The Stacks project

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

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. $\square$