Timeline for Are chain complexes over a field always injective?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 9, 2016 at 22:11 | answer | added | Qiaochu Yuan | timeline score: 2 | |
| Feb 24, 2015 at 20:50 | history | edited | Paul | CC BY-SA 3.0 |
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| Feb 24, 2015 at 20:48 | vote | accept | Paul | ||
| Feb 24, 2015 at 20:42 | answer | added | Gabriel C. Drummond-Cole | timeline score: 11 | |
| Feb 24, 2015 at 20:14 | comment | added | Paul | @darji I see now, you're correct; this part of the proof looks wrong, I'll have to think about it some more. I'm less interested in the correctness of the proof than in the answer to the question. Is there an object in Ch_{F} which is not injective? | |
| Feb 24, 2015 at 19:55 | comment | added | darij grinberg | What do you mean by "such maps should compose to give an exact sequence in each degree"? From $f_\cdot$ you don't get an exact sequence. How do you obtain $0\to \mathrm{Hom}_{\mathrm{Ch}_{\mathbb{F}}}\big(C_{m},K_{n}\big) \to \mathrm{Hom}_{\mathrm{Ch}_{\mathbb{F}}}\big(B_{m},K_{n}\big) \to \mathrm{Hom}_{\mathrm{Ch}_{\mathbb{F}}}\big(A_{m},K_{n}\big) \to \mathrm{D}_{m,n}$ ? | |
| Feb 24, 2015 at 19:53 | comment | added | Paul | @darji As you say a chain complex morphism is a collection of F-module maps which commute with the boundary operator. The fact that they commute with the boundary operator isn't really relevant at this point, all that matter is that they are linear maps. So if I have a chain map f_{.}:A_{.}->B_{.}, then I have linear maps f_{n}:A_{n}->B_{n}. By definition of image and kernel in the above proof, such maps should compose to give an exact sequence in each degree. Perhaps I'm misunderstanding your concern? | |
| Feb 24, 2015 at 19:38 | comment | added | darij grinberg | "the above sequence gives rise to sequences of $\mathbb F$-modules in each degree:" -- how so? A homomorphism of chain complexes is not just a collection of homomorphisms between their components; it has to send every component to the respective component of the image, and commute with the boundary operator. | |
| Feb 24, 2015 at 19:21 | history | asked | Paul | CC BY-SA 3.0 |