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

Lemma <strong>2.</strong>13 ([BM, Lemma <strong>2.</strong>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 <strong>2.</strong>1<strong>1.</strong>Corollary <strong>2.</strong>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 <strong>2.</strong>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 <strong>2.</strong>1<strong>2.</strong>Lemma <strong>2.</strong>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 ′ .