# Cech cohomology of coefficient in a presheaf and in its associated sheaf

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 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.

## 1 Answer

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 $$F\to aF$$).