Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...

108andsatisfying(B, y) = Coequ Fun((P χ) ◦ (δD P Q) , ε D P Q ) .A µ Q ◦ (xQ) = χ and µ B Q ◦ (Qy) = χ.Furthermore1) The morphisms cocan 1 := (A µ Q P ) ◦ (Aδ C ) : AC → QP is an isomorphism.2) The morphism cocan 1 := ( P µ B Q)◦ (δD B) : DB → P Q is an isomorphisms.3)4)5)6)(χP ) ◦ (QP δ C ) = cocan 1 ◦ (xC) and (P χ) ◦ (δ D P Q) = cocan 1 ◦ (Dy)x = ( Aε C) ◦ (cocan 1 ) −1y = ( ε D B ) ◦ (cocan 1 ) −1δ C = (A µ Q P ) ◦ (Aδ C ) ◦ (u A C)δ D = ( P µ B Q)◦ (δD B) ◦ (Du B ) .From the last equalities, we deduce that if ε C A is a regular epimorphism, so is σ Aand if Bε D is a regular epimorphism, so is σ B .Proof. Note that we are in the setting of Theorem 6.29.1) Let us check that cocan 1 is an isomorphism.( The) inverse of the functorial morphism cocan 1 is given by cocan −11 = (xC) ◦QρCP : QP → AC. Indeed we compute(xC) ◦ ( Qρ C P)◦ cocan1 ◦ (xC) = (xC) ◦ ( Qρ C P)◦( Aµ Q P ) ◦ (Aδ C ) ◦ (xC)= (xC) ◦ ( Qρ C P)◦( Aµ Q P ) ◦ (xQP ) ◦ (QP δ C )(115)= (xC) ◦ ( Qρ C P)◦ (χP ) ◦ (QP δC ) = (xC) ◦ (χP C) ◦ ( QP Qρ C P)◦ (QP δC )P rightCcol= (xC) ◦ (χP C) ◦ (QP δ C C) ◦ ( QP ∆ C) = (xC) ◦ ( w l C ) ◦ ( QP ∆ C)x coequ= (xC) ◦ (w r C) ◦ ( QP ∆ C) = (xC) ◦ ( QP ε C C ) ◦ ( QP ∆ C) C comonad= (xC) .Since xC is an epimorphism, we obtain that(xC) ◦ ( Qρ C P)◦ cocan1 = AC.On the other hand, we havecocan 1 ◦ (xC) ◦ ( ) (Qρ C P = Aµ Q P ) ◦ (Aδ C ) ◦ (xC) ◦ ( )Qρ C P= (A µ Q P ) ◦ (xQP ) ◦ (QP δ C ) ◦ ( )Qρ C (115)P = (χP ) ◦ (QP δ C ) ◦ ( )Qρ C P(118)= (χP ) ◦ (Qδ D P ) ◦ ( )Q D ρ P(113)= ( Qε D P ) ◦ ( Q D ρ P) D ρ P counital= QP