Lemma 74.11.16. The property $\mathcal{P}(f) =$“$f$ is affine” is fpqc local on the base.

**Proof.**
We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.20.3. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times _ Z X \to Z'$ is affine. Let $X'$ be a scheme representing $Z' \times _ Z X$. We obtain a canonical isomorphism

since both schemes represent the algebraic space $Z' \times _ Z Z' \times _ Z X$. This is a descent datum for $X'/Z'/Z$, see Descent, Definition 35.34.1 (verification omitted, compare with Descent, Lemma 35.39.1). Since $X' \to Z'$ is affine this descent datum is effective, see Descent, Lemma 35.37.1. Thus there exists a scheme $Y \to Z$ over $Z$ and an isomorphism $\psi : Z' \times _ Z Y \to X'$ compatible with descent data. Of course $Y \to Z$ is affine (by construction or by Descent, Lemma 35.23.18). Note that $\mathcal{Y} = \{ Z' \times _ Z Y \to Y\} $ is a fpqc covering, and interpreting $\psi $ as an element of $X(Z' \times _ Z Y)$ we see that $\psi \in \check{H}^0(\mathcal{Y}, X)$. By the sheaf condition for $X$ with respect to this covering (see Properties of Spaces, Proposition 66.17.1) we obtain a morphism $Y \to X$. By construction the base change of this to $Z'$ is an isomorphism, hence an isomorphism by Lemma 74.11.15. This proves that $X$ is representable by an affine scheme and we win. $\square$

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