Timeline for Are dg-modules over a cofibrant dg-category cofibrant?
Current License: CC BY-SA 4.0
20 events
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| Dec 13, 2020 at 13:57 | comment | added | Reid Barton | Or more simply, $\mathsf{dgMod}_C$ as an object of $\mathsf{dgCat}$ doesn't commute with colimits in $C$ in any sense because, for example, if $C$ is the initial dg-category, $\mathsf{dgMod}_C$ is not empty! | |
| Dec 13, 2020 at 4:39 | comment | added | Reid Barton | The tensor product in section 4 does not correspond to the tensor product of presentable ∞-categories which appears in formulas like $\mathsf{dgMod}_C \otimes^L \mathsf{dgMod}_D = \mathsf{dgMod}_{C \otimes^L D}$. It is closer to the ordinary product of categories. A general $C \otimes^L D$-module isn't built out of a $C$-module and a $D$-module; you need to allow formation of colimits as well. | |
| Dec 13, 2020 at 3:10 | history | edited | Stahl | CC BY-SA 4.0 |
fixed definition of fibration
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| Dec 13, 2020 at 2:28 | comment | added | Stahl | @DmitriPavlov Thanks for all your help here. I'll follow the outline you've provided! | |
| Dec 13, 2020 at 2:27 | vote | accept | Stahl | ||
| Dec 13, 2020 at 2:18 | comment | added | Dmitri Pavlov | @Stahl: For your cited fact, I suggest to observe first that dgCat_C is homotopy cocontinuous in C. Then the problem for dg-algebras reduces to the case C=D=k, the base ring, which can be shown directly. Nothing in this argument needs ∞-categories. | |
| Dec 13, 2020 at 1:38 | comment | added | Stahl | @ReidBarton Toën claims on page 10 of the paper I mentioned that Tabuada has shown that there is a model structure on $\mathsf{dgCat}$ with fibrations as defined above and whose weak equivalences are the quasi-equivalences. In section 4, Toën defines the tensor and derived tensor product of two dg-categories by $C\otimes^L D:=Q(C)\otimes D,$ where $Q:\mathsf{dgCat}\to\mathsf{dgCat}$ is a cofibrant replacement functor for this model structure which acts as the identity on objects. Perhaps I'm missing something, but nowhere have I seen a requirement that the dg-categories be cocomplete. | |
| Dec 13, 2020 at 1:26 | history | edited | Stahl | CC BY-SA 4.0 |
added 48 characters in body
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| Dec 13, 2020 at 0:18 | comment | added | Reid Barton | I'm skeptical. Don't you need to be in a model category (or 2-category) of cocomplete dg-categories for these tensor products to even make sense? In this context the question makes more sense, but then you also need to specify what the model structure is. | |
| Dec 12, 2020 at 22:58 | history | edited | Stahl | CC BY-SA 4.0 |
added original motivation
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| Dec 12, 2020 at 22:13 | comment | added | Stahl | @Dmitri I'm trying to find a more "elementary" (non-$\infty$-categorical) proof that $\mathsf{dgMod}_C\otimes^L\mathsf{dgMod}_{D}\simeq\mathsf{dgMod}_{C\otimes^L D}$, at least for $C$ and $D$ dg-algebras (I asked a different question about this and got a nice answer, but I wanted to do it without those techniques as well). However, the proof sketch I came up with required this result. | |
| Dec 12, 2020 at 18:47 | comment | added | Dmitri Pavlov | Why do you want categories of dg-modules to be cofibrant in the first place? What is the motivation behind wanting such a property? I suspect we may have an XY problem here. | |
| Dec 7, 2020 at 15:52 | answer | added | Dmitri Pavlov | timeline score: 1 | |
| S Dec 7, 2020 at 14:34 | history | suggested | gmvh |
Added top-level tag
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| Dec 7, 2020 at 11:12 | comment | added | Reid Barton | Is a dg-category equivalent (as an enriched category) to a cofibrant one again cofibrant? I'm not sure the main question makes much sense, unless the answer is "no". A priori it should depend on the choice of the category Set up to isomorphism, and not just up to equivalence. | |
| Dec 7, 2020 at 8:15 | review | Suggested edits | |||
| S Dec 7, 2020 at 14:34 | |||||
| Dec 7, 2020 at 7:24 | history | edited | Stahl | CC BY-SA 4.0 |
added 8 characters in body
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| Dec 7, 2020 at 7:24 | comment | added | Stahl | @TimPorter Yes it was, thank you! | |
| Dec 7, 2020 at 7:23 | comment | added | Tim Porter | You seem to have defined 'quasi-fully faithful ' twice. i suspect the second was intended to be the definition of 'quasi-essentially surjective'! | |
| Dec 7, 2020 at 7:01 | history | asked | Stahl | CC BY-SA 4.0 |