Lemma 76.49.1. Consider a commutative diagram

$\xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[ld]^ q \\ & Z }$

of algebraic spaces. Assume that $p$ is locally of finite type and closed. Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is unramified.

Proof. By Morphisms of Spaces, Lemma 67.38.10 there exists an open subspace $U(f) \subset X$ which is the set of points where $f$ is unramified. Moreover, formation of $U(f)$ commutes with arbitrary base change. Let $W \subset Z$ be the open subspace (see Properties of Spaces, Lemma 66.4.8) with underlying set of points

$|W| = |Z| \setminus |p|\left(|X| \setminus |U(f)|\right)$

i.e., $z \in |Z|$ is a point of $W$ if and only if $f$ is unramified at every point of $X$ above $z$. Note that this is open because we assumed that $p$ is closed. Since the formation of $U(f)$ commutes with arbitrary base change we immediately see (using Properties of Spaces, Lemma 66.4.9) that $W$ has the desired universal property. $\square$

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