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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 $X = \mathbb P^1$.

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    $\begingroup$ There seems to be absolutely no reason for this to be a sheaf, except maybe $i=0$ and $X$ irreducible. $\endgroup$
    – Alex Degtyarev
    Commented Mar 17, 2014 at 19:04

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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?

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    $\begingroup$ The derived category has few limits, so it would be a bad idea to talk about sheaves with values in the derived category. Instead we look at presheaves of chain complexes satisfying a sheaf condition up-to-homotopy. $\endgroup$
    – Zhen Lin
    Commented Mar 17, 2014 at 19:34
  • $\begingroup$ Well along the lines of this answer one might try to require distinguished triangles instead of short exact sequences... $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Mar 17, 2014 at 19:48
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    $\begingroup$ Minor correction: there is generally no $0$ at the end of the first short exact sequence. $\endgroup$
    – Keenan Kidwell
    Commented Mar 17, 2014 at 19:58
  • $\begingroup$ @KeenanKidwell Oh you're right. I'll correct that. In that case the sheaf axiom (for two open sets anyway) is indeed satisfied by $H^0$. $\endgroup$
    – Johannes Hahn
    Commented Mar 17, 2014 at 20:07

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