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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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= k ◦ (m AA U) ◦ (xA A U) ◦ (QP x A U) ◦ (Qz r P A U) .Since ( Q preserves coequalizers by assumpti<strong>on</strong>, by Lemma<str<strong>on</strong>g>2.</str<strong>on</strong>g>9 we haveQ ̂Q)(A U, Ql A U = Coequ Fun (QP xA U ◦ Qz l P A U), (QP x A U ◦ Qz r P A U) ) , so we deducethat there exists a unique functorial morphism k 1 : Q ̂Q A U → Y such that(199) k 1 ◦ (Ql A U) = k ◦ (m AA U) ◦ (xA A U) .Then we havek 1 ◦( )Qµ A Q bAU◦ (QlA A U) (150)= k 1 ◦ (Ql A U) ◦ (QP m AA U)(199)= k ◦ (m AA U) ◦ (xA A U) ◦ (QP m AA U)x= k ◦ (m AA U) ◦ (Am AA U) ◦ (xAA A U) m Aass= k ◦ (m AA U) ◦ (m A A A U) ◦ (xAA A U)m A(198)= k ◦ (A A Uλ A ) ◦ (m A A A U) ◦ (xAA A U)= k ◦ (m AA U) ◦ (AA A U A λ) ◦ (xAA A U) = x k ◦ (m AA U) ◦ (xA A U) ◦ (QP A A U A λ)((199)= k 1 ◦ (Ql A U) ◦ (QP A A U A λ) = l k 1 ◦ Q ̂Q)A U A λ ◦ (QlA A U)( )Since QlA A U is epi, we get that k 1 ◦ Qµ A Q bAU= k 1 ◦(Q ̂Q)A Uλ A . Since Q preservescoequalizers,(Q ̂Q)(A , Qpb Q= Coequ Fun Qµ A Q bAU, Q ̂Q)A Uλ A , then there exists aunique functorial morphism k 2 : Q ̂Q A → Y such that)(200) k 1 = k 2 ◦(Qpb Q.We have) ) (k 2 ◦(µ B QBU B ̂QA ◦(QBpb Q◦ Qy ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP x A U))) )y= k 2 ◦(µ B QBU B ̂QA ◦(Qy B U B ̂QA ◦(QP Qpb Q◦ (QP Ql A U) ◦ (QP QP x A U)) )(107)= k 2 ◦(χ B U B ̂QA ◦(QP Qpb Q◦ (QP Ql A U) ◦ (QP QP x A U))χ= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (χP QP A U)(200)= k 1 ◦ ◦ (Ql A U) ◦ (QP x A U) ◦ (χP QP A U)(199)= k ◦ (m AA U) ◦ (xA A U) ◦ (QP x A U) ◦ (χP QP A U)(102)= k ◦ (x A U) ◦ (χP A U) ◦ (χP QP A U)(98)= k ◦ (x A U) ◦ (χP A U) ◦ (QP χP A U)(102)= k ◦ (m AA U) ◦ (xA A U) ◦ (QP x A U) ◦ (QP χP A U)(199)= k 1 ◦ (Ql A U) ◦ (QP x A U) ◦ (QP χP A U))(200)= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (QP χP A U)167

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