Lemma 75.6.5. Let S be a scheme. Let f : X \to Y be an affine morphism of algebraic spaces over S. For E in D_\mathit{QCoh}(\mathcal{O}_ Y) we have Rf_* Lf^* E = E \otimes ^\mathbf {L}_{\mathcal{O}_ Y} f_*\mathcal{O}_ X.
Proof. Since f is affine the map f_*\mathcal{O}_ X \to Rf_*\mathcal{O}_ X is an isomorphism (Cohomology of Spaces, Lemma 69.8.2). There is a canonical map E \otimes ^\mathbf {L} f_*\mathcal{O}_ X = E \otimes ^\mathbf {L} Rf_*\mathcal{O}_ X \to Rf_* Lf^* E adjoint to the map
Lf^*(E \otimes ^\mathbf {L} Rf_*\mathcal{O}_ X) = Lf^*E \otimes ^\mathbf {L} Lf^*Rf_*\mathcal{O}_ X \longrightarrow Lf^* E \otimes ^\mathbf {L} \mathcal{O}_ X = Lf^* E
coming from 1 : Lf^*E \to Lf^*E and the canonical map Lf^*Rf_*\mathcal{O}_ X \to \mathcal{O}_ X. To check the map so constructed is an isomorphism we may work locally on Y. Hence we may assume Y and therefore X is affine. In this case the problem reduces to the case of schemes (Derived Categories of Schemes, Lemma 36.5.3) via Lemma 75.4.2 and Remark 75.6.3. \square
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