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Let $\mathcal{C}$ be a compactly generated stable $\infty$-category, linear over a field of characteristic $0$ (i.e., so that it is in particular a dg-category). Let $A$ be a co-monad acting on $\mathcal{C}$, and suppose that $A$ preserves colimits and sends compact objects to compact objects.

Let $\mathrm{coMod}_A$ be the category of $A$ co-modules in $\mathcal{C}$. I am interested in the question of when it is true that $\mathrm{coMod}_A$ is compactly generated by those co-modules that are compact in $\mathcal{C}$.

I am able to prove that all compact objects of $\mathrm{coMod}_A$ are compact in $\mathcal{C}$ (because the co-free co-module functor respects colimits in this case).

On the other hand, it seems that similar questions have a negative answer in general (e.g., co-algebras are not generally given by colimits of compact ones).

I am having trouble finding sources working in sufficient generality, but I am not very familiar with this field. Is there anything known about this property?

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  • $\begingroup$ Probably a stupid question but what definition of the $\infty$ category of comodules are you using? $\endgroup$
    – Saal Hardali
    Commented Feb 13, 2019 at 8:32
  • $\begingroup$ Opposite category of the category of modules in the opposite category of $\mathcal{C}$. $\endgroup$
    – Gal Dor
    Commented Feb 13, 2019 at 9:00
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    $\begingroup$ Aha, this is the one I know how to make sense of in this generality. Notice however that there are other useful variants which do not always agree see arxiv.org/abs/0905.2621 $\endgroup$
    – Saal Hardali
    Commented Feb 13, 2019 at 9:16
  • $\begingroup$ In this generality it is certainly false. Let $C$ be the coaugmented counital coalgebra over $\mathbb{Q}$ cogenerated by a single primitive cogenerator in (homological) degree $3$. Then $\pi_{\ast} (Comod_C(\mathbb{Q})) = \mathbb{Q}[x]$ with $|x| = 2$ so $\mathbb{Q}$ can't be compact because then $\mathbb{Q}[x^{-1}] = 0$ contradicting that $\pi_{\ast} (End_C(\mathbb{Q}))[x^{-1}] = \mathbb{Q}[x,x^{-1}] \ne 0$ $\endgroup$
    – Saal Hardali
    Commented Feb 5, 2020 at 13:54

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