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

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

98AU A can AA F = can AA F = ( CσAAF ) A ◦ (C ρ Q P AA F ) = ( Cσ A) ◦ (C ρ Q P ) = can 1 sothat also can 1 is an isomorphism.□6.6. The tame case.Definiti<strong>on</strong> 6.18. A formal dual structure M = (A, B, P, Q, σ A , σ B ) is called aMorita c<strong>on</strong>text <strong>on</strong> the categories A and B if it satisfies also the balanced c<strong>on</strong>diti<strong>on</strong>s(91) σ A ◦ ( µ B QP ) = σ A ◦ ( Q B µ P)and σ B ◦ ( P A µ Q)= σ B ◦ ( µ A P Q ) .Lemma 6.19. Let M = (A, B, P, Q, σ A , σ B ) be a Morita c<strong>on</strong>text <strong>on</strong> the categories Aand B and assume that A, B, P, Q preserve coequalizers. Hence, there exist functorialmorphisms• AB σ A BA : AQ BB P A → Id A A such that(92) AU AB σ A BA = B σ A BAwhere B σ A BA is uniquely determined by Bσ A BA ◦(Q Bp B P ) = ( A Uλ A )◦ ( Bσ A BAU )and(93) Bσ A B ◦ (p QB P ) = σ A• BA σ B AB : BP AA Q B → Id B B such that(94) BU BA σ B AB = A σ B ABwhere A σ B AB is uniquely determined by Aσ B AB ◦ (P Ap A Q) = ( B Uλ B ) ◦ ( Aσ B A BU )and(95) Aσ B A ◦ (p P A Q) = σ B .Moreover we have that(96) Bσ A BAAF = B σ A B and Aσ B ABBF = A σ B A.Proof. By definiti<strong>on</strong>, (Q BB P , p QB P ) = Coequ Fun(µBQ P, Q B µ P)and by assumpti<strong>on</strong>σ A is balanced, so that, by the universal property of the coequalizer, there exists aunique functorial morphism B σ A B : Q BBP → A such that B σ A B ◦ (p QBP ) = σ A . Now,let us c<strong>on</strong>sider the following diagramQ BB P A A UQ B µ A B P AU Q B B P A Uλ AQ BB P A UQ B p B PQ BB P ABσ A B A AUBσ A B AUBσ A BAA A Um AA UA A Uλ A A A UAUλ A AUNote that, by naturality of p Q and definiti<strong>on</strong> of B σB A we have(Bσ BAU ) A ◦ ( Q B µ A BP AU ) ◦ (p QB P A A U) = ( BσBAU ) A ◦ (p QB P A U) ◦ ( Q B Uµ A BP AU )= ( σ A AU ) ◦ ( Qµ A P AU ) (80)= (m AA U) ◦ ( σ A A A U ) = (m AA U) ◦ ( Bσ A BA A U ) ◦ (p QB P A A U)

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

Saved successfully!

Ooh no, something went wrong!