Proof. By a formal argument with base change, this reduces to the following question: A morphism $a : X' \to X$ of algebraic spaces such that $U \times _ X X' \to U$ and $Y \times _ X X' \to Y$ are isomorphisms, is an isomorphism. The family $\{ U \to X, Y \to X\}$ can be refined by an fpqc covering by Lemma 80.10.11. Hence the result follows from Descent on Spaces, Lemma 73.11.15. $\square$

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