Cech to derived spectral sequence and sheafification Let X be a topological space, let $\mathcal{U} = \{U_i\}$ be a cover of X, and let $\mathcal{F}$ be a sheaf of abelian groups on X. If X is separated, each $U_i$ is affine, and $\mathcal{F}$ is quasi-coherent, then Cech cohomology computes derived functor cohomology; in general one only gets a spectral sequence
$$
H^p(\mathcal{U},\underline{H}^q(\mathcal{F})) \Rightarrow H^{p+q}(X,\mathcal{F})
$$
where $\underline{H}^q(\mathcal{F})$ is the presheaf $U \mapsto H^q(U,\mathcal{F}|_U)$.

Question: For q > 0, $\underline{H}^q(\mathcal{F})$ sheafifies to 0.

For a quasi-coherent sheaf $\mathcal{F}$ this is clear because cohomology vanishes on affines. Is this really true in general? Brian Conrad states this in the introduction to his notes on cohomological descent.
 A: I do not think that sheaves of abelian groups need to be locally acyclic.  Let me say what I mean in an example.  Take $X=\mathbb{C}^{2}$ with the classical (metric) topology.  Let $\mathcal{F} = \mathbb{Z}_{D}$ where $D=\mathbb{C}^{\times} \times \{0\}$.  
Then for any arbitrarily small polydisk $U$ containing $(0,0)$ we have 
$H^{1}(U,\mathbb{Z}_{D}) = H^{1}(U \cap D, \mathbb{Z})$ 
is not vanishing.  It seems in this example that 
$\underline{H}^{1}(\mathcal{F})_{(0,0)}$ 
does not vanish.
A: Yes, this is true in general.
It suffices to show the stalks vanish.  Pick $x \in X$ and take an injective resolution $0 \to {\cal F} \to I^0 \to \cdots$.  For any open $U$ containing $x$, we get a chain complex
$$0 \to I^0(U) \to I^1(U) \to \cdots$$
whose cohomology groups are $H^p(U,{\cal F}|_U)$.
Taking direct limits of these sections gives the chain complex
$$0 \to I^0_x \to I^1_x \to \cdots$$
of stalks, which has zero cohomology in positive degrees because the original complex was a resolution.  However, direct limits are exact and so we find
$$0 = {\rm colim}_{x \in U} H^p(U,{\cal F}|_U) = {\underline H}^p({\cal F})_x$$
as desired.
Generally, cohomology tells you the obstructions to patching local solutions into global solutions, and this says that locally those obstructions vanish.
A: Here's another short proof: denote by $I,J$ the inclusion of sheaves on X into presheaves and sheafification respectively, then
$$
\underline{H}^p(\mathcal{F})\cong R^pI(\mathcal{F}).
$$
Since $J$ is exact,
$$
J\circ R^pI\cong R^p(J\circ I)
$$
and the later vanishes for $p>0$ as $J\circ I=id_{\mathfrak{Ab}(X)}$. So $\underline{H}^p(\mathcal{F})$ sheafifies to 0 for $p>0$.
