Lemma 15.91.24. Let A \to B be a ring map. Let I \subset A be a finitely generated ideal. If A \to B is flat and A/I \cong B/IB, then the restriction functor D(B) \to D(A) induces an equivalence D_{comp}(B, IB) \to D_{comp}(A, I).
Proof. Choose generators f_1, \ldots , f_ r of I. Denote \check{\mathcal{C}}^\bullet _ A \to \check{\mathcal{C}}^\bullet _ B the quasi-isomorphism of extended alternating Čech complexes of Lemma 15.89.4. Let K \in D_{comp}(A, I). Let I^\bullet be a K-injective complex of A-modules representing K. Since \mathop{\mathrm{Ext}}\nolimits ^ n_ A(A_ f, K) and \mathop{\mathrm{Ext}}\nolimits ^ n_ A(B_ f, K) are zero for all f \in I and n \in \mathbf{Z} (Lemma 15.91.1) we conclude that \check{\mathcal{C}}^\bullet _ A \to A and \check{\mathcal{C}}^\bullet _ B \to B induce quasi-isomorphisms
and
Some details omitted. Since \check{\mathcal{C}}^\bullet _ A \to \check{\mathcal{C}}^\bullet _ B is a quasi-isomorphism and I^\bullet is K-injective we conclude that \mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet ) \to I^\bullet is a quasi-isomorphism. As the complex \mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet ) is a complex of B-modules we conclude that K is in the image of the restriction map, i.e., the functor is essentially surjective
In fact, the argument shows that F : D_{comp}(A, I) \to D_{comp}(B, IB), K \mapsto \mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet ) is a left inverse to restriction. Finally, suppose that L \in D_{comp}(B, IB). Represent L by a K-injective complex J^\bullet of B-modules. Then J^\bullet is also K-injective as a complex of A-modules (Lemma 15.56.1) hence F(\text{restriction of }L) = \mathop{\mathrm{Hom}}\nolimits _ A(B, J^\bullet ). There is a map J^\bullet \to \mathop{\mathrm{Hom}}\nolimits _ A(B, J^\bullet ) of complexes of B-modules, whose composition with \mathop{\mathrm{Hom}}\nolimits _ A(B, J^\bullet ) \to J^\bullet is the identity. We conclude that F is also a right inverse to restriction and the proof is finished. \square
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