Switch D(A) with K(A)

THanks to ZL
https://stacks.math.columbia.edu/tag/079V#comment-8991

cohomology.tex

@@ -6566,9 +6566,9 @@ \section{Cohomology of unbounded complexes}
 \item for any additive functor
 $F : \textit{Mod}(\mathcal{O}_X) \to \mathcal{A}$
 into an abelian category $\mathcal{A}$ we consider the exact functor
-$F : K(\textit{Mod}(\mathcal{O}_X)) \to D(\mathcal{A})$ induced by $F$
+$F : K(\textit{Mod}(\mathcal{O}_X)) \to K(\mathcal{A})$ induced by $F$
 and we obtain a right derived functor
-$RF : D(\mathcal{O}_X) \to K(\mathcal{A})$.
+$RF : D(\mathcal{O}_X) \to D(\mathcal{A})$.
 \end{enumerate}
 By construction we have $RF(\mathcal{F}^\bullet) = F(\mathcal{I}^\bullet)$
 where $\mathcal{F}^\bullet \to \mathcal{I}^\bullet$ is as above.

sites-cohomology.tex

@@ -3944,9 +3944,9 @@ \section{Cohomology of unbounded complexes}
 \item for any additive functor
 $F : \textit{Mod}(\mathcal{O}) \to \mathcal{A}$
 into an abelian category $\mathcal{A}$ we consider the exact functor
-$F : K(\textit{Mod}(\mathcal{O})) \to D(\mathcal{A})$ induced by $F$
+$F : K(\textit{Mod}(\mathcal{O})) \to K(\mathcal{A})$ induced by $F$
 and we obtain a right derived functor
-$RF : D(\mathcal{O}) \to K(\mathcal{A})$.
+$RF : D(\mathcal{O}) \to D(\mathcal{A})$.
 \end{enumerate}
 By construction we have $RF(\mathcal{F}^\bullet) = F(\mathcal{I}^\bullet)$
 where $\mathcal{F}^\bullet \to \mathcal{I}^\bullet$ is as above.