Lemma 76.9.3. Let $S$ be a scheme. Let $X \subset X'$ be a thickening of algebraic spaces over $S$. For any open subspace $U \subset X$ there exists a unique open subspace $U' \subset X'$ such that $U = X \times _{X'} U'$.
Proof. Let $U' \to X'$ be the object of $X'_{spaces, {\acute{e}tale}}$ corresponding to the object $U \to X$ of $X_{spaces, {\acute{e}tale}}$ via (76.9.1.1). The morphism $U' \to X'$ is étale and universally injective, hence an open immersion, see Morphisms of Spaces, Lemma 67.51.2. $\square$
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