Timeline for An account of "Homologie nicht-additiver Funktoren. Anwendungen"'s results
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| Oct 20, 2022 at 7:42 | comment | added | Grisha Taroyan | @NicholasKuhn No, I think here symmetric powers are considered in another sense. In particular, $K(\mathbb{Z},1)$ is a simplicial abelian group and we are taking symmetric powers of this object. Although, I am not completely sure how this works. I was also somewhat confused by this fact since the topological symmetric powers of a circle are indeed just a circle. | |
| Oct 20, 2022 at 0:02 | comment | added | Nicholas Kuhn | Maybe I am misunderstanding your question about K(Z,1), but the symmetric products of the circle are all homotopy equivalent to a circle. Indeed there is an exercise in tom Dieck's book Transformation Groups that asks you to show that the nth symmetric product of the circle is a fibration over the circle with fiber an (n-1) simplex. | |
| Oct 19, 2022 at 13:05 | review | Close votes | |||
| Oct 21, 2022 at 4:07 | |||||
| Oct 19, 2022 at 12:45 | history | edited | Grisha Taroyan |
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| Oct 19, 2022 at 12:36 | history | asked | Grisha Taroyan | CC BY-SA 4.0 |