Singular cohomology as a Zariski sheaf Let $X$ be a complex algebraic variety, and consider the presheaf
$U \mapsto H^i(U^{an}, \mathbb Z)$
in the Zariski topology. Is there a theorem that says this presheaf is already a sheaf, for certain values of $i$ and maybe under some assumptions on $X$?
It seems to be true for $i = 1$ and $X = \mathbb A^1$, but it fails for $i = 2$ and $X = \mathbb P^1$.
 A: Notice that the sheaf axiom is a sort of Mayer-Vietoris property. If $X=U\cup V$ for two opens $U,V$ then the sheaf axiom for some presheaf $F$ with values in an abelian category requires
$0\to F(X)\xrightarrow{res_U^X\oplus res_V^X} F(U)\oplus F(V) \xrightarrow{res_{U\cap V}^U-res_{U\cap V}^V} F(U\cap V)$
to be exact. The situation for singular cohomology is very different, there the Mayer-Vietoris-sequence is a long exact sequence. In particular will the short sequences for $H^i$ with $i>0$ will not be exact in general because no higher cohomology vanishes in general. The analogue for cohomology would propably be that the sequence of the cochain complexes
$0 \to S^\ast_{\{U,V\}}(X) \to S^\ast(U)\oplus S^\ast(V) \to S^\ast(U\cap V) \to 0$
is exact where $S^\ast_{\{U,V\}}(X)$ denotes the cochain complex that is generated by cochains supported either in $U$ or in $V$. This is homotopy equivalent to the full cochain complex $S^\ast(X)$, so there is a "sheaf axiom up to homotopy" somewhere in here.
On the other hand long exact sequences of (co)homology are exactly what makes an distinguished triangle in a derived category. Is there maybe a notion of a derived-category-valued sheaf?
