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$\newcommand{\tot}{\operatorname{Tot}}\newcommand{\A}{\mathscr{A}}\newcommand{\L}{\mathbb{L}}\newcommand{\R}{\mathbb{R}}$Say $T$ is is a functor $\A_1\times\A_2\times\cdots\times\A_n\to\A$ of Abelian categories, additive in each variable separately. Assume the Abelian categories $\A_\bullet$ are as nice as we could wish, maybe even that they are module categories.

When defining the left or right hyperderived functors of $T$ (which Cartan-Eilenberg call instead the (co)homology invariants of $T$) we take, for some tuple $X_1,X_2,\cdots,X_n$, left or right Cartan-Eilenberg resolutions $Q_1,Q_2,\cdots,Q_n$ - which are upper half plane double complexes of (co)homological type - and according to some sign conventions also explained in the text of Cartan-Eilenberg we can form a $2n$-complex $T(Q_1,Q_2,\cdots,Q_n)$. Then we would declare the hyperderived functor of $T$ at $X_1,\cdots,X_n$ to be the homology of the totalisation of this $2n$-complex.

In the text of Cartan-Eilenberg it seems the convention is to always define $\L_\ast T(X_1,\cdots,X_n)=H_\ast(\tot^{\oplus}(T(Q_1,\cdots,Q_n))),\,\R_\ast T(X_1,\cdots,X_n)=H^\ast(\tot^{\oplus}(T(Q_1,\cdots,Q_n)))$ taking the $\oplus$-totalisation in both instances. However, in the text of Weibel (which only discusses the single variable case) the convention is to define $\L_\ast T$ by $\oplus$-totalisation and $\R_\ast T$ by $\prod$-totalisation; $\R_\ast T(X_1,\cdots,X_n):=H^\ast(\tot^{\prod}(T(X_1,\cdots,X_n)))$.

Either choice results in well-defined functors. The two notions only disagree on unbounded complexes. Are there any reasons to prefer one totalisation convention over the other, in this case? My gut tells me $\R_\ast$ is best defined with $\prod$-totalisation simply because $\oplus$ sits on the "left" and $\prod$ on the "right" when thinking about universal properties. I cannot point to any formal property of one definition over the other that would inform this choice. I suppose another benefit is that under the $\prod$-totalisation definition, $\R_\ast$ is formally dual to $\L_\ast$. But, Cartan and Eilenberg must have had a reason for doing it their way, right?

One difficulty in pinpointing reasons to prefer one definition over the other is that I am not aware of anywhere where these notions are studied in generality. The Cartan-Eilenberg texts lays out some definitions. I know of a detailed study of $\L_\ast T$ where $T$ is the tensor product in Grothendieck's EGA 3, but this is a special case.

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    $\begingroup$ Any decent source discussing cyclic homology has a section on it. Most often it is the case that one totalisation gives zero functor, but other one gives something meaningful; it's important to differentiate this situation from one where zero is actually the right answer, and using wrong totalisation gives you something meaningless. // What I mean is there's no "right" or "wrong" choice; there are two genuinely different functors, each one has its own merits. $\endgroup$
    – Denis T
    Commented Dec 24, 2023 at 18:01
  • $\begingroup$ Also not every abelian category has exact products (or coproducts). If products are exact, product totalisation of cohomological functors is cohomological; similarly for coproducts. If (co)products are not exact, there's little to no reason to call (co)product totalisation a "derived functor", since it won't be cohomological. $\endgroup$
    – Denis T
    Commented Dec 24, 2023 at 18:04
  • $\begingroup$ @DenisT Ah, but don't we show these functors are homological only in the case of bounded complexes, and if bounded complexes are involved all direct sums and products will end up finite thus there is no issue with exactness? $\endgroup$
    – FShrike
    Commented Dec 24, 2023 at 18:34
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    $\begingroup$ As you suspect, Weibel's convention is the correct one. At the level of derived ∞-categories, a left CE resolution writes a complex as the colimit of a simplicial diagram of complexes (via Dold-Kan), and taking the colimit of such a diagram corresponds to sum-totalization. A right CE resolution writes a complex as the limit of a cosimplicial diagram, and taking the limit corresponds to product-totalization. $\endgroup$
    – Marc Hoyois
    Commented Dec 27, 2023 at 12:05
  • $\begingroup$ @MarcHoyois What you're writing only works if (co)products are exact in the heart of some t-structure of derived type. Derived category of an abelian category can fail to have those properties. $\endgroup$
    – Denis T
    Commented Dec 27, 2023 at 16:45

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This is more of an extended comment rather than an answer, but the question itself is rather conversational, so...

I'd say that one reasonable definition (rather than construction) of a derived functor is as follows. Suppose that you have an additive functor $F: \mathbf C \to \mathbf A$ between abelian categories. Then the (right) derived functor $RF$ is a (right) Kan extension of $\bar{F} \circ q_{\mathbf A}$ along $q_{\mathbf C}$, where: $$\bar{F}: \mathrm{Hot^*}(\mathbf C) \to \mathrm{Hot^*}(\mathbf A)$$ is pointwise application of $F$ to a complex, and $q_{\mathbf {(-)}}$ is Verdier quotient of homotopy category of $\mathbf {(-)}$ by the subcategory of acyclic complexes (with resulting quotient category being the derived category of $\mathbf {(-)}$). $*$ can be $+, -, b$, or nothing.

This Kan extension is not guaranteed to exist! If $\mathbf C$ has enough injectives, and $* = +, b$; or $\mathbf C$ has enough projectives, and $* = -, b$, then existence of such Kan extension follows from the fact that full embedding of acyclic complexes forms a semiorthogonal decomposition. This decomposition has form $\langle \mathrm{Acyc}^+(\mathbf C), \mathrm{Hot}^+(Inj-\mathbf C) \rangle$, such that $\mathrm{Hot}^+(Inj-C) \simeq \frac{\mathrm{Hot}^+(\mathbf C)}{\mathrm{Acyc}^+(\mathbf C)} = D^+(\mathbf C)$ in first case, and $\langle \mathrm{Hot}^-(\mathrm{Proj}-\mathbf C), \mathrm{Acyc}^-(\mathbf C)\rangle$ in second. If $F$ is left exact in first case, or irght exact in second case, it will be isomorphic to zeroth cohomology of $RF$ w.r.t canonical structure on $D^*(\mathbf A)$. If $\mathbf C$ has finite projective or injective dimension, then it exists in unbouded case as well. Otherwise, you need to resort to some tricks to prove its existence.

(This treatment of derived functors is briefly mentioned in Deligne's appendix to "Residues and duality"; I'm not aware of any good text source, but you can fill in the details yourself.)

Sometimes you can apply some sort of resolutions to compute this Kan extension; this is the case when $F$ is effaceable, or there are enough injectives/projectives. But the more you think of the derived functor as a solution to an extension problem — upgrading a functor between hearts of triangulated categories to a triangulated functor between triangulated categories — the more decisively you can say which resolutions (in particular, which totalisations of degree-wise constructions) make sense, and which do not.

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  • $\begingroup$ I was recently thinking (with regards to the Kan extension perspective and whether or not it is guaranteed to exist): in the special case of modules over a ring, it can be shown that all complexes - even unbounded ones - admit “semifree” and “semiinjective” resolutions, these being surjective and injective quasiisomorphisms into complexes satisfying a certain lifting property. Do you know then if we can use that to say any left/right exact functor between module categories has a Kan extension downstairs on $D(R)\to D(S)$? Ie with $\ast=\text{all the complexes}$? $\endgroup$
    – FShrike
    Commented Mar 24 at 16:18
  • $\begingroup$ I realise this maybe has nothing to do with my original question but since you mentioned related things in your answer.. $\endgroup$
    – FShrike
    Commented Mar 24 at 16:18
  • $\begingroup$ @FShrike Indeed, for any additive functor (not necessarily exact on any side; this condition plays a role only when you want to check whether a morphism from F to/from the zeroth right/left derived of F is iso) from a module category all derived functors exist. They only depend on left/right "exactification" of F, which is sort of sheafification (construction of this adjoint for small categories is in Freyd's "Abelian categories", for locally presentable I do not know a good reference). $\endgroup$
    – Denis T
    Commented Mar 26 at 12:49
  • $\begingroup$ Precise condition for existence of derived functor as a Kan extension for triangulated categories along a Verder quotient by some C is in the beginning of SGA4T3. I append (c) to a morphism name to denote that its cone is in C. So, condition is: for any object X there exists f(c), such that for all g(c) exists h(c): $f: X \to Y$, $g: Y \to Z$, $h: Z \to W$, with a condition that $F(gh): F(Y) \to F(W)$ is iso. Then RF(X) is F(Y). // Infinity-categorical version of this statement is thoroughly covered in Cisinski's "Higher categories and homotopical algebra". $\endgroup$
    – Denis T
    Commented Mar 26 at 13:00
  • $\begingroup$ This condition (which is more or less just a repackaging of existence of pointwise Kan extension) is weaker that existence of enough injectives in abelian category (and dual is weaker than enough projectives); it is always satisfied, of course, when homotopy category is equivalent to the derived category (which is the case when abelian category has finite Yoneda global dimension). One can compare this to Spaltenstein's "homotopy resolutions", which you mentioned; their existence is sufficient, but also not necessary. $\endgroup$
    – Denis T
    Commented Mar 26 at 13:15

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