Injective resolution for right derived functor This question is base on my previous question, 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.
 A: You are certainly allowed to do this and you will get a right derived functor because you are looking at resolutions $G^\bullet \to I^\bullet$ (note how the arrow is pointing to the right!). This works because any two injective resolutions $I^\bullet$ of $G^\bullet$ are homotopy equivalent, so you can use this to compute the right derived functor of absolutely any functor.
However, the derived tensor product is defined as a left derived functor.
Maybe this will help: the derived tensor product can be computed by taking a K-flat resolution in either variable. Thus you can first take a K-flat resolution $F^\bullet \to F$ (if $F$ is a sheaf this can be just your usual bounded above complex of flat modules resolving $F$) and then $Tot(F^\bullet \otimes G^\bullet)$ computes the derived tensor product. You don't even have to replace $G^\bullet$; in fact, since now $F^\bullet$ is K-flat you can reoplace $G^\bullet$ by any quasi-isomorphic complex and you get the same answer in the derived category.
A: I think you have a little confusion between right and left derived functors. Let me repeat the definition using the language of Kan extensions. In the following for any abelian category $A$ we will write $D(A)$ for its derived category (i.e. the localization of the category of complexes to weak equivalences). Let also $\iota_A:A\to D(A)$ be the natural inclusion.
If $A,B$ are abelian categories and $F:A\to B$ is an additive functor we say that a functor $RF:D(A)\to D(B)$ equipped with a natural transformation $\iota_B \circ F \to RF \circ \iota_A$ is the right derived functor if it is universal among those functors.
We can also define its left derived functor $LF$ in the same way, but with a map $LF \circ \iota_A \to \iota_B \circ F$. There is a deep difference between those two concepts. You seem to think that if a functor has both a left derived functor and a right derived functor those are equal, but I do not see of any obvious reason why this should be true.
In fact left and right "derived" functors is, in my opinion, a bad terminology (a better one would be maybe left and right "approximations") with which we are unfortunately stuck.
The main difference here between a flat resolution and an injective resolution is that flat resolutions map into your object, while injective resolutions map out of your object.
Moreover I really don't think that it is true that if $I^\bullet$ is an acyclic complex of injective modules then $I^\bullet\otimes M$ is acyclic for all $M$. If it is I will be very surprised (I tried for a while to come up with a counterexample but I wasn't able to).
