Does $H^i(X,F)\cong H^i(Y,f_*F)$ hold for $X\to Y$ finite but $F$ not necessarily quasi-coherent? Let $X\to Y$ be a finite morphism between schemes,$F$ be a sheaf of abelian groups but not necesarily quasi-coherent. Does $H^i(X,F)\cong H^i(Y,f_*F)$ still hold for sheaf cohomology with Zariski topology? (It holds for etale cohomology)
 A: Great question! Here is a counterexample:
Let $(A,\mathfrak m)$ be a DVR, and let $A \subseteq B$ be a finite extension of domains such that $B$ has exactly two primes above $\mathfrak m$. For example, $A = \mathbb Z_{(5)}$, and $B = \mathbb Z_{(5)}[i]$, with the primes $(1+2i), (1-2i)$ lying above $(5)$.
Then $Y = \operatorname{Spec} A$ is the space $\{x,\eta\}$ in which $x$ is closed and $\eta$ is open, and $X = \operatorname{Spec} B$ is the space $\{x, y,\eta\}$ in which $x$ and $y$ are closed, but $\eta$ is not. The map sends both $x$ and $y$ to $x$, and $\eta$ to $\eta$.
Lemma 1. Let $\mathcal F$ be a sheaf on $Y$. Then $H^i(Y, \mathcal F) = 0$ for all $i > 0$.
Proof. We will show that taking global sections is exact. Given a surjection $\mathcal F \to \mathcal G$ on $Y$, and a global section $s \in \mathcal G(Y)$, there exists a covering on which $s$ comes from $\mathcal F$. But every covering of $Y$ has to include a copy of $Y$ itself, so $s$ comes from $\mathcal F(Y)$. $\square$
Remark. The lemma actually generalises (with the exact same proof!) to arbitrary local rings. Indeed, the only open that contains the closed point of $Y$ is $Y$ itself.
It now suffices to construct a sheaf $\mathcal F$ on $X$ whose first cohomology is nontrivial. To do this, recall from Hartshorne Exercise II.1.19 the exact sequence
$$0 \to j_! (\mathcal F|_U) \to \mathcal F \to i_* (\mathcal F|_Z) \to 0,$$
where $j \colon U \to X$ and $i \colon Z \to X$ are a complementary open and closed set in a topological space $X$.
Lemma 2. Let $U = \{\eta\}$, $Z = \{x,y\}$, and $\mathcal F = \mathbb Z$ (the constant sheaf). Then $\mathcal F(X) \to i_*(\mathcal F|_Z) (X)$ is not surjective. Consequently (since the constant sheaf $\mathbb Z$ on the irreducible space $X$ is flasque),
$$H^1(X, j_!(\mathcal F|_U)) \neq 0.$$
Proof. Since $X$ is irreducible, it is connected, so $\mathcal F(X) = \mathbb Z$. On the other hand, $Z$ is a discrete space of two points, so $\underline{\operatorname{Sh}}(Z) \cong \underline{\operatorname{Ab}} \times \underline{\operatorname{Ab}}$, given by
$$\mathcal G \mapsto (\mathcal G_x, \mathcal G_y).$$
We use that $\mathcal F|_Z = i^{-1} \mathcal F$ is the sheafification of the presheaf
$$V \mapsto \operatorname*{colim}_{W \supseteq i(V)} \mathcal F(W).$$
Since sheafification doesn't alter stalks, and $\{x,\eta\}$ is the unique minimal open set containing $x$, we see that
$$(\mathcal F|_Z)_x = \mathcal F|_Z(\{x\}) = \mathcal F(\{x,\eta\}) = \mathbb Z,$$
and similarly $(\mathcal F|_Z)_y = \mathbb Z$. Thus, we see that
$$i_*(\mathcal F|_Z)(X) = \mathcal F|_Z(U) = \mathbb Z \oplus \mathbb Z.$$
Thus, the map $\mathcal F(X) \to (i_*\mathcal F|_Z)(X)$ can never be surjective, as $\mathbb Z$ does not surject onto $\mathbb Z \oplus \mathbb Z$. $\square$
Remark. The proof of the vanishing of higher direct images for étale cohomology for a finite morphism relies on the fact that strictly Henselian local rings (the local rings for the étale topology) have no higher cohomology, and that a finite covering of a strictly Henselian local ring is again a finite product of strictly Henselian local rings. However, as we saw above, finite coverings of local rings for the Zariski topology do have higher cohomology. Maybe we should see this as a hint that the Zariski topology does not have a good local theory in the same way that the étale topology does.
