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I'm learning homological algebra using several references books. But I find two definitions of derived functor in general triangulated category. I wonder to know which definition is more generally accepted.

Let $\mathcal{D}$ be a triangulated category and $\mathcal{C}$ a strictly full triangulated subcategory of $\mathcal{D}$ with the quotient $Q:\mathcal{D}\to \mathcal{D}/\mathcal{C}$. Given a triangulated category $\mathcal{T}$ and a triangulated functor $F:\mathcal{D}\to \mathcal{T}$, I have seen two different definitions of derived functor of $F$ with respect to $\mathcal{C}$.

Version 1 A right derived functor of $F$ with respect to $\mathcal{C}$ is a pair consisting of a triangulated functor $\mathcal{R}F:\mathcal{D}/\mathcal{C}\to \mathcal{T}$ and a trinatural transformation $\eta:F\Rightarrow (\mathcal{R}F)Q$ such that: given any triangulated functor $G:\mathcal{D}/\mathcal{C}\to \mathcal{T}$ and trinatural transformation $\xi:F\Rightarrow G Q$, there is a unique trinatural transformation $\chi:\mathcal{R}F\Rightarrow G$ such that $\xi=(\chi Q)\eta$.

For example, this definition occurs in note1 (page 198) and note2 (page 72).

Version 2 A right derived functor of $F$ with respect to $\mathcal{C}$ is just a left Kan extension $(\text{Lan}_QF,\eta)$. Also, some references require $\text{Lan}_QF$ to be triangulated without $\eta$ being compatible with translations.

For example, this definition occurs in page 253 of "Categories and Sheaves" by Masaki Kashiwara (essentially the same) and some books in other languages.

I think these two definition are not equivalent. A ordinary Kan extension $\text{Lan}_QF$ has no reason to be triangulated, and its natural transformation $\eta$ also has no reason to be compatible with translations. On the converse, the left Kan extension in the sense of triangulated categories (version 1) can not be an ordinary left Kan extension.

I wonder to know which definition is more generally accepted.

Z. He
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  • Are you sure that the ordinary Kan extension is not automatically triangulated in this situation? It's not exactly the same, and I haven't checked whether it can be adapted to prove this, so I might be missing an important difference, but the Stacks Project seems to prove something similar (see in particular Lemma 13.14.6). – Jeremy Rickard Feb 26 '24 at 07:36
  • @JeremyRickard I have roughly checked that Lemma. Since I'm not familiar with the language of ind-object and essentially constant used there, I still have no idea why the ordinary Kan extesion implies the Kan extension in 2-cateogory of triangulated category. I'll read the notes in the Stacks Project more carefully when time makes that possible. – Z. He Feb 27 '24 at 03:13
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    $\def\T{\mathcal{T}}$The Version 2 is missing the property of $(\operatorname{Lan}_QF,\eta)$ being an “absolute left Kan extension,” i.e., it must be preserved by any functor $H:\T\to\T'$ in the sense that $(H\circ\operatorname{Lan}_QF,H\eta)$ is the left Kan extension of $H\circ F$ along $Q$, for all such $H$. This is actually what Kashiwara and Schapira require in Definition 10.3.1 (the one in p. 253 from their book that you mentioned). There is a discussion regarding Version 1 vs Version 2 in Görtz, Wedhorn, Algebraic Geometry II, Remark F.168. – Elías Guisado Villalgordo Sep 20 '24 at 08:33

1 Answers1

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$\def\Triang{\mathbf{Triang}} \def\Cat{\mathbf{Cat}} \def\C{\mathcal{C}} \def\A{\mathcal{A}} \def\S{\mathcal{S}} \def\D{\mathcal{D}} \def\Trans{\mathbf{Trans}}$Write $\Cat$ for the 2-category of categories, $\Trans$ for the 2-category of categories with translation and $\Triang$ for its 2-subcategory of triangulated categories (the latter ones are defined here). Given objects $\C,\A,\D$ belonging to some 2-category and 1-morphisms $F:\C\to\A$ and $K:\C\to\D$, we write $K^\dagger F$ for the left extension of $F$ along $K$.¹ The left extension $K^\dagger F$ is said to be absolute if it is preserved by any 1-morphism, i.e., $G\cdot K^\dagger F=K^\dagger(GF)$, for any 1-morphism $G$ with domain $\A$.

[Der, Theorem 4.1]. Let $F:\C\to\A$ be a triangulated functor of triangulated categories, and let $\S$ be a saturated multiplicative system compatible with the triangulated structure on $\C$ [SP, 05R2]. Let $Q:\C\to\S^{-1}\C$ be the localization functor. Consider the following assertions:

(a) $\exists Q^\dagger F$ in $\Triang$,

(b) $\exists Q^\dagger F$ in $\Triang$ and is absolute,

(c) $\exists Q^\dagger F$ in $\Cat$, and

(d) $\exists Q^\dagger F$ in $\Cat$ and is absolute.

Then (d) implies all other assertions.

Assertions (a) and (d) correspond respectively with your versions 1 and 2. These are equivalent respectively to Verdier's and Deligne's definition for the derived functor, given in [V, p. 38, Définition 1.2] and [D, 1.2.1]. What about the converse? The short answer is: we don't seem to know. According to [GW, Remark F.168]:

It is not clear (to us), whether a derived functor as in Definition F.164 (i.e., as in (a)) is necessarily a derived functor in the sense of [KaSh] (i.e., as in (d)). In practice this distinction is not important because the results for the existence of derived functors in [KaSh] show that all techniques we use to construct derived functors (which all rely on Theorem F.173) also ensure the existence of the right derived functor in their stronger sense.

(Italics are mine; Theorem F.173 is just the general result to compute derived functors in terms of resolutions.)

If you want to see a proof of the theorem above, I wrote one in [Der]. The proof consists of two parts:

  1. One first shows that if $F$ is a triangulated functor of triangulated categories and (d) holds, then $Q^\dagger F$ is triangulated. This part is [Der, Corollary 2.2] and its proof is summarized here.

  2. One afters check that the universal comparison transformation $F\Rightarrow Q^\dagger F\cdot Q$ is trinatural, and concludes by the following result.

[Der, Proposition 4.3]. Let $F:\C\to\A$, $K:\C\to\D$ be 1-morphisms in $\Trans$. If $(K^\dagger F,\eta:F\Rightarrow K^\dagger F\cdot K)$ exists in $\Cat$, then there is a unique isomorphism $\chi:K^\dagger F\cdot T\cong T\cdot K^\dagger F$ such that $\eta$ becomes a trinatural transformation. Moreover, $((K^\dagger F,\chi),\eta)$ is the left extension in $\Trans$ of $F$ along $K$.


¹A left extension in $\Cat$ is just a left Kan extension. In general, left extensions can be defined in any 2-category [L, Sect. 2.2].

References

[Der] ―, Derived Triangulated Functors are Triangulated, https://eliasguisado.wordpress.com/work/

[SP] The Stacks Project Authors, The Stacks Project

[L] S. Lack. “A 2-Categories Companion.” In: Towards Higher Categories. 2009. https://arxiv.org/abs/math/0702535.

[V] J.-L. Verdier. “Catégories Dérivées (état 0)”. In: SGA4½, Cohomologie étale. Ed. by P. Deligne. Vol. 569. Lecture Notes in Mathematics. Springer-Verlag, 1977, pp. 262–311

[D]. P. Deligne. “Cohomologie à supports propres”. In: SGA4, Théorie de Topos et Cohomologie Étale des Schémas. Vol. III, Exp. XVII. Lecture Notes in Math. 305. Springer, 1971, pp. 250-461

[GW] U. Görtz, T. Wedhorn, Algebraic Geometry II: Cohomology of Schemes, Springer, 2023

[KaSh] M. Kashiwara and P. Schapira. Categories and Sheaves. Grundlehren der mathematischen Wissenschaften. Springer Berlin Heidelberg, 2005.