Lemma 76.12.3. In Situation 76.12.1 let $K$ be as in Lemma 76.12.2. For any perfect object $E$ of $D(\mathcal{O}_ X)$ we have

$M = R\Gamma (X, K \otimes ^\mathbf {L} E)$ is a perfect object of $D(A)$ and there is a canonical isomorphism $R\Gamma (X_ n, \mathcal{F}_ n \otimes ^\mathbf {L} E|_{X_ n}) = M \otimes _ A^\mathbf {L} A_ n$ in $D(A_ n)$,

$N = R\mathop{\mathrm{Hom}}\nolimits _ X(E, K)$ is a perfect object of $D(A)$ and there is a canonical isomorphism $R\mathop{\mathrm{Hom}}\nolimits _{X_ n}(E|_{X_ n}, \mathcal{F}_ n) = N \otimes _ A^\mathbf {L} A_ n$ in $D(A_ n)$.

In both statements $E|_{X_ n}$ denotes the derived pullback of $E$ to $X_ n$.

**Proof.**
Proof of (2). Write $E_ n = E|_{X_ n}$ and $N_ n = R\mathop{\mathrm{Hom}}\nolimits _{X_ n}(E_ n, \mathcal{F}_ n)$. Recall that $R\mathop{\mathrm{Hom}}\nolimits _{X_ n}(-, -)$ is equal to $R\Gamma (X_ n, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (-, -))$, see Cohomology on Sites, Section 21.36. Hence by Derived Categories of Spaces, Lemma 74.25.8 we see that $N_ n$ is a perfect object of $D(A_ n)$ whose formation commutes with base change. Thus the maps $N_ n \otimes _{A_ n}^\mathbf {L} A_{n - 1} \to N_{n - 1}$ coming from $\varphi _ n$ are isomorphisms. By More on Algebra, Lemma 15.97.3 we find that $R\mathop{\mathrm{lim}}\nolimits N_ n$ is perfect and that its base change back to $A_ n$ recovers $N_ n$. On the other hand, the exact functor $R\mathop{\mathrm{Hom}}\nolimits _ X(E, -) : D_\mathit{QCoh}(\mathcal{O}_ X) \to D(A)$ of triangulated categories commutes with products and hence with derived limits, whence

\[ R\mathop{\mathrm{Hom}}\nolimits _ X(E, K) = R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathrm{Hom}}\nolimits _ X(E, \mathcal{F}_ n) = R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathrm{Hom}}\nolimits _ X(E_ n, \mathcal{F}_ n) = R\mathop{\mathrm{lim}}\nolimits N_ n \]

This proves (2). To see that (1) holds, translate it into (2) using Cohomology on Sites, Lemma 21.48.4.
$\square$

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