Lemma 73.11.14. The property $\mathcal{P}(f) =$“$f$ is an open immersion” is fpqc local on the base.

**Proof.**
We will use Lemma 73.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 66.12.1. Consider a cartesian diagram

of algebraic spaces over $S$ where $Z' \to Z$ is a surjective flat morphism of affine schemes, and $X' \to Z'$ is an open immersion. We have to show that $X \to Z$ is an open immersion. Note that $|X'| \subset |Z'|$ corresponds to an open subscheme $U' \subset Z'$ (isomorphic to $X'$) with the property that $\text{pr}_0^{-1}(U') = \text{pr}_1^{-1}(U')$ as open subschemes of $Z' \times _ Z Z'$. Hence there exists an open subscheme $U \subset Z$ such that $X' = (Z' \to Z)^{-1}(U)$, see Descent, Lemma 35.13.6. By Properties of Spaces, Proposition 65.17.1 we see that $X$ satisfies the sheaf condition for the fpqc topology. Now we have the fpqc covering $\mathcal{U} = \{ U' \to U\} $ and the element $U' \to X' \to X \in \check{H}^0(\mathcal{U}, X)$. By the sheaf condition we obtain a morphism $U \to X$ such that

is commutative. On the other hand, we know that for any scheme $T$ over $S$ and $T$-valued point $T \to X$ the composition $T \to X \to Z$ is a morphism such that $Z' \times _ Z T \to Z'$ factors through $U'$. Clearly this means that $T \to Z$ factors through $U$. In other words the map of sheaves $U \to X$ is bijective and we win. $\square$

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