Timeline for Cup product with arbitrary coefficients
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2012 at 20:32 | history | edited | Ralph | CC BY-SA 3.0 |
cohomology is dual space of homology
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| Apr 13, 2012 at 1:47 | comment | added | Ralph | I've restricted (*) to pid's since even a formulation for hereditary rings doesn't apply to Hatcher's 3.9. I think it will be easier for the OP this way. | |
| Apr 13, 2012 at 1:38 | history | edited | Ralph | CC BY-SA 3.0 |
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| Apr 13, 2012 at 1:22 | comment | added | Mariano Suárez-Álvarez | :) Also, a semi-hereditary ring has $Tor$ dimension at most $1$, but over $\mathbb Z/P^r$ with $r>1$ we have $\mathrm{Tor}_n(\mathbb Z/p,\mathbb Z/p)\neq0$ for all $n$. | |
| Apr 13, 2012 at 1:16 | history | edited | Ralph | CC BY-SA 3.0 |
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| Apr 13, 2012 at 1:11 | comment | added | Ralph | @Mariano: You beat me by a few seconds. | |
| Apr 13, 2012 at 1:07 | history | edited | Ralph | CC BY-SA 3.0 |
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| Apr 13, 2012 at 1:05 | comment | added | Mariano Suárez-Álvarez | $\mathbb Z/m$ is hereditary iff $m$ is square free (and then it is semisimple) | |
| Apr 13, 2012 at 0:57 | comment | added | Ralph | @David: The OP defines a map $\tilde{h}$ and asks in b), c) about the image and kernel of $\tilde{h}$. And that are the questions I answer. Where's the problem ? | |
| Apr 13, 2012 at 0:39 | comment | added | David White | I thought he wanted to take the construction of $\tilde{h}$, which he understands, and try to make it work to get a map like $h$, i.e. a map $H^k(X;G)\rightarrow Hom(H_k(X),G)$. That's how I read the ``with $G$ in place of $R$'' part of his question. If I'm right, then this answer is probably not what he's looking for, though I still think it's cool. | |
| Apr 13, 2012 at 0:24 | history | answered | Ralph | CC BY-SA 3.0 |