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

$\xymatrix{ X' \ar[r] \ar[d] & X \ar[d] \\ Z' \ar[r] & Z }$

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

$\xymatrix{ U' \ar[r] \ar[d]^{\cong } \ar@/_3ex/[dd] & U \ar[d] \ar@/^3ex/[dd] \\ X' \ar[r] \ar[d] & X \ar[d] \\ Z' \ar[r] & Z }$

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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