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In his Cohomological methods in intersection theory, Cisinski writes:

"[...] note however that, by Balzin’s work [Bal19, Theorem 2], it is clear that one can go back and forth between the language of fibred $\infty$-categories and the one of algebraic derivators."

By Bal19 he intends Balzin's Reedy model structures in families. Let me quote the abovementioned Theorem 2.

Thm. 2 in Bal19.

Let $\mathsf{E} \to \mathsf{R}$ be a left model Reedy fibration. Then the induced $\infty$-functor $\mathsf{L}\text{Sect}(\mathsf{R}, \mathsf{E}) \to \text{Sect}(\mathsf{R}, \mathsf{LE})$ is an equivalence of quasicategories.

Even though I sense a connection between derivators, fibred $\infty$-categories and this statement, I cannot make it precise. Could someone help me spelling out what Cisinski means precisely?

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    $\begingroup$ Ask him! You'll get the best possible answer. $\endgroup$ Commented Feb 5, 2020 at 9:43

1 Answer 1

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I am no Denis-Charles but given the other paper you quoted let me think of a sketch, perhaps you will be able to make the right out of it.

Let $\mathcal E \to \mathcal C$ be a Quillen presheaf (model categorical fibres, Quillen pairs as transition functors) where each $\mathcal E(c)$ is stable. This would imply, if we had the cofibrant generation in fibres for example, that for any functor $F: \mathcal D \to \mathcal C$, the category of sections $Sect(\mathcal D,F^*\mathcal E)$ is a stable model category. You can then introduce a $2$-functor $\mathbb D: (Cat/ \mathcal C)^{op} \to CAT$ that sends $F: \mathcal D \to \mathcal C$ to the homotopy category $Ho (Sect(\mathcal D,F^*\mathcal E))$ and study its properties. Something like that could be called a relative (pre)derivator (maybe it already is, forgive my ignorance, then) valued in triangulated categories.

We can also start with the infinity-localisation $L \mathcal E \to \mathcal C$ and study its infinity-categorical sections $Sect(\mathcal D,F^* L \mathcal E)$ for functors $F: \mathcal D \to \mathcal C$. Since a localisation of a stable model category is a stable infinity-category (a statement true even without cofibrant generation due to Proposition 3 mentioned below), $L \mathcal E \to \mathcal C$ is a bicartesian fibration with stable fibres and using Section 5 of HTT (co/limits of sections computed fibrewise) we can verify that each $Sect(\mathcal D,F^* L \mathcal E)$ is stable. Taking their homotopy categories would again yield another $2$-functor $\mathbb D':(Cat/ \mathcal C)^{op} \to CAT$ valued in triangulated categories.

Proposition 3 of the Reedy paper means however that $Ho (Sect(\mathcal D,F^*\mathcal E)) \cong Ho (Sect(\mathcal D,F^* L \mathcal E))$ where the latter is the homotopy category of an infinity-category and the former is that of a model category. In other words, the relative derivators $\mathbb D$ and $\mathbb D'$ are equivalent; the equivalence comes from the fairly canonical map $L Sect(\mathcal D,F^*\mathcal E)) \to Sect(\mathcal D,F^* L \mathcal E)$ (universality of localisation is used here) so it should be trackable along base changes. Thus you can study your relative derivators any way you want, via the model-categorical presentation or the infinity-categorical one.

I suppose you get an algebraic derivator if you replace $\mathcal C$ by $Sch$ (no restriction on the size of $\mathcal C$ was made above) and consider something like $QCoh \to Sch$ for the Quillen presheaf, but you will have to take it from here to see if that fits. Maybe I should add something about such relative derivators to the Reedy paper if life permits.

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    $\begingroup$ This seems a bit debatable to me. In this answer you start from some kind of model category and you show that you can construct a derivator out of it, or an infinity category. These two object somehow compute the same homotopy theory. I completely agree, but you are STARTING from the wrong data. If I am given a derivator, how do I construct an infinity category? If I am given an infinity category, how do I construct a derivator? $\endgroup$ Commented Oct 13, 2021 at 15:06
  • $\begingroup$ I don't claim I answered anything, so there is not much to debate. One could try proving a statement similar to arXiv:math/0603339 for relative derivators that are presentable biCartesian (to avoid the example of sections of a Cartesian, and not coCartesian, fibration over a simplex that can happen to be presentable). If that was done, then perhaps you could start with an algebraic derivator, check its local presentability so that it comes from a model-categorical bifibration and then use the above to not worry about anything. It seems like a lot of work, maybe something else was meant by DCC. $\endgroup$
    – Edouard
    Commented Oct 19, 2021 at 13:10

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