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

4need some material related to Gabriel-Popescu theorem which is contained in theappendix.The last part is the first outcome of a joint work with J. Gomez-Torrecillas.It is devoted to the introduction of the bicategory of balanced bimodule functorsBIM (C). First we fix some notation and terminology about 2-categories and bicategories.Then we define the bicategory of balanced bimodule functors and finallywe study how it can be related to entwined modules and comodules.<strong>2.</strong> <strong>Preliminaries</strong><strong>2.</strong><strong>1.</strong> <strong>Some</strong> <strong>results</strong> on equalizers and coequalizers. In the following, most ofthe computations are justified. We denote by the name of a functorial morphism,its naturality property.Definitions <strong>2.</strong><strong>1.</strong> Let α : B → C be a functorial morphism. We say that α is• a functorial monomorphism, or simply a monomorphism, if for every β, γ :A → B such that α ◦ β = α ◦ γ we have β = γ.• a functorial regular monomorphism, or simply a regular monomorphism, ifα is the equalizer of two functorial morphisms.• a functorial epimorphism, or simply an epimorphism, if for every β, γ : C →D such that β ◦ α = γ ◦ α we have β = γ.• a regular epimorphism, or simply a regular epimorphism, if α is the coequalizerof two functorial morphisms.Definition <strong>2.</strong><strong>2.</strong> A parallel pair α, β : F → F ′ is said to be reflexive if the twoarrows have a common right inverse δ : F ′ → F.Definition <strong>2.</strong>3. A reflexive equalizer is an equalizer of a reflexive parallel pair.Definition <strong>2.</strong>4. A reflexive coequalizer is a coequalizer of a reflexive parallel pair.Lemma <strong>2.</strong>5. Let F, G, H be functors and let f : F → G, g : G → H and h : F → Hbe functorial morphisms such that h = g ◦ f. Assume that f is a functorial isomorphism.Then h is a regular epimorphism if and only if g is a regular epimorphism.Proof. First, let us assume that g is a regular epimorphism, i.e.(H, g) = Coequ Fun (α, β). Then we haveh ◦ f −1 ◦ α = g ◦ f ◦ f −1 ◦ α = g ◦ f ◦ f −1 ◦ β = h ◦ f −1 ◦ β.Now, let χ : F → X be a functorial morphism such that χ ◦ f −1 ◦ α = χ ◦ f −1 ◦ β.By the universal property of the coequalizer (H, g) = Coequ Fun (α, β), there existsa unique functorial morphism χ : H → X such that χ ◦ g = χ. Then, by composingto the right with f we getχ ◦ h = χ ◦ g ◦ f = χ ◦ f −1 ◦ f = χ.Moreover, let χ ′ be another functorial morphism such that χ ′ ◦ h = χ. Since we alsohave χ ◦ h = χ we haveχ ′ ◦ g ◦ f = χ ′ ◦ h = χ = χ ◦ h = χ ◦ g ◦ f