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256 U. Bunke, T. Schick, M. Spitzweck, and A. Thom3.3.11 We consider the functor which associates to F 2 Sh Ab S the coneC.F/ WD Cone Hom ShAb S=Y .F jY ;I jY / ! Hom ShAb S=Y .F jY ;J/ :In order to show that the marked map in (15) is a quasi isomorphism we must show thatC.F/ is acyclic for all F 2 Sh Ab S.The restriction functor .:::/ jY is exact by Lemma 3.8. For H 2 Sh Ab S=Y thefunctor Hom ShAb S=Y .:::;H/is a right-adjoint. As a contravariant functor it transformscolimits into limits and is left exact. In particular, the functorsHom ShAb S=Y ..:::/ jY ;I jY /;Hom ShAb S=Y ..:::/ jY ;J/transform coproducts of sheaves into products of complexes. It follows that F ! C.F/also transforms coproducts of sheaves into products of complexes. If P 2 Sh Ab S is acoproduct of representable sheaves, then by Lemma 3.13 C.A/ is a product of acycliccomplexes and hence acyclic.We claim that F ! C.F/ transforms short exact sequences of sheaves to shortexact sequences of complexes. Since J is injective and .:::/ jY is exact, the functorF 7! Hom ShAb S=Y .F jY ;J/is a composition of exact functors and thus has this property.Furthermore we have Hom ShAb S=Y .F jY ;I jY / Š Hom ShAb S .F; I / jY . Since I isinjective, the functor F ! Hom ShAb S=Y .F jY ;I jY / has this property, too. This impliesthe claim.We now argue by induction. Let n 2 N and assume that we have already shownthat H i .C.F // Š 0 for all F 2 Sh Ab S and i