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I am studying derived functors and I came across a problem: given an additive left exact functor $F$ between two abelian categories $\mathcal{C}, \mathcal{D}$ with enough injectives one defines its right derived functor $RF : D^{-}(\mathcal{C}) \rightarrow D^{-}(\mathcal{D})$ by means of the universal property. Moreover one can define the classical derived functors $R^{i}F(X) = H^{i}(RF(\mathcal{I}^{\bullet}_{X}))$ where by $\mathcal{I}^{\bullet}_{X}$ I mean an injective resolution of $X$.

As far as I understood $R^{i}F$ is meant to be a functor between $\mathcal{C}$ and $\mathcal{D}$, but here comes my question: if I consider $\mathcal{C} = \mathcal{D} = \{ \text{ sheaves of abelian groups on a topological space } \}$ then the cohomology groups of a complex of abelian sheaves need not to be abelian sheaves but only presheaves, so we need to sheafify. Is my reasoning correct or I am losing something?

Thank you.

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You need to take cohomology in the category of sheaves of abelian groups, where the cokernel of a map of sheaves is by definition the sheafification of the cokernel in the category of presheaves.

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