The Stacks project

Proof. Fully faithfulness. Suppose we have $i \in I$ and objects $X_ i$, $Y_ i$ of $\mathcal{C}_{S_ i}$. Denote $X = X_ i \times _{S_ i} S$ and $Y = Y_ i \times _{S_ i} S$. Suppose given a morphism $f : X \to Y$ over $S$. We can choose a finite affine open covering $Y_ i = V_{i, 1} \cup \ldots \cup V_{i, m}$ such that $V_{i, j} \to Y_ i \to S_ i$ maps into an affine open $W_{i, j}$ of $S_ i$. Denote $Y = V_1 \cup \ldots \cup V_ m$ the induced affine open covering of $Y$. Since $f : X \to Y$ is quasi-compact (Schemes, Lemma 26.21.14) after increasing $i$ we may assume that there is a finite open covering $X_ i = U_{i, 1} \cup \ldots \cup U_{i, m}$ by quasi-compact opens such that the inverse image of $U_{i, j}$ in $Y$ is $f^{-1}(V_ j)$, see Lemma 32.4.11. By Lemma 32.10.1 applied to $f|_{f^{-1}(V_ j)}$ over $W_ j$ we may assume, after increasing $i$, that there is a morphism $f_{i, j} : V_{i, j} \to U_{i, j}$ over $S$ whose base change to $S$ is $f|_{f^{-1}(V_ j)}$. Increasing $i$ more we may assume $f_{i, j}$ and $f_{i, j'}$ agree on the quasi-compact open $U_{i, j} \cap U_{i, j'}$. Then we can glue these morphisms to get the desired morphism $f_ i : X_ i \to Y_ i$. This morphism is unique (up to increasing $i$) because this is true for the morphisms $f_{i, j}$.

To show that the functor is essentially surjective we argue in exactly the same way. Namely, suppose that $X$ is an object of $\mathcal{C}_ S$. Pick $i \in I$. We can choose a finite affine open covering $X = U_1 \cup \ldots \cup U_ m$ such that $U_ j \to X \to S \to S_ i$ factors through an affine open $W_{i, j} \subset S_ i$. Set $W_ j = W_{i, j} \times _{S_ i} S$. This is an affine open of $S$. By Lemma 32.10.1, after increasing $i$, we may assume there exist $U_{i, j} \to W_{i, j}$ of finite presentation whose base change to $W_ j$ is $U_ j$. After increasing $i$ we may assume there exist quasi-compact opens $U_{i, j, j'} \subset U_{i, j}$ whose base changes to $S$ are equal to $U_ j \cap U_{j'}$. Claim: after increasing $i$ we may assume the image of the morphism $U_{i, j, j'} \to U_{i, j} \to W_{i, j}$ ends up in $W_{i, j} \cap W_{i, j'}$. Namely, because the complement of $W_{i, j} \cap W_{i, j'}$ is closed in the affine scheme $W_{i, j}$ it is affine. Since $U_ j \cap U_{j'} = \mathop{\mathrm{lim}}\nolimits U_{i, j, j'}$ does map into $W_{i, j} \cap W_{i, j'}$ we can apply Lemma 32.4.9 to get the claim. Thus we can view both

as schemes over $W_{i, j'}$ whose base changes to $W_{j'}$ recover $U_ j \cap U_{j'}$. Hence after increasing $i$, using Lemma 32.10.1, we may assume there are isomorphisms $U_{i, j, j'} \to U_{i, j', j}$ over $W_{i, j'}$ and hence over $S_ i$. Increasing $i$ further (details omitted) we may assume these isomorphisms satisfy the cocycle condition mentioned in Schemes, Section 26.14. Applying Schemes, Lemma 26.14.1 we obtain an object $X_ i$ of $\mathcal{C}_{S_ i}$ whose base change to $S$ is isomorphic to $X$; we omit some of the verifications. $\square$