This question is base on my previous [question](https://mathoverflow.net/questions/159547/why-is-the-derived-tensor-product-only-defined-for-bounded-above-derived-categor/159549#comment408017_159549), 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"](http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN496067613&DMDID=DMDLOG_0011&LOGID=LOG_0011&PHYSID=PHYS_0057), 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.



Injective resolution for right derived functor
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.


