Lemma 15.90.23. Let $A \to B$ be a ring map. Let $I \subset A$ be an ideal. The inverse image of $D_{comp}(A, I)$ under the restriction functor $D(B) \to D(A)$ is $D_{comp}(B, IB)$.

Proof. Using Lemma 15.90.2 we see that $L \in D(B)$ is in $D_{comp}(B, IB)$ if and only if $T(L, f)$ is zero for every local section $f \in I$. Observe that the cohomology of $T(L, f)$ is computed in the category of abelian groups, so it doesn't matter whether we think of $f$ as an element of $A$ or take the image of $f$ in $B$. The lemma follows immediately from this and the definition of derived complete objects. $\square$

Comment #6421 by Peng DU on

Need add "in" after "is" in the statement.

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