Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...
Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...
Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...
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220• 2-cells: m<strong>on</strong>ad functor transformati<strong>on</strong>s in C.Remark 10.5. We denote by (X, 1 X ) in C the trivial m<strong>on</strong>ad <strong>on</strong> the object X withtrivial multiplicati<strong>on</strong> and unit m 1X : 1 1X · 1 1X → 1 1X and u 1X : 1 X → 1 1X .Definiti<strong>on</strong> 10.6. Let C, C ′ be two 2-categories and let G : C → C ′ be a pseudofunctor.Then the pseudofunctor Mnd (G) : Mnd (C) → Mnd (C ′ ) is defined asfollows:• Mnd (G) (X, A) = (G (X) , G (A))• Mnd (G) (Q, φ) = (G (Q) , G (φ))• Mnd (G) (σ) = G (σ) .Remark 10.7. Note that Cmd (C) = Mnd (C ∗ ) where C ∗ is the dual reversing2-cells of C.1<str<strong>on</strong>g>1.</str<strong>on</strong>g> C<strong>on</strong>structi<strong>on</strong> of BIM (C)The idea of defining this bicategory goes back to the strict m<strong>on</strong>oidal category ofbalanced bimodule functors that we defined in Subsecti<strong>on</strong> 3.<str<strong>on</strong>g>2.</str<strong>on</strong>g> We observed that,c<strong>on</strong>sidering bimodule functors with respect to the same m<strong>on</strong>ad <strong>on</strong> both sides, wehave a unit and a compositi<strong>on</strong>, so that they form a strict m<strong>on</strong>oidal category. In thecase we c<strong>on</strong>sider a bimodule with respect to two different m<strong>on</strong>ads, the unit objectfails and the compositi<strong>on</strong> between them is no l<strong>on</strong>ger inside the class of objects of thecategory. The way to solve this problem is to look at balanced bimodule functorsas 0-cells of a bicategory, changing the definiti<strong>on</strong> of their product.Definiti<strong>on</strong> 1<str<strong>on</strong>g>1.</str<strong>on</strong>g><str<strong>on</strong>g>1.</str<strong>on</strong>g> Let X be a 0-cell and let (Y, B) be a m<strong>on</strong>ad in C. A left B-modulein C (or simply a left B-module) is a pair (Q, λ Q ) where Q : X → Y is a 1-cell andλ Q : B · Q → Q is a 2-cell in C, satisfying the associativity and unitality propertieswith respect to the m<strong>on</strong>ad B, i.e.λ Q (m B · 1 Q ) = λ Q (1 B · λ Q ) and λ Q (u B · 1 Q ) = 1 Q .Definiti<strong>on</strong> 1<str<strong>on</strong>g>1.</str<strong>on</strong>g><str<strong>on</strong>g>2.</str<strong>on</strong>g> Let (X, A) and (Y, 1 Y ) be m<strong>on</strong>ads in C. A right A-module in C(or simply a right A-module) is a m<strong>on</strong>ad functor in C ∗ , i.e. a 1-cell Q : X → Yand a 2-cell ρ Q : Q · A → 1 Y · Q = Q in C, satisfying the associativity and unitalityproperties with respect to the m<strong>on</strong>ad A, i.e.ρ Q (1 Q · m A ) = ρ Q (ρ Q · 1 A ) and ρ Q (1 Q · u A ) = 1 Q .Definiti<strong>on</strong> 1<str<strong>on</strong>g>1.</str<strong>on</strong>g>3. Let (X, A) and (Y, B) be m<strong>on</strong>ads in C. A B-A-bimodule in C (orsimply a B-A-bimodule) is a triple (Q, λ Q , ρ Q ) where• (Q, λ Q ) is a left B-module in C• (Q, ρ Q ) is a right A-module in C• the compatibility c<strong>on</strong>diti<strong>on</strong> holdsλ Q (1 B · ρ Q ) = ρ Q (λ Q · 1 A ) .Lemma 1<str<strong>on</strong>g>1.</str<strong>on</strong>g>4. Let (X, A) , (Y, B) be m<strong>on</strong>ads in C and let (Q, λ Q ) be a left A-moduleand (Q, ρ Q ) be a right B-module. Then (Q, λ Q ) = Coequ C (m A · 1 Q , 1 A · λ Q ) and(Q, ρ Q ) = Coequ C (1 Q · m B , ρ Q · 1 B ).