Lemma 60.18.4. Let $\mathcal{C}$ be a category endowed with the chaotic topology. Let $X$ be an object of $\mathcal{C}$ such that every object of $\mathcal{C}$ has a morphism towards $X$. Assume that $\mathcal{C}$ has products of pairs. Then for every abelian sheaf $\mathcal{F}$ on $\mathcal{C}$ the total cohomology $R\Gamma (\mathcal{C}, \mathcal{F})$ is represented by the complex

$\mathcal{F}(X) \to \mathcal{F}(X \times X) \to \mathcal{F}(X \times X \times X) \to \ldots$

associated to the cosimplicial abelian group $[n] \mapsto \mathcal{F}(X^ n)$.

Proof. Note that $H^ q(X^ p, \mathcal{F}) = 0$ for all $q > 0$ as any presheaf is a sheaf on $\mathcal{C}$. The assumption on $X$ is that $h_ X \to *$ is surjective. Using that $H^ q(X, \mathcal{F}) = H^ q(h_ X, \mathcal{F})$ and $H^ q(\mathcal{C}, \mathcal{F}) = H^ q(*, \mathcal{F})$ we see that our statement is a special case of Cohomology on Sites, Lemma 21.13.2. $\square$

Comment #3625 by shanbei on

second line of this proof:

It should be "$H^q(X, \mathcal{F}) = H^q(h_X, \mathcal{F})$"?

Comment #3655 by BB on

The statement of the lemma assumes that C has products, but the proof only uses finite non-empty products. In potential applications, I think one does not want to assume C has a final object (as implied by the existence of products, at least under some conventions). So I think it might be good to replace "has products" with "has finite non-empty products".

Comment #5445 by Hao on

In the proof, "sheaves are presheaves" is somehow confusing. Maybe "any presheaf is a sheaf" is better.

What is the definition of "total complex" in the statement?

Comment #5668 by on

THanks and fixed here. The terminology "total cohomology" is just a way of referring to $R\Gamma(-, -)$.

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