Lemma 29.8.5. Let $f : X \to S$ be a dominant morphism of schemes. Let $g : S' \to S$ be an open morphism of schemes, and let $f' : X' \to S'$ be the base change of $f$ by $g$. Then $f'$ is dominant.
Proof. If $f'$ is not dominant, then $f'(X')$ is contained in some closed subset $B$ not equal to $S'$. Then $S'-B$ is open in $S'$ and nonempty. Since $g$ is open, $g(S'-B)$ is nonempty and open in $S$. Since $f(X)$ is dense in $S$, there is a point $x\in X$ with $f(x) \in g(S'-B)$. Let $W = \mathop{\mathrm{Spec}}(\kappa (f(x)))$, and write $X_ W$, $S_ W$, $S'_ W$ for the pullbacks of these schemes from $S$ to $W$. Then $X_ W$ and $(S'-B)_ W$ are nonempty, and so $X_ W \times _ W (S'-B)_ W$ is nonempty, as $W$ is the spectrum of a field. This contradicts that $f'(X \times _ S S')$ is contained in $B \subset S'$. So in fact $f'$ is dominant. $\square$
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