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

i.e. Hom B (Y, iX) equalizes Hom B (Y, αX) and Hom B (Y, γX) , for every X ∈ A andY ∈ B. Let now ζ : Z → Hom B (Y, F X) be a map such that Hom B (Y, αX) ◦ ζ =Hom B (Y, γX) ◦ ζ. Then, for every X ∈ A, Y ∈ B and for every z ∈ Z we have(αX) ◦ ζ (z) = Hom B (Y, αX) (ζ (z)) = Hom B (Y, γX) ◦ (ζ (z))= (γX) ◦ ζ (z) .Then, for every X ∈ A and Y ∈ B there exists a unique morphism θ z : Y → EX inB such that(iX) ◦ θ z = ζ (z)i.e.Hom B (Y, iX) (θ z ) = ζ (z) .The assignment z ↦→ θ z defines a map θ : Z → Hom B (Y, EX) such that Hom B (Y, iX)◦θ = ζ.□<strong>2.</strong><strong>2.</strong> Contractible Equalizers and Coequalizers.11Definition <strong>2.</strong>18. Let C be a category.eightuple (Z, X, Y, d, d 0 , d 1 , s, t) whereA contractible ( or split) equalizer is asuch thatZ d Xsd 0td 1 Yt ◦ d 0 = Id Xs ◦ d = Id Zt ◦ d 1 = d ◦ sd 0 ◦ d = d 1 ◦ d.Proposition <strong>2.</strong>19. Let C be a category and let (Z, X, Y, d, d 0 , d 1 , s, t) be a contractibleequalizer. Then (Z, d) = Equ C (d 0 , d 1 ) .Proof. Let ξ : L → X be such thatthenLetso thatd 0 ◦ ξ = d 1 ◦ ξξ = Id X ◦ ξ = t ◦ d 0 ◦ ξ = t ◦ d 1 ◦ ξ = d ◦ (s ◦ ξ) .ξ ′ = s ◦ ξ : L → Zξ = d ◦ ξ ′ .Let now ξ ′′ : L → Z be such that d ◦ ξ ′′ = ξ. Thenξ ′′ = Id Z ◦ ξ ′′ = s ◦ d ◦ ξ ′′ = s ◦ ξ = ξ ′ .□