Timeline for Two functors from Grp to Grp?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 5, 2009 at 20:41 | comment | added | Theo Johnson-Freyd | So just to clarify, your functor is trivial on any morphism that does not involve G_0. | |
| Nov 5, 2009 at 20:02 | comment | added | Steven Gubkin | It could really go either way couldn't it? It would be interesting to see a proof that there was no such functor. | |
| Nov 5, 2009 at 18:19 | comment | added | user717 | This is a funny question and a funny answer and it would be really funny if there would be a T which is not isomorphic to the identity. If somebody finds out, please post it! | |
| Nov 5, 2009 at 18:15 | comment | added | Reid Barton | Ah, I was thinking that the condition "T(G) = G" was evil, but actually it's the conclusion of whether two functors are equal that's the relevant evil in my example (we should instead ask for an isomorphism in the functor category). | |
| Nov 5, 2009 at 17:55 | comment | added | Andrew Critch | Actually, I think it's quite satisfying. It's important to be aware that the generality of functors and natural transformations still allows them to be fairly "unnatural", and I think this is the simplest example to illustrate that (especially since it works for any object with a non-trivial automorphism in any category). | |
| Nov 5, 2009 at 17:53 | comment | added | Steven Gubkin | Excellent! I thought I had considered this solution (or something very much like it) at some point or another, but it hadn't worked out. I guess I was remembering incorrectly. Thanks! | |
| Nov 5, 2009 at 17:51 | vote | accept | Steven Gubkin | ||
| Nov 5, 2009 at 17:45 | history | answered | Reid Barton | CC BY-SA 2.5 |