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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8Since the second row is a coequalizer by assumption, there is a unique morphismy : C ′ → X such that(1) y ◦ i ′ = x ◦ e ′ .We calculatey ◦ n ′′ ◦ i = y ◦ i ′ ◦ n ′ (1)= x ◦ e ′ ◦ n ′ = x ◦ e ′ ◦ m ′ (1)= y ◦ i ′ ◦ m ′= y ◦ m ′′ ◦ iand since i is an epimorphism we get thaty ◦ n ′′ = y ◦ m ′′ .Since the third column is a coequalizer, there exists a unique morphism z : C ′′ → Xsuch thatThenz ◦ e ′′ = y.z ◦ i ′′ ◦ e ′ = z ◦ e ′′ ◦ i ′ = y ◦ i ′ = x ◦ e ′so we get that z ◦ i ′′ = x. Since e ′′ ◦ i ′ = i ′′ ◦ e ′ and e ′′ , e ′ , i ′ are epimorphism, wededuce that i ′′ is epimorphism and hence z is unique with respect to z ◦ i ′′ = x. □Corollary ong>2.ong>1ong>2.ong> Let F , F ′ : A → B be functors and α, β : F → F ′ be functorialmorphisms. Assume that, for every X ∈ A, B has coequalizers of αX and βXhence there exists (Q, q) = Coequ Fun (α, β), cf. Lemma ong>2.ong>7. Assume that (P, p) =Coequ A (f, g) of morphisms f, g : X → Y in A and that both F and F ′ preserveCoequ A (f, g). Then also Q preserves Coequ A (f, g).Proof. The following diagram (in B) is serially commutative by naturalityF XαXβXF ′ XqXQXF fF gF ′ fF ′ gQfQgF YαYβYF ′ YqY QYF pF ′ pQpF PαPβPF ′ PqP QPThe columns are coequalizers by Lemma ong>2.ong>7. The first and second rows are coequalizersby the assumption that F and F ′ preserve coequalizers. Thus the third row isa coequalizer by Lemma ong>2.ong>1ong>1.ong>□

Lemma ong>2.ong>13 ([BM, Lemma ong>2.ong>5]). Consider the following serially commutative diagramin an arbitrary category KAiBee ′ e ′′A ′ i ′ B ′ f ′ C ′g ′n mn ′m ′n ′′ m ′′A ′′ i ′′ B ′′ f ′′ g ′′ C ′′Assume that all columns are equalizers and also the second and third rows are equalizers.Then also the first row is an equalizer.Proof. Dual to Lemma ong>2.ong>1ong>1.ong>Corollary ong>2.ong>14. Let G,G ′ : C → K be functors and γ, θ : G → G ′ be functorialmorphisms. Assume that, for every X ∈ C, K has equalizers of γX and θX hencethere exists (E, e) = Equ Fun (γ, θ), cf. Lemma ong>2.ong>8. Assume that (I, i) = Equ C (f, g)of morphisms f, g : X → Y in C and that both G and G ′ preserve Equ C (f, g). Thenalso E preserves Equ C (f, g).Proof. Dual to Corollary ong>2.ong>1ong>2.ong>Lemma ong>2.ong>15. Let Z, Z ′ , W, W ′ : A → B be functors, let a, b : Z → W and a ′ , b ′ :Z ′ → W ′ be functorial morphisms, let ϕ : Z → Z ′ and ψ : W → W ′ be functorialisomorphisms such thatψ ◦ a = a ′ ◦ ϕ and ψ ◦ b = b ′ ◦ ϕ.Assume that there exist (E, i) = Equ Fun (a, b) and (E ′ , i ′ ) = Equ Fun (a ′ , b ′ ). Then ϕinduces an isomorphism ̂ϕ : E → E ′ such that ϕ ◦ i = i ′ ◦ ̂ϕ.fgC9□□EˆϕiZϕE ′ i ′Z ′ b ′aWProof. Let us define ̂ϕ. Let us computebψa ′ W ′a ′ ◦ ϕ ◦ i = ψ ◦ a ◦ i defi= ψ ◦ b ◦ i = b ′ ◦ ϕ ◦ iand since (E ′ , i ′ ) = Equ Fun (a ′ , b ′ ) there exists a unique functorial morphism ̂ϕ : E →E ′ such thati ′ ◦ ̂ϕ = ϕ ◦ i.Note that ̂ϕ is mono since so are i and i ′ and ϕ is an isomorphism. Considerϕ −1 : Z ′ → Z and ψ −1 : W ′ → W . Then we havea ◦ ϕ −1 = ψ −1 ◦ a ′ and b ◦ ϕ −1 = ψ −1 ◦ b ′ .

8Since the sec<strong>on</strong>d row is a coequalizer by assumpti<strong>on</strong>, there is a unique morphismy : C ′ → X such that(1) y ◦ i ′ = x ◦ e ′ .We calculatey ◦ n ′′ ◦ i = y ◦ i ′ ◦ n ′ (1)= x ◦ e ′ ◦ n ′ = x ◦ e ′ ◦ m ′ (1)= y ◦ i ′ ◦ m ′= y ◦ m ′′ ◦ iand since i is an epimorphism we get thaty ◦ n ′′ = y ◦ m ′′ .Since the third column is a coequalizer, there exists a unique morphism z : C ′′ → Xsuch thatThenz ◦ e ′′ = y.z ◦ i ′′ ◦ e ′ = z ◦ e ′′ ◦ i ′ = y ◦ i ′ = x ◦ e ′so we get that z ◦ i ′′ = x. Since e ′′ ◦ i ′ = i ′′ ◦ e ′ and e ′′ , e ′ , i ′ are epimorphism, wededuce that i ′′ is epimorphism and hence z is unique with respect to z ◦ i ′′ = x. □Corollary <str<strong>on</strong>g>2.</str<strong>on</strong>g>1<str<strong>on</strong>g>2.</str<strong>on</strong>g> Let F , F ′ : A → B be functors and α, β : F → F ′ be functorialmorphisms. Assume that, for every X ∈ A, B has coequalizers of αX and βXhence there exists (Q, q) = Coequ Fun (α, β), cf. Lemma <str<strong>on</strong>g>2.</str<strong>on</strong>g>7. Assume that (P, p) =Coequ A (f, g) of morphisms f, g : X → Y in A and that both F and F ′ preserveCoequ A (f, g). Then also Q preserves Coequ A (f, g).Proof. The following diagram (in B) is serially commutative by naturalityF XαXβXF ′ XqXQXF fF gF ′ fF ′ gQfQgF YαYβYF ′ YqY QYF pF ′ pQpF PαPβPF ′ PqP QPThe columns are coequalizers by Lemma <str<strong>on</strong>g>2.</str<strong>on</strong>g>7. The first and sec<strong>on</strong>d rows are coequalizersby the assumpti<strong>on</strong> that F and F ′ preserve coequalizers. Thus the third row isa coequalizer by Lemma <str<strong>on</strong>g>2.</str<strong>on</strong>g>1<str<strong>on</strong>g>1.</str<strong>on</strong>g>□

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