Timeline for If a $\otimes$-idempotent object has a dual, must it be self-dual?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 9, 2017 at 16:14 | vote | accept | Tim Campion | ||
| Aug 8, 2017 at 21:18 | history | edited | MTyson | CC BY-SA 3.0 |
Removed incorrect answer
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| Aug 8, 2017 at 21:17 | comment | added | MTyson | @ToddTrimble, that makes sense. I was wondering why no one else pointed out that example. | |
| Aug 8, 2017 at 21:11 | comment | added | Tim Campion | Good point..... | |
| Aug 8, 2017 at 21:08 | comment | added | Todd Trimble | The countable infinite direct sum of copies of $k$ is not dualizable. Sure, it has a dual, but that's not the same thing. In fact for modules over a commutative ring, dualizable means the same as finitely generated projective. | |
| Aug 8, 2017 at 21:07 | vote | accept | Tim Campion | ||
| Aug 8, 2017 at 21:10 | |||||
| Aug 8, 2017 at 21:07 | comment | added | Tim Campion | Wow, thanks, that's great! And such a simple counterexample to (1), I surely should have thought of that! | |
| Aug 8, 2017 at 21:03 | history | undeleted | MTyson | ||
| Aug 8, 2017 at 21:03 | history | edited | MTyson | CC BY-SA 3.0 |
Fixed gap in proof, answered question 1
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| Aug 8, 2017 at 1:15 | history | deleted | MTyson | via Vote | |
| Aug 7, 2017 at 23:59 | comment | added | Anton Fetisov | I don't understand why do you drop out the disjoint loops. Looks like you proved that those 2 morphisms are inverse up to multiplication by $\mathrm{dim}\, X$, which is not invertible in general. | |
| Aug 7, 2017 at 23:11 | review | First posts | |||
| Aug 7, 2017 at 23:19 | |||||
| Aug 7, 2017 at 23:09 | history | answered | MTyson | CC BY-SA 3.0 |