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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