Lemma 88.25.3. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Z \to Y$ be morphisms of algebraic spaces. Let $T \subset |X|$ be closed. Assume that

$X$ is locally Noetherian,

$g$ is a monomorphism and locally of finite type,

$f|_{X \setminus T} : X \setminus T \to Y$ factors through $g$, and

$f_{/T} : X_{/T} \to Y$ factors through $g$,

then $f$ factors through $g$.

**Proof.**
Consider the fibre product $E = X \times _ Y Z \to X$. By assumption the open immersion $X \setminus T \to X$ factors through $E$ and any morphism $\varphi : X' \to X$ with $|\varphi |(|X'|) \subset T$ factors through $E$ as well, see Formal Spaces, Section 87.14. By More on Morphisms of Spaces, Lemma 76.20.3 this implies that $E \to X$ is étale at every point of $E$ mapping to a point of $T$. Hence $E \to X$ is an étale monomorphism, hence an open immersion (Morphisms of Spaces, Lemma 67.51.2). Then it follows that $E = X$ since our assumptions imply that $|X| = |E|$.
$\square$

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