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Tag 01JR

Chapter 25: Schemes > Section 25.17: Fibre products of schemes

Lemma 25.17.3. 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. $\square$

    The code snippet corresponding to this tag is a part of the file schemes.tex and is located in lines 3040–3050 (see updates for more information).

    \begin{lemma}
    \label{lemma-open-fibre-product}
    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 $p^{-1}(V) \cap q^{-1}(W)$.
    \end{lemma}
    
    \begin{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
    \ref{categories-section-fibre-products}.
    \end{proof}

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