Note: Below, cohomology means Čech cohomology.
(Liu, exercise 5.3.13) Let $X,S$ be locally Noetherian schemes, $f:X\to S$ a projective morphism and $\mathcal{F,G}$ quasicoherent sheaves on $X$ with a surjective morphism $\mathcal{F}\to\mathcal{G}$. Suppose that the fibers of $f$ are of dimension $\leq d$. I would like to show that $$H^d(X,\mathcal{F})\to H^d(X,\mathcal{G})$$ is surjective, and here's what I've managed to prove so far.
First off, let $\mathcal{K}$ be the kernel of the morphism $\mathcal{F}\to\mathcal{G}$. Then $\mathcal{K}$ is quasicoherent. Proposition 5.2.34 tells us that our assumptions ($f$ projective, $Y$ locally Noetherian, $\dim X_s\leq d$) imply the vanishing of the $p$:th higher direct images of any quasicoherent sheaf on $X$ for $p\geq d+1$. In particular, $$R^pf_*\mathcal{K}=0,\ p\geq d+1.$$ In other words, for every open affine $V\subseteq S$, we have $$(R^pf_*\mathcal{K})(V)=H^p(f^{-1}(V),\mathcal{K}|_{f^{-1}(V)})=0,\ p\geq d+1\tag{$\star$}.$$ Now, from the short exact sequence $$0\to \mathcal{K} \to \mathcal{F} \to \mathcal{G} \to 0$$ and quasicoherence of $\mathcal{G}$, we obtain a long exact sequence in cohomology containing the following piece: $$H^d(X,\mathcal{F})\to H^d(X,\mathcal{G})\to H^{d+1}(X,\mathcal{K})$$ The surjectivity of the first arrow is equivalent to the vanishing of the last group, but is it possible to conclude that from $(\star)$?
If $[k]\in H^{d+1}(X,\mathcal{K})$, so $k=(k_\alpha)\in C^{d+1}(\mathcal{U},\mathcal{K})$ for some cover $\mathcal{U}$, I don't see how I can find a refinement of $\mathcal{U}$ for which $(\star)$ would be applicable.
Edit: It seems like the exercise is false as stated. Let $X=S=\mathbb{P}^1_k$, $\mathcal{F}=\mathcal{O}_X$ and $\mathcal{G}=k(x) \oplus k(y)$ the direct sum of scyscraper sheafs supported at $x\neq y$. There is a natural surjective morphism of sheaves $\mathcal{F}\to\mathcal{G}$. Now, if $f:X\to X$ is the identity morphism (which is projective) then every fiber has dimension $0$ and the claim would imply that we have a surjection $$k=H^0(X,\mathcal{O}_X)\to H^0(X, k(x) \oplus k(y))=k(x) \oplus k(y),$$ which is impossible.