How is the canonical morphism $V \times_U W \to X \times_S Y$ an open immersion?

Question

I don't see where in this proof below, (from Stacks project), it is shown that $V \times_U W \to X \times_S Y$ is an open immersion.

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Answer 1

Clearly, $p^{-1}(V) \cap q^{-1}(W) \to X \times_S Y$ is an open immersion. By the uniqueness of fiber products one has $p^{-1}(V) \cap q^{-1}(W) \cong V \times_U W$, and the two maps $p^{-1}(V) \cap q^{-1}(W) \to X \times_S Y$ and $V \times_U W \to X \times_S Y$ are the same under that identification. Hence also $V \times_U W \to X \times_S Y$ is an open immersion.