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

20By composing with P u A Y we getk ◦ (P Az) ◦ (P u A Y ) = h ◦ ( µ A P Y ) ◦ (P u A Y )and thusk ◦ (P u A Z) ◦ (P z) = h.Let l := k ◦ (P u A Z) : P Z → H. Then we havel ◦ (P z) = k ◦ (P u A Z) ◦ (P z) u A= k ◦ (P Az) ◦ (P u A Y )(3)= h ◦ ( µ A P Y ) ◦ (P u A Y ) = h.Let l ′ : P Z → H be another morphism such thatThen we havel ′ ◦ (P z) = h.l ◦ ( µ A P Z ) ◦ (P Az) = l ◦ (P z) ◦ ( µ A P Y ) = h ◦ ( µ A P Y )= l ′ ◦ (P z) ◦ ( µ A P Y ) = l ′ ◦ ( µ A P Z ) ◦ (P Az) .Since P A preserves coequalizers, we have that P Az is an epimorphism. Since µ A P Zis also an epimorphism, we deduce that l = l ′ . Therefore we obtain that (P Z, P z) =Coequ B (P f, P g) . The second statement can be proved similarly. We consider theabove coequalizerXfg Yin the category B and assume that AQ preserves it. By applying to it the functorsAQ and Q we get the following diagram in Au A QXAQXQXA µ Q XAQfAQgQfQgu A QYAQYQYA µ Q YzAQzQz Zu A QZAQZQZA µ Q ZBy assumption, the first row is a coequalizer. Assume that there exists a morphismh : QY → H such thath ◦ (Qf) = h ◦ (Qg) .Then, by composing with A µ Q X we geth ◦ (Qf) ◦ (A µ Q X ) = h ◦ (Qg) ◦ (A µ Q X )and since A µ Q is a functorial morphism we obtainh ◦ (A µ Q Y ) ◦ (AQf) = h ◦ (A µ Q Y ) ◦ (AQg) .Since (AQZ, AQz) = Coequ B (AQf, AQg), there exists a unique morphism k :AQZ → H such that(4) k ◦ (AQz) = h ◦ (A µ Q Y ) .By composing with u A QY we getk ◦ (AQz) ◦ (u A QY ) = h ◦ (A µ Q Y ) ◦ (u A QY )