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In an abelian category $\mathcal{A}$, for a system $\{F_i,\phi_{ij}\}$ we have an exact sequence

$0\to \lim F_i\to \prod F_i \to \prod F_i$

where the second map is given by $id-\prod\phi_{ij}$. Is there a version of this for stable $\infty$-categories? Meaning, if $\mathcal{C}$ is a stable $\infty$-category and $F_i$ is a system in $\mathcal{C}$, then is there a fiber sequence in $\mathcal{C}$ given by

$\lim F_i\to \prod F_i \to \prod F_i$?

In case $\mathcal{C}=\mathcal{D}^+(A)$ where $A$ is an abelian category with exact products, enough invectives and the system is countable, then this is true, see for example Stacks Project Lemma 0BK7. I was wondering if one can conclude the same for a more general $\mathcal{C}$

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In the case of an $\mathbb N^{op}$-indexed system specifically, the answer is yes (note that this is implicit in the Stacks project link you gave); in fact if you replace "fiber sequence" by "equalizer", this holds in an arbitrary $\infty$-category with the appropriate limits (namely products and equalizers). The description in terms of fibers does not hold in general though (in an arbitrary $\infty$-category with limits, the "fiber" without specifying a basepoint does not even make sense)

There are several ways a proof could go. One approach is to use the fact that $\mathbb N$, as an $\infty$-category, is the infinite pushout $[1]\coprod_{[0]}[1]\coprod_{[0]}[1] \dots$. Another approach is to use the Yoneda lemma to reduce to the case of the $\infty$-category of spaces, and there, use explicit models for homotopy limits (this is probably simpler to actually write down, if not as conceptual).

For a general filtered poset $I$, however, homotopy limits over $I^{op}$ can be more complicated.

There is always the general "Bousfield-Kan formula", which expresses $\lim_{I^{op}}$ in the form of the totalization of a cosimplicial object (the $\infty$-analogue of an equalizer): namely, $\lim_{I^{op}}F$ can be described as the limit over $\Delta$ of a functor that looks like $[n]\mapsto \prod_{i_0\to ... \to i_n} F(i_0)$ .

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  • $\begingroup$ Thank you very much for your answer! Ok I had the feeling that this was probably too optimistic. $\endgroup$
    – user197402
    Commented Nov 1, 2022 at 18:34
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    $\begingroup$ Could you elaborate on that? I do not know how to prove this in spaces... Using the description of $\mathbb N$ as an infinite pushout, I only know how to produce $lim F_i$ as a pullback of the diagonal of the product of the $F_i$'s. In a pointed situation, that gives a fiber sequence $\prod_i\Omega F_i\to lim F_i\to\prod_i F_i$. In a stable context, of course, we can deloop and get the fiber sequence expressing the limit as a fiber. $\endgroup$
    – D.-C. Cisinski
    Commented Nov 1, 2022 at 21:17
  • $\begingroup$ @D.-C.Cisinski : doesn't this pullback exactly describe an equalizer ? (note that at the beginning of my answer I said "if you replace 'fiber sequence' by 'equalizer'", so I am not claiming that we get a fiber sequence in spaces, for that I agree with your comment) $\endgroup$
    – Maxime Ramzi
    Commented Nov 1, 2022 at 21:45
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    $\begingroup$ Yes, the pullback does precisely that. I just wanted to insist on the fact that the description of the limit as a fiber is only true in a stable context. $\endgroup$
    – D.-C. Cisinski
    Commented Nov 1, 2022 at 22:38
  • $\begingroup$ Right, maybe I should stress that more than what I wrote. I agree with this $\endgroup$
    – Maxime Ramzi
    Commented Nov 2, 2022 at 10:10

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