12.07.2015 Views

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 ...

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

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 ).

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!