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I am reading Jean-Luc Brylinski's book on loop spaces. In his book, he claims that:

Suppose that for any presheaf $F$ of Abelian group such that its associated sheaf $aF$ is $0$, the groups $\check{H}^q(X, F)$ are all zero. Then for arbitrary presheaf $A$ of Abelian group and its associated sheaf $aA$, $$\check{H}^q(X,A)\cong \check{H}^q(X,aA). $$

On the level of cochains, I know that taking direct limit gives the map $$\prod_{i_0<...<i_q} A(U_{i_0...i_q}) \ \to \prod_{i_0<...<i_q} aA(U_{i_0...i_q})$$

I think this claim is a purely algebraic one, but I do not know how to construct this isomorphism. Any help is appreciated.

PS. I cheat a little bit. The definition of cochains in his book does not require ordering among indices.

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Step 1 : Consider the exact sequences of presheaves $$ 0\to\ker(F\to aF)\to F\to\operatorname{im}(F\to aF)\to 0$$ $$ 0\to \operatorname{im}(F\to aF)\to aF\to\operatorname{coker}(F\to aF)\to 0$$

The presheaves $K=\ker(F\to aF)$ and $C=\operatorname{coker}(F\to aF)$ have the property that $aK=0$ and $aC=0$. This is because the functor $F\mapsto aF$ commutes with limits and colimits, so $aK=\ker(aF\to aF)=0$ and similarly for $C$.

Step 2 : Since infinite products are exact in the category of abelian groups, an exact sequence of presheaves $$0\to F'\to F\to F''\to 0$$ yields an exact sequence of complexes : $$0\to \check{C}(\mathcal{U},F')\to\check{C}(\mathcal{U},F)\to\check{C}(\mathcal{U},F'')\to 0$$ and taking cohomology gives a long exact sequence : $$...\to\check{H}^p(\mathcal{U},F')\to\check{H}^p(\mathcal{U},F)\to\check{H}^p(\mathcal{U},F'')\to\check{H}^{p+1}(\mathcal{U},F')\to....$$ Since filtered colimit are exact in the category of abelian groups, taking the colimits over all coverings gives a long exact sequence $$...\to\check{H}^p(X,F')\to\check{H}^p(X,F)\to\check{H}^p(X,F'')\to\check{H}^{p+1}(X,F')\to....$$ Step 3 Combine Step 1 and Step 2 together. The two short exact sequences from Step 1 gives long exact sequences. Since $aK=0$ and $aC=0$, by assumption, $\check{H}^q(X,K)=0$ and $\check{H}^q(X,C)=0$. From the long exact sequences $$\check{H}^q(X,F)\to\check{H}^q(X,\operatorname{im}(F\to aF))\to\check{H}^q(X,aF)$$ are both isomorphisms. Hence the result. (Note also that by functoriality, this composition is just the morphism induced in Cech cohomology by the natural morphism $F\to aF$).

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