Lemma 73.49.6. 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 presentation,

2. $p$ is flat,

3. $p$ is universally closed,

4. $q$ is locally of finite type,

5. $q$ is closed, and

6. $q$ 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 an isomorphism.

Proof. By Lemma 73.49.5 there exists an open subspace $W_1 \subset Z$ such that $f_{Z'}$ is surjective and flat if and only if $Z' \to Z$ factors through $W_1$. By Lemma 73.49.3 there exists an open subspace $W_2 \subset Z$ such that $f_{Z'}$ is a closed immersion if and only if $Z' \to Z$ factors through $W_2$. We claim that $W = W_1 \cap W_2$ works. Certainly, if $f_{Z'}$ is an isomorphism, then $Z' \to Z$ factors through $W$. Hence it suffices to show that $f_ W$ is an isomorphism. By construction $f_ W$ is a surjective flat closed immersion. In particular $f_ W$ is representable. Since a surjective flat closed immersion of schemes is an isomorphism (see Morphisms, Lemma 29.25.1) we win. (Note that actually $f_ W$ is locally of finite presentation, whence open, so you can avoid the use of this lemma if you like.) $\square$

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