Lemma 109.23.1. Let S be a scheme and s \in S a point. Let f : X \to S and g : Y \to S be families of curves. Let c : X \to Y be a morphism over S. If c_{s, *}\mathcal{O}_{X_ s} = \mathcal{O}_{Y_ s} and R^1c_{s, *}\mathcal{O}_{X_ s} = 0, then after replacing S by an open neighbourhood of s we have \mathcal{O}_ Y = c_*\mathcal{O}_ X and R^1c_*\mathcal{O}_ X = 0 and this remains true after base change by any morphism S' \to S.
Proof. Let (U, u) \to (S, s) be an étale neighbourhood such that \mathcal{O}_{Y_ U} = (X_ U \to Y_ U)_*\mathcal{O}_{X_ U} and R^1(X_ U \to Y_ U)_*\mathcal{O}_{X_ U} = 0 and the same is true after base change by U' \to U. Then we replace S by the open image of U \to S. Given S' \to S we set U' = U \times _ S S' and we obtain étale coverings \{ U' \to S'\} and \{ Y_{U'} \to Y_{S'}\} . Thus the truth of the statement for the base change of c by S' \to S follows from the truth of the statement for the base change of X_ U \to Y_ U by U' \to U. In other words, the question is local in the étale topology on S. Thus by Lemma 109.4.3 we may assume X and Y are schemes. By More on Morphisms, Lemma 37.72.7 there exists an open subscheme V \subset Y containing Y_ s such that c_*\mathcal{O}_ X|_ V = \mathcal{O}_ V and R^1c_*\mathcal{O}_ X|_ V = 0 and such that this remains true after any base change by S' \to S. Since g : Y \to S is proper, we can find an open neighbourhood U \subset S of s such that g^{-1}(U) \subset V. Then U works. \square
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