Lemma 36.29.3. In Situation 36.29.1 the category of perfect objects of $D(\mathcal{O}_ S)$ is the colimit of the categories of perfect objects of $D(\mathcal{O}_{S_ i})$.

**Proof.**
For every open $U_0 \subset S_0$ consider the condition $P$ that the functor

is an equivalence where ${}_{perf}$ indicates the full subcategory of perfect objects and 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. First, we observe that we already know the functor is fully faithful by Lemma 36.29.2. Thus it suffices to prove essential surjectivity.

We first check condition (2) of the induction principle. Thus suppose that we have $S_0 = U_0 \cup V_0$ and that $P$ holds for $U_0$, $V_0$, and $U_0 \cap V_0$. Let $E$ be a perfect object of $D(\mathcal{O}_ S)$. We can find $i \geq 0$ and $E_{U, i}$ perfect on $U_ i$ and $E_{V, i}$ perfect on $V_ i$ whose pullback to $U$ and $V$ are isomorphic to $E|_ U$ and $E|_ V$. Denote

the maps adjoint to the isomorphisms $Lf_ i^*E_{U, i} \to E|_ U$ and $Lf_ i^*E_{V, i} \to E|_ V$. By fully faithfulness, after increasing $i$, we can find an isomorphism $c : E_{U, i}|_{U_ i \cap V_ i} \to E_{V, i}|_{U_ i \cap V_ i}$ which pulls back to the identifications

Apply Cohomology, Lemma 20.45.1 to get an object $E_ i$ on $S_ i$ and a map $d : E_ i \to Rf_{i, *}E$ which restricts to the maps $a$ and $b$ over $U_ i$ and $V_ i$. Then it is clear that $E_ i$ is perfect and that $d$ is adjoint to an isomorphism $Lf_ i^*E_ i \to E$.

Finally, we check condition (1) of the induction principle, in other words, we check the lemma holds when $S_0$ is affine. Say $S_0 = \mathop{\mathrm{Spec}}(A_0)$, $S_ i = \mathop{\mathrm{Spec}}(A_ i)$, and $S = \mathop{\mathrm{Spec}}(A)$. Using Lemmas 36.3.5 and 36.10.7 we see that we have to show that

This is clear from the fact that perfect complexes over rings are given by finite complexes of finite projective (hence finitely presented) modules. See More on Algebra, Lemma 15.74.17 for details. $\square$

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