Isomorphism passing to the derived category

Suppose to have an additive right exact functor $F: \mathcal A \rightarrow \mathcal B$ between Abelian categories and suppose that $F(A)=B$ for an object $A$ in $\mathcal A$. Denote with $D(\mathcal A)$, $D(\mathcal B)$the derived categories of chain complexes (if you want, bounded), and denote with $Q$ the localization functor.

Suppose that $F$ has a right total derived functor $\textbf R F: D(\mathcal A) \rightarrow D(\mathcal B)$. In general, I would not expect that $Q(B) \simeq (\textbf R F)(Q(A))$, when $F$ is not exact.

Then I can't understand why this fact seems to be used roundabout in some cases. To say, I am working on Examples 5.4 of Huybrecht's Fourier-Mukai transform in algebraic geometry and it seems to be used in the following argument:

Let $X$ a smooth projective variety, $p,q: X \times X \rightarrow X$ the two projections and $i: X \rightarrow X \times X$ the diagonal morphism. The structure sheaf $\mathcal O _\Delta$ of the diagonal is canonically isomorphic to $i_* \mathcal O _X$. Regarded as complexes concentrated in degree 0, $i_* \mathcal O _X$ and $\mathcal O _\Delta$ define objects in $D^b(X)$ (the usual derived category of $X$). If $\mathcal E$ is any object in $D^b(X)$, the integral transform can be defined. Huybrechts write:

$\Phi_{\mathcal O _\Delta}(\mathcal E) = \textbf R p_* ( q^* \mathcal E \otimes^\textbf L \mathcal O _\Delta) = \textbf R p_* ( q^* \mathcal E \otimes^\textbf L \textbf R i_* \mathcal O _X)$

where $\mathcal O _\Delta = \textbf R i_* \mathcal O _X$ seems to be used ($q^*$ is not derived since $q$ is flat, hence $q^*$ is exact).

• For a closed embedding of varieties/schemes $i\colon X\to Y$, the direct image functor $i_*$ (acting from the category of quasi-coherent sheaves on $X$ to the category of quasi-coherent sheaves on $Y$, or from the category of sheaves of $\mathcal O_X$-modules to the category of sheaves of $\mathcal O_Y$-modules) is exact. So there is no difference between $i_*$ and $\mathbf R i_*$, or in other words, the functor $\mathbf R i_*$ does actually form a commutative square with the functor $i_*$ and the localization functors $Q$. – Leonid Positselski Jun 16 '16 at 11:19