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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #6221 by Yuto Masamura on

Comment #6357 by Johan on

There are also: