5
$\begingroup$

Let $R$ be a commutative ring. Let $\{A_i\}_{i \in I}$ be a diagram of $R$-linear $1$-categories, indexed by a finite poset $I$. (If this matters, assume that the $A_i$ have finitely many objects).

Given an $R$-linear $1$-category $A$, let $D(A)$ be the infinity category $Fun(A, ch(R))$, localized at the quasi-isomorphisms. The reason for this notation is that if we assume $A$ has finitely many objects, then we can view $A$ equivalently as a semi-simple $R$-algebra with idempotents indexed by the objects, and $D(-)$ is the ordinary unbounded derived category of the abelian category of modules over said algebra.

Question: do we have $D(colim A_i)= holim D(A_i)$? (Note that I am distinguishing here between the $1$-categorical colimit and the infinity-categorical homotopy limit.)

Improve this question
$\endgroup$
4
  • $\begingroup$ Indexed by a finite set $I$ or a finite category ? $\endgroup$
    – Maxime Ramzi
    Commented Oct 3, 2023 at 16:02
  • $\begingroup$ @MaximeRamzi A finite poset, sorry for the confusion. $\endgroup$
    – Laurent Cote
    Commented Oct 3, 2023 at 16:38
  • 2
    $\begingroup$ It essentially going to boils down to the question of wether the colimit of the $A_i$ is a homotopy colimits of dg-categories. So sometime yes, but in general no. $\endgroup$
    – Simon Henry
    Commented Oct 3, 2023 at 18:29
  • 2
    $\begingroup$ A línear category with one object is the same as an algebra, but not necessarily semi-simple. $\endgroup$
    – Fernando Muro
    Commented Oct 3, 2023 at 21:05

0

Reset to default

You must log in to answer this question.