Lemma 74.11.7. The property $\mathcal{P}(f) =$“$f$ is universally injective” 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 67.9.5. Let $Z' \to Z$ be a flat surjective morphism of affine schemes over $S$ and let $f : X \to Z$ be a morphism from an algebraic space to $Z$. Assume that the base change $f' : X' \to Z'$ is universally injective. Let $K$ be a field, and let $a, b : \mathop{\mathrm{Spec}}(K) \to X$ be two morphisms such that $f \circ a = f \circ b$. As $Z' \to Z$ is surjective there exists a field extension $K'/K$ and a morphism $\mathop{\mathrm{Spec}}(K') \to Z'$ such that the following solid diagram commutes
\[ \xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[rrd] \ar@{-->}[rd]_{a', b'} \ar[dd] \\ & X' \ar[r] \ar[d] & Z' \ar[d] \\ \mathop{\mathrm{Spec}}(K) \ar[r]^{a, b} & X \ar[r] & Z } \]
As the square is cartesian we get the two dotted arrows $a'$, $b'$ making the diagram commute. Since $X' \to Z'$ is universally injective we get $a' = b'$. This forces $a = b$ as $\{ \mathop{\mathrm{Spec}}(K') \to \mathop{\mathrm{Spec}}(K)\} $ is an fpqc covering, see Properties of Spaces, Proposition 66.17.1. Hence $f$ is universally injective as desired. $\square$
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