Is the composite of absolute derived functors a derived functor?

Question

Let me recall the following definition. Let $F: C \to D$ be a functor between homotopical categories. Denote by $\gamma_C: C \to \mathrm{Ho} C$ the localization and similary for $D$. A total left derived functor $\mathbf{L}F: \mathrm{Ho}C \to \mathrm{Ho}D$ for $F$ is a right Kan extension of $\gamma_D \circ F$ along $\gamma_C$. This will be called an absolute derived functor if it is absolute as a Kan extension.

It is well-know that the following two properties of derived functors:

(a) if $F: C \to D$ and $F': D \to E$ are functors between homotopical categories that admit total left derived functors $\mathbf{L}F$ and $\mathbf{L}F'$, then $\mathbf{L}F' \circ \mathbf{L}F$ is a total left derived functor for $F' \circ F$.

(b) if $F : C \to D$ is left adjoint to $G: D \to C$ and $F$ and $G$ admit left and right total derived functors $\mathbf{L}F $ and $\mathbf{R}G$ respectively, then $\mathbf{L}F $ is left adjoint to $\mathbf{R} G$.

are in fact true if these functors are constructed via deformations (or, in particular, via (co)fibrant replacements in model categories), but are no more true in general, meaning, for derived functors that plainly satisfy the definition above.

Anyway, it is known, by a theorem of Maltsiniotis, that (b) holds true whenever the involved derived functors are not simply Kan extensions, but absolute ones.

I do not know if this is true also for (a), and I can not find any reference in the literature. For instance, Riehl in various places in her books or articles mentions the theorem by Maltsiniotis for adjunctions, but never says a word about composition of absolute derived functors. Cisinski in his book on Higher categories and homotopical algebra seems to suggest (Corollary 2.3.4 and Proposition 2.3.6) that (a) is in fact true for derived functors which are absolute Kan extensions, but does not give a proof.

So the question is: is property (a) true if $\mathbf{L}F $ and $\mathbf{L}F'$ are absolute Kan extensions?

Answer 1

Here is a somewhat degenerate example that illustrates what can go wrong.

Let $\textbf{Ab}$ be the category of abelian groups, considered as a homotopical category where the weak equivalences are the isomorphisms, and let $\textbf{Ch}$ be the category of chain complexes of abelian groups, considered as a homotopical category where the weak equivalences are the quasi-isomorphisms. The obvious functor $\textbf{Ab} \to \textbf{Ch}$ sending each abelian group to the corresponding chain complex concentrated in degree 0 is homotopical. On the other hand, if $M$ is not a flat abelian group, then $M \otimes_\mathbb{Z} {-}$ is not a homotopical functor $\textbf{Ch} \to \textbf{Ch}$. Nonetheless, the obvious functor $\textbf{Ab} \to \textbf{Ch}$ followed by $M \otimes_\mathbb{Z} {-}$ is homotopical, so is its own (absolute) derived functor. On the other hand, if $N$ is an abelian group such that $\textrm{Tor}^1_\mathbb{Z} (M, N)$ is non-zero, then $M \otimes^\textbf{L}_\mathbb{Z} N$ (i.e. the value of the (absolute) left derived functor of $M \otimes_\mathbb{Z} {-}$ at $N$) is not quasi-isomorphic to $M \otimes_\mathbb{Z} N$ (considered as a chain complex concentrated in degree 0). Thus, we have a homotopical functor $\textbf{Ab} \to \textbf{Ch}$ and a left deformable functor $\textbf{Ch} \to \textbf{Ch}$ such that the composite of the (absolute) left derived functors is different from the (absolute) left derived functor of the composite.

What is true is that the universal property of absolute Kan extensions gives you a comparison between the composite of absolute derived functors and the absolute derived functor of the composite.