Timeline for Exact triple yields a distinguished triangle in derived category
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| May 16, 2014 at 14:16 | comment | added | David E Speyer | Edited to sketch more of the argument. | |
| May 16, 2014 at 1:27 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| May 15, 2014 at 18:41 | comment | added | David E Speyer | You shouldn't be able to fit that into a commuting diagram of quasi-isomorphisms with the distinguished triangle coming from the cone and cylinder. I am pretty sure I worked out at some point that $M \overset{0}{\to} K[1]$ is is quasi-isomorphic to the triangle coming from the cone/cylinder construction if and only if the underlying sequence of modules is split in $A$, and this one isn't. | |
| May 15, 2014 at 18:21 | comment | added | user50838 | Is $A$ the category of $R$-modules in your example? Isn't it the case that we can take $M \stackrel 0\rightarrow K[1]$? It seems, that corresponding long exact sequence works fine then. | |
| May 15, 2014 at 12:23 | history | answered | David E Speyer | CC BY-SA 3.0 |