Timeline for Cocomplete but not complete abelian category
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 29, 2022 at 19:37 | comment | added | Jeremy Rickard | @TimCampion I'm not sure, although I think it probably is. However, if you change the definition of $\alpha$-good to mean that, for every $\beta>\alpha$, $V_\alpha\otimes_{k_\alpha}k_\beta\to V_\beta$ is an isomorphism (rather than just surjective), then I'm sure that gives a moderate category, and more or less the same proof shows that it is cocomplete but not complete. | |
| Jul 29, 2022 at 15:40 | comment | added | Tim Campion | Is $\mathfrak C$ moderate, in the sense of having no more isomorphism classes of objects than the size of the universe? | |
| Oct 20, 2014 at 9:28 | vote | accept | Simone Virili | ||
| Oct 15, 2014 at 16:22 | comment | added | Fernando Muro | This is an example from the book! | |
| Oct 15, 2014 at 15:30 | comment | added | Jeremy Rickard | @ZhenLin I think that's a nice way of thinking about what's happening. The abelian categories in the $\mathbf{On}$-indexed sequence are all complete and cocomplete. The functors between them preserve coproducts, and so the coproduct of any set of objects can be constructed in any of the categories that contains all the objects, and survives in the direct limit. But the functors don't preserve products, and the product of a set of objects gets bigger as we go through the ordinals and gets pushed out of the top of the category (technical term) as we take the direct limit. | |
| Oct 15, 2014 at 15:14 | comment | added | Zhen Lin | This looks like the direct limit of an $\mathbf{On}$-indexed sequence of cocomplete abelian categories along colimit-preserving exact functors, so it should be just abstract nonsense that it is a cocomplete abelian category. | |
| Oct 15, 2014 at 14:54 | comment | added | Jeremy Rickard | @ToddTrimble Yes, exactly. In fact, it would be enough for $k_\beta$ to be an infinite extension of $k_\alpha$ for some $\beta>\alpha$. | |
| Oct 15, 2014 at 14:05 | comment | added | Todd Trimble | I'm not seeing a thing wrong with this, and it's carefully written. Congratulations on this ingenious answer. Just a small note that $k_\beta \otimes_{k_\alpha} -$ preserves arbitrary products iff $k_\beta$ is a finite extension of $k_\alpha$ (being canonically isomorphic to $\hom_{k_\alpha-\text{Vect}}(k_\beta, -)$), in which case the canonical map $k_\beta \otimes_{k_\alpha} W(\alpha) \to W(\beta)$ would be an isomorphism. That's why we needed $k_\beta$ to be an infinite extension of $k_\alpha$. | |
| Oct 15, 2014 at 11:31 | history | answered | Jeremy Rickard | CC BY-SA 3.0 |