Timeline for Standard name of "atomic morphisms"?
Current License: CC BY-SA 2.5
19 events
| when toggle format | what | by | license | comment | |
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| Jan 13, 2010 at 11:31 | vote | accept | Hans-Peter Stricker | ||
| Jan 4, 2010 at 22:17 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
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| Jan 4, 2010 at 16:17 | answer | added | David E Speyer | timeline score: 4 | |
| Jan 4, 2010 at 16:01 | answer | added | Mike Shulman | timeline score: 8 | |
| Jan 4, 2010 at 15:19 | answer | added | Reid Barton | timeline score: 7 | |
| Jan 4, 2010 at 12:49 | comment | added | Qiaochu Yuan | If the category contains an object C isomorphic, but not identical, to either A or B then again no morphisms satisfy the conditions you give. I think that you should at least adjust for that. But I guess you are free to choose your own definitions. | |
| Jan 4, 2010 at 11:33 | comment | added | Hans-Peter Stricker | @José: I want to use the notion of "prime morphism" (sounds good!) in a follow-up question concerning "reconstruction" of the inner structure of an object from its "position" in a category. | |
| Jan 4, 2010 at 11:29 | comment | added | Hans-Peter Stricker | Ok, now I understand what you mean. But that's a statement: if f:A->B is "atomic" (in the above sense), then neither A nor B have nontrivial automorphisms. Why should I rephrase the definition? | |
| Jan 4, 2010 at 11:24 | comment | added | José Figueroa-O'Farrill | I am curious now: could you perhaps elaborate on why you find them interesting? | |
| Jan 4, 2010 at 11:24 | comment | added | José Figueroa-O'Farrill | I think that the definition makes sense. A better name would be "prime", since the defining property is that they should not admit a nontrivial factorisation. Alas, googling "prime morphism" yields an even smaller number of hits. As Qiaochu says they are bound to be rare. | |
| Jan 4, 2010 at 11:15 | comment | added | Qiaochu Yuan | I still don't think you're phrasing this definition correctly. If either A or B has nontrivial automorphisms then no morphisms satisfy the conditions you give. | |
| Jan 4, 2010 at 10:45 | comment | added | Hans-Peter Stricker | @Qiaochu: I tried to be more explicit, but did not really change anything. | |
| Jan 4, 2010 at 10:42 | history | edited | Hans-Peter Stricker | CC BY-SA 2.5 |
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| Jan 4, 2010 at 10:28 | comment | added | Kim Morrison |
@Qiaochu, why don't those statements make sense? With your names for sources and targets, clearly g=f additionally means that A=C, etc.
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| Jan 4, 2010 at 10:21 | comment | added | Qiaochu Yuan | If you don't specify that f is an endomorphism, none of the statements g = f, h = id, g = id, h = f make sense. | |
| Jan 4, 2010 at 10:04 | comment | added | Hans-Peter Stricker | I want to say that f cannot be non-trivially decomposed. What is wrong with my formulation? | |
| Jan 4, 2010 at 9:45 | comment | added | Qiaochu Yuan | Because that doesn't make sense with what you've written, and neither does C = A or C = B. So you must mean A = B = C, and you should say that. | |
| Jan 4, 2010 at 9:44 | comment | added | Qiaochu Yuan | When you say f = gh, do you mean f : A \to B, h : A \to C, g : C \to B for an arbitrary choice of C? | |
| Jan 4, 2010 at 9:40 | history | asked | Hans-Peter Stricker | CC BY-SA 2.5 |