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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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113Proof. Since (C U, ) and (D U, D F ) are adjuncti<strong>on</strong>s, (C U C Q D , D P CC F ) = ( Q D , D P )and (D U D P C , C Q DD F ) = ( P C , C Q ) are also adjuncti<strong>on</strong>s.□6.9. Coherds and M<strong>on</strong>ads. In this subsecti<strong>on</strong> we prove that in the case whenthere exist coequalizers in the base categories and all functors occurring in a formalcodual structure preserve them, we establish an equivalence between coherds <strong>on</strong><strong>on</strong>e hand, and m<strong>on</strong>ads <strong>on</strong> the other hand, together with two natural isomorphismgenerating a Galois map.Theorem 6.36. Let A and B be categories in both of which the coequalizer of anypair of parallel morphisms exists. Let X = (C, D, P, Q, δ C , δ D ) be a formal codualstructure <strong>on</strong> A and B. Then we have(1) If A = (A, m A , u A ) is a m<strong>on</strong>ad <strong>on</strong> the category A and ( Q, A µ Q : AQ → Q )is a left A-module functor such that(i) the functorial morphism cocan 1 := (A µ Q P ) ◦ (Aδ C ) : AC → QP is anisomorphism(ii) the functorial morphism cocan 2 := (A µ Q D ) ◦ ( AρQ) D : AQ → QD is anisomorphismthen χ := A µ Q ◦ ( Aε C Q ) ◦ ( cocan −11 Q ) : QP Q → Q is a copretorsor andthus a coherd.(2) If B = (B, m B , u B ) is a m<strong>on</strong>ad <strong>on</strong> the category B and ( Q, µ B Q : QB → Q) isa right B-module functor such that(i) the functorial morphism cocan 1 := ( P µ Q) B ◦ (δD B) : DB → P Q is anisomorphism(ii) the functorial morphism cocan 2 := ( (Cµ Q) B ◦ Cρ Q B ) : QB → CQ is anisomorphismthen χ := µ B Q ◦ ( Qε D B ) ◦ (Qcocan −1 1 ) : QP Q → Q is a copretorsor andthus a coherd.Proof. Let us prove 1), the other is similar. We have to prove thatχ := A µ Q ◦ ( Aε C Q ) ◦ ( cocan −11 Q )satisfies (111) , (112) , (113) . We compute( A µ Q ass)=cocan 1 ◦ (m A C) = (A µ Q P ) ◦ (Aδ C ) ◦ (m A C)= (A µ Q P ) ◦ (m A QP ) ◦ (AAδ C )( Aµ Q P ) ◦ ( A A µ Q P ) ◦ (AAδ C ) = (A µ Q P ) ◦ (Acocan 1 )so we obtaincocan 1 ◦ (m A C) = (A µ Q P ) ◦ (Acocan 1 )i.e.(131) (m A C) ◦ Acocan −11 = ( ) (cocan −11 ◦ Aµ Q P ) .We computeχ ◦ (QP χ)= A µ Q ◦ ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (QP A µ Q )◦ ( QP Aε C Q ) ◦ ( QP cocan −11 Q )

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