Lemma 26.17.3. Let $f : X \to S$ and $g : Y \to S$ be morphisms of schemes with the same target. Let $X \times _ S Y$, $p$, $q$ be the fibre product. Suppose that $U \subset S$, $V \subset X$, $W \subset Y$ are open subschemes such that $f(V) \subset U$ and $g(W) \subset U$. Then the canonical morphism $V \times _ U W \to X \times _ S Y$ is an open immersion which identifies $V \times _ U W$ with $p^{-1}(V) \cap q^{-1}(W)$.
Proof. Let $T$ be a scheme. Suppose $a : T \to V$ and $b : T \to W$ are morphisms such that $f \circ a = g \circ b$ as morphisms into $U$. Then they agree as morphisms into $S$. By the universal property of the fibre product we get a unique morphism $T \to X \times _ S Y$. Of course this morphism has image contained in the open $p^{-1}(V) \cap q^{-1}(W)$. Thus $p^{-1}(V) \cap q^{-1}(W)$ is a fibre product of $V$ and $W$ over $U$. The result follows from the uniqueness of fibre products, see Categories, Section 4.6. $\square$
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