# Derived functor defined with an adapted subcategory

I'm currently reading the book "Fourier-Mukai Transforms in Algebraic Geometry" written by Daniel Huybrechts, and I'm facing a problem that I can't find answers on the internet. Here is the context : (cf. section 2.2 remark 2.51)

Consider $\mathcal{A}$ and $\mathcal{B}$ two abelian categories, and suppose we have an $\textit{exact}$ functor $$F : K^+(\mathcal{A}) \to K(\mathcal{B})$$ (notice that the functors needs not to come from a functor between abelian categories), where $K(\mathcal{A})$ denotes the homotopy category of complexes of objects in $\mathcal{A}$ and the $"+"$ means that we take bounded below complexes. My goal is to understant in which case we can define a derived functor $RF : D^+(\mathcal{A}) \to D(\mathcal{B})$ between the derived categories.

The remark 2.51 in the book tells the following :

Suppose there exists a triangulated subcategory $\mathcal{K}_F$ of $K^+(\mathcal{A})$ which is $\textit{adapted}$ to $F$, $i.e.$ which satisfies :

If $A^\bullet$ is acyclic ($i.e$ $H^n(A^\bullet)=0$ for all $n$), then $F(A^\bullet)$ is acyclic.

Any $A^\bullet \in K^+(\mathcal{A})$ is quasi-isomorphic to a complex in $\mathcal{K}_F$.

Then there exists a derived functor $RF : D^+(\mathcal{A}) \to D(\mathcal{B})$ satisfying classic properties of derived functor.

Here are my attempts : for any complex $A^\bullet$ in $K^+(\mathcal{A})$ we can find a complex $T_{A^\bullet} \in \mathcal{K}_F$ and a quasi-isomorphism $A^\bullet \to T_{A^\bullet}$, thus it is natural to define $RF(A^\bullet) := F(T_{A^\bullet})$ as we usually do when we consider abelian categories with enough injectives. Now my problem comes when I want to define $RF$ on arrows. I know, by the first hypothesis, that $F$ preserves quasi-isomorphisms $\textit{within}$ $\mathcal{K_F}$.

Consider an arrow $A^\bullet \leftarrow C^\bullet \rightarrow B^\bullet$ from $A^\bullet$ to $B^\bullet$ in $D^+(\mathcal{A})$, where the left arrow is a quasi-isomorphism. I would like to obtain a roof $$F(T_{A^\bullet}) \leftarrow F(T_{C^\bullet}) \rightarrow F(T_{B^\bullet})$$ but I can't figure out how to get a morphism $F(T_{C^\bullet}) \to F(T_{A^\bullet})$. I tried multiple way, for instance to use axioms of triangulated categories to fill a morphism of distinguished triangle as : $\require{AMScd}$

\begin{CD} C^\bullet @>>> T_{C^\bullet} @>>> C(j_{C^\bullet}) @>>> C^\bullet \\ @VVV@.@VVV@VVV\\ A^\bullet @>>> T_{A^\bullet} @>>> C(j_{A^\bullet}) @>>> A^\bullet \end{CD}

where $C(f)$ means the $\textit{mapping cone}$ of $f$ and $j_{A^\bullet}$ is the quasi-isomorphism $A^\bullet \to T_{A^\bullet}$, but I failed to find a good complex morphism $C(j_{C^\bullet}) \to C(j_{A^\bullet})$.

Now looking to other references ("Derived categories of coherent sheaves and equivalences between them", D. O. Orlov, lemma 1.1.6 ; "Residues and duality", Harthshorne, proposition 3.3 & theorem 5.1), I think that the thing I should show is that $\mathcal{K}_F$ is $\textit{left cofinal}$ in $K^+(\mathcal{A})$ with respect to the quasi-isomorphisms, that it for any quasi-isomorphism $$s : X^\bullet \to Y^\bullet$$ with $Y^\bullet \in \mathcal{K}_F$, there is a morphism $f : Z^\bullet \to X^\bullet$ with $Z \in \mathcal{K}_F$ and such that $s\circ f$ is a quasi-isomorphism.

I'm not an expert about localization in categories, and I'm probably missing some simplier homological algebra arguments, but if anyone could give me some help on this, I would really appreciate.

• I think this will be easier if you use the "dual" description of maps in the derived category: i.e., as diagrams $A^\bullet\rightarrow C^\bullet\leftarrow B^\bullet$ where the second arrow is a quasi-isomorphism. – Jeremy Rickard Apr 6 '18 at 10:20
• By the way, I'd interpret "quasi-isomorphic" to simply mean isomorphic in the derived category (it ought to be an equivalence relation) rather than that there is a quasi-isomorphism in one specific direction. But for this result I think you really do need that stronger condition for $\mathcal{K}_F$ (Hartshorne requires it), so I think Huybrechts' statement is a bit misleading. – Jeremy Rickard Apr 6 '18 at 10:25
• Thanks to your comments, I think I find a solution to my problem in Hartshorne's "Residues & Duality". I'm new to Stackexchange, should I write a answer or explaining in comments (or maybe delete my question) ? – Dominique MATTEI Apr 6 '18 at 15:29
• I’d suggest you write a brief answer, as it might be useful for somebody else at some point. And (maybe after waiting a while for comments) it’s quite OK to accept your own answer. – Jeremy Rickard Apr 6 '18 at 16:29

## 1 Answer

I follow the previous notations. According to Hartshorne's $\textit{Residues and duality}$ (Chapter I, Proposition 4.1), we can complete the diagram $T_{A^\bullet} \leftarrow C^\bullet \rightarrow T_{C^\bullet}$ into a commutative diagram : \begin{CD} C^\bullet @>>> T_{C^\bullet} \\ @VVV@VVV\\ T_{A^\bullet} @>>> D_A^\bullet, \end{CD} where all arrows are quasi-isomorphisms. Composing with a quasi-isomorphism $D_A^\bullet \to T_{D_A^\bullet}$ and doing the same with $B^\bullet$ (careful here, only horizontal arrows of the previous diagram would be quasi-isomorphism in the case of $B^\bullet$), we obtain a roof $T_{D_A^\bullet} \leftarrow T_{C^\bullet} \rightarrow T_{D_B^\bullet}$, and since $T_{D_A^\bullet}$ is quasi-isomorphic $\textit{within}$ $\mathcal{K}_F$ to $T_{A^\bullet}$ (same for $B^\bullet)$, we can apply $F$ and get the roof we originally wanted.