Timeline for Completeness of the Infinity Category of A-infinity Categories
Current License: CC BY-SA 4.0
9 events
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| Oct 15, 2023 at 17:52 | comment | added | Simon Henry | To be clear this is meant as a toy model for the whole category of $A_\infty$ categories that has the advantages of being covered by standard theorem about operads. I don't claim it answers your questions - it just shows that the argument saying there is no such model structure is invalid as it would apply equally to this toy model were we know there is a model structure (the counter example ni the paper you cite only use one object $A_\infty$-categories). | |
| Oct 15, 2023 at 17:42 | comment | added | Simon Henry | We fix a set $X$ and we consider The $\infty$-category of $A_\infty$-categories with $X$ as set of objects (and identity on objects functors between them). | |
| Oct 15, 2023 at 17:09 | comment | added | TheWildCat | @SimonHenry By "fixed set of objects" do you mean we consider the infinity category of $A_{\infty}$-categories with objects in this set? | |
| Oct 15, 2023 at 13:50 | comment | added | Simon Henry | ... the right notion homotopically speaking. but the correct notion of "weak functor" appears when you look at morphism out of a cofibrant replacement. I have never really looked at the theory of $A_\infty$-categories itself. But for $A_\infty$-categories with fixed set of objects, it is well known there is a model structure (they are algebras for a cofibrant operad) and I would be extremely surprised if varying the set of objects would create any problems. | |
| Oct 15, 2023 at 13:46 | comment | added | Simon Henry | $C$ is a cofibrant replacement comonad (probably the bar resolution). So the thing the paper looks at is a CoKleisli category, so it is not surprising it doesn't have all limits. Once you have the model structure the correct notion of morphism you use to compute the homotopy category are these "weak" or $A_\infty$ functor but the model structure itself is meant to be on the category of strict algebraic morphisms. The situation is similar to the Bergner model structure on simplicial categories. The model structure is on the category of strict functors, which are obviously not (...) | |
| Oct 15, 2023 at 13:41 | comment | added | Simon Henry | Ok. The mathematical content of the paper is not wrong, but I think there is a big misunderstanding of how model structures are supposed to work: (strictly unital or non-unital) $A_\infty$ categories are an essentially algebraic notion so the category of $A_\infty$-categories has all limits and colimits. The problem is that the category the paper considers use a different notion of morphism, the $A_\infty$ functors, corresponding to "weak functors". These are not meant to be the morphism in the model category. They corresponds to arrows $CX \to Y$ where C is (...) | |
| Oct 15, 2023 at 7:22 | comment | added | TheWildCat | @SimonHenry The 1-category of $A_{\infty}$-categories does not have a model structure, and in fact there're simple counterexamples showing that the category of $A_{\infty}$-algebras does not have arbitrary limits and colimits (with reference: Localizations of the category of $A_{\infty}$-categories and internal Homs). However, the 1-categorical description is a "fake category", which means we should really look at the infinity category, which might have a model structure but I cannot find in the literature. | |
| Oct 13, 2023 at 20:19 | comment | added | Simon Henry | Note that the notion of $A_\infty$ only makes sense 1-categorically. From the $\infty$-category theoretic point of view, the $A_\infty$ operads is equivalent to the Assoc operads. Given the way you write your question it feels like you have a very precise concrete definition of $A_\infty$-categories in mind. In this case you probably have a model structure for those, and the $\infty$-category associated to a model category is always complete and co-complete. | |
| Oct 13, 2023 at 16:31 | history | asked | TheWildCat | CC BY-SA 4.0 |