Lemma 29.9.4. The base change of a surjective morphism is surjective.

Proof. Let $f: X \to Y$ be a morphism of schemes over a base scheme $S$. If $S' \to S$ is a morphism of schemes, let $p: X_{S'} \to X$ and $q: Y_{S'} \to Y$ be the canonical projections. The commutative square

$\xymatrix{ X_{S'} \ar[d]_{f_{S'}} \ar[r]_ p & X \ar[d]^{f} \\ Y_{S'} \ar[r]^{q} & Y. }$

identifies $X_{S'}$ as a fibre product of $X \to Y$ and $Y_{S'} \to Y$. Let $Z$ be a subset of the underlying topological space of $X$. Then $q^{-1}(f(Z)) = f_{S'}(p^{-1}(Z))$, because $y' \in q^{-1}(f(Z))$ if and only if $q(y') = f(x)$ for some $x \in Z$, if and only if, by Lemma 29.9.3, there exists $x' \in X_{S'}$ such that $f_{S'}(x') = y'$ and $p(x') = x$. In particular taking $Z = X$ we see that if $f$ is surjective so is the base change $f_{S'}: X_{S'} \to Y_{S'}$. $\square$

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).