Lemma 88.25.4. Let $S$ be a scheme. Let $X$, $W$ be algebraic spaces over $S$ with $X$ locally Noetherian. Let $T \subset |X|$ be a closed subset. Let $a, b : X \to W$ be morphisms of algebraic spaces over $S$ such that $a|_{X \setminus T} = b|_{X \setminus T}$ and such that $a_{/T} = b_{/T}$ as morphisms $X_{/T} \to W$. Then $a = b$.
Proof. Let $E$ be the equalizer of $a$ and $b$. Then $E$ is an algebraic space and $E \to X$ is locally of finite type and a monomorphism, see Morphisms of Spaces, Lemma 67.4.1. Our assumptions imply we can apply Lemma 88.25.3 to the two morphisms $f = \text{id} : X \to X$ and $g : E \to X$ and the closed subset $T$ of $|X|$. $\square$
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