This question is base on my previous questionquestion, and I repeat it here:
Suppose $X$ is a projective variety and $D^{+}(X)$ is the derived cateogry of bounded below complexes of sheaf of $\mathcal{O}_X$-modules. Let $F$ be a sheaf, and I want to define a derived tensor $\otimes F: D^{+}(X) \to D^{+}(X)$ as follows:
Suppose $G^{\bullet} \in D^{+}(X)$, and I lift it to the homotopy category $K^{+}(X)$ (also denoted by $G^{\bullet}$), let $I^{\bullet} \in K^{+}(X)$ be a complex of injectives which is quasi-isomorphic to $G^{\bullet}$(this can always obtain because $G^{\bullet}$ is bounded below). Then tensoring $F$ to $I^{\bullet}$, we have a complex $F\otimes I^{\bullet}$. Finally, map this complex to $D^{+}(X)$. My question is, is this the correct way to define the derived tensor $\otimes F$?
I know this is weird, because $\otimes F$ is right exact and one supposed to use flat resolution. However, here is my reason why I got the above procedure:
By Chapter I Corollary 5.3 (page 56) of the book "Residues and Duality", if $A,B$ are two abelian categories, and $F:A \to B$ is an additive functor, and assume $A$ has enough injectives, then the derived functor $\mathbf{R}^{+}F$ exists.
The construction can be found in Theorem 5.1 loc.cit. The main fact is although injective objects are not $\otimes F$-acyclic, if a complex of injectives $I^{\bullet}\in K^{+}(X)$ is acyclic (i.e. $H^i(I^{\bullet})=0$ for all $i$), then $F \otimes I^{\bullet}$ is also acyclic (i.e. $H^i(F \otimes I^{\bullet})=0$ for all $i$).
Please correct me if anything is wrong.
