Timeline for What is the opposite category of the category of modules (or Hopf algebra representations)?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Mar 9 at 17:02 | comment | added | Kubrick | @Greg Stevenson sorry for kind of necroposting, I wonder if you can elaborate a little on how can the Eilenberg swindle be used after deducing that iso. | |
| Jun 26, 2010 at 0:57 | history | edited | Greg Stevenson | CC BY-SA 2.5 |
added some stuff
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| Jun 25, 2010 at 13:06 | comment | added | Akhil Mathew | @Tom Goodwillie: Thanks! That clarifies things as well. | |
| Jun 25, 2010 at 4:28 | comment | added | Yemon Choi | @Greg: thanks, that clarifies things. | |
| Jun 25, 2010 at 2:16 | comment | added | Greg Stevenson | @Yemon: I just included the requirement for generators/cogenerators for completeness - you are right that a one dimensional vector space is both a generator and a cogenerator in k-Mod and that is not a problem. The problem arises in trying to require that both filtered limits and colimits be simultaneously exact. | |
| Jun 25, 2010 at 1:53 | comment | added | Tom Goodwillie | More simply: If a category has products and coproducts, and if the same object is both initial and final, then you get a map from the coproduct of any set of objects to their product. In R-Mod this map is always a monomorphism. Therefore in R-Mod^{op} this map is always an epimorphism. But in S-Mod it is never an epimorphism in the case of infinitely many nontrivial objects | |
| Jun 25, 2010 at 1:37 | comment | added | Yemon Choi | Greg, a naive question here (just betraying my misunderstanding I suspect) - in the category of vector spaces over a field $k$, why isn't the one-dimensional space both a generator and cogenerator? And isn't this category the same as $k$-mod? | |
| Jun 25, 2010 at 0:44 | comment | added | Greg Stevenson | The proof is not so hard I don't think - consider the canonical map from the countable coproduct of copies of $M$ to the countable product in $R$-Mod and its opposite. Then deduce that it is an isomorphism and use an Eilenberg swindle to show $M = 0$. | |
| Jun 25, 2010 at 0:36 | vote | accept | Akhil Mathew | ||
| Jun 25, 2010 at 0:35 | history | answered | Greg Stevenson | CC BY-SA 2.5 |