The sheafification of taking cohomology is trivial?

Question

Consider the Nisnevich site of a noetherian scheme $S$ of finite Krull dimension (the objects are schemes $U$ smooth and of finite type over $S$), let $A$ be a sheaf of abelian groups on this site. I want to know:

Is the Nisnevich sheafification of the presheaf $$U\mapsto\mathrm{H}_{\mathrm{Nis}}^{n}(U, A)$$ trivial for $n>0$ (or does this presheaf have trivial stalks)?

One can also assume $S=\mathrm{Spec}(k)$ for $k$ a field if needed.

The case of $n=1$ is trivial since $\mathrm{H}_{\mathrm{Nis}}^{1}$ classifies torsors, which are Nisnevich locally trivial.

Answer 1

This is true (in any site), you are deriving the identity functor, which is exact, so you get zero.

More generally, for a map of sites $$ \varepsilon\colon C \to D $$ (which by definition is a functor $\varepsilon^{-1}$ in the opposite direction satisfying some properties), the higher direct image functors $R^i \varepsilon_* \mathscr{F}$ coincide with the sheafification of $U\mapsto H^i(\varepsilon^{-1}(U), \mathscr{F})$. If $C=D$ and $\varepsilon = {\rm id}$, we have $R^i \varepsilon_* = 0$ for $i>0$.