Timeline for Example of an additive functor admitting no right derived functor
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 21, 2017 at 19:37 | vote | accept | Pierre-Yves Gaillard | ||
| Mar 21, 2017 at 19:24 | history | edited | Yonatan Harpaz | CC BY-SA 3.0 |
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| Mar 21, 2017 at 19:23 | comment | added | Yonatan Harpaz | Actually the formula in my previous comment is inaccurate. There are only ${\bf N}$ worth of copies of ${\bf Z}/2$ in the end, not ${\bf Z}$ worth. Also I may have confused the positive and negative grading conventions. | |
| Mar 21, 2017 at 19:16 | comment | added | Yonatan Harpaz | @Pierre-YvesGaillard, there are several approached to derived functors. I admit I am less familiar with the definition via universal colimits, but on chain complexes of (not necessarily finite dimensional) vector spaces there are also various model structures which one can use to compute derived functors. The definition with universal colimits seems a bit hard to verify in general (I suppose in the book they give some useful criteria). | |
| Mar 21, 2017 at 19:11 | comment | added | Yonatan Harpaz | @Pierre-YvesGaillard, Since we can identify $F(-)$ with ${\rm Hom}_{{\cal C}}({\bf Z}/2,-)$ and since $X$ can be identified with the product of ${\bf Z}/2[n]$ we get that $R_0 F(X) \cong \prod_{n \in {\bf Z}} R_0{\rm Hom}({\bf Z/2},{\bf Z/2}[n]) \cong \prod_{n \in {\bf Z}} R_0{\rm Ext^{-n}}({\bf Z/2},{\bf Z/2}) \cong \prod_{n \in {\bf Z}} {\bf Z}/2$. | |
| Mar 21, 2017 at 19:09 | comment | added | Pierre-Yves Gaillard | [When you say "$RF(X)$ will be the derived functor there" you don't take into account, it seems to me, the condition on the arbitrary functor $G$ in the definition of $RF(X)$ given in the question.] | |
| Mar 21, 2017 at 18:53 | comment | added | Pierre-Yves Gaillard | It seems to me we can argue as follows. Assume by contradiction that $RF$ exists. The proof of Theorem 13.4.1 p. 337 in Kashiwara and Schapira's book shows $$R_0F(X)\cong\prod{\rm Hom}_{{\cal D}({\cal C})}({\bf Z}/2[0],{\bf Z}/2[n]),$$ which is infinite dimensional over ${\bf Z}/2$, a contradiction. What do you think? | |
| Mar 21, 2017 at 18:38 | history | edited | Yonatan Harpaz | CC BY-SA 3.0 |
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| Mar 20, 2017 at 23:24 | history | answered | Yonatan Harpaz | CC BY-SA 3.0 |