This question is related to a question I asked a few days ago. Since there seems to be no (at least for me) satisfying reference for cohomology and base change as stated by Vakil in his script in exercise 28.2.M (or below), I would like to record a proof in this post. But I am no expert and not sure about some steps. Let me start with the statement.

Let $\pi \colon X \to Y$ be a proper finitely presented morphism, $\mathscr F$ coherent over $X$ and flat over $Y$ and the base change map $\phi^p_y$ be surjective at $y \in Y$. Then the following hold.

- There is an open neighborhood $U$ of $y$ such that for any $\phi \colon Z \to U$, the base change $\phi^p_Z$ is an isomorphism.
- Furthermore, $\phi^{p-1}_y$ is surjective if and only if $R^p\pi_*\mathscr F$ is locally free in some neighborhood of $y$.

The idea of the proof (as indicated by Vakil) is to reduce to the case of a locally Noetherian base using techniques from Grothendieck. As the statement is local we can assume that $Y = \operatorname{Spec} R$ is affine. Now we get the following Cartesian diagram $\require{AMScd}$ \begin{CD} X @>\alpha>> X_0\\ @V \pi V V @VV \pi_0 V\\ Y @>>\alpha_0> Y_0 \end{CD} where $Y_0$ is Noetherian and affine (in fact a $\mathbb Z$-subalgebra $R_0$ of $R$) and $\pi_0$ is proper. Moreover there is a coherent sheaf $\mathscr F_0$ on $X_0$ flat over $Y_0$ such that $\mathscr F$ is the pullback of $\mathscr F_0$ by $\alpha$. Now let $y \in Y$ be a point (we identify $y$ with $\operatorname{Spec} k(y)$) and let $y_0$ denote the image under $\alpha_0$. We get the two Cartesian squares $\require{AMScd}$ \begin{CD} X_y @>>> X_{0,y_0} @>>> X_0\\ @VVV @VVV @VV\pi_0V\\ y @>>> y_0 @>>> Y_0 \end{CD} Since the map at the bottom left is flat, the base change map of the left square is an isomorphism. Hence the base change map of the right square is an isomorphism if and only the base change map of the outer diagram is an isomorphism. This should imply (here I am not sure) that the morphism $\phi_y$ in the statement of the theorem is surjective / an isomorphism if and only if $\phi_{y_0}$ is surjective / an isomorphism (this would certainly be the case if it is true that $\phi_y = \phi_{y_0} \otimes k(y)$).

Hence the assumption of the theorem holds at $y$ if and only if it holds at $y_0$ and we assume now that this is the case. By cohomology and base change for a Noetherian base there is an open subset $U$ of $Y_0$ such that the base change map with any morphism mapping to $U$ is an isomorphism. Then the same statement is certainly true for $\alpha_0^{-1}(U)$ and after possibly shrinking $Y$ and $Y_0$ we may assume that the property holds for those open sets.

It is left to show that the second assertion of the theorem holds. Similarly to before we get that $\phi_y^{p-1}$ is surjective if and only if $\phi_{y_0}^{p-1}$ is surjective. Since the base change map associated with $\alpha_0$ is an isomorphism we see that $R^p\pi_*\mathscr F$ is locally free if $R^p\pi_{0*}\mathscr F_0$ is locally free (since it is a pullback of the last sheaf). But I have no idea how I would get the reverse implication.

I would be really grateful if someone could fill in the missing steps (or point out where I did something completely wrong). I am especially insecure about what I am doing since Conrad in *Conrad, Brian*, **Grothendieck duality and base change**, Lecture Notes in Mathematics. 1750. Berlin: Springer. x, 296 p. (2000). ZBL0992.14001. seems to use more sophisticated arguments to arrive at weaker statement in section 5.1