Timeline for Is the category $\operatorname{sVect}$ an "algebraic closure" of $\operatorname{Vect}$?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 24, 2020 at 15:59 | vote | accept | Tim Campion | ||
| Apr 23, 2020 at 21:45 | history | edited | Theo Johnson-Freyd | CC BY-SA 4.0 |
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| Apr 23, 2020 at 21:36 | comment | added | Theo Johnson-Freyd | @TimCampion I wrote my answer without really reading your question. I apologize deeply for that. I'll add some comments in the question above. | |
| Apr 23, 2020 at 2:50 | comment | added | Tim Campion | @TheoJohnson-Freyd This is great, thanks! I don't suppose you know anything about the relative version of being weakly terminal? That is, the property of every map $A \to sVect$ extending along various $A \to B$'s to yield $B \to sVect$'s? | |
| Apr 22, 2020 at 20:12 | comment | added | Theo Johnson-Freyd | @Todd I can’t argue with that | |
| Apr 22, 2020 at 20:11 | comment | added | Theo Johnson-Freyd | @LSpice I meant “emit”: I want to quantify over all (nonzero) symmetric monoidal functors with specified domain. | |
| Apr 22, 2020 at 19:38 | comment | added | Todd Trimble | Side comment: even in ordinary algebra, I think a lot of people would consider "the" algebraic closure to be something of an abuse of language, because there is no preferred way of comparing two such. Too many automorphisms. | |
| Apr 22, 2020 at 17:34 | comment | added | LSpice | Do you mean 'admit' or 'emit' in "which does not emit a symmetric monoidal functor"? Both makes sense to me, but I'd expect the former. | |
| Apr 22, 2020 at 17:33 | history | edited | LSpice | CC BY-SA 4.0 |
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| Apr 22, 2020 at 17:22 | history | edited | Theo Johnson-Freyd | CC BY-SA 4.0 |
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| Apr 22, 2020 at 17:15 | comment | added | Theo Johnson-Freyd | Of course, you should add appropriate words to my answer like linear, cocontinuous, or what have you, to specify your ambient "categorified linear algebra". The details of how you add those words don't matter too much, because you expect (co)completion operations that move you between worlds. When defining "separable extension", you should demand very strong finiteness (i.e. dualizability) conditions. With those conditions in place, the different worlds for categorified linear algebra all match; see the appendix A bestiary of 2-vector spaces. | |
| Apr 22, 2020 at 17:10 | history | answered | Theo Johnson-Freyd | CC BY-SA 4.0 |