The Stacks project

Lemma 35.23.8. The property $\mathcal{P}(f) =$“$f$ is universally injective” is fpqc local on the base.

Proof. A base change of a universally injective morphism is universally injective (this is formal). Being universally injective is Zariski local on the base; this is clear from the definition. Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ 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 $S' \to S$ is surjective and by the discussion in Schemes, Section 26.13 there exists a field extension $K'/K$ and a morphism $\mathop{\mathrm{Spec}}(K') \to S'$ such that the following solid diagram commutes

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[rrd] \ar@{-->}[rd]_{a', b'} \ar[dd] \\ & X' \ar[r] \ar[d] & S' \ar[d] \\ \mathop{\mathrm{Spec}}(K) \ar[r]^{a, b} & X \ar[r] & S } \]

As the square is cartesian we get the two dotted arrows $a'$, $b'$ making the diagram commute. Since $X' \to S'$ is universally injective we get $a' = b'$, by Morphisms, Lemma 29.10.2. Clearly this forces $a = b$ (by the discussion in Schemes, Section 26.13). Therefore Lemma 35.22.4 applies and we win.

An alternative proof would be to use the characterization of a universally injective morphism as one whose diagonal is surjective, see Morphisms, Lemma 29.10.2. The lemma then follows from the fact that the property of being surjective is fpqc local on the base, see Lemma 35.23.7. (Hint: use that the base change of the diagonal is the diagonal of the base change.) $\square$

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