By the universal property of coproducts and thenition of the composition in A R B , it is enough to prove that

|xxu| x u x ( A B ) u

xy xa|xxu|by x Ax Ru B y 1x 1x x Au B y = xy , (18)


foabno|x

Thfa

w

fopr

enReth

foanid

Coth

Ifda

PrgeisOco(Lerpr

(2fa

ReTuccar all u S . If x Au is an initial object we have nothing to prove, as the domains of the sides of theove equation are also initial objects (recall that x Au B y = if x Au = ). Let us suppose that x Au ist an initial object. Then by the denition of simple twisting systems (the third condition) we getxu| = u and

x|xxu|y xax|xxu|buy x Ax Rxu B y

(1Ax 1Bx x Au B y

)= x uy xaxubuy (1Ax x Au 1Bu u B y).us Eq. (18) immediately follows by the fact 1Ax and 1

Bu are the identity morphisms of x and u. The

ct that 1x is a right identity map of x can be proved analogously.We now claim that { xy}x,yS is an M-functor. Taking into account the denition of and xcyz

e must prove that

x|yyz|z xax|yyz|azz x Ay R yz Az

(x Ay 1Ay y Az 1Az

)= x zz ( xayz 1Az ), (19)r all x, y and z in S . Once again, if y Az = we have nothing to prove. In the other case, one canoceed as in the proof of (18) to get this equation. Similarly, is an M-functor.It remains to prove the fact that A R B factorizes through A and B . As a matter of fact, for thisriched category, we shall show that xy is the identity map of x(A R B)y , for all x and y in S .call that xy is the unique map such that xy x uy = xcuy xuy , for all u S . Hence to concludee proof of the theorem it is enough to obtain the following relation

x|uuu|y xau|uuu|buy x Au Ruu B y

(x Au 1Bu 1Au u Bz

)= x uy , (20)r all u S . We may suppose that x Au is not initial object. Thus |uuu| = u and we can take x = ud y = u in (16). Hence, using the same reasoning as in the proof of (18), we deduce the requiredentity. rollary 2.15. Let A and B be enriched categories over an S-distributivemonoidal category M . Let us supposeat for all x, y, z and t in S the function | | : S3 S satises the equations

xy|yzt|= |xzt|, |xyz|zt= |xyt|, |xxy| = y and |xyy| = x. (21){ x R yz }x,y,zS is a family of maps which satises the identities (13)(16) for all x, y, z and t in S, then theta in 2.11 dene an M-category A R B that factorizes through A and B .

oof. Let x, y, u, v , z, w and t be arbitrary elements in S . By using the rst two identities in (21) wet |uyq| = |pvq| = |pzw|, where p = |uyv| and q = |vzw|. Hence the rst statement in Lemma 2.13true. In particular, the squares (R) and (L) in the diagram from Lemma 2.13(2) are well dened.n the other hand, under the assumptions of the corollary, the relation (17) hold. Therefore we canntinue as in the proof of the second part of Lemma 2.13 to show that (R) is commutative. Similarly,) is commutative too. It follows that the outer square of is commutative too. By the universal prop-ty of the coproduct we deduce that the composition is associative, see the rst paragraph of theoof of Theorem 2.14.Furthermore, the relations in (21) together with the identities (13)(16) imply Eqs. (18), (19) and

0). Proceeding as in the proof of Theorem 2.14 we conclude that A R B is an M-category thatctorizes through A and B . mark 2.16. Throughout this remark we assume that M is a T -distributive monoidal category, whereis an arbitrary set. In other words, any family of objects in M has a coproduct and the tensor prod-t is distributive over all coproducts. It was noticed in [RW, 2.1 and 2.2] that, for such a monoidal

tegory M , one can dene a bicategory M-mat as follows.


frta1-

Thso

Lethhkhoce

Sabprfa

cathdeinitth

It

sttw

3.

mgoFo

sa

LebeanBy construction, its 0-cells are arbitrary sets. If I and J are two sets, then the 1-cells in M-matom I to J are the J I-indexed families of objects in M . A 2-cell with source {X ji}( j,i) JI andrget {Y ji}( j,i) JI is a family { f ji}( j,i) JI of morphisms f ji : X ji Y ji . The composition of thecells {Xkj}(k, j)K J and {Y ji}( j,i) JI is the family {Zki}(k,i)KI , where

Zki :=j J

Xkj Y ji .

e vertical composition in M-mat of { f ji}( j,i) JI and {g ji}( j,i) JI makes sense if and only if theurce of f ji and the target of g ji are equal for all i and j. If it exists, then it is dened by

{ f ji}( j,i) JI {g ji}( j,i) JI = { f ji g ji}( j,i) JI .

t { f ji}( j,i) JI and { f kj}(k, j)K J be 2-cells such that f ji : X ji Y ji and f kj : X kj Y kj . Bye universal property of coproducts, for each (k, i) K I , there exists a unique morphismi : j J X kj X ji j J Y kj Y ji whose restriction to X kj X ji is f kj f ji . By denition, therizontal composition of { f kj}(k, j)K J and { f ji}( j,i) JI is the family {hki}(k,i)KI . The identity 1-lls and 2-cells in M-mat are the obvious ones.As pointed out in [RW], a monad on a set S in M-mat is an M-category with the set of objects

, and conversely. In particular, given two M-categories with the same set of objects, one may speakout distributive laws between the corresponding monads in M-mat . In our terminology, they areecisely the twisting systems. In view of [RW, 3.1], factorizable enriched categories generalize strictctorization systems.In conclusion, the Theorem 2.3 may be regarded as a version of [RW, Proposition 3.3] for enrichedtegories. For a simple twisting system (R, | |) between B and A, the enriched category A R Bat we constructed in Theorem 2.14 can also be described in terms of monads. Let : B A A Bnote the distributive law associated to (R, | |), where (A,mA,1A) and (B,mB ,1B) are the monadsM-mat corresponding to A and B, respectively. By the general theory of monads in a bicategory,follows that A B is a monad in M-mat with respect to the multiplication and the unit given bye formulae:

m := (mA mB) (IdA IdB) and 1 := 1A 1B .

is not dicult to show that A R B is the M-category associated to (A B,m,1).By replacing Set-mat with a suitable bicategory, one obtains similar results for other algebraic

ructures, such as PROs and PROPs; see [La]. We also would like to note that distributive laws be-een pseudomonads are investigated in [Mar].We are indebted to the referee for pointing the papers [La,Mar,RW] out to us.

Matched pairs of enriched categories

Throughout this section (M ,,1,) denote a braided category and we take M to be theonoidal category Coalg(M ). Our aim is to characterize simple twisting systems between two cate-ries that are enriched over M . We start by investigating some properties of the morphisms in M .r the moment, we impose no conditions on M .A slightly more general version of the following lemma is stated in [La, Proposition 3.2]. For the

ke of completeness we include a proof of it.

mma 3.1. Let (C,C , C ), (D1,D1 , D1 ) and (D2,D2 , D2 ) be coalgebras in M . Let f : C D1 D2

a morphism of coalgebras. Then f1 := (D1 D2 ) f and f2 := (D1 D2) f are coalgebra morphisms

d the following relations hold:


Co( f

PrDth

Htr

Bypl

CoLef2

Ta

Th

Thi( f1 f2) C = f , (22)( f2 f1) C = D1,D2 ( f1 f2) C . (23)

nversely, let f1 : C D1 and f2 : C D2 be coalgebra morphisms such that (23) holds. Then f :=1 f2) C is a coalgebra map such that

(D1 D2) f = f1 and (D1 D2) f = f2. (24)oof. Let us assume that f : C D1 D2 is a coalgebra morphism. Let i := Di , for i = 1,2. Clearly,1 D2 and D1 D2 are coalgebra morphisms. In conclusion f1 and f2 are morphisms in M . One other hand, as f is a morphism in M we have

(D1 D1,D2 D2) (D1 D2) f = ( f f ) C . (25)ence, using the denition of f1 and f2, the relation (25), the fact that the braiding is a naturalansformation and the compatibility relation between the comultiplication and the counit we get

( f1 f2) C = (D1 2 1 D2) ( f f ) C= [D1 ((2 1) D1,D2 D2) (D1 D2)] f= (D1 1 2 D2) (D1 D2) f = f .

applying 1 D2 D1 2 to (25) and using once again the compatibility between the comulti-ication and the counit we obtain

( f2 f1) C = (1 D2 D1 2) ( f f ) C= (1 D2 D1 2) (D1 D1,D2 D2) (D1 D2) f= D1,D2 (1 D1 D2 2) (D1 D2) f = D1,D2 f .

nversely, let us assume that f1 : C D1 and f2 : C D2 are morphisms in M such that (23) holds.t f := ( f1 f2) C . By the denition of the comultiplication on D1 D2 and the fact that f1 andare morphisms in M, we get

D1D2 f = (D1 D1,D2 D2) (D1 D2) ( f1 f2) C= (D1 D1,D2 D2) ( f1 f1 f2 f2) (C C ) C .

king into account (23) and the fact that the comultiplication is coassociative, it follows that

D1D2 f =[f1

(D1,D2 ( f1 f2) C

) f2] (C C ) C= [ f1 (( f2 f1) C ) f2] (C C ) C= [(( f1 f2) C ) (( f1 f2) C )] C = ( f f ) C .

e formula that denes f together with i f i = C yield

(1 2) f = (1 f1 2 f2) C = (C C ) C = C .us f is a morphism of coalgebras, so the lemma is proved. The equations in (24) are obvious, as

f i = C .


Rear

3.el( xan

Lex R

In

Cothmantw

Le

(1

(2

(3mark 3.2. Let f , f : C D1 D2 be coalgebra morphisms. By the preceding lemma, f and f e equal if and only if

(1 D2) f = (1 D2) f and (D1 2) f = (D1 2) f .

3. The morphisms xyz and xyz . Let A and B denote two M-categories whose objects are theements of a set S . The hom-objects of A and B are coalgebras, which will be denoted byAy, xAy , x

Ay ) and ( xB y, x

By , x

By ). By denition, the composition and the identity maps in A

d B are coalgebra morphisms. Note that the comultiplication of xB y Az is given by

x B y Az = ( xB y x B y ,y Az y Az) xByAz .

t | | : S3 S be a function and let R denote an S3-indexed family of coalgebra morphismsyz : xB y Az x A|xyz|Bz . We dene xyz : xB y Az x A|xyz| and xyz : xB y Az |xyz|Bz by

xyz := x A|xyz|Bz x R yz and xyz :=(x

A|xyz|Bz) x R yz . (26)

view of Lemma 3.1, xyz and xyz are coalgebra morphisms and they satisfy the relations(x yz xyz

) x B y Az = x R yz , (27)(x yz xyz

) x B y Az = x A|xyz|,|xyz|Bz ( x yz xyz ) x B y Az . (28)nversely, if one starts with := { xyz }x,y,zS and := { xyz }x,y,zS , two families of coalgebra mapsat satisfy (28), then by formula (27) we get the set R := { x R yz }x,y,zS whose elements are coalgebraaps, cf. Lemma 3.1. Therefore, there is an one-to-one correspondence between the couples (,)d the sets R as above. Our goal is to characterize those couples (,) that corresponds to a simpleisting system in M .

mma 3.4. The statements below are true.

) If |xy|yzt|| = |xzt| then the relation (13) is equivalent to the following equations:

x zt xbyz At = |xzt|b|yzt|t xy|yzt|Bt (xB y y zt yzt

) ( xB y y Bz At ), (29)x zt xbyz At = x y|yzt| xB y zt . (30)

) If |xyz|zt|| = |xyt| then the relation (14) is equivalent to the following equations:

x yt xB yazt = xa|xyz||xyt| x A|xyz| zt (x yz x yz z At

) (x B y Az z At), (31)x yt xB yazt = |xyz| zt x yz At . (32)

) If |xyy| = x then the relation (15) is equivalent to the following equations:

x xy (1Bx x Ay

)= xAy 1By , (33)x (1B A )= A . (34)x y x x y x y


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Prinan) If |xxy| = y then the relation (16) is equivalent to the following equations:

x yy (xB y 1Ay

)= xB y, (35)x yy

(xB y 1Ay

)= 1Ax xBy . (36)oof. In order to prove the rst statement we apply the Remark 3.2 to

f := x R zt xbyz At and f := x A|xzt|b|yzt|t x R y|yzt|Bt xB y Rzt .

ote that f is well dened and its target is x A|xzt|Bt , since the codomain of x R y|yzt|Bt xB y Rzt is|xy|yzt||Bt and |xy|yzt|| = |xzt|. Clearly, f and f are coalgebra morphisms, since the composited the tensor product of two morphisms in M remain in M . An easy computation, based on Eq. (27)d the formulae of xyz and xyz , yields us

xA|xzt|Bt f = x zt xbyz At,

x A|xzt|Bt f = x zt xbyz At,x

A|xzt|Bt f = |xzt|b|yzt|t x y|yzt| Bt (xB y y zt yzt

) ( xB y y Bz At ).king into account that xb

yz is a coalgebra morphism and using the denition of xyz we get

x A|xzt|Bt f = x A|xzt|B|yzt| x R y|yzt| (xB y

(y A|yzt|Bt y Rzt

))= x y|yzt| xB y zt .view of the Remark 3.2, we have f = f if and only if

x A|xzt|Bt f = x A|xzt|Bt f and xA|xzt|Bt f = xA|xzt|Bt f .

us, if |xy|yzt|| = |xzt|, then (13) is equivalent to (29) together with (30). We omit the proof of thecond statement, being similar.To prove the third part of the lemma we reiterate the above reasoning. We now take f and f tothe coalgebra morphisms

f := x Rxy (1Bx x Ay

)and f := x Ay 1By .

nce |xxy| = y both f and f target in x A y B y . It is easy to see that (33) together with (34) areuivalent to (15). Similarly, one shows that the fourth statement is true. eorem 3.5. We keep the notation and the assumptions from 3.3. The set R is a simple twisting system in if and only if the families and satisfy the following conditions:

i) If xB y Az is not an initial object then x A|xyz|Bz is not an initial object as well.i) If xB y Bz At is not an initial object in M , then |xy|yzt|| = |xzt| and Eqs. (29) and (30) hold.i) If xB y Az At is not an initial object in M , then ||xyz|zt| = |xyz| and Eqs. (31) and (32) hold.) If x A y is not an initial object in M , then |xxy| = y and Eqs. (33) and (34) hold.) If xB y is not an initial object in M , then |xyy| = x and Eqs. (35) and (36) hold.

oof. The condition (i) is a part of the denition of simple twisting systems. If xB y Bz At is not anitial object in M then we may assume that |xy|yzt|| = |xzt|. Thus, by Lemma 3.4, the relation (13)

d Eqs. (29) and (30) are equivalent. To conclude the proof we proceed in a similar way.


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ofX6. Matched pairs and the bicrossed product. Let :={ xyz }x,y,zS and := { xyz }x,y,zS be twomilies of maps as in 3.3. We shall say that the quintuple (A, B,,, | |) is a matched pair of M-tegories if and only if and satisfy the conditions (i)(v) from the above theorem. For a matchedir (A, B,,, | |) we have just seen that (R, | |) is a simple twisting system in M , where:= { x R yz }x,y,zS is the set of coalgebra morphisms which are dened by the formula (27). Hence,pposing that M is an S-distributive domain, we may construct the twisted tensor product A R B ,hich is an enriched category over M . We shall call it the bicrossed product of (A, B,,, | |) ande shall denote it by A B .oposition 3.7. The bicrossed product of a matched pair (A, B,,, | |) is enriched over the monoidaltegory M := Coalg(M ).

oof. Let {Ci}ii be a family of coalgebras in M . Let us assume that the underlying family of objectss a coproduct C := xSCi in M . Let {i}iI by the set of canonical inclusions into C . There areique maps : C C C and : C 1 such that

i = (i i) i and i = i,

r all i I , where i and i are the comultiplication and the counit of Ci . It is easy to see that,,) is a coalgebra in M . Note that, by the construction of the coalgebra structure on C , theclusion i is a coalgebra map, for any i I . Furthermore, let f i : Ci D be a coalgebra morphismr every i I . By the universal property of the coproduct there is a unique map f : C D in M ch that f i = f i , for all i I . It is not dicult to see that f is a morphism of coalgebras, so, {i}iI ) is the coproduct of {Ci}iI in M .In particular, x Au B y =

uS x Au B y has a unique coalgebra structure such that the inclusion x uz :

u B y x Au B y is a coalgebra map, for all x, y and u in S . Recall that for the construction of themposition map xc

yz : x Au B y Av Bz x Aw Bz one applies the universal property of the coproduct to

u,v}u,vS2 , where

fu,v = x |uyv|z xa|uyv|bz x Au R yv Bz.

nce A and B are M-categories and u Ryv is a coalgebra map, in view of the foregoing remarks,

follows that xcyz is a morphism in M , for all x, y, z S . The identity map of x in A B is the

algebra map x xx (1Ax 1Bx ). In conclusion, A B is enriched over M . Examples

In this section we give some examples of (simple) twisting systems. We start by considering these of Set-categories, that is usual categories.

1. Simple twisting systems of enriched categories over Set . The category Set is a braided monoidaltegory with respect to the cartesian product, its unit object being {}. Clearly, the empty set is theitial object in Set, and this category is an S-distributive domain, for any set S . We have alreadyticed that the () hypothesis holds in Set.Let C be an enriched category over Set. Thus, by denition, C is a category in the usual sense, thatxC y is a set for all x, y S . An element in xC y is regarded as a function from y to x.It is easy to see that a given set X can be seen in a unique way as a coalgebra in Set. As a matterfact the comultiplication and the counit of this coalgebra are given by the diagonal map : X X and the constant map : X {},(x) = x x and (x) = .


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Thapthbviously, any function f : X Y is morphism of coalgebras in Set. Consequently, any category Cay be seen as an enriched category over Coalg(Set),Our aim is to describe the simple twisting systems between two categories B and A. In view of

e foregoing discussion and of our results in the previous section, for any simple twisting system:= { x R yz }x,y,zS there is a unique matched pair (A, B,,, | |), and conversely. These structurese related each other by the formulae (26) and (27).Since A is an usual category, the composition of morphisms will be denoted in the traditional way

g , for any g x A y and g y Az (recall that the domain and the codomain of g are y and x,spectively). The same notation will be used for B . On the other hand, for any f xB y and g y Aze shall write

f g := x yz ( f , g) and f g = xyz ( f , g).nce the comultiplication in this case is always the diagonal map, and the counit is the constant map{}, the conditions of Theorem 3.5 and the following ones are equivalent.

i) If xB y Az is not empty then x A|xyz|Bz is not empty as well.i) For any ( f , f , g) xB y Bz At we have |xy|yzt|| = |xzt|, and(

f f ) g = f ( f g) and ( f f ) g = [ f ( f g)] ( f g).i) For any ( f , g, g) xB y A z At we have ||xyz|zt| = |xyt|, and

f (g g)= ( f g) g and f (g g)= ( f g) [( f g) g].v) For any g x A y we have |xxy| = y, and

1Bx g = g and 1Bx g = 1By .v) For any f xB y we have |xyy| = x, and

f 1Ay = 1Bx and f 1Ay = f .In this case the bicrossed product A B is the category whose hom-sets are given by

x(A B)y =uS

x Au B y .

e identity of x in A B is (1Ax ,1Bx ). For (g, f ) x Au B y and (g, f ) y A v Bz we have(g, f ) (g, f )= (g ( f g), ( f g) f ).

mark 4.2. R. Resebrugh and R.J. Wood showed that every twisting systems between two Set-tegories B and A is completely determined by a left action of B on A and a right action A on B . More precisely, given a twisting system R = { xR yz }x,y,zS and the morphisms f xB y and

y Az , then xR yz ( f , g) is an element in x Au Bz , where u is a certain element of S . Hence, there areique morphisms f g x Au and f g u Bz such that

xRyz (g, f ) = ( f g, f g).

e actions and must satisfy several compatibility conditions, which are similar to those thatpear in the above characterization of simple twisting systems. For details the reader is referred to

e second section of [RW].


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4.to3. The bicrossed product of two groupoids. We now assume that (A, B,,,| |) is a matchedir of groupoids. Recall that a groupoid is a category whose morphisms are invertible. We claimat A B is also a groupoid. Indeed, as in the case of monoids, one can show that a category is aoupoid if and only if every morphism is right invertible (or left invertible). Since

x(A B)y =uS

x Au B y,

is enough to prove that (g, f ) is right invertible, where g x Au and f u B y are arbitrary mor-isms. Therefore, we are looking for a pair (g, f ) y Av v Bx such that

g ( f g)= 1Ax and ( f g) f = 1Bx .nce g is an invertible morphism in x Au we get that f g = g1 u Ax . Since f is invertible too,

g = 1Ay g =(f 1 f ) g = f 1 ( f g)= f 1 g1.

s g y Av and f 1 g1 y A|yux| we must have v = |yux|. Thus we can now take

f = [ f ( f 1 g1)]1 |yux|Bx.4. The smash product category. We take M to be the monoidal category K-Mod, where K is a com-utative ring. Hence in this case we work with K-linear categories. Let H be a K-bialgebra. We deneenriched category H over K-Mod by setting xHx = H and xH y = 0, for x = y. The composition of

orphisms in H is given by the multiplication in H and the identity of x is the unit of H . For themultiplication of H we shall use the -notation

(h) =

h(1) h(2).

t A denote an H-module category, i.e. a category enriched in H-Mod. Thus H acts on x A y , for anyy S , and the composition and the identity maps in A are H-linear morphisms. Obviously, A isK-linear category. Our aim is to associate to A a simple twisting system R = { x R yz }x,y,zS . First wene | | : S3 S by |xyz| = z. Then, using the actions : H x Az x Az , we dene

x Rxz : H x Az x Az H, x Rxz(h f ) =

h(1) f h(2).

r x = y we take x R yz = 0. It is easy to see that R is a simple twisting system of K-linear categories.early K-Mod is S-distributive, for any set S . If K is a eld then K-Mod is a domain, so in this casee twisted tensor product of A and H with respect to R makes sense, cf. Theorem 2.14. It is callede smash product of A by H , and it is denoted by A#H . By denition, x(A#H)y = x A y H and

( f h) ( f h)= f (h(1) f ) h(2)h, (37)r any f x A y , f y Az and h,h H .

5. The semidirect product. Let A be a category. Let us suppose that (B, ,1) is a monoid that actsthe left on each x A y via : B x Ay x Ay .


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bee dene the category B so that xBx = B and xB y = , for x = y. The composition of morphisms isven by the multiplication in B . To this data we associate a matched pair (A, B, | |,,), setting g = f for any ( f , g) xBx Ay , and dening the function | | : S3 S by |xyz| = z. Note thatx = y then xB y Az is empty, so xyz and xyz coincide with the empty function. One shows that, B,,, | |) is a matched pair if and only if for any (g, g) x A y Az and f B

f (g g)= ( f g) ( f g),f 1Ax = 1Ax .

e corresponding bicrossed product will be denoted in this case by A B . If |S| = 1 then A can beentied with a monoid and A B is the usual semidirect product of two monoids. For this reasone shall call A B the semidirect product of A with B . Note that x(A B)y = x A y B . For any, f B and (g, g) x A y Az , the composition of morphisms in A B is given by

(g, f ) (g, f )= (g ( f g), f f ).6. Twisting systems between algebras in M . We now consider a twisting system R between two M-tegories B and A with the property that S = {x0}. Obviously, M is S-distributive. We shall use thetation A = x0 Ax0 and B = x0 Bx0 . The composition map a := x0ax0x0 and 1A := 1Ax0 dene an algebraructure on A. A similar notation will be used for the algebra corresponding to the M-category B .t R be a morphism from B A to A B .Since x0

x0x0 is the identity map of B A, by Proposition 2.5, we deduce that x0 Rx0x0 = R denes

twisting system between B and A if and only if R satises the relations (13)(16) with respect toe unique map | | : S3 S . In turn, they are equivalent to the following identities

R (b A) = (A b) (R B) (B R), (38)R (B a) = (a B) (A R) (R A), (39)

R (1B A)= A 1B , (40)R (B 1A) = 1A B. (41)

conclusion, in the case when |S| = 1, to give a twisting system between B and A is equivalent tove a twisting map between the algebras B and A, that is a morphism R which satises (38)(41).By applying Corollary 2.15 to a twisting map R : B A A B (viewed as a twisting systemtween two M-categories with one object) we get the twisted tensor algebra A R B . The unit of thisgebra is 1A 1B and its multiplication is given by

m = (a b) (A R B).

ote that, in view of the foregoing remarks, an algebra C in M factorizes through A and B if and onlyit is isomorphic to a twisted tensor algebra A R B , for a certain twisting map R .An algebra in the monoidal category K-Mod is by denition an associative and unital K-algebra.isted tensor K-algebras were investigated for instance in [Ma1,Tam,CSV,CIMZ,LPoV,JLPvO].Coalgebras over a eld K are algebras in the monoidal category (K-Mod)o . Hence a twisting maptween two coalgebras (A,A, A) and (B,B , B) is a K-linear mapR : A K B B K A


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anhich satises the equations that are obtained from (38)(41) by making the substitutions a := A ,:= B , 1A := A and 1B := B , and reversing the order of the factors with respect to the compositionM . For example (38) should be replaced with

(B K A) R = (B K R) (R K B) (A K B).bviously A R B is the K-coalgebra (A K B,,), where := A B and

= (A R B) (A K B).An algebra in -Mod- is called a -ring. Specializing M to -Mod- we nd the denition of

e twisted tensor -ring. Dually, -corings are algebras in (-Mod-)o . Thus in this particular casee are led to the construction of the twisted tensor -coring.By denition a monad on a category A is an algebra in [A, A]. If (F ,F , F ) and (G,G , G) are

onads in M , then a natural transformation

: G F F Gtises the relations (38)(41) if and only if is a distributive law, cf. [Be]. Let F 2 := F F . For everystributive law we get a monad (F G,, ), where := F G G = F G F and

:= F G F 2G FG = FG F G2 FG.istributive laws between comonads can be dened similarly, or working in [A, A]o .Finally, twisting maps in Opmon(M) have been considered in [BV]. In [BV] the authors denebimonad in M as an algebra in Opmon(M). Hence a twisting map between two bimonads is anmonoidal distributive law between the underlying monads. For any opmonoidal distributive law tween the bimonads G and F , there is a canonical bimonad structure on the endofunctor F G . SeeV, Section 4] for details.

7. Matched pairs of algebras in Coalg(M ). Let M denote the category of coalgebras in a braidedonoidal category (M ,,1,). By denition, a bialgebra in M is an algebra in M . We x two bial-bras (A,a,1A,A, A) and (A,b,1B ,B , B) in M and take R : B A A B to be a morphismM . By Lemma 3.1, there are the coalgebra maps : B A A and : B A B such that

R = ( ) BA and A,B ( ) BA = ( ) BA . (42)e have seen that R is a twisting map in M if and only if it satises the relations (38)(41). In viewLemma 3.4, these equations are equivalent to the fact that (A,) is a left B-module and (B,) isright A-module such that the following identities hold:

(b A) = b ( B) (B ) (B BA), (43) (B a) = a (A ) ( A) (BA A), (44)

(1B A) = A 1B , (45) (B 1A) = 1A B . (46)

denition, a matched pair of bialgebras in M consists of a left B-action (A,) and a right A-action,) in M such that the second equation in (42) and the relations (43)(46) hold. For a matchedir of bialgebras we shall use the notation (A, B,,).Summarizing, there is an one to-one-correspondence between twisting maps of bialgebras in M

d matched pairs of bialgebras in M . If (A, B,,) is a matched pair of bialgebras and R is the


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forresponding twisting map, then A R B will be called the bicrossed product of the bialgebras A and, and it will be denoted by A B . Note that A B is an algebra in M . Thus the bicrossed productA and B is a bialgebra in M . The unit of this bialgebra is 1A 1B and the multiplication is given

m = (a b) (A B) (A BA B).

a coalgebra A B is the tensor product coalgebra of A and B . We also conclude that a bialgebrain M factorizes through the sub-bialgebras A and B if and only if C = A B .As a rst application, let us take M to be the category of sets, which is braided with respect to

e braiding given by (X, Y ) (Y , X) and ( f , g) (g, f ), for any sets X, Y and any functions f , g .e have already noticed that there is a unique coalgebra structure on a given set X

(x) = (x, x), (x) = ,

here denotes the empty set; recall that the unit object in Set is {}. Hence an ordinary monoid,. an algebra in Set, has a natural bialgebra structure in this braided category. Moreover any twistingap R : B A A B between two monoids (A, ,1A) and (B, ,1B) is a twisting map of bialgebrasSet. Let (A, B,,) be the corresponding matched pair. One easily shows that the second condition(42) is always true. By notation, the functions and map ( f , g) B A to f g and f g ,spectively. Hence Eqs. (43)(46) are equivalent to the following ones:

(f f ) g = [ f ( f g)] ( f g),f (g g)= ( f g) [( f g) g],1B g = 1B and f 1A = 1A .

nce R( f , g) = ( f g, f g) the product of the monoid A B is dened by the formula(g, f ) (g, f )= (g( f g), ( f g) f ).

conclusion, a monoid C factorizes through A and B if and only if C = A B .We have seen that the bicrossed product of two groupoids is a groupoid. Thus, if A and B are

oups, then A B is a group as well. This result was proved by Takeuchi who introduced theatched pairs of groups in [Tak].We now consider the braided category K-Mod, whose braiding is the usual ip map. An algebraM , the monoidal category of K-coalgebras, is a bialgebra over the ring K, and conversely. Proceed-g as in the previous case, one shows that a twisting map R : B K A A K B of bialgebras isiquely determined by the coalgebra maps : B K A A and : B K A B via the formula

R( f g) =

( f(1) g(1)) ( f(2) g(2)).

sing the -notation, the second equation in (42) is true if and only if

( f(1) g(1)) ( f(2) g(2)) =

( f(2) g(2)) ( f(1) g(1)),r any f B and g A. On the other hand, Eqs. (43)(46) hold if and only if


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((i(i(gg) f =[g (g(1) f(1))](g(2) f(2)),

g ( f f )= (g(1) f(1))[(g(2) f(2)) f ],f 1A = B( f )1A,1B g = A(g)1B ,

r any f , f B and any g, g A. Thus, we rediscover the denition of matched pairs of bialgebrasd the formula for the multiplication of the double cross product, see [Ma2, Theorem 7.2.2]. Namely,

( f g)( f g)= f (g(1) f (1)) (g(2) f (2))g.8. Twisting systems between thin categories. Our aim now is to investigate the twisting systemstween two thin categories B and A. Recall that a category is thin if there is at most one morphismtween any couple of objects. Thus, for any x and y in S we have that either x A y = { x gy} or x A y ise empty set. Clearly, if x A y = { x gy} and y Az = {y gz} then x gy y gz = x gz . The identity morphismx is x gx . Similarly, if xB y is not empty then xB y = { x f y}.We x a twisting system R between B and A. It is dened by a family of maps

xRyz : xB y y Az

uS

Au Bz

at render commutative the diagrams in Fig. 3. We claim that R is simple. We need a function | : S3 S such that the image of xR yz is included into x A|xyz|Bz for all (x, y, z) S3. Let T S3note the set of all triples such that xB y Az = xB y y Az is not empty. Of course, if (x, y, z) ist in T then xR

yz is the empty function, so we can take |xyz| to be an arbitrary element in S . For

, y, z) T there exists |xyz| S such that

xRyz ( x f y, y gz) = ( xg|xyz|, |xyz| f z). (47)

ence xRyz is a function from xB y Az to x A|xyz|Bz . Note that x A|xyz|Bz is not empty in this case. For

y (x, y, z) S3 we set x R yz := xR yz . By Proposition 2.5 and Corollary 2.7 it follows that R is simple.e would like now to rewrite the conditions from the denition of simple twisting systems in anuivalent form, that only involves properties of T and | |. For instance let us show that the rstndition from Corollary 2.7 is equivalent to:

i) If (y, z, t) T and (x, y, |yzt|) T then |xy|yzt|| = |xzt|.

Indeed, if xB y Bz At is not empty then y Bz At = , so (y, z, t) T . We have already noticed thatA|yzt|Bt is not empty, provided that y Bz At is so. Since xB y and y A|yzt| are not empty it followsat xB y A|yzt| has the same property, that is (x, y, |yzt|) T . Therefore, if xB y Bz At is not empty then, z, t) T and (x, y, |yzt|) T . It is easy to see that the reversed implication is also true. Thus itmains to prove that Eq. (13) holds in the case when xB y Bz At is not empty. But this is obvious,

xRzt xbyz At and ( x A|xy|yzt||b|yzt|t ) xR y|yzt|Bt xB y Rzt have the same source xB y Bz At and the samerget x A|xzt|Bt = x A|xy|xyz||Bt . Both sets are singletons, so the above two morphisms must be equal.Proceeding in a similar way, we can prove that the other three conditions from Corollary 2.7 are

spectively equivalent to:

ii) If (x, y, z) T and (|xyz|, z, t) T then ||xyz|zt| = |xyt|.ii) If (x, x, y) T then |xxy| = y.

v) If (x, y, y) T then |xyy| = x.


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v) If (x, y, z) T then x A|xyz|Bz is not empty.

Hence for a twisting system R the function | | satises the above ve conditions. Conversely,t | | : S3 S denote a function such that the above ve conditions hold. Let xR yz be the emptynction, if (x, y, z) is not in T . Otherwise we dene xR

yz by the formula (47). In view of the foregoing

marks it is not dicult to see that R := { xR yz }x,y,zS is a simple twisting system.Clearly two functions | | and | | induce the same twisting system if and only if their restric-n to T are equal. Summarizing, we have just proved the theorem below.

eorem 4.9. Let A and B be thin categories. Let T denote the set of all triples (x, y, z) S3 such that xB y Aznot empty. If R is a twisting system between B and A then there exists a function | | : S3 S such thate conditions (i)(v) from the previous subsection hold, and conversely. Two functions | | and | | inducee same twisting system R if and only if their restriction to T are equal.

10. The twisted tensor product of thin categories. Let R be a twisting system between two thintegories B and A. By the preceding theorem, R is simple and there are T and || : S3 S suchat the conditions (i)(v) hold. In particular, the twisted tensor product of these categories exists. Bynition, we have x(A R B)y =

uS( x Au u Bz). We can identify this set with

x S y := {u S | x Au B y = }.

r u x S y and v y Sz we have (u, y, v) T . Thus u R yv ( u f y, y gv) = ( u g|uyv|, |uyv| f v), so the com-sition in A R B is given by

( xgu, u f y) ( y gv , v f z) = ( xgu u g|uyv|, |uyv| f v v f z) = ( xg|uyv|, |uyv| f z).

t C(S, T , | |) be the category whose objects are the elements of S . By denition, the hom-set(S, T , | |)y is x S y , the identity map of x S is x itself and the composition is given by

: x S y y Sz x Sz, u v = |uyv|.

erefore, we have just proved that A R B and C(S, T , | |) are isomorphic.

mark 4.11. Let C be a small category. Let S denote the set of objects in C . The category C factorizesrough two thin categories if and only if there are T S and | | : S3 S as in the previousbsection such that C is isomorphic to C(S, T , | |).

12. Twisting systems between posets. Any poset is a thin category, so we can apply Theorem 4.9 toaracterize a twisting system R between two posets B := (S,) and A := (S,). In this setting therresponding set T contains all (x, y, z) S3 such that x y and y z. For simplicity, we shall writeis condition as x y z. A similar notation will be used for arbitrarily long sequences of elementsS . For instance, x y z t u means that x y, y z, z t and t u. The function | | musttises the following conditions:

i) If x y z then x |xyz| z.i) If x y z t then |xy|yzt|| = |xzt|.i) If x y z t then ||xyz|zt| = |xyt|.v) If x y then |xxy| = y.

v) If x y then |xyy| = x.


sa

4.{1

NThxBThha

soItbea

A

1

Bympi

Nm

A

ThIDIn the case when the posets and are identical, an example of function | | : S3 S thattises the above conditions is given by |xyz| = z, if y = z, and |xyz| = x, otherwise.

13. Example of twisting map between two groupoids. Let A be a groupoid with two objects, S =,2}. The hom-sets of A are the following:

1A2 = {u}, 2A1 ={u1

}, 1A1 = {Id1}, 2A2 = {Id2}.

ote that A is thin. We set B := A and we take R to be a twisting system between B and A. Byeorem 4.9 there are T and | | : S3 S that satises the conditions (i)(v) in 4.8. Since all setsy Az = xB y y Az are nonempty it follows that T = S . Thus |xxy| = y and |xyy| = x, for all x, y S .ere are two triples (x, y, z) S3 such that x = y and y = z, namely (1,2,1) and (2,1,2). Hence weve to compute |121| and |212|. If we assume that |121| = 1, then

1 = |221| = 21|121|= |211| = 2,we get a contradiction. Thus |121| = 2, and proceeding in a similar way one proves that |212| = 1.is easy to check that | | satises the required conditions, so there is only one twisting map Rtween A and itself. Since A is a groupoid, the corresponding bicrossed product C := A A isgroupoid as well, see Subsection 4.3. By denition,

1C1 =

x{1,2}1Ax x A1 =

{(Id1, Id1),

(u,u1

)}.

nalogously one shows that

C2 ={(Id1,u), (u, Id2)

}, 2C1 =

{(Id2,u

1), (u1, Id1)} and 2C2 = {(Id2, Id2), (u1,u)}.construction of the twisting map R we get 1R21(u,u

1) 1A|121| |121|A1 = {(u,u1)}. The otheraps xR

yz can be determined analogously. The complete structure of this groupoid is given in the

cture below, where we used the notation f := (u,u1) and g := (Id1,u).

ote that f 2 = Id1 and g1 = (Id2,u1). Now we can say easily which arrow corresponds to a givenorphism in C , as in each home-set we have identied at least one element.

cknowledgments

The rst named author was nancially supported by the funds of the Contract POSDRU/6/1.5/S/12.e second named author was nancially supported by CNCSIS, Contract 560/2009 (CNCSIS code

_69).


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Factorizable enriched categories and applicationsIntroduction1 Preliminaries and notation2 Factorizable M-categories and twisting systems3 Matched pairs of enriched categories4 ExamplesAcknowledgmentsReferences

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