Lemma 89.3.3. Let X, x_ i, U_ i \to X, u_ i be as in (89.3.0.1). If f : Y \to X corresponds to g_ i : Y_ i \to U_ i under F, then Y_{x_ i} \cong (Y_ i)_{u_ i} as algebraic spaces.
Proof. This is clear because u_ i \to x_ i is an isomorphism. \square
Lemma 89.3.3. Let X, x_ i, U_ i \to X, u_ i be as in (89.3.0.1). If f : Y \to X corresponds to g_ i : Y_ i \to U_ i under F, then Y_{x_ i} \cong (Y_ i)_{u_ i} as algebraic spaces.
Proof. This is clear because u_ i \to x_ i is an isomorphism. \square
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