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

228(241)= (1 Q • B l Q ′) ζ Q,B,Q ′p Q•B B,Q ′ (p Q,B · 1 Q ′)so that, since p Q•B B,Q ′ (p Q,B · 1 Q ′) is an epimorphism, we getr Q • B 1 Q ′ = (1 Q • B l Q ′) ζ Q,B,Q ′.Proposition 1<strong>1.</strong>1<strong>1.</strong> Let (X, A) , (Y, B) be monads in C, let (P, λ P , ρ P ) , (Q, λ Q , ρ Q )be A-B-bimodules in C, let (P ′ , λ P ′, ρ P ′) , (Q ′ , λ Q ′, ρ Q ′) be B-C-bimodules in C andlet f : P → Q, f ′ : P ′ → Q ′ be bimodule morphisms in C. Then there exists a uniqueA-C-bimodule morphism f • B f ′ : P • B P ′ → Q • B Q ′ .Proof. Since f is an A-B-bimodule morphism, we have that(245) λ Q (1 A · f) = fλ P and ρ Q (f · 1 B ) = fρ P .Since f ′ is a B-C-bimodule morphism, we have that(246) λ Q ′ (1 B · f ′ ) = f ′ λ P ′ and ρ Q ′ (f ′ · 1 C ) = f ′ ρ P ′.Let us consider the following diagramP · B · P ′ ρ P ·1 P ′ 1 P ·λ P ′f·1 B·f ′ P · P ′ p P,P ′f·f ′P • B P ′f• B f ′Q · B · Q ′ ρ Q·1 Q ′ Q · Q ′ p Q,Q ′ Q • B Q ′1 Q·λ Q ′Note that the left square serially commutes, in factandThus, we get that(f · f ′ ) (ρ P · 1 P ′) = (f · 1 Q ′) (1 P · f ′ ) (ρ P · 1 P ′)ρ P= (f · 1 Q ′) (ρ P · 1 Q ′) (1 P · 1 B · f ′ )(245)= (ρ Q · 1 Q ′) (f · 1 B · 1 Q ′) (1 P · 1 B · f ′ ) = (ρ Q · 1 Q ′) (f · 1 B · f ′ )(f · f ′ ) (1 P · λ P ′) = (f · 1 Q ′) (1 P · f ′ ) (1 P · λ P ′)(246)= (f · 1 Q ′) (1 P · λ Q ′) (1 P · 1 B · f ′ )f= (1 Q · λ Q ′) (f · 1 B · 1 Q ′) (1 P · 1 B · f ′ )= (1 Q · λ Q ′) (f · 1 B · f ′ ) .p Q,Q ′ (f · f ′ ) (ρ P · 1 P ′) = p Q,Q ′ (f · f ′ ) (1 P · λ P ′)and since (P • B P ′ , p P,P ′) = Coequ C (ρ P · 1 P ′, 1 P · λ P ′) we deduce that there exists aunique 2-cell f • B f ′ : P • B P ′ → Q • B Q ′ such that(247) (f • B f ′ ) p P,P ′ = p Q,Q ′ (f · f ′ ) .□