Timeline for Any exact faithful functor is represented by a unique projective generator
Current License: CC BY-SA 4.0
12 events
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| May 11, 2020 at 19:22 | history | edited | Phil Tosteson | CC BY-SA 4.0 |
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| May 11, 2020 at 19:07 | comment | added | S.Farr | @PhilTosteson, good, then it makes sense. I think you also need to change the 'left exact' to 'right exact' in the last line. | |
| May 11, 2020 at 19:05 | vote | accept | S.Farr | ||
| May 11, 2020 at 19:03 | comment | added | Phil Tosteson | @S.Farr Sorry, I wrote it backwards-- I am using left exactness in the argument. | |
| May 11, 2020 at 19:03 | history | edited | Phil Tosteson | CC BY-SA 4.0 |
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| May 11, 2020 at 19:01 | comment | added | S.Farr | Why is the Hom-functor tautologically right exact? I thought since Hom is right adjoint to the tensor functor, it usually just is left exact. | |
| May 11, 2020 at 15:28 | comment | added | Phil Tosteson | @JeremyRickard Thanks, I was missing a dual. I edited the post to fix the mistake-- I think it is correct now. | |
| May 11, 2020 at 15:26 | history | undeleted | Phil Tosteson | ||
| May 11, 2020 at 15:26 | history | edited | Phil Tosteson | CC BY-SA 4.0 |
Fixed the mistake pointed out by Jeremy Rickard
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| May 11, 2020 at 13:11 | history | deleted | Phil Tosteson | via Vote | |
| May 11, 2020 at 7:02 | comment | added | Jeremy Rickard | Assuming that by $G(R)^*$ you mean the vector space dual, then $P=G(R)^*$ is not usually projective. Even when $G=\text{Hom}_R(R,-)$ is the forgetful functor, $G(R)^*\cong R^*$ which is not projective unless $R$ is self-injective. | |
| May 10, 2020 at 21:45 | history | answered | Phil Tosteson | CC BY-SA 4.0 |