Lemma 76.49.3. Consider a commutative diagram

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

of algebraic spaces. Assume that

1. $p$ is locally of finite type,

2. $p$ is universally closed, and

3. $q : Y \to Z$ is separated.

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 a closed immersion.

Proof. We will use the characterization of closed immersions as universally closed, unramified, and universally injective morphisms, see Lemma 76.14.9. First, note that since $p$ is universally closed and $q$ is separated, we see that $f$ is universally closed, see Morphisms of Spaces, Lemma 67.40.6. It follows that any base change of $f$ is universally closed, see Morphisms of Spaces, Lemma 67.9.3. Thus to finish the proof of the lemma it suffices to prove that the assumptions of Lemma 76.49.2 are satisfied. The projection $\text{pr}_0 : X \times _ Y X \to X$ is universally closed as a base change of $f$, see Morphisms of Spaces, Lemma 67.9.3. Hence $X \times _ Y X \to Z$ is universally closed as a composition of universally closed morphisms (see Morphisms of Spaces, Lemma 67.9.4). This finishes the proof of the lemma. $\square$

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