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.

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    $\begingroup$ Ok, I still don't understand, but my advice is really to revisit some basic points. Why do you think this is a RIGHT derived functor????? derived tensor product is a LEFT derived functor. $\endgroup$
    – user36931
    Mar 7, 2014 at 14:18
  • $\begingroup$ I see where I was wrong. But why derived tensor has to be a left derived functor? It could well be a right derived functor as I defined -- this make sense, but maybe meaningless -- I don't have any idea about this... $\endgroup$
    – Li Yutong
    Mar 7, 2014 at 23:11
  • $\begingroup$ Well, yes, in this way you the right derived functors of tensor products. But since Tensor products are right exact, its $i$-th right derived functor is identically zero, save for $i = 0$. $\endgroup$ Mar 9, 2014 at 16:26

2 Answers 2


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.

  • $\begingroup$ Thank you very much! I see where I was wrong: $\otimes F$ can be used to define a right derived functor, but we usually define it as a left derived functor, and they two may not the same. $\endgroup$
    – Li Yutong
    Mar 7, 2014 at 23:04

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).

  • $\begingroup$ Thank you for your answer! I see where I was wrong. However, for the statement $I^{\bullet}$ is acyclic then $I^{\bullet} \otimes M$ is also acyclic, you can prove by using "Residues and Duality" Chapt I, Lemma 4.5 Page 41. $\endgroup$
    – Li Yutong
    Mar 7, 2014 at 23:07
  • $\begingroup$ If $I^\bullet$ is an acyclic complex of injective objects, bounded below, it is contractible, i.e. homotopy equivalent to the zero complex. Since any additive functor (exact or not) has to preserve homotopy equivalences, it follows that in such a case $I^\bullet \otimes M$ is acyclic as well. $\endgroup$ Feb 23, 2015 at 17:53
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    $\begingroup$ For unbounded complexes, the claim is false. Consider the acyclic complex $\cdots \stackrel{2}{\to} \mathbb{Z}/4 \stackrel{2}{\to} \cdots$ of $\mathbb{Z}/4$-modules. This is a complex of projective modules (so, in the dual category, it's a complex of injective objects). But tensoring it with $\mathbb{Z}/2$, one obtains a complex which has cohomology in every degree. (This counterexample is in Gelfand/Manin.) $\endgroup$ Feb 23, 2015 at 17:56

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