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
$$\prod_{i_0<...<i_q} A(U_{i_0...i_q}) \ \to \prod_{i_0<...<i_q} aA(U_{i_0...i_q})$$
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.
 A: 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 $F\to aF$).
