Timeline for Colimit density and monads
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 12, 2015 at 13:30 | comment | added | Tim Campion | @ZhenLin Ah, I see what I did wrong. I thought the coproduct of the objects of the generator would form a generator. But this won't be true generically because even though an object $X$ admits a map from some object of the generator, it need not admit a map from every object of the generator, so it need not admit a map from the coproduct. It is interesting that this construction does work for simplicial sets. | |
| Jan 12, 2015 at 8:39 | comment | added | Todd Trimble | @TimCampion But I think you're right that simplicial sets are monadic, as the sum $H$ of the representables is (regular) projective and is a dense generator (in this example, every representable is a retract of $H$). I hadn't realized that before. | |
| Jan 12, 2015 at 8:29 | comment | added | Zhen Lin | @TimCampion But it's not true. A Grothendieck topos need not contain a generating object, let alone one that is projective. | |
| Jan 12, 2015 at 7:56 | vote | accept | arsmath | ||
| Jan 12, 2015 at 7:56 | comment | added | arsmath | By "colimit-dense" I mean that every object is a colimit of some diagram, not necessarily the canonical diagram, but your example works for that as well. I knew Pos wasn't monadic over Set, but I assumed that you couldn't get the 3-element chain as a colimit from the 2-element chain, though now I see that you can. I think I overgeneralized the proof that Pos is not regular (that you link to) to show that the 3-element chain was not a colimit, though it clearly doesn't follow. | |
| Jan 12, 2015 at 3:42 | comment | added | Tim Campion | The converse direction of the characterization theorem in the linked-to note of Vitale does show that if a cocomplete category with a dense generator is exact, then it is monadic over $\mathsf{Set}$. This surprises me: it implies that every Grothendieck topos is monadic over $\mathsf{Set}$. In particular, contrary to my claim, simplicial sets are monadic over $\mathsf{Set}$. | |
| Jan 12, 2015 at 1:38 | history | answered | Todd Trimble | CC BY-SA 3.0 |