ag.algebraic geometry - equivalence of Grothendieck-style versus Cech-style sheaf cohomology

Let $X$ be a topological space, and $T$ its category of open sets with the usual Grothendieck topology. Let $T'$ be any sieve of $T$ (a subcategory of $T$ such that if $U$ is in $T'$ then any subset of $U$ is also in $T'$). For example, $T'$ might be the collection of open subsets subordinate to the open subsets in a cover $mathcal{U}$. Any sheaf on $T$ induces a functor on $T'$ which can be viewed as a sheaf on $T'$ if $T'$ is given the minimal topology (the only covers are the identity maps). This determines a morphism of topoi $f : T rightarrow T'$, hence a spectral sequence

$H^p(T', R^q f_ast F) Rightarrow H^{p+q}(T, F)$ .

(One could surely also convince oneself that such a spectral sequence exists without any reference to topoi.)

The Cech cohomology of $F$ with respect to some covering family $mathcal{U}$ is

$H^p(mathcal{U}, F) = H^p(T', f_ast F)$

where $T' = T'(U)$ is the sieve associated to the cover $mathcal{U}$. The Cech cohomology is then the filtered colimit

$check{H}^p(T, F) = varinjlim_{(T',f)} H^p(T', f_ast F)$

taken over the projections $f : T rightarrow T'$ associated as above to covering families $mathcal{U}$.

One evidently has edge homomorphisms

$check{H}^p(T, F) rightarrow H^p(T, F)$

from the spectral sequence, and the question is when these induce an isomorphism. If we could somehow eliminate the $R^p f_ast F$, $p > 0$, by passing to a "small enough" cover we would have equality. This condition already holds in many cases; the following condition is more general (but I haven't checked carefully that it actually works!):

For every cover $mathcal{U}$ of $X$, every $U_1, ldots, U_n in mathcal{U}$, and every class in $alpha in H^p(U_1 mathop{times}_X cdots mathop{times}_X U_n, F)$, $p > 0$, there exists a refinement $mathcal{U}'$ of $mathcal{U}$ such that the restriction of $alpha$ under the map

$H^p(U_1 mathop{times}_X cdots mathop{times}_X U_n, F) rightarrow H^p(U'_1 mathop{times}_X cdots mathop{times}_X U'_n, F)$

is zero.

To make sense of this, one must use some convention for the covers $mathcal{U}$ and $mathcal{U}'$ to ensure there is a map as above. For example, one could work only with covers indexed by the points of $X$ (a cover is then a collection of neighborhoods of each point of $X$).

A more refined version of the above condition would say that Cech cohomology equals cohomology in degrees at most $q$ if the above condition holds for $p leq q$. Since it always holds for $p = 0,1$ this implies that

$check{H}^1(T, F) = H^1(T, F)$

in general.

Edit in response to David's comment:

The Cech complex always computes cohomology correctly in a presheaf category (i.e., when the topology is "chaotic": an object has no covers by anything except itself). Trying to compute cohomology in an arbitrary site using the Cech complex is (heuristically) something like trying to approximate the site by a presheaf category.

Here is how Cech cohomology computes cohomology of presheaves. Consider any category $T'$. If $F$ is a presheaf of groups on $T'$ then the sheaf cohomology groups of $F$ are the derived functors of the inverse limit for diagrams of shape $T'$. They are also computed as

$Ext(mathbf{Z}, F)$

where $mathbf{Z}$ is the constant sheaf associated to the integers. Remarkably, in a presheaf category, $mathbf{Z}$ has a canonical projective resolution associated to any cover of the final presheaf. A cover of the final presheaf is a collection of objects $U$ of $T'$ such that every object of $T'$ has a map to at least one object of $U$. The $i$-th term of this complex is the direct sum, over all choices of $i$ elements $U_1, ..., U_i$ of $U$, of the groups $mathbf{Z}_{U_1 times cdots times U_i}$. (You can check this is projective by noting it is the extension by $0$ of $mathbf{Z}$ from the slice category $T' / U_1 times cdots times U_i$ and extension by $0$ preserves projectives (since it has an exact right adjoint) and $mathbf{Z}$ is projective on the slice category since all higher cohomology of all sheaves vanishes (since it has a final object). It's also easy to check by a direct calculation.)

Denote this complex by $K$. Since this is a projective resolution of $mathbf{Z}$, $mathrm{Hom}(K, F)$ computes the cohomology of $F$. But it is also easy to see that this is just the Cech complex of $F$.