Lemma 86.9.3. Let $X_\lambda , \lambda \in \Lambda$ and $X = \mathop{\mathrm{colim}}\nolimits X_\lambda$ be as in Definition 86.9.1. Then $X_\lambda \to X$ is representable and a thickening.

Proof. The statement makes sense by the discussion in Spaces, Section 64.3 and 64.5. By Lemma 86.9.2 the morphisms $X_\lambda \to X$ are representable. Given $U \to X$ where $U$ is a scheme, then the discussion following Definition 86.9.1 shows that Zariski locally on $U$ the morphism factors through some $X_\mu$ with $\lambda \leq \mu$. In this case $U \times _ X X_\lambda = U \times _{X_\mu } X_\lambda$ so that $U \times _ X X_\lambda \to U$ is a base change of the thickening $X_\lambda \to X_\mu$. $\square$

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