Timeline for Splitting of $H\mathbb{Z}$-module spectra
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 11, 2019 at 22:18 | answer | added | John Rognes | timeline score: 7 | |
| Jan 11, 2019 at 19:50 | comment | added | Dylan Wilson | What is your definition of $H\mathbb{Z}$ and the unit map? (It seems hard to have a definition of those two things without also having a proof that the unit map is an isomorphism on $\pi_0$...) | |
| Jan 11, 2019 at 19:39 | history | edited | Arun Debray |
added some tags
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| Jan 11, 2019 at 16:56 | comment | added | Denis Nardin | @user09127 What are you starting from? One quick way to see that the map $\mathbb{S}→H\mathbb{Z}$ is an iso on π_0 is that it is a map of rings and both rings are $\mathbb{Z}$. Or maybe what you're missing is that $π_*(H\mathbb{Z}∧X)\cong H_*X$? | |
| Jan 11, 2019 at 16:30 | comment | added | user09127 | @JohnRognes That's what I don't quite get. I assume that the way to prove this is to show that the unit map $S\rightarrow H\mathbb{Z}$ induces an isomorphism on $\pi_0$ (which has to be true), but how would your prove that? I guess a description of the action of the product of $H\mathbb{Z}$ on homotopy groups would do the trick. But, once again, that's not a precise argument. | |
| Jan 11, 2019 at 16:06 | history | edited | user09127 | CC BY-SA 4.0 |
added 8 characters in body
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| Jan 6, 2019 at 10:48 | comment | added | Denis Nardin | Maybe the missing observation here is that $H\mathbb{Z}\wedge SG\cong HG$ for all abelian groups $G$? | |
| Jan 6, 2019 at 10:47 | comment | added | John Rognes | Because $\tilde \alpha$ is the wedge sum of maps $\Sigma^k H(\pi_k M) \to M$ inducing isomorphism on $\pi_k$. | |
| Jan 6, 2019 at 8:49 | history | asked | user09127 | CC BY-SA 4.0 |