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

• 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 1<strong>1.</strong>5. Moreover,such horizontal composition is weakly associative and unital by Propositions1<strong>1.</strong>7 and 1<strong>1.</strong>6.]• 2-cells are bimodule morphisms in C.Example 1<strong>1.</strong>14. Let us consider the bicategory SetMat as defined in [RW, <strong>2.</strong>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