Lemma 87.25.5. Let $S$ be a scheme. Let $X$, $Y$ be locally Noetherian algebraic spaces over $S$. Let $T \subset |X|$ and $T' \subset |Y|$ be closed subsets. Let $a, b : X \to Y$ be morphisms of algebraic spaces over $S$ such that $a|_{X \setminus T} = b|_{X \setminus T}$, such that $|a|(T) \subset T'$ and $|b|(T) \subset T'$, and such that $a_{/T} = b_{/T}$ as morphisms $X_{/T} \to Y_{/T'}$. Then $a = b$.

**Proof.**
Consequence of the more general Lemma 87.25.4.
$\square$

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