Timeline for The definition of a group object is wrong?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| May 31, 2011 at 16:17 | comment | added | Ryan Reich | I meant $F(i) i$. | |
| May 31, 2011 at 16:16 | comment | added | Ryan Reich | I think the "virtual group object" setting is the only appropriate one in which to interpret this question, because it emphasizes that the inverse only exists in $\mathbf{D}$. In that sense, $i^2$ is the identity in $\mathbf{D}$ like for any group object, and thus lifts to $\mathbf{C}$ (as the identity, since $S$ is faithful). If you happen to be working with some kind of inversion $F$, and if $F^2 = I$, then you could also say $s(i)i = \mathrm{id}$, but I wonder if it is possible to have, say, an $F$ of order 3, so that we only have "$i^3 = \mathrm{id}$". An example would be nice here. | |
| May 31, 2011 at 16:03 | comment | added | Qiaochu Yuan | Oh, ha again. I thought I said something different, but that really is the last comment in your answer. Okay, before I say anything else, can someone tell me whether there's any sense in which $i^2$ must be a morphism? | |
| May 31, 2011 at 15:52 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
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| May 31, 2011 at 15:50 | comment | added | Ryan Reich | Ha, comment conflict. What you just edited in is also what I was suggesting. | |
| May 31, 2011 at 15:49 | comment | added | Ryan Reich | Actually, it sounds like you are really motivating my last comment to my answer. So you do think it is meaningful to consider an inversion map which simply has no analogue in $\mathbf{C}$? | |
| May 31, 2011 at 15:46 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
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| May 31, 2011 at 15:31 | history | answered | Qiaochu Yuan | CC BY-SA 3.0 |