Suppose we have Grothendieck abelian categories $\mathcal{A}, \mathcal{B}$. Suppose also we have given an exact functor of triangulated categories $$ F \colon D(\mathcal{A}) \to D(\mathcal{B}) $$ where $D(\mathcal{A})$ and $D\mathcal({B})$ denote appropriate derived categories of complexes (possibly bounded below or both ways). Suppose that

  • for any injective $I \in \mathcal{A}$ we have a functorial maps which are quasi-isomorphisms $F(I) \cong h^0F(I)$, and

  • $F(D^{\geq 0}(\mathcal{A})) \subseteq D^{\geq 0}(\mathcal{B})$ ($F$ is $t$-left exact for the standard $t$-structure).

Is it then true that $F$ is the right derived functor of is zeroth cohomology? I.e. $F \cong R(h^0F)$. If not, are there known counterexamples? Or which approporiate additional assumptions are needed so that such a statement holds?

I believe that the above setup yields that the $i$th cohomology of $F$ is canonically isomorphic to the $i$th right derived functor of $h^0F$. However, I don't see how to extend the given quasi-isomorphism $F(I) \cong h^0F(I)$ for injectives in $\mathcal{A}$ to a natural transformation of functors $F \circ Q \to h^0F$ where $Q$ denotes the natural functor from the homotopy category $K(\mathcal{A})$ to $D(\mathcal{A})$. Once one has this, the universal porperty of the right derived functor should give the result.

Some Background: In the situation I am interested in the functor $F$ arises as a composition of a left and right derived functor in a much bigger ambient category, whose restriction to $D(\mathcal{A})$ happens to be left exact and satisfy the condition $F(I) \cong h^0F(I)$ for injective objects of $\mathcal{A}$. I could imagine that this is not an uncommon situation...

The result I.Proposition 7.4 in Residues and Duality (Hartshorne) is a statement of the type I am looking for. it says that under similar assumptions as above a right derived functor is (a shift of) the left derived functor of its highest non-vanishing cohomology...


2 Answers 2


Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories with enough injective objects. Let me use the notation $D^+(\mathcal{A})$ and $D^+(\mathcal{B})$ to denote the stable $\infty$-categories whose homotopy categories are the (cohomologically bounded below) derived categories of $\mathcal{A}$ and $\mathcal{B}$, respectively (you can also consider unbounded derived categories, but the situation is a bit more subtle).

Let $\mathcal{C} \subseteq \mathrm{Fun}( D^{+}( \mathcal{A} ), D^{+}( \mathcal{B}) )$ be the full subcategory spanned by those functors which are exact, left t-exact, and carry injective objects of $\mathcal{A}$ into the heart of $D^{+}( \mathcal{B} )$. Then the construction $$F \in \mathcal{C} \mapsto h^0 F|_{ \mathcal{A} }$$ determines an equivalence from $\mathcal{C}$ to the category of left exact functors from $\mathcal{A}$ to $\mathcal{B}$. The inverse of this equivalence is "taking the right derived functor".

Consequently, one can answer your question as follows: given a functor of triangulated categories $G: hD^{+}(\mathcal{A}) \rightarrow hD^{+}(\mathcal{B})$, it arises as a right derived functor (of a left exact functor of abelian categories) if and only if

a) The functor $G$ lifts to an exact functor of stable $\infty$-categories $D^{+}(\mathcal{A}) \rightarrow D^{+}(\mathcal{B})$ (anything that you build by composing derived functors will have this property).

b) The functor $G$ is left t-exact and carries injective objects of $\mathcal{A}$ into the heart of $hD^{+}(\mathcal{B})$.

  • $\begingroup$ That is exactly the type of statement I was looking for. Thank you. Could you point me to a reference of the equivalence of functor categories you state in your answer. Also, if possible, a reference to your statement that "anything built up by composing derived functors will lift to the stable infinity categories". $\endgroup$
    – mabli
    Jul 20, 2016 at 6:17
  • $\begingroup$ See Thm of Lurie's Higher Algebra. As for the other statement: derived functors themselves lift (example so just compose the lifts. $\endgroup$ Jul 20, 2016 at 17:43

Here's a counterexample for unbounded derived categories (this doesn't answer the revised question with the "$t$-left exact" condition).

Suppose $\mathcal{A}$ is a Grothendieck category with enough projectives, where projectives and injectives coincide: for example, the module category of $k[x]/(x^2)$ for a field $k$.Let $F$ be the left derived functor of a right exact, but not exact, functor.

Added (19 July 2016):

I don't have a definitive answer for the general revised question, but here's a reason that I think that (if the answer is "yes") then it might be hard without extra conditions.

I believe that it is even still an open problem whether there could be a self-equivalence $F$ of the derived category of a module category $\text{Mod}(A)$ such that the restriction of $F$ to $\text{Mod}(A)$ is the identity, $F(X)\cong X$ for every object $X$ of the derived category, but $F$ is not isomorphic to the identity functor. If such a functor existed, then it would of course be a counterexample.

This is something that annoyed me for several weeks when I was doing my PhD thirty-odd years ago. I made some comments in Section 7 of my paper "Morita theory for derived categories", although I'm afraid the comments don't say much more than "I don't know".

The difficulty is probably just an artefact of the axioms for an abstract triangulated category not being rigid enough, and it's likely that the problem is easier for functors that arise in nature.

Also, I don't rule out the possibility of simpler, more natural counterexamples for the question asked in the OP, although I haven't been able to think of any.

  • $\begingroup$ Yes indeed, very good point. It reminds me that I forgot to add an assumption on $F$, namely that $F$ is left exact in the sense that $F(D^{\geq 0}(\mathcal{A})) \subseteq D^{\geq 0}(\mathcal{B}). I edited the question accordingly... $\endgroup$
    – mabli
    Jul 19, 2016 at 10:01
  • 1
    $\begingroup$ In fact, this has annoyed me for quite a bit longer than a few weeks already. Seeing the answer of @Jacob Lurie it seems you are right in your assessment that this is an artefact of the shortcomings of triangulated categories which are resolved in a better context... $\endgroup$
    – mabli
    Jul 20, 2016 at 6:15

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