The Bicategory of Open Functors
aa r X i v : . [ m a t h . C T ] F e b The Bicategory of Open Functors
Alexandre Fernandez, Luidnel Maignan, and Antoine Spicher
Univ Paris Est Creteil, LACL, 94000, Creteil, France
February 17, 2021
Abstract
We want to replace categories, functors and natural transformationsby categories, open functors and open natural transformations. In analogywith open dynamical systems, the adjective open is added here to meanthat some external information is taken into account. For the particularuse of the authors, such an open functor is described by two components:a presheaf representing the possible external influences for each input, anda classical functor from the category of elements of this presheaf to thecategory of results. Considering the appropriate notion of compositionthen leads to a bicategory. This report describes this bicategory with aslittle auxiliary constructions as possible and gives all the details of allthe proofs needed to establish the bicategory, as explicitly as possible.Subsequent reports will give other presentations of this bicategory andcompare it to other existing constructions, e.g. spans, fibrations, pseudo-adjunctions, Kleisli bicategories of pseudo-monads, and profunctors (ordistributors).
We want to replace categories, functors and natural transformations by cate-gories, open functors and open natural transformations. In analogy with open dynamical systems, the adjective open is added here to mean that some externalinformation is taken into account.The origin of this will is that we found ourselve in a case where we neededto generalize the powerset monad P :
Set → Set in a setting where sets arereplaced by categories. More precisely, the goal was to represent a “local def-inition to global definition” phenomenon as a pointwise Kan extension as weusually do when thinking of spatially extended dynamical systems in terms of global transformations (Maignan and Spicher (2015); Fernandez et al. (2019)),except that, this time, some non-determinism came into play. The first trieswas to replace objects by sets of objects and morphisms by some sort of sets ofmorphisms. However, we found convincing examples showing that multiplicitiesare important, and these multiplicities come from the different ways and reasonswhy a result could be obtained. These “reasons” form the aforementioned ex-ternal information, not present in the input, but nonetheless taken into accountto produce the result. This external information can be the result of a choice,1ither from a random source, from an external scheduler, or simply from an-other part of the system that is not modeled but just taken as secondary inputinstead (Maignan and Gruau (2008, 2009, 2010); Arrighi et al. (2013)).Although it seemed clear that some settings should already exist to modelthis, there were questions on small details in each possible setting. In such acase, it is a good idea to first formalize the intuition as directly as possible, andonly then to try to map it in other settings in a precise formal way to fix thosedetails. This is exactly the purpose of this work. In this report, we only presentopen functors and open natural transformations as directly as possible, with aslittle reference to other auxiliary constructions as possible. Proofs are also madeat a very elementary and explicit level for various reasons, including the checkingof some details, the use of some parts as exercises, and the accessibility for someintented audience. Also we do not spend time on any foundational issues aboutsets of sets or categories of categories. The rest of the work should come asadditional reports hopefully, and we will try to update this report to link thedifferent reports together. A short description of the remaining content is givenat the end.
Let us compile here all the categorical definitions manipulated in this report,in order to make the proofs easier to check and to fix the notations. Note thatwe did our best to indicate every time a definition or proposition of the reportis used. So if a step of proof as no indication, its justification should be inthis section. For more information, one can consult MacLane (2013); Borceux(2008); B´enabou (1967).
We denote by
Cat the strict 2-category of categories, functors and naturaltransformations.By default, the composition of two arrows f : c → c ′ and g : c ′ → c ′′ in acategory C is denoted g · f : c → c ′′ , and the identity arrow of an object c ∈ C isdenoted id[ c ] : c → c . The collection of arrows from an object c to an object c ′ is denoted [ c, c ′ ]. In these notations, the category C is left implicit but can beretrieve from the context.The application of a functor F on an object c and an arrow f are denoted F ( c ) and F ( f ) respectively. Also, Id[ C ] : C → C is the identity functor of thecategory C and G ◦ F : C → E the composition of two functors F : C → D and G : D → E .The component at c ∈ C of a natural transformation θ : F = ⇒ G : C → D isdenoted θ [ c ] : F ( c ) → G ( c ). We also use the “arrow component” notation, i.e. for any f : c → c ′ , θ [ f ] : F ( c ) → G ( c ′ ) is defined as θ [ f ] = G ( f ) · θ [ c ] = θ [ c ′ ] · F ( f ).Also, ι [ F ] : F = ⇒ F : C → D is the identity natural transformation of a functor F : C → D , φ • θ : F → H : C → D the vertical composition of θ : F = ⇒ G : C →D and φ : G = ⇒ H : C → D , and φ ◦ θ : G ◦ F → G ′ ◦ F ′ : C → E the horizontalcomposition of θ : F = ⇒ F ′ : C → D and φ : G = ⇒ G ′ : D → E . Withour notations, the components of these compositions are ( φ • θ )[ c ] = φ [ c ] · θ [ c ]and ( φ ◦ θ )[ c ] = φ [ θ [ c ]] for any c ∈ C . Also ( φ • θ )[ g · f ] = φ [ g ] · θ [ f ] and2 φ ◦ θ )[ f ] = φ [ θ [ f ]] for any f : c → c ′ ∈ C . We finally recall the exchange law( φ ′ • φ ) ◦ ( θ ′ • θ ) = ( φ ′ ◦ θ ′ ) • ( φ ◦ θ ) and its special case ( φ ′ • φ ) ◦ ι [ F ] = ( φ ′ ◦ ι [ F ]) • ( φ ◦ ι [ F ]). Recall that the exchange law only expresses that compositionof functors is functorial.Note that we use brackets where the literature typically uses subscripts be-cause we need lots of nestings, e.g. ψ [ φ [ θ [ c ]]] instead of ψ φ θc (and this is a simpleinstance). Also, the “arrow component” notation is used as it is practical notto choose between equal compositions, e.g. for some θ : F = ⇒ F ′ , φ : G = ⇒ G ′ , ψ : H = ⇒ H ′ , we write ψ [ φ [ θ [ c ]]] instead of ψ [ G ′ ( F ′ ( c ))] · H ( φ [ F ′ ( c )]) · H ( G ( θ [ c ]))or H ′ ( φ [ F ′ ( c )]) · ψ [ G ( F ′ ( c ))] · H ( G ( θ [ c ])) or ψ [ G ′ ( F ′ ( c ))] · H ( G ′ ( θ [ c ])) · H ( φ [ F ( c )])or H ′ ( G ′ ( θ [ c ])) · φ [ G ′ ( F ( c ))] · H ( φ [ F ( c )]) or H ′ ( φ [ F ′ ( c )]) · H ′ ( G ( θ [ c ])) · ψ [ G ( F ( c ))]or H ′ ( G ′ ( θ [ c ])) · H ′ ( φ [ F ( c )]) · ψ [ G ( F ( c ))]. Diagrammatically, this particular ex-ample amounts to consider the naturality cube induced by the three naturaltransformations and to link directly the initial corner to the terminal corner bynaming the diagonal arrow, thanks to the arrow component notation, instead ofexpressing it by one of the six paths along the edges. This notation also makessome reasoning easier to write, e.g. the definition of horizontal composition ofnatural transformations. This lengthy paragraph is required as it prepare forthe proofs below. We denote by
Set the category of sets and functions, and denote by g ◦ f thecomposition of two functions f : X → Y and g : Y → Z . Its opposite category isdenoted Set op , its objects are simple sets, but an arrow f from Y to X in Set op is denoted f : Y X and its associated function f po : X → Y . This is doneto reduce the risk of a disordering. Given a functor F : C →
Set op , we denoteby ∫ F its category of elements whose objects are of the form ( c, x ) with c ∈ C and x ∈ F ( c ), and arrows of the form ( f, x ′ ) : ( c, F ( f ) po ( x ′ )) → ( c ′ , x ′ ) with f : c → c ′ and x ′ ∈ F ( c ′ ). The composition in ∫ F is ( g, x ′′ ) · ( f, x ′ ) := ( g · f, x ′′ )(with x ′ = F ( g ) po ( x ′′ ) for arrows to be composable) and the identity arrowsid[ c, x ] := (id[ c ] , x ). Given another functor G : C →
Set op and a naturaltransformation θ : F = ⇒ G , we denote by ( ∫ θ ) po : ∫ G → ∫ F the functordefined as ( ∫ θ ) po ( c, y ) = ( c, θ [ c ] po ( y )) and ( ∫ θ ) po ( f, y ′ ) = ( f, θ [ c ′ ] po ( y ′ )) with f : c → c ′ . We recall also that for any fixed category C , categories of elementsand functors of elements form a functor ∫ − from [ C , Set op ] to Cat op . C Set op FG ∫ F ∫ Gθ ( ∫ θ ) po ( c, θ [ c ] po ( y )) ( c ′ , θ [ c ′ ] po ( y ′ ))( f, θ [ c ′ ] po ( y ′ ))( c, y ) ( c ′ , y ′ )( f, y ′ )( ∫ θ ) po ( ∫ θ ) po ( ∫ θ ) po Note that we write id[ c, x ] and not id[( c, x )], and we write F ( c, x ) and not F (( c, x )) (as done for ( ∫ θ ) po ( c, y ) above). This shorthand is only used for anyapplication on objects or arrows from a category of elements. For an arbitrarypair, we will denote h x, y i and write F ( h x, y i ). To make things more readable,we rarely recall the domains and codomains of any arrow ( f, x ′ ). Our experience3s that these cumbersome data obscure things with too much details that areeasier to retrieve than to read. Finally, the common usage is to use [ C op , Set ]instead of [ C , Set op ] but we find it easier to switch convention for the particularneeds of this work, so that the arrows of C are always used in the same directionduring composition of open functors below, where the codomain of an openfunctor is also used as the domain of the following functor. A bicategory, a notion introduced by B´enabou (1967), is almost like a category,except that between two objects there is not just a collection of arrows but acategory of arrows. Similarly, the composition is not simply a function but afunctor. If it is still associative, left unital and right unital for the identities,we have a strict 2-category, e.g.
Cat . But since arrows form categories, we canhave different but isomorphic arrows, allowing to consider compositions that areassociative, left unital, and right unital only up to isomorphisms, as long as thechosen isomorphisms are coherent. Here, coherence means that the two ways totransform ( i ◦ ( h ◦ ( g ◦ f ))) into ((( i ◦ h ) ◦ g ) ◦ f ) should be equal, and similarlyfor the two transformations from ( g ◦ (id[ c ′ ] ◦ f )) to g ◦ f .More formally, a bicategory is composed of (1) a collection of 0-cells, (2)for each pair of 0-cells c, c ′ a category ( c → c ′ ) whose objects are called 1-cells,and arrows 2-cells (3) for each 0-cell c an identity 1-cell i [ c ] : c → c , (4) for eachtriplet of 0-cell c, c ′ , c ′′ a composition functor ◦ c,c ′ ,c ′′ from ( c ′ → c ′′ ) × ( c → c ′ ) to( c → c ′′ ), (5) for each pair of 0-cell c, c ′ a natural isomorphism lu c,c ′ , called leftunitor, from the functor ( i [ c ′ ] ◦ − ) to the identity functor ( − ) both from ( c → c ′ )to ( c → c ′ ) (6) for each pair of 0-cell c, c ′ a natural isomorphism ru c,c ′ , calledright unitor, from functor ( − ◦ i [ c ]) to identity functor ( − ) both from ( c → c ′ )to ( c → c ′ ) (7) for each quadruplet of 0-cell c, c ′ , c ′′ , c ′′′ a natural isomorphism a c,c ′ ,c ′′ ,c ′′′ , called associator, from functor ( − ◦ ( − ◦ − )) to functor (( − ◦ − ) ◦ − )both from ( c ′′ → c ′′′ ) × ( c ′ → c ′′ ) × ( c → c ′ ) to ( c → c ′′′ ). These data shouldbe coherent in the sense that this pentagon diagram in ( c → c ′′′′ ) and trianglediagram in ( c → c ′′ ) should commute. i ◦ ( h ◦ ( g ◦ f ))( i ◦ h ) ◦ ( g ◦ f ) i ◦ (( h ◦ g ) ◦ f )( i ◦ ( h ◦ g )) ◦ f (( i ◦ h ) ◦ g ) ◦ f a [ g ◦ f, h, i ] a [ f, g, i ◦ h ] id[ i ] ◦ a [ f, g, h ] a [ f, h ◦ g, i ] a [ i, h, g ] ◦ id[ f ] g ◦ ( i [ c ] ◦ f ) ( g ◦ i [ c ]) ◦ fg ◦ f a [ f, i [ c ] , g ]id[ g ] ◦ lu [ f ] ru [ g ] ◦ id[ f ]4he vocabulary “0-cell”, “1-cell”, “2-cell” is only used here to make the previousparagraph readable. Below, 0-cells are categories, 1-cells are open functors and2-cells are open natural transformations. Remark . Given two categories C and D , an open functor F from C to D needs to describe for each input c ∈ C the set of possible external interactionsavailable for c . In all examples considered up to now by the authors, it is thecase that for each f : c → c ′ , there is an associated “restriction” map in theother direction, namely from the set of interactions available for c ′ into the setof those available for c . The reason is that, in the considered examples, c can beviewed as a part of c ′ through f , so that each external interaction on the wholeof c ′ can be restricted to an interaction c alone. We therefore have a functor F α : C →
Set op describing these interactions and restrictions and its categoryof elements ∫ F α represents the complete information, inputs and their externalinformation/interactions. From this “complete” category, the data of a classicalfunctor F β : ∫ F α → D finishes the description of the open functor F . Definition 2.
Given two categories C and D , an open functor F from C to D ,denoted F : C ◦−→ D , is given by two functors F α : C →
Set op and F β : ∫ F α →D , as in the following diagram. C Set op D F α ∫ F α F β We denote J C , D K the collection of all open functors from C to D .Remark . Given two objects c, c ′ ∈ C , an arrow f : c → c ′ , and an element x ′ ∈ F α ( c ′ ), the arrow ( f, x ′ ) : ( c, F α ( f ) po ( x ′ )) → ( c ′ , x ′ ) in the category ofelements ∫ F α is sent to F β ( f, x ′ ) : F β ( c, F α ( f ) po ( x ′ )) → F β ( c ′ , x ′ ). We do notrecall systematically those domains and codomains below. Definition 4.
The identity open functor Id[ C ] : C ◦−→ C of a category C is givenby Id[ C ] α = ( c
7→ { ⋆ } , f id[ { ⋆ } ] op ) , and Id[ C ] β = (( c, ⋆ ) c, ( f, ⋆ ) f ) .Remark . More generally, any classical functor can be turned into an openfunctor by setting no external information, i.e. a singleton set for every input,as done in this last definition for the classical identity functor. We leave thisidea for a later report.Given two open functors F and G , their composition G ◦ F has a G ◦ F α com-ponent corresponding of pairs of the form h x, y i with x taken in F α and y takenin G α in the appropriate manner so that ( G ◦ F ) β ( c, h x, y i ) = G β ( F β ( c, x ) , y ). Definition 6.
Given three categories C , D and E and two open functors F : C ◦−→ D and G : D ◦−→ E , as in the following diagram, C Set op D Set op E F α ∫ F α F β G α ∫ G α G β heir composition G ◦ F : C ◦−→ D is given by: ( G ◦ F ) α ( c ) = a x ∈ F α ( c ) ( G α ◦ F β )( c, x ) , ( G ◦ F ) α ( f ) po ( h x ′ , y ′ i ) = h F α ( f ) po ( x ′ ) , ( G α ◦ F β )( f, x ′ ) po ( y ′ ) i , ( G ◦ F ) β ( c, h x, y i ) = G β ( F β ( c, x ) , y ) , and ( G ◦ F ) β ( f, h x ′ , y ′ i ) = G β ( F β ( f, x ′ ) , y ′ ) for any c, c ′ ∈ C , f : c → c ′ ∈ C , x ∈ F α ( c ) , x ′ ∈ F α ( c ′ ) such that x = F α ( f ) po ( x ′ ) , y ∈ ( G α ◦ F β )( c, x ) and y ′ ∈ ( G α ◦ F β )( c, x ′ ) such that y = G α ( F β ( f, x ′ )) po ( y ′ ) . Proposition 7.
All identity open functors and all open functor compositionsare indeed open functors.Proof.
Although obvious, one should still remember to check this.It is now easy to see that this composition is neither strictly associative,strictly left unital, nor strictly right unital. Indeed, the more we compose, themore we have pairs of pairs, even when we compose with open identity functors.In fact, the nesting of the pairs mimics the nesting of the compositions. Thisis why we have a bicategory and not a 2-category. So let us now introducethe open natural transformations first, in order to state the properties of thecomposition and its relations with identity open functors up to open naturalisomorphisms.
Remark . The following definitions of open natural transformations can bejustified as being intuitively induced by the diagrams. It becomes even obviousin other presentations of open functors, but this is out of the scope of this report.We only propose a comparison in Remark 15 below as an additional clue.
Definition 9.
Given two categories C and D and two open functors F, G : C ◦−→D , an open natural transformation θ from F to G , denoted θ : F ◦ = ⇒ G : C ◦−→D , is given by two natural transformations θ α : F α = ⇒ G α : C →
Set op and θ β : F β ◦ ( ∫ θ α ) po = ⇒ G β : ∫ G α → D , as in the following diagram. C Set op D F α ∫ F α F β G α ∫ G α G β θ α ( ∫ θ α ) po θ β We denote by J F, G K the collection of all open natural transformations betweenthem. Definition 10.
Given two categories C and D and an open functor F : C ◦−→ D ,its identity open natural transformation ι [ F ] : F ◦ = ⇒ F is the open naturaltransformation given by ι [ F ] α = ι [ F α ] and ι [ F ] β = ι [ F β ] . Set op D F α ∫ F α F β F α ∫ F α F β ι [ F ] (Id[ ∫ F α ]) po ι [ F β ] Definition 11.
Given two categories C and D , three open functors F, G, H : C ◦−→ D , and two open natural transformations θ : F ◦ = ⇒ G : C ◦−→ D and φ : G ◦ = ⇒ H : C ◦−→ D , their open vertical composition φ • θ : F ◦ = ⇒ H : C ◦ = ⇒ D is the open natural transformation given by ( φ • θ ) α = φ α • θ α and ( φ • θ ) β = φ β • ( θ β ◦ ι [ ∫ ( φ α ) po ]) . C Set op D F α ∫ F α F β G α ∫ G α G β H α ∫ H α H β θ α ( ∫ θ α ) po θ β φ α ( ∫ φ α ) po φ β Proposition 12.
All identity open natural transformations and all open verticalcompositions are indeed open natural transformations.Proof.
Although obvious, one should still remember to check this.
Proposition 13.
Given two categories C and D , the collection J C , D K of allopen functors forms a category where the collection of arrows between two openfunctors F, G : C ◦−→ D is given by the collection J F, G K of all open natural trans-formation, the composition by the open vertical composition • and the identityarrows by the identity open natural transformations.Proof. We need to prove that the open vertical composition is associative andhas the identity open natural transformations as neutral elements. For theassociativity, let us consider four open functors
F, G, H, I : C ◦−→ D , three opennatural transformations θ : F ◦ = ⇒ G , φ : G ◦ = ⇒ H , and ψ : H ◦ = ⇒ I , and provethat ( ψ • ( φ • θ )) = (( ψ • φ ) • θ ).( ψ • ( φ • θ )) α = ψ α • ( φ α • θ α ) (by Def. 11 two times)= ( ψ α • φ α ) • θ α = (( ψ • φ ) • θ ) α (by Def. 11 two times)( ψ • ( φ • θ )) β = ψ β • (( φ β • ( θ β ◦ ι [( ∫ φ α ) po ])) ◦ ι [( ∫ ψ α ) po ]) (by Def. 11 two times)= ψ β • (( φ β ◦ ι [( ∫ ψ α ) po ]) • (( θ β ◦ ι [( ∫ φ α ) po ]) ◦ ι [( ∫ ψ α ) po ]))= ( ψ β • ( φ β ◦ ι [( ∫ ψ α ) po ])) • ( θ β ◦ ( ι [( ∫ φ α ) po ] ◦ ι [( ∫ ψ α ) po ]))= ( ψ β • ( φ β ◦ ι [( ∫ ψ α ) po ])) • ( θ β ◦ ι [( ∫ φ α ) po ◦ ( ∫ ψ α ) po ])= ( ψ β • ( φ β ◦ ι [( ∫ ψ α ) po ])) • ( θ β ◦ ι [ ∫ ( ψ α • φ α ) po ])= (( ψ • φ ) • θ ) β (by Def. 11 two times)7or the neutrality, consider two open functors F, G : C ◦−→ D and an opennatural transformation θ : F ◦ = ⇒ G . Let us prove that θ • ι [ F ] = θ .( θ • ι [ F ]) α = θ α • ι [ F ] α (by Def. 11)= θ α • ι [ F α ] (by Def. 10)= θ α ( θ • ι [ F ]) β = θ β • ( ι [ F ] β ◦ ι [( ∫ θ α ) po ]) (by Def. 11)= θ β • ( ι [ F β ] ◦ ι [( ∫ θ α ) po ]) (by Def. 10)= θ β • ι [ F β ◦ ( ∫ θ α ) po ]= θ β Finally, let us prove that ι [ F ] • θ = θ .( ι [ G ] • θ ) α = ι [ G ] α • θ α (by Def. 11)= ι [ G α ] • θ α (by Def. 10)= θ α ( ι [ G ] • θ ) β = ι [ G ] β • ( θ β ◦ ι [( ∫ ι [ G ] α ) po ]) (by Def. 11)= ι [ G β ] • ( θ β ◦ ι [( ∫ ι [ G α ]) po ]) (by Def. 10)= θ β ◦ ι [(Id[ ∫ G α ]) po ]= θ β Definition 14.
Given three categories C , D and E , four open functors F, F ′ : C ◦−→ D and
G, G ′ : D ◦−→ E and two open natural transformations θ : F ◦ = ⇒ F ′ : C ◦−→ D and φ : G ◦ = ⇒ G ′ : D ◦−→ E , their open horizontal com-position φ ◦ θ : G ◦ F ◦ = ⇒ G ′ ◦ F ′ : C ◦−→ E is the open natural transfor-mation given by ( φ ◦ θ ) α [ c ] po ( h x ′ , y ′ i ) = h θ α [ c ] po ( x ′ ) , ( φ α ◦ θ β )[ c, x ′ ] po ( y ′ ) i and ( φ ◦ θ ) β [ c, h x ′ , y ′ i ] = φ β [ θ β [ c, x ′ ] , y ′ ] . C Set op D Set op E F α ∫ F α F β F ′ α ∫ F ′ α F ′ β θ α ( ∫ θ α ) po θ β G α ∫ G α G β G ′ α ∫ G ′ α G ′ β φ α ( ∫ φ α ) po φ β Remark . The reader is invited to compare these formulas with those of( G ◦ F ) α ( f ) po ( h x ′ , y ′ i ) and ( G ◦ F ) β ( f, h x ′ , y ′ i ) in Def 6, as a small justificationof the previous definition. Proposition 16.
All open horizontal compositions are indeed open naturaltransformations.Proof.
Although obvious, one should still remember to check this.
Proposition 17.
Given any three categories C , D and E , the composition ofopen functors together with the horizontal composition of open natural transfor-mations form a functor from J D , E K × J C , D K to J C , E K .Proof. We need to prove that identities and compositions are preserved. Forthe identities, let us consider two functors F : C ◦−→ D and G : D ◦−→ E . The8dentity open natural transformation of (
G, F ) in J D , E K × J C , D K is ( ι [ G ] , ι [ F ]).We want to show that ι [ G ] ◦ ι [ F ] = ι [ G ◦ F ].( ι [ G ] ◦ ι [ F ]) α [ c ] po ( h x, y i )= h ι [ F ] α [ c ] po ( x ) , ( ι [ G ] α ◦ ι [ F ] β )[ c, x ] po ( y ) i (by Def. 14)= h ι [ F α ][ c ] po ( x ) , ( ι [ G α ] ◦ ι [ F β ])[ c, x ] po ( y ) i (by Def. 10 three times)= h id[ F α ( c )] po ( x ) , id[( G α ◦ F β )( c, x )] po ( y ) i = h x, y i = id[( G ◦ F ) α ( c )] po ( h x, y i )= ι [( G ◦ F ) α ][ c ] po ( h x, y i )= ι [ G ◦ F ] α [ c ] po ( h x, y i ) (by Def. 10)( ι [ G ] ◦ ι [ F ]) β [ c, h x, y i ]= ι [ G ] β [ ι [ F ] β [ c, x ] , y ] (by Def. 14)= ι [ G β ][ ι [ F β ][ c, x ] , y ] (by Def. 10 two times)= ι [ G β ][id[ F β ( c, x )] , y ]= ι [ G β ][ F β ( c, x ) , y ]= id[ G β ( F β ( c, x ) , y )]= id[( G ◦ F ) β ( c, h x, y i )] (by Def. 6)= ι [ G ◦ F ] β [ c, h x, y i ] (by Def. 10)Consider six functors F, F ′ , F ′′ : C ◦−→ D and
G, G ′ , G ′′ : D ◦−→ E and fouropen natural transformations θ : F ◦ = ⇒ F ′ , θ ′ : F ′ ◦ = ⇒ F ′′ , φ : G ◦ = ⇒ G ′ and φ ′ : G ′ ◦ = ⇒ G ′′ , as in the following diagram. C Set op D Set op E F α ∫ F α F β F ′ α ∫ F ′ α F ′ β F ′′ α ∫ F ′′ α F ′′ β θ α ( ∫ θ α ) po θ β θ ′ α ( ∫ θ ′ α ) po θ ′ β G α ∫ G α G β G ′ α ∫ G ′ α G ′ β G ′′ α ∫ G ′′ α G ′′ β φ α ( ∫ φ α ) po φ β φ ′ α ( ∫ φ ′ α ) po φ ′ β We need to show that ( φ ′ • φ ) ◦ ( θ ′ • θ ) = ( φ ′ ◦ θ ′ ) • ( φ ◦ θ ). The equations beinglengthy, we start by showing that both terms have ( − ) α part which are pairs.We develop the ( − ) α part of the first.(( φ ′ • φ ) ◦ ( θ ′ • θ )) α [ c ] po ( h x ′′ , y ′′ i )= h ( θ ′ • θ ) α [ c ] po ( x ′′ ) , (( φ ′ • φ ) α ◦ ( θ ′ • θ ) β )[ c, x ′′ ] po ( y ′′ ) i (by Def. 14)9ow let us proceed similarly for the ( − ) α part of second term.(( φ ′ ◦ θ ′ ) • ( φ ◦ θ )) α [ c ] po ( h x ′′ , y ′′ i )= (( φ ′ ◦ θ ′ ) α • ( φ ◦ θ ) α )[ c ] po ( h x ′′ , y ′′ i ) (by Def. 11)= ( φ ◦ θ ) α [ c ] po (( φ ′ ◦ θ ′ ) α [ c ] po ( h x ′′ , y ′′ i ))= ( φ ◦ θ ) α [ c ] po ( h θ ′ α [ c ] po ( x ′′ ) , ( φ ′ α ◦ θ ′ β )[ c, x ′′ ] po ( y ′′ ) i ) (by Def. 14 here and below)= h θ α [ c ] po ( θ ′ α [ c ] po ( x ′′ )) , ( φ α ◦ θ β )[ c, θ ′ α [ c ] po ( x ′′ )] po (( φ ′ α ◦ θ ′ β )[ c, x ′′ ] po ( y ′′ )) i Now let us show that both pairs have the same first and second components.( θ ′ • θ ) α [ c ] po ( x ′′ )= ( θ ′ α • θ α )[ c ] po ( x ′′ ) (by Def. 11)= θ α [ c ] po ( θ ′ α [ c ] po ( x ′′ ))(( φ ′ • φ ) α ◦ ( θ ′ • θ ) β )[ c, x ′′ ] po ( y ′′ )= (( φ ′ α • φ α ) ◦ ( θ ′ β • ( θ β ◦ ι [( ∫ θ ′ α ) po ])))[ c, x ′′ ] po ( y ′′ ) (by Def. 14)= (( φ ′ α ◦ θ ′ β ) • ( φ α ◦ ( θ β ◦ ι [( ∫ θ ′ α ) po ])))[ c, x ′′ ] po ( y ′′ )= (( φ α ◦ ( θ β ◦ ι [( ∫ θ ′ α ) po ]))[ c, x ′′ ] po ◦ ( φ ′ α ◦ θ ′ β )[ c, x ′′ ] po )( y ′′ )= ( φ α ◦ θ β )[ c, θ ′ α [ c ] po ( x ′′ )] po (( φ ′ α ◦ θ ′ β )[ c, x ′′ ] po ( y ′′ ))Now, the ( − ) β part of both terms are compositions.(( φ ′ • φ ) ◦ ( θ ′ • θ )) β [ c, h x ′′ , y ′′ i ]= ( φ ′ • φ ) β [( θ ′ • θ ) β [ c, x ′′ ] , y ′′ ] (by Def. 14)= ( φ ′ β • ( φ β ◦ ι [( ∫ φ ′ α ) po ]))[( θ ′ β • ( θ β ◦ ι [( ∫ θ ′ α ) po ]))[ c, x ′′ ] , y ′′ ] (by Def. 11 two times)= ( φ ′ β • ( φ β ◦ ι [( ∫ φ ′ α ) po ]))[ θ ′ β [ c, x ′′ ] · θ β [ c, θ ′ α [ c ] po ( x ′′ )] , y ′′ ]= ( φ ′ β • ( φ β ◦ ι [( ∫ φ ′ α ) po ]))[( θ ′ β [ c, x ′′ ] , y ′′ ) · ( θ β [ c, θ ′ α [ c ] po ( x ′′ )] , G ′′ α ( θ ′ β [ c, x ′′ ]) po ( y ′′ ))]= φ ′ β [ θ ′ β [ c, x ′′ ] , y ′′ ] · ( φ β ◦ ι [( ∫ φ ′ α ) po ])[ θ β [ c, θ ′ α [ c ] po ( x ′′ )] , G ′′ α ( θ ′ β [ c, x ′′ ]) po ( y ′′ )](( φ ′ ◦ θ ′ ) • ( φ ◦ θ )) β [ c, h x ′′ , y ′′ i ]= (( φ ′ ◦ θ ′ ) β • (( φ ◦ θ ) β ◦ ι [( ∫ ( φ ′ ◦ θ ′ ) α ) po ]))[ c, h x ′′ , y ′′ i ] (by Def. 11)= ( φ ′ ◦ θ ′ ) β [ c, h x ′′ , y ′′ i ] · (( φ ◦ θ ) β ◦ ι [( ∫ ( φ ′ ◦ θ ′ ) α ) po ])[ c, h x ′′ , y ′′ i ]Let us show that both compositions have the same first and second members. φ ′ β [ θ ′ β [ c, x ′′ ] , y ′′ ]= ( φ ′ ◦ θ ′ ) β [ c, h x ′′ , y ′′ i ] (by Def. 14)( φ β ◦ ι [( ∫ φ ′ α ) po ])[ θ β [ c, θ ′ α [ c ] po ( x ′′ )] , G ′′ α ( θ ′ β [ c, x ′′ ]) po ( y ′′ )]= φ β [ θ β [ c, θ ′ α [ c ] po ( x ′′ )] , φ ′ α [ F ′ β [ c, θ ′ α [ c ] po ( x ′′ )]] po ( G ′′ α ( θ ′ β [ c, x ′′ ]) po ( y ′′ ))]= φ β [ θ β [ c, θ ′ α [ c ] po ( x ′′ )] , ( φ ′ α [ F ′ β [ c, θ ′ α [ c ] po ( x ′′ )]] po ◦ G ′′ α ( θ ′ β [ c, x ′′ ] po ))( y ′′ )]= ( φ ◦ θ ) β [ c, h θ ′ α [ c ] po ( x ′′ ) , φ ′ α [ θ ′ β [ c, x ′′ ]] po ( y ′′ ) i ]= ( φ ◦ θ ) β [ c, h θ ′ α [ c ] po ( x ′′ ) , ( φ ′ α ◦ θ ′ β )[ c, x ′′ ] po ( y ′′ ) i ] (by Def. 14)= ( φ ◦ θ ) β [ c, ( φ ′ ◦ θ ′ ) α [ c ] po ( h x ′′ , y ′′ i )] (by Def. 14)= (( φ ◦ θ ) β ◦ ι [( ∫ ( φ ′ ◦ θ ′ ) α ) po ])[ c, h x ′′ , y ′′ i ]10 efinition 18. Given two categories C and D , their left unitor is the (clas-sical) natural isomorphism lu CD : (Id[ D ] ◦ − ) = ⇒ ( − ) : J C , D K → J C , D K with component lu CD F : Id[ D ] ◦ F ◦ = ⇒ F defined as lu CD F α [ c ] po ( x ) = h x, ⋆ i and lu CD F β [ c, x ] = id[ F β ( c, x )] . Similarly, their right unitor is the (classi-cal) natural isomorphism ru CD : ( − ◦ Id[ C ]) = ⇒ ( − ) : J C , D K → J C , D K withcomponent ru CD F : F ◦ Id[ C ] ◦ = ⇒ F defined as ru CD F α [ c ] po ( x ) = h ⋆, x i and ru CD F β [ c, x ] = id[ F β ( c, x )] . Proposition 19.
These are indeed a natural isomorphisms.Proof.
These maps are trivially isomorphism. Now, consider some open naturaltransformation θ : F ◦ = ⇒ G : C ◦−→ D . We need to prove that the naturalitysquare commutes, i.e. lu CD G • ( ι [Id[ D ]] ◦ θ ) = θ • lu CD F . For each component( − ) α and ( − ) β , we show that both terms reduce to the same expression.( θ • lu CD F ) α [ c ] po ( y ) = ( θ α • lu CD F α )[ c ] po ( y ) (by Def. 11)= ( lu CD F α [ c ] po ◦ θ α [ c ] po )( y )= lu CD F α [ c ] po ( θ α [ c ] po ( y ))= h θ α [ c ] po ( y ) , ⋆ i (by Def. 18)( lu CD G • ( ι [Id[ D ]] ◦ θ )) α [ c ] po ( y )= ( lu CD Gα • ( ι [Id[ D ]] ◦ θ ) α )[ c ] po ( y ) (by Def. 11)= (( ι [Id[ D ]] ◦ θ ) α [ c ] po ◦ lu CD Gα [ c ] po )( y )= ( ι [Id[ D ]] ◦ θ ) α [ c ] po ( h y, ⋆ i ) (by Def. 18)= h θ α [ c ] po ( y ) , ( ι [Id[ D ]] α ◦ θ β )[ c, y ] po ( ⋆ ) i (by Def. 14)= h θ α [ c ] po ( y ) , ⋆ i ( θ • lu CD F ) β [ c, y ] = ( θ β • ( lu CD F β ◦ ι [( ∫ θ α ) po ]))[ c, y ] (by Def. 11)= θ β [ c, y ] · lu CD F β [ c, θ α [ c ] po ( y )]= θ β [ c, y ] · id[ F β ( c, θ α [ c ] po ( y ))] (by Def. 18)= θ β [ c, y ]( lu CD G • ( ι [Id[ D ]] ◦ θ )) β [ c, y ]= ( lu CD Gβ • (( ι [Id[ D ]] ◦ θ ) β ◦ ι [( ∫ lu CD Gα ) po ]))[ c, y ] (by Def. 11)= lu CD Gβ [ c, y ] · ( ι [Id[ D ]] ◦ θ ) β [ c, lu CD Gα [ c ] po ( y )]= id[ G β ( c, y )] · ( ι [Id[ D ]] ◦ θ ) β [ c, h y, ⋆ i ] (by Def. 18 two times)= ( ι [Id[ D ]] ◦ θ ) β [ c, h y, ⋆ i ]= ι [Id[ D ] β ][ θ β [ c, y ] , ⋆ ] (by Def. 14 and 10)= θ β [ c, y ] (by Def. 4)Similarly for the right unitor, we need to prove that the naturality square com-mutes, i.e. ru CD G • ( θ ◦ ι [Id[ D ]]) = θ • ru CD F .( θ • ru CD F ) α [ c ] po ( y ) = ( θ α • ru CD F α )[ c ] po ( y ) (by Def. 11)= ( ru CD F α [ c ] po ◦ θ α [ c ] po )( y )= ru CD F α [ c ] po ( θ α [ c ] po ( y ))= h ⋆, θ α [ c ] po ( y ) i (by Def. 18)11 ru CD G • ( θ ◦ ι [Id[ D ]])) α [ c ] po ( y )= ( ru CD Gα • ( θ ◦ ι [Id[ D ]]) α )[ c ] po ( y ) (by Def. 11)= (( θ ◦ ι [Id[ D ]]) α [ c ] po ◦ ru CD Gα [ c ] po )( y )= ( θ ◦ ι [Id[ D ]]) α [ c ] po ( h ⋆, y i ) (by Def. 18)= h ι [Id[ D ]] α [ c ] po ( ⋆ ) , ( θ α ◦ ι [Id[ D ]] β )[ c, y ] po ( ⋆ ) i (by Def. 14)= h ⋆, θ α [ c ] po ( y ) i (by Def. 4 and 10)( θ • ru CD F ) β [ c, y ] = ( θ β • ( ru CD F β ◦ ι [( ∫ θ α ) po ]))[ c, y ] (by Def. 11)= θ β [ c, y ] · ru CD F β [ c, θ α [ c ] po ( y )]= θ β [ c, y ] · id[ F β ( c, θ α [ c ] po ( y ))] (by Def. 18)= θ β [ c, y ]( ru CD G • ( θ ◦ ι [Id[ D ]])) β [ c, y ]= ( ru CD Gβ • (( θ ◦ ι [Id[ D ]]) β ◦ ι [( ∫ ru CD Gα ) po ]))[ c, y ] (by Def. 11)= ru CD Gβ [ c, y ] · ( θ ◦ ι [Id[ D ]]) β [ c, ru CD Gα [ c ] po ( y )]= id[ G β ( c, y )] · ( θ ◦ ι [Id[ D ]]) β [ c, h ⋆, y i ] (by Def. 18 two times)= ( θ ◦ ι [Id[ D ]]) β [ c, h ⋆, y i ]= θ β [ ι [Id[ D ] β ][ c, ⋆ ] , y ] (by Def. 14 and 10)= θ β [Id[ D ] β ( c, ⋆ ) , y ]= θ β [ c, y ] (by Def. 4) Definition 20.
Given four open categories B , C , D and E , and three openfunctors F : B ◦−→ C , G : C ◦−→ D , and H : D ◦−→ E , their associator is the (classical) natural isomorphism a BCDE : ( − ◦ ( − ◦ − )) → (( − ◦ − ) ◦ − ) : J D , E K × J C , D K × J B , C K → J B , E K with component a BCDE [ F, G, H ] : H ◦ ( G ◦ F ) ◦−→ ( H ◦ G ) ◦ F : B ◦−→ E defined as ( a BCDE [ F, G, H ]) α [ b ] po ( h x, h y, z ii ) = hh x, y i , z i and ( a BCDE [ F, G, H ]) β [ b, h x, h y, z ii ] = id[ H β ( G β ( F β ( b, x ) , y ) , z )] (or equivalently ( a BCDE [ F, G, H ]) β = ι [(( H ◦ G ) ◦ F ) β ] ). Proposition 21.
This is indeed a natural isomorphism.Proof.
These maps are trivially isomorphism. Now, consider three open naturaltransformations θ : F ◦ = ⇒ F ′ : B ◦−→ C , φ : G ◦ = ⇒ G ′ : C ◦−→ D , and ψ : H ◦ = ⇒ H ′ : D ◦−→ E . We need to prove that the naturality square commutes, i.e. a ′ • ( ψ ◦ ( φ ◦ θ )) = (( ψ ◦ φ ) ◦ θ ) • a where a stands for a BCDE [ F, G, H ] and a ′ a BCDE [ F ′ , G ′ , H ′ ].( a ′ • ( ψ ◦ ( φ ◦ θ ))) α [ b ] po ( h x ′ , h y ′ , z ′ ii )= (( ψ ◦ ( φ ◦ θ )) α [ b ] po ◦ a ′ α [ b ] po )( h x ′ , h y ′ , z ′ ii ) (by Def. 11)= ( ψ ◦ ( φ ◦ θ )) α [ b ] po ( a ′ α [ b ] po ( h x ′ , h y ′ , z ′ ii ))= ( ψ ◦ ( φ ◦ θ )) α [ b ] po ( hh x ′ , y ′ i , z ′ i ) (by Def. 20)= h ( φ ◦ θ ) α [ b ] po ( h x ′ , y ′ i ) , ( ψ α ◦ ( φ ◦ θ ) β )[ b, h x ′ , y ′ i ] po ( z ′ ) i (by Def. 14)= h ( φ ◦ θ ) α [ b ] po ( h x ′ , y ′ i ) , ψ α [( φ ◦ θ ) β [ b, h x ′ , y ′ i ]] po ( z ′ ) i = hh θ α [ b ] po ( x ′ ) , ( φ α ◦ θ β )[ b, x ′ ] po ( y ′ ) i , ψ α [ φ β [ θ β [ b, x ′ ] , y ′ ]] po ( z ′ ) i (by Def. 14)= hh θ α [ b ] po ( x ′ ) , φ α [ θ β [ b, x ′ ]] po ( y ′ ) i , ψ α [ φ β [ θ β [ b, x ′ ] , y ′ ]] po ( z ′ ) i ((( ψ ◦ φ ) ◦ θ ) • a ) α [ b ] po ( h x ′ , h y ′ , z ′ ii )= ( a α [ b ] po ◦ (( ψ ◦ φ ) ◦ θ ) α [ b ] po )( h x ′ , h y ′ , z ′ ii ) (by Def. 11)= a α [ b ] po ((( ψ ◦ φ ) ◦ θ ) α [ b ] po ( h x ′ , h y ′ , z ′ ii ))= a α [ b ] po ( h θ α [ b ] po ( x ′ ) , (( ψ ◦ φ ) α ◦ θ β )[ b, x ′ ] po ( h y ′ , z ′ i ) i ) (by Def. 14)= a α [ b ] po ( h θ α [ b ] po ( x ′ ) , ( ψ ◦ φ ) α [ θ β [ b, x ′ ]] po ( h y ′ , z ′ i ) i ) (and Def. 14 gives)= a α [ b ] po ( h θ α [ b ] po ( x ′ ) , h φ α [ θ β [ b, x ′ ]] po ( y ′ ) , ( ψ α ◦ φ β )[ θ β [ b, x ′ ] , y ′ ] po ( z ′ ) ii )= a α [ b ] po ( h θ α [ b ] po ( x ′ ) , h φ α [ θ β [ b, x ′ ]] po ( y ′ ) , ψ α [ φ β [ θ β [ b, x ′ ] , y ′ ]] po ( z ′ ) ii )= hh θ α [ b ] po ( x ′ ) , φ α [ θ β [ b, x ′ ]] po ( y ′ ) i , ψ α [ φ β [ θ β [ b, x ′ ] , y ′ ]] po ( z ′ ) i (by Def. 20)( a ′ • ( ψ ◦ ( φ ◦ θ ))) β [ b, h x ′ , h y ′ , z ′ ii ]= ( a ′ β • (( ψ ◦ ( φ ◦ θ )) β ◦ ι [( ∫ a ′ α ) po ]))[ b, h x ′ , h y ′ , z ′ ii ] (by Def. 11)= a ′ β [ b, h x ′ , h y ′ , z ′ ii ] · (( ψ ◦ ( φ ◦ θ )) β ◦ ι [( ∫ a ′ α ) po ])[ b, h x ′ , h y ′ , z ′ ii ]= id[ . . . ] · (( ψ ◦ ( φ ◦ θ )) β ◦ ι [( ∫ a ′ α ) po ])[ b, h x ′ , h y ′ , z ′ ii ] (by Def. 20)= (( ψ ◦ ( φ ◦ θ )) β ◦ ι [( ∫ a ′ α ) po ])[ b, h x ′ , h y ′ , z ′ ii ]= ( ψ ◦ ( φ ◦ θ )) β [ b, a ′ α [ b ] po ( h x ′ , h y ′ , z ′ ii )]= ( ψ ◦ ( φ ◦ θ )) β [ b, hh x ′ , y ′ i , z ′ i ] (by Def. 20)= ψ β [( φ ◦ θ ) β [ b, h x ′ , y ′ i ] , z ′ ] (by Def. 14)= ψ β [ φ β [ θ β [ b, x ′ ] , y ′ ] , z ′ ] (by Def. 14)((( ψ ◦ φ ) ◦ θ ) • a ) β [ b, h x ′ , h y ′ , z ′ ii ]= ((( ψ ◦ φ ) ◦ θ ) β • ( a β ◦ ι [( ∫ (( ψ ◦ φ ) ◦ θ ) α ) po ]))[ b, h x ′ , h y ′ , z ′ ii ] (by Def. 11)= (( ψ ◦ φ ) ◦ θ ) β [ b, h x ′ , h y ′ , z ′ ii ] · ( a β ◦ ι [( ∫ (( ψ ◦ φ ) ◦ θ ) α ) po ])[ b, h x ′ , h y ′ , z ′ ii ]= (( ψ ◦ φ ) ◦ θ ) β [ b, h x ′ , h y ′ , z ′ ii ] · a β [ b, (( ψ ◦ φ ) ◦ θ ) α [ b ] po ( h x ′ , h y ′ , z ′ ii )]= (( ψ ◦ φ ) ◦ θ ) β [ b, h x ′ , h y ′ , z ′ ii ] · id[ . . . ] (by Def. 20)= (( ψ ◦ φ ) ◦ θ ) β [ b, h x ′ , h y ′ , z ′ ii ]= ( ψ ◦ φ ) β [ θ β [ b, x ′ ] , h y ′ , z ′ i ] (by Def. 14)= ψ β [ φ β [ θ β [ b, x ′ ] , y ′ ] , z ′ ] (by Def. 14)13 roposition 22. The collection of all categories together with open functorsand open natural transformations, in the precise sense defined up to now, forma bicategory.Proof.
The only remaining tasks are to prove the coherence of the associatorvia the pentagon identity, and the coherence of the left and right unitors via thetriangle identity. Let us begin by the pentagon identity, which amounts to thecommutativity of the following diagram for any categories A , B , C , D , and E andany open functors F : A ◦−→ B , G : B ◦−→ C , H : C ◦−→ D , I : D ◦−→ E . I ◦ ( H ◦ ( G ◦ F ))( I ◦ H ) ◦ ( G ◦ F ) I ◦ (( H ◦ G ) ◦ F )( I ◦ ( H ◦ G )) ◦ F (( I ◦ H ) ◦ G ) ◦ F a [ G ◦ F, H, I ] a [ F, G, I ◦ H ] ι [ I ] ◦ a [ F, G, H ] a [ F, H ◦ G, I ] a [ I, H, G ] ◦ ι [ F ]For the ( − ) α components first and then for the ( − ) β components, let us showthat both paths reduces to the same expression. For any k ∈ A and e ∈ (( I ◦ H ) ◦ G ) ◦ F ) α we have the following for the three step path.(( a [ G, H, I ] ◦ ι [ F ]) • a [ F, H ◦ G, I ] • ( ι [ I ] ◦ a [ F, G, H ])) α [ k ] po ( e )= ( a [ G, H, I ] ◦ ι [ F ]) α • a [ F, H ◦ G, I ] α • ( ι [ I ] ◦ a [ F, G, H ]) α )[ k ] po ( e ) (by Def. 11)= ( ι [ I ] ◦ a [ F, G, H ]) α [ k ] po ◦ a [ F, H ◦ G, I ] α [ k ] po ◦ ( a [ G, H, I ] ◦ ι [ F ]) α [ k ] po ( e )Now e has the form h w, h x, h y, z iii for some w ∈ F α , x ∈ G α , y ∈ H α , z ∈ I α .Let us show that the functions in the last line above simply does what we expect.( a [ G, H, I ] ◦ ι [ F ]) α [ k ] po ( h w, h x, h y, z iii )= h ι [ F ] α [ k ] po ( w ) , ( a [ G, H, I ] α ◦ ι [ F ] β )[ k, w ] po ( h x, h y, z ii ) i ) (by Def. 14)= h ι [ F ] α [ k ] po ( w ) , a [ G, H, I ] α [ ι [ F ] β [ k, w ]] po ( h x, h y, z ii ) i )= h ι [ F ] α [ k ] po ( w ) , a [ G, H, I ] α [id[ F β ( k, w )]] po ( h x, h y, z ii ) i ) (by Def. 10)= h ι [ F ] α [ k ] po ( w ) , a [ G, H, I ] α [ F β ( k, w )] po ( h x, h y, z ii ) i )= h ι [ F ] α [ k ] po ( w ) , hh x, y i , z ii (by Def. 20)= h w, hh x, y i , z ii (by Def. 10) a [ F, H ◦ G, I ] α [ k ] po ( h w, hh x, y i , z ii )= hh w, h x, y ii , z i (by Def. 20)14 ι [ I ] ◦ a [ F, G, H ]) α [ k ] po ( hh w, h x, y ii , z i )= h a [ F, G, H ] α [ k ] po ( h w, h x, y ii ) , ( ι [ I ] α ◦ a [ F, G, H ] β )[ k, h w, h x, y ii ] po ( z ) i (by Def. 14)= hhh w, x i , y i , ( ι [ I ] α ◦ a [ F, G, H ] β )[ k, h w, h x, y ii ] po ( z ) i (by Def. 20)= hhh w, x i , y i , ι [ I ] α [ a [ F, G, H ] β [ k, h w, h x, y ii ]] po ( z ) i = hhh w, x i , y i , ι [ I ] α [id[ H β ( G β ( F β ( k, w ) , x ) , y )]] po ( z ) i (by Def. 20)= hhh w, x i , y i , ι [ I ] α [ H β ( G β ( F β ( k, w ) , x ) , y )] po ( z ) i = hhh w, x i , y i , z i (by Def. 10)The ( − ) α component of the two step path indeed leads to the same result.( a [ F, G, I ◦ H ] • a [ G ◦ F, H, I ]) α [ k ] po ( h w, h x, h y, z iii )= ( a [ F, G, I ◦ H ] α • a [ G ◦ F, H, I ] α )[ k ] po ( h w, h x, h y, z iii ) (by Def. 11)= ( a [ G ◦ F, H, I ] α [ k ] po ◦ a [ F, G, I ◦ H ] α [ k ] po )( h w, h x, h y, z iii )= a [ G ◦ F, H, I ] α [ k ] po ( a [ F, G, I ◦ H ] α [ k ] po ( h w, h x, h y, z iii ))= a [ G ◦ F, H, I ] α [ k ] po ( hh w, x i , h y, z ii ) (by Def. 20)= hhh w, x i , y i , z i (by Def. 20)Now let us compute the ( − ) β component of the three step path, step by step,using freely the previously established equations.(( a [ G, H, I ] ◦ ι [ F ]) • θ ) β [ k, h w, h x, h y, z iii ] (and Def. 11 gives)= (( a [ G, H, I ] ◦ ι [ F ]) β • ( θ β ◦ ι [( ∫ ( a [ G, H, I ] ◦ ι [ F ]) α ) po ]))[ k, h w, h x, h y, z iii ]= ( a [ G, H, I ] ◦ ι [ F ]) β [ k, h w, h x, h y, z iii ] · θ β [ k, h w, hh x, y i , z ii ]= a [ G, H, I ] β [ ι [ F ] β [ k, w ] , h x, h y, z ii ]] · θ β [ k, h w, hh x, y i , z ii ] (by Def. 14)= a [ G, H, I ] β [id[ F β ( k, w )] , h x, h y, z ii ]] · θ β [ k, h w, hh x, y i , z ii ] (by Def. 10)= a [ G, H, I ] β [ F β ( k, w ) , h x, h y, z ii ]] · θ β [ k, h w, hh x, y i , z ii ]= id[ I β ( H β ( G β ( F β ( k, w ) , x ) , y ) , z )] · θ β [ k, h w, hh x, y i , z ii ] (by Def. 20)= θ β [ k, h w, hh x, y i , z ii ]( a [ F, H ◦ G, I ] • θ ′ ) β [ k, h w, hh x, y i , z ii ]= ( a [ F, H ◦ G, I ] β • ( θ ′ β ◦ ι [( ∫ a [ F, H ◦ G, I ] α ) po ]))[ k, h w, hh x, y i , z ii ] (by Def. 11)= a [ F, H ◦ G, I ] β [ k, h w, hh x, y i , z ii ] · θ ′ β [ k, hh w, h x, y ii , z i ]= id[ I β (( H ◦ G ) β ( F β ( k, w ) , h x, y i , z )] · θ ′ β [ k, hh w, h x, y ii , z i ] (by Def. 20)= θ ′ β [ k, hh w, h x, y ii , z i ]( ι [ I ] ◦ a [ F, G, H ]) β [ k, hh w, h x, y ii , z i ]= ι [ I ] β [ a [ F, G, H ] β [ k, h w, h x, y ii ] , z ] (by Def. 14)= ι [ I ] β [id[ H β ( G β ( F β ( k, w ) , x ) , y )] , z ]] (by Def. 20)= ι [ I ] β [ H β ( G β ( F β ( k, w ) , x ) , y ) , z ]= id[ I β ( H β ( G β ( F β ( k, w ) , x ) , y ) , z )] (by Def. 10)15ombining these three equations by letting θ of the first equation be the secondequation and θ ′ of the second equation be the third equation, we obtain thefollowing one.(( a [ G, H, I ] ◦ ι [ F ]) • ( a [ F, H ◦ G, I ] • ( ι [ I ] ◦ a [ F, G, H ]))) β [ k, h w, h x, h y, z iii ]= ( a [ F, H ◦ G, I ] • ( ι [ I ] ◦ a [ F, G, H ])) β [ k, h w, hh x, y i , z ii ]= ( ι [ I ] ◦ a [ F, G, H ]) β [ k, hh w, h x, y ii , z i ]= id[ I β ( H β ( G β ( F β ( k, w ) , x ) , y ) , z )]Now let us use the same strategy for the ( − ) β components of the two step path.( a [ F, G, I ◦ H ] • θ ) β [ k, h w, h x, h y, z iii ]= ( a [ F, G, I ◦ H ] β • ( θ β ◦ ι [( ∫ a [ F, G, I ◦ H ] α ) po ]))[ k, h w, h x, h y, z iii ] (by Def. 11)= a [ F, G, I ◦ H ] β [ k, h w, h x, h y, z iii ] · θ β [ k, hh w, x i , h y, z ii ]= id[( I ◦ H ) β ( G β ( F β ( k, w ) , x ) , h y, z i )] · θ β [ k, hh w, x i , h y, z ii ] (by Def. 20)= θ β [ k, hh w, x i , h y, z ii ] a [ G ◦ F, H, I ] β [ k, hh w, x i , h y, z ii ]= id[ I β ( H β ( G β ( F β ( k, w ) , x ) , y ) , z )] (by Def. 20)Combining these two steps we obtain the desired equation. We are left witha similar task for the triangle identity, i.e. the condition that the followingdiagram commutes. G ◦ (Id[ C ] ◦ F ) ( G ◦ Id[ C ]) ◦ FG ◦ F a [ F, Id[ C ] , G ] ι [ G ] ◦ lu [ F ] ru [ G ] ◦ ι [ F ]Let us start with the ( − ) α components of the two-steps path.( ru [ G ] ◦ ι [ F ]) α [ k ] po ( h x, y i )= h ι [ F ] α [ k ] po ( x ) , ( ru [ G ] α ◦ ι [ F ] β )[ k, x ] po ( y ) i (by Def. 14)= h x, ( ru [ G ] α ◦ ι [ F ] β )[ k, x ] po ( y ) i (by Def. 10)= h x, ru [ G ] α [ ι [ F ] β [ k, x ]] po ( y ) i = h x, ru [ G ] α [Id[ F ( k, x ) β ]] po ( y ) i (by Def. 10)= h x, ru [ G ] α [ F ( k, x ) β ] po ( y ) i = h x, h ⋆, y ii (by Def. 18) a [ F, Id[ C ] , G ] α [ k ] po ( h x, h ⋆, y ii )= hh x, ⋆ i , y i (by Def. 20)16ow the ( − ) α components of the one step path.( ι [ G ] ◦ lu [ F ]) α [ k ] po ( h x, y i )= h lu [ F ] α [ k ] po ( x ) , ( ι [ G ] α ◦ lu [ F ] β )[ k, x ] po ( y ) i (by Def. 14)= h lu [ F ] α [ k ] po ( x ) , ι [ G ] α [ lu [ F ] β [ k, x ]] po ( y ) i = h lu [ F ] α [ k ] po ( x ) , ι [ G ] α [id[ F ( k, x )]] po ( y ) i (by Def. 18)= h lu [ F ] α [ k ] po ( x ) , ι [ G ] α [ F ( k, x )] po ( y ) i = h lu [ F ] α [ k ] po ( x ) , y i (by Def. 10)= hh x, ⋆ i , y i (by Def. 18)Both paths agree on the ( − ) α component. Now let us calculate the ( − ) β com-ponent of the two-steps path.(( ru [ G ] ◦ ι [ F ]) • θ ) β [ k, h x, y i ]= (( ru [ G ] ◦ ι [ F ]) β • ( θ β ◦ ι [( ∫ ( ru [ G ] ◦ ι [ F ]) α ) po ]))[ k, h x, y i ] (by Def. 11)= ( ru [ G ] ◦ ι [ F ]) β [ k, h x, y i ] · θ β [ k, h x, h ⋆, y ii ]= ru [ G ] β [ ι [ F ] β [ k, x ] , y ]] · θ β [ k, h x, h ⋆, y ii ] (by Def. 14)= ru [ G ] β [id[ F β ( k, x )] , y ]] · θ β [ k, h x, h ⋆, y ii ] (by Def. 10)= ru [ G ] β [ F β ( k, x ) , y ]] · θ β [ k, h x, h ⋆, y ii ]= id[ G β ( F β ( k, x ) , y )] · θ β [ k, h x, h ⋆, y ii ] (by Def. 18)= θ β [ k, h x, h ⋆, y ii ] a [ F, Id[ C ] , G ] β [ k, h x, h ⋆, y ii ]= id[ G β (Id[ C ] β ( F β ( k, x ) , ⋆ ) , y )] (by Def. 20)= id[ G β ( F β ( k, x ) , y )] (by Def. 4)Now the one step path.( ι [ G ] ◦ lu [ F ]) β [ k, h x, y i ]= ι [ G ] β [ lu [ F ] β [ k, x ] , y ] (by Def. 14)= ι [ G ] β [id[ F β ( k, x )] , y ] (by Def. 18)= ι [ G ] β [ F β ( k, x ) , y ]= id[ G β ( F β ( k, x ) , y )] (by Def. 10) This report presents all the formal details of the proof that the open functors,described as directly as possible, form a bicategory. The subsequent reports willestablish that (1) this bicategory can be presented as the Kleisli bicategory ofsome pseudo-monad on
Cat , (2) this pseudo-monad arise from a particularly17imple adjunction linking this bicategory with
Cat , (3) the formal definition inthis report can be made easier to manipulate by the use of discrete fibrations in-stead of categories of elements through the so called Grothendieck construction,and doing so presents this bicategory as a particular bicategory of spans, (4) thisbicategory is a sub-bicategory of the bicategory of profunctors ( a.k.a. distribu-tors), (5) many manipulations in this report can be modularized because thereis a factorization system separated the ( − ) α part and the ( − ) β part. (6) Kanextensions in this bicategory have a particular form in terms of the underlyingstrict 2-category Cat . Note that the many presentations of this bicategory isstrongly related to the many possible presentations one can do on the notion of“relation”: powerset monad for (1) and (2) above, spans for (3), and character-istic functions of the relation for (4). These results are those that are alreadydrafted and only need to be cleaned up and made available.
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