Timeline for Universal property of module categories over monads
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 8, 2014 at 9:00 | vote | accept | Martin Brandenburg | ||
| Apr 22, 2014 at 20:13 | comment | added | Mike Shulman | Well, at least it requires proof. | |
| Apr 21, 2014 at 19:11 | comment | added | Martin Brandenburg | @Mike Shulman: Well, I think that a functor on $\mathsf{Mod}(T)$ which preserves reflexive coequalizers and coproducts of free modules, is automatically cocontinuous. Is this not correct? | |
| Apr 20, 2014 at 5:03 | answer | added | Mike Shulman | timeline score: 6 | |
| Apr 19, 2014 at 21:55 | comment | added | Mike Shulman | @MartinBrandenburg, that shows that Mod(T) has coproducts, but not that $\tilde{G}$ preserves them. | |
| Apr 19, 2014 at 7:05 | comment | added | Martin Brandenburg | @Daniel: Coproducts may be reduced to reflexive coequalizers of coproducts of free modules (see Linton's paper). | |
| Apr 19, 2014 at 0:48 | comment | added | Daniel Schäppi | Is it clear that $\tilde{G}$ preserves coproducts? If the categories or the monad are not additive, then coproducts in the category of modules (algebras) are usually quite different from coproducts in the base category (e.g. groups vs. sets). If $T$ is additive and finitary (preserves filtered colimits), this is certainly not a problem. | |
| Apr 17, 2014 at 10:51 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Apr 17, 2014 at 9:10 | comment | added | Mariano Suárez-Álvarez | @MartinBrandenburg, well, modules are algbras in the sense of universal algebra (with one unary operator per element in the ring, satisfyign the relations you know) That is where the name $T$-algebra comes from, I guess. | |
| Apr 17, 2014 at 6:58 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Apr 17, 2014 at 6:43 | comment | added | Martin Brandenburg | Exactly, Tom. My motivation for the terminology "modules" is the example $T=A \otimes -$ for some algebra $A$ in a monoidal category, here $T$-modules are left $A$-modules (not $A$-algebras). In my understanding an algebra should carry some sort of associative binary operation, but a $T$-action is a morphism $Tx \to x$ satisfying two conditions, which really abstracts the definition of a left module. | |
| Apr 17, 2014 at 1:14 | comment | added | Tom Leinster | @Dimitri: I think Martin means the category of algebras, which some people call the category of modules. | |
| Apr 17, 2014 at 0:18 | comment | added | Dimitri Chikhladze | But I am not sure what you mean by Mod(T). | |
| Apr 17, 2014 at 0:17 | comment | added | Dimitri Chikhladze | One way to actually construct the Kleisli object in the 2-category of cocomplete categories and cocontinuous functors I imagine is to somehow make the Kleisli category of T cocomplete. I dont't know how this would work. | |
| Apr 17, 2014 at 0:08 | comment | added | Dimitri Chikhladze | Forgetting colimits the category with that universal property is the Kleisli category for the monad. One can define a Kleisli object in any 2-category. If your T is cocontinuous, then the universal category that you are after would be the Kleisli object of T in the 2-category of cocomplete categories and cocontinuous functors. This is purely formal statement of course. | |
| Apr 16, 2014 at 18:01 | history | asked | Martin Brandenburg | CC BY-SA 3.0 |