Lemma 66.22.3. Let $S$ be a scheme. Let $f : X \to Y$ be a birational morphism of algebraic spaces over $S$ which are decent and have finitely many irreducible components. If $y \in Y$ is the generic point of an irreducible component, then the base change $X \times _ Y \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}) \to \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y})$ is an isomorphism.

**Proof.**
Let $X' \subset X$ and $Y' \subset Y$ be the maximal open subspaces which are representable, see Lemma 66.20.4. By Lemma 66.21.3 the fibre of $f$ over $y$ is consists of points of codimension $0$ of $X$ and is therefore contained in $X'$. Hence $X \times _ Y \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}) = X' \times _{Y'} \mathop{\mathrm{Spec}}(\mathcal{O}_{Y', y})$ and the result follows from Morphisms, Lemma 29.49.3.
$\square$

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