The Stacks project

Proof. After replacing $Z$ by the open subspace found in Lemma 76.49.1 we may assume that $f$ is already unramified; note that this does not destroy assumption (2) or (3). By Morphisms of Spaces, Lemma 67.38.9 we see that $\Delta _{X/Y} : X \to X \times _ Y X$ is an open immersion. This remains true after any base change. Hence by Morphisms of Spaces, Lemma 67.19.2 we see that $f_{Z'}$ is universally injective if and only if the base change of the diagonal $X_{Z'} \to (X \times _ Y X)_{Z'}$ is an isomorphism. Let $W \subset Z$ be the open subspace (see Properties of Spaces, Lemma 66.4.8) with underlying set of points

i.e., $z \in |Z|$ is a point of $W$ if and only if the fibre of $|X \times _ Y X| \to |Z|$ over $z$ is in the image of $|X| \to |X \times _ Y X|$. Then it is clear from the discussion above that the restriction $p^{-1}(W) \to q^{-1}(W)$ of $f$ is unramified and universally injective.

Conversely, suppose that $f_{Z'}$ is unramified and universally injective. In order to show that $Z' \to Z$ factors through $W$ it suffices to show that $|Z'| \to |Z|$ has image contained in $|W|$, see Properties of Spaces, Lemma 66.4.9. Hence it suffices to prove the result when $Z'$ is the spectrum of a field. Denote $z \in |Z|$ the image of $|Z'| \to |Z|$. The discussion above shows that

is surjective. By Properties of Spaces, Lemma 66.4.3 in the commutative diagram

the vertical arrows are surjective. It follows that $z \in |W|$ as desired. $\square$