Base change requires its actual map and hypotheses

Assume the Axiom of Choice (The Axiom of Choice) and the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$-indexed chain) as required by the cited cohomology-and-base-change theorem.

The cohomology and base-change map of Cohomology and base-change map is a comparison between the fibre of a higher direct image and the cohomology of a fibre of $f$: for $f:X\to S$, an ${\mathcal{O}}_X$-module $\mathcal{F}$, a point $s\in S$ and $q\ge 0$ it is the $\kappa (s)$-linear map ${\phi }_s^q\text{\hspace{0.17em}⁣}:(R^q{f}_{\ast }\mathcal{F})(s)⟶H^q(X_s,{\mathcal{F}}_s),$ where $(R^q{f}_{\ast }\mathcal{F})(s)=(R^q{f}_{\ast }\mathcal{F}{)}_s{\otimes }_{{\mathcal{O}}_{S,s}}\kappa (s)$ is the fibre of the higher direct image at $s$ (Fibre of a module sheaf at a point). It is not a licence to commute cohomology with arbitrary base change, and its hypotheses are exactly those of Cohomology and base change for proper flat coherent families: $f$ proper of finite presentation over an arbitrary base $S$, and $\mathcal{F}$ coherent and flat over $S$ (Flat and faithfully flat modules and ring homomorphisms). Under them the theorem states:

(i) ${\phi }_s^q$ is surjective if and only if it is an isomorphism, and then all base changes of $R^q{f}_{\ast }\mathcal{F}$ over a neighbourhood of $s$ are isomorphisms, so the criterion is checked in the degree $q$ whose fibre dimension is being computed;

(ii) assuming ${\phi }_s^q$ is surjective, $R^q{f}_{\ast }\mathcal{F}$ is locally free of finite rank in a neighbourhood of $s$ if and only if the adjacent map ${\phi }_s^{q-1}$ is surjective. The adjacent condition is automatic for $q=0$; local freeness alone does not imply surjectivity of ${\phi }_s^q$.

In particular the fibre dimension $h^q(s)={\dim }_{\kappa (s)}{H}^q(X_s,{\mathcal{F}}_s)$ equals the dimension of the fibre of $R^q{f}_{\ast }\mathcal{F}$ at $s$ whenever the corresponding surjectivity holds; equality of these dimensions alone does not imply surjectivity; the rank of a locally free $R^q{f}_{\ast }\mathcal{F}$ is not by itself a formula for $h^q$, and the degree shift in the local-freeness criterion (ii) must be respected.

The hypotheses are not automatic. Even a proper flat family with a coherent sheaf flat over the base can have jumping fibre dimensions. Let $k$ be a field, let $S=\mathrm{Spec}k[a]$ (The underlying space of an affine spectrum), let $X={\mathbb{P}}_S^1$ be the relative projective line (Relative projective space from standard charts) with twisting sheaves ${\mathcal{O}}_X(d)$ (Twisting sheaf on Proj) and projection $f:X\to S$, which is proper, flat and of finite presentation, and let $\mathcal{E}$ be the rank-two finite locally free ${\mathcal{O}}_X$-module (Locally free sheaves of finite rank) given by the extension $0\to {\mathcal{O}}_X(-2)\to \mathcal{E}\to {\mathcal{O}}_X\to 0$ whose extension class is $a$ times the generator $[1\mathrm{/}(x_0{x}_1)]$ of $H^1({\mathbb{P}}_k^1,\mathcal{O}(-2))=k$ (Generator cocycle for H1 of O(-2), Cohomology of O(d) on projective space). The construction of $\mathcal{E}$ and the computations of the fibre dimensions are carried out in the companion example An upper jump of h0 in a flat projective family of this frontier's examples page, where it is shown that $h^0({\mathcal{E}}_{(a)})=1\text{and}{h}^0({\mathcal{E}}_u)=0\text{ for every }u\mathrm{\ne }(a)\in S\mathrm{.}$ Now suppose that ${\phi }_{(a)}^0$ were surjective. Then by the theorem, (i) and (ii) with the degree $-1$ condition automatic for $q=0$, the sheaf $R^0{f}_{\ast }\mathcal{E}=f_{\ast }\mathcal{E}$ would be locally free of finite rank on a neighbourhood $U$ of the origin, with ${\phi }_u^0$ an isomorphism for every $u\in U$; the dimension of the fibre of $f_{\ast }\mathcal{E}$ at $u\in U$ is the locally constant rank, so $h^0({\mathcal{E}}_u)={\dim }_{\kappa (u)}(f_{\ast }\mathcal{E})(u)$ would be constant after shrinking $U$ around the origin to a constant-rank neighbourhood. This contradicts the displayed jump $h^0({\mathcal{E}}_{(a)})=1$, $h^0({\mathcal{E}}_u)=0$ for $u\mathrm{\ne }(a)$. Consequently ${\phi }_{(a)}^0$ is not surjective, and in particular not an isomorphism: properness and flatness of the family do not by themselves make the base-change map an isomorphism, and the rank of a locally free $R^q{f}_{\ast }\mathcal{F}$ cannot in general be used to compute $h^q$ without checking the comparison map.