Timeline for A covariant functor on a given abelian category and comparison of homology in target and source
Current License: CC BY-SA 4.0
13 events
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| Nov 11, 2020 at 7:03 | comment | added | Ali Taghavi | @WilleLiou thanks for your comment. | |
| Nov 11, 2020 at 0:39 | comment | added | Wille Liu | The point is that for $\mathrm{Hom}(G, C_*(X))\cong \mathrm{RHom}(G, C_*(X))$ to hold, you need to require that $\mathrm{Ext}^{> 0}(G, C_k) = 0$ for all $k$. Otherwise, the most we can say is that there is a spectral sequence connecting $\mathrm{Hom}(G, C_*(X))$ and $\mathrm{RHom}(G, C_*(X))$. | |
| Nov 11, 2020 at 0:33 | comment | added | Wille Liu | @AliTaghavi $\mathrm{R} \mathrm{Hom}(-, -)$ is the derived functor of $\mathrm{Hom}$. See Ch 3 of Methods of Homological Algebras by Gelfand and Manin for a nice textbook on homological algebra. | |
| Nov 10, 2020 at 23:15 | history | edited | Ali Taghavi |
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| Nov 10, 2020 at 23:14 | comment | added | Ali Taghavi | @DenisNardin By the first part of your comment do you mean"it does not distinguish quasi isomorphic complexs? What is the reason? | |
| Nov 10, 2020 at 23:11 | comment | added | Ali Taghavi | @DenisNardin may you ellaborate your comment?what is $\mathbb{R}Hom(G,C_*(X)$? | |
| Nov 10, 2020 at 10:24 | comment | added | Denis Nardin | This is not good because it doesn't identify quasi-isomorphic complexes. Maybe you want something like $\mathbb{R}\operatorname{Hom}(G,C_*(X))$? It can be computed by taking a dg-injective replacement for $C_*(X)$ so it's less explicit. | |
| Nov 10, 2020 at 8:11 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 60 characters in body
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| Nov 10, 2020 at 7:48 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 33 characters in body
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| Nov 9, 2020 at 21:49 | comment | added | M.G. | In fact, the complex is also boring if $G$ is the additive group of any field regardless if its char. | |
| Nov 9, 2020 at 21:41 | history | edited | M.G. | CC BY-SA 4.0 |
corrected spelling
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| Nov 9, 2020 at 21:31 | comment | added | M.G. | If $C$ is the chain complex of a finite CW-complex $X$, then the $C_n$-s are simply finite direct sums of copies of $\mathbb{Z}$. So, this boils down to $\mathrm{Hom}(G,\mathbb{Z})$. If $G$ is a finitely generated abelian group, then your complex discards any torsion of $G$, and you get a complex of finite direct sums of $\mathbb{Z}$. In particular, your complex does not care about coefficients like $\mathbb{Z} / p \mathbb{Z)$ :-) | |
| Nov 9, 2020 at 20:59 | history | asked | Ali Taghavi | CC BY-SA 4.0 |