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

and since p QB P A A U is an epimorphism, we get that ( Bσ A BAU ) ◦ ( Q B µ A BP AU ) =(m AA U) ◦ ( Bσ A B A AU ) . Moreover, by using naturality of p Q , definition of B σ A B , naturalityof σ A we have(Bσ A BAU ) ◦ (Q BB P A Uλ A ) ◦ (p QB P A A U) = ( Bσ A BAU ) ◦ (p QB P A U) ◦ (Q B U B P A Uλ A )= ( σ A AU ) ◦ (Q B U B P A Uλ A ) = (A A Uλ A ) ◦ ( σ A AU A F A U )= (A A Uλ A ) ◦ ( Bσ A BA A U ) ◦ (p QB P A A U)and since p QB P A A U is an epimorphism, we get that ( Bσ A BAU ) ◦ (Q BB P A Uλ A ) =(A A Uλ A ) ◦ ( Bσ A B A AU ) so that the left square serially commutes. Since B, P, Q preservecoequalizers, by Corollary <strong>2.</strong>12, also Q BB P = Coequ Fun(µBQ BU B P, Q B Uλ BB P )preserves them so that both the rows are coequalizers. Hence, there exists a uniquefunctorial morphism B σ A BA : Q BBP A → A U such that(97) Bσ A BA ◦ (Q B p B P ) = ( A Uλ A ) ◦ ( Bσ A BAU ) .Now, by using naturality of A µ QB , definition of B σ A BA , definition of Bσ A B , coequalizingproperty of A Uλ A , we computeBσ A BA ◦ (A µ QB BP A)◦ (AQB p B P ) ◦ (Ap QB P A U)= B σBA A ◦ (Q B p B P ) ◦ (A µ QB BP A U ) ◦ (Ap QB P A U)(7)= ( A Uλ A ) ◦ ( BσBAU ) A ◦ (p QB P A U) ◦ (A µ QB U B P A U )= ( A Uλ A ) ◦ ( σ A AU ) ◦ (A µ QB U B P A U )(80)= ( A Uλ A ) ◦ (m AA U) ◦ ( Aσ A AU ) = ( A Uλ A ) ◦ (A A Uλ A ) ◦ ( Aσ A AU )= ( A Uλ A ) ◦ (A A Uλ A ) ◦ ( A B σ A BAU ) ◦ (Ap QB P A U)= ( A Uλ A ) ◦ ( A B σ A BA)◦ (AQB p B P ) ◦ (Ap QB P A U)and since (AQ B p B P ) ◦ (Ap QB P A U) is an epimorphism, we get B σBA A ◦ (A )µ QB BP A =( A Uλ A )◦ ( A B σBA) A so that B σBA A induces a functorial morphism ABσBA A : AQ BB P A →Id A A such that A U AB σBA A = BσBA A . Similarly, one can prove that there exists a uniquefunctorial morphism A σA B : P AAQ → B such that A σA B ◦ (p P AQ) = σ B and it inducesa unique functorial morphism BA σAB B : BP AA Q B → Id B B such that B U BA σAB B = AσABBwhere A σAB B is uniquely determined by AσAB B ◦ (P Ap A Q) = ( B Uλ B ) ◦ ( AσA B BU ) .Finally we compute(Bσ BAAF ) A ◦ (Q B p B P AF ) ◦ (p QB P A) (97)= ( A Uλ AA F ) ◦ ( BσBAU A A F ) ◦ (p QB P A)= m A ◦ ( BσBA ) A ◦ (p QB P A) (93)= m A ◦ ( σ A A ) (80)= σ A ◦ ( )Qµ A P(93)= B σB A ◦ (p QB P ) ◦ ( )Qµ A (11)P = B σB A ◦ (p QB P ) ◦ ( )Q B Uµ A BPp Q=B σ A B ◦ ( Q B µ A BP)◦ (pQB P A) (13),(14)= B σ A B ◦ (Q B p B P AF ) ◦ (p QB P A) .Since B and Q preserve coequalizers, by Corollary <strong>2.</strong>12, also Q B preserves them sothat ( Q B µ A BP)◦ (pQB P A) is epi and we deduce thatBσ A BAAF = B σ A B.99