Timeline for Are all vector-space valued functors on sets free?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 14, 2014 at 20:26 | comment | added | Andreas Blass | @ChrisHeunen The linear relation would have to be essentially the one in the answer, in order for $J$ to be a functor. You need $J(f)$ to make sense, in particular for $f$ a permutation. | |
| Jul 14, 2014 at 20:06 | vote | accept | Chris Heunen | ||
| Jul 14, 2014 at 20:05 | comment | added | Chris Heunen | Very nice, thanks! Would imposing any linear relation on $G$ do the trick? | |
| Jul 14, 2014 at 16:52 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
Changed A to J: A(n) wasn't the best notation when I was using S(n) for the symmetric group!
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| Jul 14, 2014 at 16:44 | history | undeleted | Jeremy Rickard | ||
| Jul 14, 2014 at 16:43 | history | deleted | Jeremy Rickard | via Vote | |
| Jul 14, 2014 at 16:33 | history | answered | Jeremy Rickard | CC BY-SA 3.0 |