2
$\begingroup$

Suppose I have two diagrams $\{\mathcal{C}_i\}_{i\in \mathcal{I}}$ and $\{\mathcal{D}_i\}_{i\in \mathcal{I}}$ in the $\infty$-category of DG-categories over a field $k$ with continuous functors (i.e. colimit-preserving functors), indexed by the same small category $\mathcal{I}$. More generally, one can replace DG-categories with presentable stable $\infty$-categories.

Let $F$ be a not necessarily continuous functor between the two diagrams. Let $\{\mathcal{C}_i\}_{i\in \mathcal{I}^{op}}$ and $\{\mathcal{D}_i\}_{i\in \mathcal{I}^{op}}$ be the opposite diagrams in the $\infty$-category of DG-categories from taking right adjoints. Assume that $F$ is also a well-defined functor between the two opposite diagrams, i.e. the functors $F_i:\mathcal{C}_i\rightarrow \mathcal{D}_i$ commute with the right adjoints (in the usual sense).

Let $\text{DGCat}_{\text{cont}}$ denote the category of DG categories with colimits, and morphisms are $k$-linear functors that commute with homotopy colimits.

By 1.3.2 of Gaitsgory's notes, $\text{colim}_{i\in \mathcal{I}}\mathcal{C}_i$ in $\text{DGCat}_{\text{cont}}$ is equivalent to $\lim_{i\in \mathcal{I}^{op}}\mathcal{C}_i$ in DGCat. So the functor $F$ gives a canonical functor $\widetilde{F}: \widetilde{\mathcal{C}}:=\lim_{i\in \mathcal{I}^{op}}\mathcal{C}_i\rightarrow \widetilde{\mathcal{D}}:=\lim_{i\in \mathcal{I}^{op}}\mathcal{D}_i$ from the universal property of limits. Let $\text{ev}_i^{\mathcal{C}}: \widetilde{\mathcal{C}}\rightarrow \mathcal{C}_i$ be the tautological evaluation functor, whose left adjoint is denoted by $\text{ins}_i^{\mathcal{C}}$. Set similar notations for $\mathcal{D}_i$. By definition, we have canonical isomorphisms $F_i\circ \text{ev}^{\mathcal{C}}_i\simeq \text{ev}_i^{\mathcal{D}}\circ\widetilde{F}$.

My questions are:

(1) Is it true in general that $\widetilde{F}\circ {\text{ins}_i^{\mathcal{C}}}\simeq \text{ins}_i^{\mathcal{D}}\circ F_i$?

(2) If we assume that each $F_i$ is continuous, does (1) follow easily? In this case, $F$ determines a canonical functor $\widetilde{F}':\text{colim}_{i\in \mathcal{I}}\mathcal{C}_i\rightarrow \text{colim}_{i\in \mathcal{I}}\mathcal{D}_i$ from the universal property of colimits. The question is whether this agrees with $\widetilde{F}$ under the natural equivalence $\text{colim}_{i\in \mathcal{I}}\mathcal{C}_i\rightarrow \widetilde{C}$ (and also for the diagram $\{\mathcal{D}_i\}_{i\in I}$) induced by $\text{ins}_i^{\mathcal{C}}$.

$\endgroup$

1 Answer 1

1
$\begingroup$

The answer to (2) is yes, by a nice result of Gaitsgory plus an easy categorical argument. To spell it out, the situation of the OP is the following, where functors going down are $ev_i$ and functors going up are their left adjoints $ins_i$:

$$ \begin{array}{rrrr} \widetilde{D} & \stackrel{\widetilde{F}}{\leftrightarrows} & \widetilde{C}\\\ \uparrow \downarrow & & \uparrow \downarrow \\\ D_i & \stackrel{F_i}{\leftrightarrows} & C_i \end{array} $$

Here we know $\widetilde{F}$ has a left adjoint by Lemma 2.6.4 in Chapter 1 of the book by Gaitsgory and Rozenblyum, which also tells us that the two ways of going around by right adjoints are isomorphic. Following their notation, let $F^R$ denote the right adjoint of $\widetilde{F}$ and $F_i^R$ denote the right adjoint of $F_i$). Lemma 2.6.4 points out that $\text{ev}^{\mathcal{C}}_i \circ F^R \simeq F_i^R \circ \text{ev}_i^{\mathcal{D}}$ as functors from $\widetilde{D}$ to $C_i$. Be warned that Gaitsgory and Rozenblyum start with $\Phi: D\to C$ rather than $F: C\to D$, so I've had to swap $C$s and $D$s from their notation.

Now, when you have a square of adjunctions, and the two ways of going around by right adjoints commute, then the two ways of going around by left adjoints also commute. This argument works just as well in an $\infty$-categorical context, a DG-category context, or a 1-categorical context, and in our case proves that $\widetilde{F}\circ {\text{ins}_i^{\mathcal{C}}}\simeq \text{ins}_i^{\mathcal{D}}\circ F_i$, as desired.

$\endgroup$
1
  • $\begingroup$ Thanks a lot! That answers my question (2). $\endgroup$
    – mikeS
    Commented Jul 1, 2022 at 3:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .