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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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30Since, by Propositi<strong>on</strong> 3.30, (Q B , p Q ) = Coequ Fun(µBQ BU,Q B Uλ B), we get thatAU ( A Q) B= Q B and A Up A Q = p Q .MoreoverAUλ A ( A Q) B: A A U ( A Q) B= AQ B → A U ( A Q) B= Q B .By Propositi<strong>on</strong> 3.30, we know that ( )Q B , A µ QB is a left A-module functor andAUλ AA (Q B ) = A µ QB . Hence we geti.e.AUλ A ( A Q) B= A µ QB(8)= A Uλ AA (Q B )( A Q) B= A (Q B ) .Notati<strong>on</strong> 3.3<str<strong>on</strong>g>2.</str<strong>on</strong>g> Let A = (A, m A , u A ) be a m<strong>on</strong>ad over a category A and letB = (B, m B , u B ) be a m<strong>on</strong>ad over a category B. Assume that both A and B havecoequalizers and A preserves them. Let Q : B → A be an A-B-bimodule functor. Inview of Propositi<strong>on</strong> 3.31, we setAQ B = ( A Q) B= A (Q B ) .Lemma 3.33. Let B = (B, m B , u B ) be a m<strong>on</strong>ad over a category B and assume that Bhave coequalizers. Let ( Q : B → A, µ B Q)be a right B-module functor. With notati<strong>on</strong>sof Propositi<strong>on</strong> 3.30 we have that(13) Q B λ BB F = µ B Q.Furthermore, if we assume that the functors Q, B preserve coequalizers we also have(14) Q B λ BB P = p Q BP .Proof. Let us c<strong>on</strong>sider the following diagramQB B U B F B U B Fµ B QBU B F B U B F Q B Uλ B B F B U B FQ B U B F B U B F p QBF B U B F Q BB F B U B F□QB B Uλ B B FQ B Uλ B B FQ B λ B B FQB B U B Fµ B QBU B FQ B Uλ B B F Q B U B Fp QB F Q BB FNote that QB B Uλ BB F = QBm B and Q B Uλ BB F = Qm B so that the left squareserially commutes because of the associativity of m B and of µ B Q . Both the rows arecoequalizers in view of the dual versi<strong>on</strong> of Lemma <str<strong>on</strong>g>2.</str<strong>on</strong>g>10 so that, by the universal propertyof coequalizers, there exists a unique functorial morphism ζ : Q BB F B U B F →Q BB F such that ζ ◦ (p QB F B U B F ) = (p QB F ) ◦ (Q B Uλ BB F ). Since p Q : Q B U → Q B isa functorial morphism, we know that Q B λ BB F makes the right square be commutative,but since by (15) we have p QB F = µ B Q we also have that µB Q makesthe right square commute. Therefore, we deduce that ζ = Q B λ BB F = µ B Q . Assumingthat Q and B preserve coequalizers, by Lemma 3.21, we get that B Ualso preserves coequalizers so that, in view of Corollary <str<strong>on</strong>g>2.</str<strong>on</strong>g>12 we also have that

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