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

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230On the other hand, we can first consider the canonical vertical composites gf :P → W and g ′ f ′ : P ′ → W ′ , which are still bimodule morphisms, and then we cancompose them horizontally gettingWe have to prove thatLet us consider the following diagramsand(gf) • B (g ′ f ′ ) : P • B P ′ → W • B W ′ .(g • B g ′ ) (f • B f ′ ) = (gf) • B (g ′ f ′ ) .P · B · P ′f·1 B·f ′ ρ P ·1 P ′ 1 P ·λ P ′Q · B · Q ′ ρ Q·1 Q ′g·1 B·g ′ (gf)·1 B·(g ′ f ′ )1 Q·λ Q ′P · P ′ p P,P ′f·f ′Q · Q ′ p Q,Q ′g·g ′P • B P ′f• B f ′Q • B Q ′g• B g ′W · B · W ′ρ W ·1 W ′W · W ′ p W,W ′ W • B W ′P · B · P ′1 W ·λ W ′ρ P ·1 P ′ 1 P ·λ P ′P · P ′ p P,P ′(gf)·(g ′ f ′ )P • B P ′W · B · W ′ρ W ·1 W ′W · W ′ p W,WW •′ B W ′1 W ·λ W ′(gf)• B (g ′ f ′ )We have to prove that (gf) • B (g ′ f ′ ) makes the external square of the first diagramcommutative. Since Bim (C) is a bicategory, in particular we have that (gf)·(g ′ f ′ ) =(g · g ′ ) (f · f ′ ) so that, by the commutativity of the first diagram, we deduce thatalso the left square of the second one commutes, i.e.[(gf) · (g ′ f ′ )] (ρ P · 1 P ′) = (ρ W · 1 W ′) [(gf) · 1 B · (g ′ f ′ )][(gf) · (g ′ f ′ )] (1 P · λ P ′) = (1 W · λ W ′) [(gf) · 1 B · (g ′ f ′ )] .Therefore, the exists the unique 2-cell (gf) • B (g ′ f ′ ) : P • B P ′ → W • B W ′ such thatThen we have[(gf) • B (g ′ f ′ )] p P,P ′ = p W,W ′ [(gf) · (g ′ f ′ )] .[(gf) • B (g ′ f ′ )] p P,P ′ = p W,W ′ [(gf) · (g ′ f ′ )] = p W,W ′ [(g · g ′ ) (f · f ′ )]= p W,W ′ (g · g ′ ) (f · f ′ ) = (g • B g ′ ) (f • B f ′ ) p P,P ′and since p P,P ′ is an epimorphism, we get that(gf) • B (g ′ f ′ ) = (g • B g ′ ) (f • B f ′ ) .Definition 1ong>1.ong>13. The bicategory BIM (C) consists of• 0-cells are monads in C□

• 1-cells are bimodules in C together with their horizontal composition definedas follows. Let (X, A) , (Y, B) and (W, C) be monads in C and let Q : Y → Xand Q ′ : W → Y be respectively an A-B-bimodule with (Q, λ Q , ρ Q ) and aB-C-bimodule in C with (Q ′ , λ Q ′, ρ Q ′). Then the horizontal composition ofthe two bimodules is given by (Q • B Q ′ , p Q,Q ′) = Coequ C (ρ Q · 1 Q ′, 1 Q · λ Q ′)[Note that Q • B Q ′ is an A-C-bimodule in C by Proposition 1ong>1.ong>5. Moreover,such horizontal composition is weakly associative and unital by Propositions1ong>1.ong>7 and 1ong>1.ong>6.]• 2-cells are bimodule morphisms in C.Example 1ong>1.ong>14. Let us consider the bicategory SetMat as defined in [RW, ong>2.ong>1].The objects of this bicategory are sets, denoted by A, B, . . .. An arrow (1-cell)M : A → B is a set valued matrix which, to fix notation ,has entries M (a, b) forevery a ∈ A and b ∈ B. A 2-cell f : M → N : A → B is a matrix of functionsf (a, b) : M (a, b) → N (a, b). Moreover, for A −→ MB −→ LC we have L · M : A → Cdefined by(L · M) (a, c) = ∑ b∈BL (b, c) × M (a, b) .A monad in SetMat on an object A is thus a pair (A, M) where A is a set andM : A → A is a matrix whose entries are M (a, b) for every a, b ∈ A, i.e. itis a small category with set of objects A. Hence, a monad functor is a functorF : (A, M) → (B, N) where A and B are the sets of objects of the small categoriesM and N. Note that, since F is a functor between categories, F is just a mapF : A → B at the level of objects. This map induces a 1-cell Q F : A → B definedas follows{}∅ if b ≠ F (a)Q F (a, b) =.{(a, F (a))} if b = F (a)Moreover, we can consider the following 2-cell φ F : Q F A → BQ F defined, for every(a, b) ∈ A × B, by the mapnote thatQ F A (a, b) = ∑ a ′ ∈Aφ F (a, b) : Q F A (a, b) → BQ F (a, b) .Q F (a ′ , b) × A (a, a ′ ) =⋃a ′ ∈F ← (b)where F ← (b) = {a ∈ A | F (a) = b}. Similarly we have{(a ′ , F (a ′ ))} × A (a, a ′ )BQ F (a, b) = ∑ b ′ ∈BB (b ′ , b) × Q F (a, b ′ ) = B (F (a) , b) × {(a, F (a))} .We can identify the set⋃{(a ′ , F (a ′ ))} × A (a, a ′ ) = Q F A (a, b) =anda ′ ∈F ← (b)⋃a ′ ∈F ← (b)B (F (a) , b) × {(a, F (a))} = BQ F (a, b) = B (F (a) , b)A (a, a ′ )231

230On the other hand, we can first c<strong>on</strong>sider the can<strong>on</strong>ical vertical composites gf :P → W and g ′ f ′ : P ′ → W ′ , which are still bimodule morphisms, and then we cancompose them horiz<strong>on</strong>tally gettingWe have to prove thatLet us c<strong>on</strong>sider the following diagramsand(gf) • B (g ′ f ′ ) : P • B P ′ → W • B W ′ .(g • B g ′ ) (f • B f ′ ) = (gf) • B (g ′ f ′ ) .P · B · P ′f·1 B·f ′ ρ P ·1 P ′ 1 P ·λ P ′Q · B · Q ′ ρ Q·1 Q ′g·1 B·g ′ (gf)·1 B·(g ′ f ′ )1 Q·λ Q ′P · P ′ p P,P ′f·f ′Q · Q ′ p Q,Q ′g·g ′P • B P ′f• B f ′Q • B Q ′g• B g ′W · B · W ′ρ W ·1 W ′W · W ′ p W,W ′ W • B W ′P · B · P ′1 W ·λ W ′ρ P ·1 P ′ 1 P ·λ P ′P · P ′ p P,P ′(gf)·(g ′ f ′ )P • B P ′W · B · W ′ρ W ·1 W ′W · W ′ p W,WW •′ B W ′1 W ·λ W ′(gf)• B (g ′ f ′ )We have to prove that (gf) • B (g ′ f ′ ) makes the external square of the first diagramcommutative. Since Bim (C) is a bicategory, in particular we have that (gf)·(g ′ f ′ ) =(g · g ′ ) (f · f ′ ) so that, by the commutativity of the first diagram, we deduce thatalso the left square of the sec<strong>on</strong>d <strong>on</strong>e commutes, i.e.[(gf) · (g ′ f ′ )] (ρ P · 1 P ′) = (ρ W · 1 W ′) [(gf) · 1 B · (g ′ f ′ )][(gf) · (g ′ f ′ )] (1 P · λ P ′) = (1 W · λ W ′) [(gf) · 1 B · (g ′ f ′ )] .Therefore, the exists the unique 2-cell (gf) • B (g ′ f ′ ) : P • B P ′ → W • B W ′ such thatThen we have[(gf) • B (g ′ f ′ )] p P,P ′ = p W,W ′ [(gf) · (g ′ f ′ )] .[(gf) • B (g ′ f ′ )] p P,P ′ = p W,W ′ [(gf) · (g ′ f ′ )] = p W,W ′ [(g · g ′ ) (f · f ′ )]= p W,W ′ (g · g ′ ) (f · f ′ ) = (g • B g ′ ) (f • B f ′ ) p P,P ′and since p P,P ′ is an epimorphism, we get that(gf) • B (g ′ f ′ ) = (g • B g ′ ) (f • B f ′ ) .Definiti<strong>on</strong> 1<str<strong>on</strong>g>1.</str<strong>on</strong>g>13. The bicategory BIM (C) c<strong>on</strong>sists of• 0-cells are m<strong>on</strong>ads in C□

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