Lemma 36.29.2. In Situation 36.29.1. Let $E_0$ and $K_0$ be objects of $D(\mathcal{O}_{S_0})$. Set $E_ i = Lf_{i0}^*E_0$ and $K_ i = Lf_{i0}^*K_0$ for $i \geq 0$ and set $E = Lf_0^*E_0$ and $K = Lf_0^*K_0$. Then the map
\[ \mathop{\mathrm{colim}}\nolimits _{i \geq 0} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_{S_ i})}(E_ i, K_ i) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ S)}(E, K) \]
is an isomorphism if either
$E_0$ is perfect and $K_0 \in D_\mathit{QCoh}(\mathcal{O}_{S_0})$, or
$E_0$ is pseudo-coherent and $K_0 \in D_\mathit{QCoh}(\mathcal{O}_{S_0})$ has finite tor dimension.
Proof.
For every open $U_0 \subset S_0$ consider the condition $P$ that the canonical map
\[ \mathop{\mathrm{colim}}\nolimits _{i \geq 0} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_{U_ i})}(E_ i|_{U_ i}, K_ i|_{U_ i}) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(E|_ U, K|_ U) \]
is an isomorphism, where $U = f_0^{-1}(U_0)$ and $U_ i = f_{i0}^{-1}(U_0)$. We will prove $P$ holds for all quasi-compact opens $U_0$ by the induction principle of Cohomology of Schemes, Lemma 30.4.1. Condition (2) of this lemma follows immediately from Mayer-Vietoris for hom in the derived category, see Cohomology, Lemma 20.33.3. Thus it suffices to prove the lemma when $S_0$ is affine.
Assume $S_0$ is affine. Say $S_0 = \mathop{\mathrm{Spec}}(A_0)$, $S_ i = \mathop{\mathrm{Spec}}(A_ i)$, and $S = \mathop{\mathrm{Spec}}(A)$. We will use Lemma 36.3.5 without further mention.
In case (1) the object $E_0^\bullet $ corresponds to a finite complex of finite projective $A_0$-modules, see Lemma 36.10.7. We may represent the object $K_0$ by a K-flat complex $K_0^\bullet $ of $A_0$-modules. In this situation we are trying to prove
\[ \mathop{\mathrm{colim}}\nolimits _{i \geq 0} \mathop{\mathrm{Hom}}\nolimits _{D(A_ i)}(E_0^\bullet \otimes _{A_0} A_ i, K_0^\bullet \otimes _{A_0} A_ i) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{D(A)}(E_0^\bullet \otimes _{A_0} A, K_0^\bullet \otimes _{A_0} A) \]
Because $E_0^\bullet $ is a bounded above complex of projective modules we can rewrite this as
\[ \mathop{\mathrm{colim}}\nolimits _{i \geq 0} \mathop{\mathrm{Hom}}\nolimits _{K(A_0)}(E_0^\bullet , K_0^\bullet \otimes _{A_0} A_ i) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{K(A_0)}(E_0^\bullet , K_0^\bullet \otimes _{A_0} A) \]
Since there are only a finite number of nonzero modules $E_0^ n$ and since these are all finitely presented modules, this map is an isomorphism.
In case (2) the object $E_0$ corresponds to a bounded above complex $E_0^\bullet $ of finite free $A_0$-modules, see Lemma 36.10.2. We may represent $K_0$ by a finite complex $K_0^\bullet $ of flat $A_0$-modules, see Lemma 36.10.4 and More on Algebra, Lemma 15.66.3. In particular $K_0^\bullet $ is K-flat and we can argue as before to arrive at the map
\[ \mathop{\mathrm{colim}}\nolimits _{i \geq 0} \mathop{\mathrm{Hom}}\nolimits _{K(A_0)}(E_0^\bullet , K_0^\bullet \otimes _{A_0} A_ i) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{K(A_0)}(E_0^\bullet , K_0^\bullet \otimes _{A_0} A) \]
It is clear that this map is an isomorphism (only a finite number of terms are involved since $K_0^\bullet $ is bounded).
$\square$
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