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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 ...

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and since C preserves coequalizers, Ck is an epimorphism, so that we get(Cz ′ ) ◦ ( ∆ C K ) = C ρ Z ◦ z ′ .Lemma 4.2<str<strong>on</strong>g>2.</str<strong>on</strong>g> Let C = ( C, ∆ C , ε C) be a com<strong>on</strong>ad over a category A, let L, N :B → A be functors and let ρ : L → CL be a coassociative and counital functorialmorphism, that is (L, ρ) is a left C-comodule functor. Let u : N → L and letφ : N → CN be functorial morphisms such that(31) ρ ◦ u = (Cu) ◦ φ.If CCu and u are m<strong>on</strong>omorphisms, then φ is coassociative and counital, that is(N, φ) is a left C-comodule functor.Proof. Let us prove that φ is coassociative(CCu) ◦ (Cφ) ◦ φ (31)= (Cρ) ◦ (Cu) ◦ φ (31)= (Cρ) ◦ ρ ◦ uρcoass= ( ∆ C L ) ◦ ρ ◦ u (31)= ( ∆ C L ) ◦ (Cu) ◦ φ ∆C= (CCu) ◦ ( ∆ C N ) ◦ φ.Since CCu is a m<strong>on</strong>omorphism we get thatLet us prove that φ is counital(Cφ) ◦ φ = ( ∆ C N ) ◦ φ.u ◦ ( ε C N ) ◦ φ εC = ( ε C L ) ◦ (Cu) ◦ φ (31)= ( ε C L ) ◦ ρ ◦ u ρcounit= u.Since u is a m<strong>on</strong>omorphism we c<strong>on</strong>clude.4.<str<strong>on</strong>g>1.</str<strong>on</strong>g> Lifting of comodule functors. This subsecti<strong>on</strong> collects the dual <str<strong>on</strong>g>results</str<strong>on</strong>g> forliftings of module functors so that <strong>on</strong>e can skip reading all the proofs we keep herein order to give details of the <str<strong>on</strong>g>results</str<strong>on</strong>g> we use in the following.Propositi<strong>on</strong> 4.23 ([W] 3.5). Let C = ( C, ∆ C , ε C) be a com<strong>on</strong>ad <strong>on</strong> a category A,let D = ( D, ∆ D , ε D) be a com<strong>on</strong>ad <strong>on</strong> a category B and let T : A → B be a functor.Then there is a bijecti<strong>on</strong> between the following collecti<strong>on</strong>s of dataF functors ˜T : C A → D B that are liftings of T (i.e. D U ˜T = T C U)M functorial morphisms Ξ : T C → DT such that(∆ D T ) ◦ Ξ = (DΞ) ◦ (ΞC) ◦ ( T ∆ C) (and ε D T ) ◦ Ξ = T ε Cgiven by( )a : F → M where a ˜T = (D U D F T ε C) ◦( )D Uγ D ˜T C Fb : M → F where D Ub (Ξ) = T C U and D Uγ D b (Ξ) = Ξ ◦ ( T C Uγ C) i.e.b (Ξ) (( X, C ρ X))=(T X, (ΞX) ◦(T C ρ X))and b (Ξ) (f) = T (f) .Proof. Let ˜T : C A → D B be a lifting of the functor T : A → B (i.e. D U ˜T = T C U).Define a functorial morphism ξ : ˜T C F → D F T as the compositeξ := (D F T ε C) ( )◦ γ D ˜T C F51□□

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