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}](http://latex.mathoverflow.net/png?%5Cmathcal%7BU%7D).  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'](http://latex.mathoverflow.net/png?f%20%3A%20T%20%5Crightarrow%20T%27), hence a spectral sequence

![H^p(T', R^q f\_\ast F) \Rightarrow H^{p+q}(T, F)](http://latex.mathoverflow.net/png?H%5Ep%28T%27%2C%20R%5Eq%20f%5F%5Cast%20F%29%20%5CRightarrow%20H%5E%7Bp%2Bq%7D%28T%2C%20F%29) .

(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}](http://latex.mathoverflow.net/png?%5Cmathcal%7BU%7D) is 

![H^p(\mathcal{U}, F) = H^p(T', f\_\ast F)](http://latex.mathoverflow.net/png?H%5Ep%28%5Cmathcal%7BU%7D%2C%20F%29%20%3D%20H%5Ep%28T%27%2C%20f%5F%5Cast%20F%29)

where ![T' = T'(U)](http://latex.mathoverflow.net/png?T%27%20%3D%20T%27%28U%29) is the sieve associated to the cover ![\mathcal{U}](http://latex.mathoverflow.net/png?%5Cmathcal%7BU%7D).  The Cech cohomology is then the filtered colimit

![\check{H}^p(T, F) = \varinjlim\_{(T',f)} H^p(T', f\_\ast F)](http://latex.mathoverflow.net/png?%5Ccheck%7BH%7D%5Ep%28T%2C%20F%29%20%3D%20%5Cvarinjlim%5F%7B%28T%27%2Cf%29%7D%20H%5Ep%28T%27%2C%20f%5F%5Cast%20F%29)

taken over the projections ![f : T \rightarrow T'](http://latex.mathoverflow.net/png?f%20%3A%20T%20%5Crightarrow%20T%27) associated as above to covering families ![\mathcal{U}](http://latex.mathoverflow.net/png?%5Cmathcal%7BU%7D).

One evidently has edge homomorphisms

![\check{H}^p(T, F) \rightarrow H^p(T, F)](http://latex.mathoverflow.net/png?%5Ccheck%7BH%7D%5Ep%28T%2C%20F%29%20%5Crightarrow%20H%5Ep%28T%2C%20F%29)

from the spectral sequence, and the question is when these induce an isomorphism.  If we could somehow eliminate the ![R^p f\_\ast F](http://latex.mathoverflow.net/png?R%5Ep%20f%5F%5Cast%20F), 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}](http://latex.mathoverflow.net/png?%5Cmathcal%7BU%7D) of X, every ![U\_1, \ldots, U\_n \in \mathcal{U}](http://latex.mathoverflow.net/png?U%5F1%2C%20%5Cldots%2C%20U%5Fn%20%5Cin%20%5Cmathcal%7BU%7D), and every class in ![\alpha \in H^p(U\_1 \mathop{\times}\_X \cdots \mathop{\times}\_X U\_n, F)](http://latex.mathoverflow.net/png?%5Calpha%20%5Cin%20H%5Ep%28U%5F1%20%5Cmathop%7B%5Ctimes%7D%5FX%20%5Ccdots%20%5Cmathop%7B%5Ctimes%7D%5FX%20U%5Fn%2C%20F%29), p > 0, there exists a refinement ![\mathcal{U}'](http://latex.mathoverflow.net/png?%5Cmathcal%7BU%7D%27) of ![\mathcal{U}](http://latex.mathoverflow.net/png?%5Cmathcal%7BU%7D) such that the restriction of ![\alpha](http://latex.mathoverflow.net/png?%5Calpha) 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)](http://latex.mathoverflow.net/png?H%5Ep%28U%5F1%20%5Cmathop%7B%5Ctimes%7D%5FX%20%5Ccdots%20%5Cmathop%7B%5Ctimes%7D%5FX%20U%5Fn%2C%20F%29%20%5Crightarrow%20H%5Ep%28U%27%5F1%20%5Cmathop%7B%5Ctimes%7D%5FX%20%5Ccdots%20%5Cmathop%7B%5Ctimes%7D%5FX%20U%27%5Fn%2C%20F%29)

is zero.

To make sense of this, one must use some convention for the covers ![\mathcal{U}](http://latex.mathoverflow.net/png?%5Cmathcal%7BU%7D) and ![\mathcal{U}'](http://latex.mathoverflow.net/png?%5Cmathcal%7BU%7D%27) 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](http://latex.mathoverflow.net/png?p%20%5Cleq%20q).  Since it always holds for p = 0,1 this implies that

![\check{H}^1(T, F) = H^1(T, F)](http://latex.mathoverflow.net/png?%5Ccheck%7BH%7D%5E1%28T%2C%20F%29%20%3D%20H%5E1%28T%2C%20F%29) 

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.