Lemma 38.28.3. In Situation 38.28.1 let K be as in Lemma 38.28.2. 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, Section 20.44. Hence by Derived Categories of Schemes, Lemma 36.30.7 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, Lemma 20.50.5.
\square
Comments (0)