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

151that is(168)Similarly, fromwe deduce that( ) (Qχ ◦ δD QQ ) ◦ ( D (hQ) −1) = (hQ) −1 ◦ ( ) (P µ ) B Q ◦ (jB) ◦ DσB.(ε D P Q ) ◦ (DhQ) = (hQ) ◦ ( ε D QQ )(169) (hQ) −1 ◦ ( ε D P Q ) = ( ε D QQ ) ◦ ( D (hQ) −1) .Thus we haveso thaty ′ ◦ (hQ) −1 ◦ ( P µ B Q)◦ (jB) ◦(DσB ) (168)= y ′ ◦ ( Qχ ) ◦ ( δ D QQ ) ◦ ( D (hQ) −1)defy ′= y ′ ◦ ( ε D QQ ) ◦ ( D (hQ) −1) (169)= y ′ ◦ (hQ) −1 ◦ ( ε D P Q )y ′ ◦ (hQ) −1 ◦ ( P µ B Q)◦ (jB) ◦(DσB ) = y ′ ◦ (hQ) −1 ◦ ( ε D P Q ) .(( ) (Since (Z, π Z ) = Coequ ) Fun P µBQ ◦ (jB) ◦ DσB, ε D P Q ) , by the universal propertyof coequalizers, there exists a unique functorial morphism ν Z′ : Z → B′ suchthat(170) ν ′ Z ◦ π Z = y ′ ◦ (hQ) −1 .Now we want to prove that ν ′ Z is the two-sided inverse of ν Z. Let us computeand since y ′ is an epimorphism we getMoreoverν ′ Z ◦ ν Z ◦ y ′ (167)= ν ′ Z ◦ π Z ◦ (hQ)(170)= y ′ ◦ (hQ) −1 ◦ (hQ) = y ′ν ′ Z ◦ ν Z = Id B ′.ν Z ◦ ν ′ Z ◦ π Z(170)= ν Z ◦ y ′ ◦ (hQ) −1 (167)= π Z ◦ (hQ) ◦ (hQ) −1 = π Zand since π Z is an epimorphism we deduce thatν Z ◦ ν ′ Z = Id Z .Thus ν Z is a functorial isomorphism between B ′ and Z with inverse ν Z ′ . Now we wantto construct an isomorphism between B and Z. Consider the parallel pairDP Q(P µ S Q )◦(jS)◦(Dσ S) (ε D P Q) P Qand computeσ B ◦ ( ) (P µ ) B Q ◦ (jB) ◦ DσB (81),j= m B ◦ ( σ B B ) ◦ ( P Qσ B) ◦ (jP Q)σ= B m B ◦ ( Bσ B) ◦ ( σ B P Q ) ◦ (jP Q) (67)= m B ◦ ( Bσ B) ◦ (u B P Q) ◦ ( ε D P Q )u B= m B ◦ (u B B) ◦ σ B ◦ ( ε D P Q ) Bmonad= σ B ◦ ( ε D P Q ) .