Lemma 37.72.4. Let $f : X \to Y$ be a proper morphism of schemes. Assume

1. for all $y \in Y$ we have $\dim (X_ y) \leq 1$ and $H^1(X_ y, \mathcal{O}_{X_ y}) = 0$, and

2. $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$ is surjective.

Then $\mathcal{O}_{Y'} \to f'_*\mathcal{O}_{X'}$ is surjective for any base change $f' : X' \to Y'$ of $f$.

Proof. We may assume $Y$ and $Y'$ affine. Then we can choose a closed immersion $Y' \to Y''$ with $Y'' \to Y$ a flat morphism of affines. By flat base change (Cohomology of Schemes, Lemma 30.5.2) we see that the result holds for $X'' \to Y''$. Thus we may assume $Y'$ is a closed subscheme of $Y$. Let $\mathcal{I} \subset \mathcal{O}_ Y$ be the ideal cutting out $Y'$. Then there is a short exact sequence

$0 \to \mathcal{I}\mathcal{O}_ X \to \mathcal{O}_ X \to \mathcal{O}_{X'} \to 0$

where we view $\mathcal{O}_{X'}$ as a quasi-coherent module on $X$. By Lemma 37.72.3 we have $H^1(X, \mathcal{I}\mathcal{O}_ X) = 0$. It follows that

$H^0(Y, \mathcal{O}_ Y) \to H^0(Y, f_*\mathcal{O}_ X) = H^0(X, \mathcal{O}_ X) \to H^0(X, \mathcal{O}_{X'})$

is surjective as desired. The first arrow is surjective as $Y$ is affine and since we assumed $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$ is surjective and the second by the long exact sequence of cohomology associated to the short exact sequence above and the vanishing just proved. $\square$

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