Lemma 32.22.3. In Situation 32.22.1. Let $X \to S$ be quasi-separated and of finite type. Given $i \in I$ and a diagram
as in (32.22.2.1) for $i' \geq i$ let $X_{i'}$ be the scheme theoretic image of $X \to S_{i'} \times _{S_ i} W$. Then $X = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} X_{i'}$.
Proof. Since $X$ is quasi-compact and quasi-separated formation of the scheme theoretic image of $X \to S_{i'} \times _{S_ i} W$ commutes with restriction to open subschemes (Morphisms, Lemma 29.6.3). Hence we may and do assume $W$ is affine and maps into an affine open $U_ i$ of $S_ i$. Let $U \subset S$, $U_{i'} \subset S_{i'}$ be the inverse image of $U_ i$. Then $U$, $U_{i'}$, $S_{i'} \times _{S_ i} W = U_{i'} \times _{U_ i} W$, and $S \times _{S_ i} W = U \times _{U_ i} W$ are all affine. This implies $X$ is affine because $X \to S \times _{S_ i} W$ is a closed immersion. This also shows the ring map
is surjective. Let $I$ be the kernel. Then we see that $X_{i'}$ is the spectrum of the ring
where $I_{i'}$ is the inverse image of the ideal $I$ (see Morphisms, Example 29.6.4). Since $\mathcal{O}(U) = \mathop{\mathrm{colim}}\nolimits \mathcal{O}(U_{i'})$ we see that $I = \mathop{\mathrm{colim}}\nolimits I_{i'}$ and we conclude that $\mathop{\mathrm{colim}}\nolimits \mathcal{O}(X_{i'}) = \mathcal{O}(X)$. $\square$
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