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

and thus we can rewrite the above relation([ρ (P ·B)•B (Q·A) (p P ·B,Q·A · 1 A ) = ρ (P ·B)•B (Q·A) h −1 (1 P · p B,Q·A ) ] )· 1 Aandso that= ρ (P ·B)•B (Q·A)(h−1 · 1 A)(1P · p B,Q·A · 1 A )p P ·B,Q·A (1 P · 1 B · ρ Q·A ) = h −1 (1 P · p B,Q·A ) (1 P · 1 B · ρ Q·A )(237)= h −1 ( 1 P · ρ B•B (Q·A))(1P · p B,Q·A · 1 A )ρ (P ·B)•B (Q·A)(h−1 · 1 A)(1P · p B,Q·A · 1 A ) = h −1 ( 1 P · ρ B•B (Q·A))(1P · p B,Q·A · 1 A ) .Since 1 P · p B,Q·A · 1 A is epi, we get( )ρ (P ·B)•B (Q·A) h−1 · 1 ( )A = h−11 P · ρ B•B (Q·A)and thusso that we get thatρ (P ·B)•B (Q·A) = h −1 ( 1 P · ρ B•B (Q·A))(h · 1A )ρ (P ·B)•B (Q·A) ≃ 1 P · ρ B•B (Q·A) ≃ 1 P · ρ Q·A = ρ P ·Q·A .1<strong>2.</strong> Entwined modules and comodulesLet (X, 1 X ) , (Y, B) be monads in C and let us compute the categoryMnd (C) ((X, 1 X ) , (Y, B)) . Note that C (X, B) : C (X, Y ) → C (X, Y ) is a monadover the category C (X, Y ) . In fact, we set multiplication and unit of the monadto be C (X, m B ) = m B (−) : C (X, B · B) → C (X, B) and C (X, u B ) = u B (−) :C (X, 1 Y ) → C (X, B). In fact we haveandC (X, m B ) C (X, m B · 1 B ) = m B (m B · 1 B ) = m B (1 B · m B )= C (X, m B ) C (X, 1 B · m B )C (X, m B ) C (X, u B · 1 B ) = m B (u B · 1 B ) = 1 B = m B (1 B · u B )= C (X, m B ) C (X, 1 B · u B )The objects of such category are the monad functors (Q, φ) from (X, 1 X ) to (Y, B) ,i.e. the 1-cells Q : X → Y together with the 2-cells φ : B · Q = C (X, B) Q → Qsatisfying the following conditionsφ (1 B · φ) = φ (m B · 1 Q )φ (u B · 1 Q ) = 1 Qwhich says that φ gives a structure of left C (X, B)-module to the 1-cell Q : X → Y .Therefore, we can conclude thatMnd (C) ((X, 1 X ) , (Y, B)) = C(X,B) C (X, Y ).Now, following [St, pg. 158], we define the bicategory of comonads as follows:Cmd (C) = Mnd (C ∗ ) ∗where (−) ∗denotes the bicategory obtained by reversing235□