Cohomology and Base Change - Degree 0 Sanity Check

Question

Both Vakil and Hartshorne describe Cohomology and Base Change in the following way:

Suppose $f:X \rightarrow Y$ is a projective (in Vakil, proper) morphism of Noetherian schemes, $F$ a coherent sheaf on $X$ flat over $Y.$ If the natural map $\phi^i(y): R^if_*(F) \rightarrow H^i(X_y,F_y)$ is surjective, then $\phi^{i-1}(y)$ is also surjective if and only if $R^if_*(F)$ is locally free near $y$.

An important special case would be when $i=0$. I'm nervous about the fact that neither author singles out this special case, and my silly question is: does the hypothesis "$\phi^{-1}(y)$ is surjective" hold vacuously when $i=0$? In other words, is surjectivity of $\phi^0(y)$ enough to conclude local freeness of $f_*F$ near $y$? Thanks.

Answer 1

Since the posting of this question, Vakil's text has been amended to address this exact issue. Here's the wording of the theorem in the November 18, 2017 version, which is the most recent as of the posting of this answer:

28.1.6 Cohomology and Base Change Theorem. Suppose $\pi:X\to Y$ is proper, $Y$ is locally Noetherian, $\mathcal{F}$ is coherent over $X$ and flat over $Y$, and $\phi_q^p: R^p\pi_*\mathcal{F}\otimes\kappa(q)\to H^p(X_q,\mathcal{F}|_{X_q})$ is surjective. Then the following hold.

(i) There is an open neighborhood $U$ of $q$ such that for any $\psi:Z\to U$, $\phi_Z^p$ is an isomorphism. In particular, $\phi_q^p$ is an isomorphism.

(ii) Furthermore, $\phi_q^{p-1}$ is surjective (hence an isomorphism by (i)) if and only if $R^p\pi_*\mathcal{F}$ is locally free in some open neighborhood of $q$ (or equivalently, $(R^p\pi_*\mathcal{F})_q$ is a free $\mathcal{O}_{Y,q}$-module, Exercise 13.7.F). This in turn implies that $h^p$ is constant in an open neighborhood of $q$.

(Proofs of Theorems 28.1.5 and 28.1.6 will be given in §28.2. Note in (ii) that if $p=0$, $\phi_q^{p-1}$ is automatically surjective, as the codomain of $\phi_q^{-1}$ is $H^{-1}(X_q,\mathcal{F}|_{X_q})$, which is $0$ by definition.)

(My emphasis added on the final sentence.)