Timeline for An abstract nonsense proof of the Hurewicz theorem
Current License: CC BY-SA 4.0
15 events
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| Sep 3, 2018 at 15:03 | history | edited | Jeff Strom | CC BY-SA 4.0 |
equivalence level increased
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| Oct 13, 2017 at 1:06 | comment | added | Dylan Wilson | In this case I used spectrum as a shorthand. What I mean is that we have a sequence of spaces $K_n$ and homotopy equivalences $\Omega K_n \cong K_{n-1}$. The Eilenberg-Steenrod axioms then become the fact that $[-, X]$ takes cofiber sequences to exact sequences, and the suspension axiom is the loops-suspension adjunction. So I think that's all I've used. The only thing I can think of is that maybe it's not obvious that the $K_n$ have no homotopy below dimension $n$? But again, if we're allowed cellular approximation, then this is okay from the explicit cell structure on geometric realization | |
| Oct 13, 2017 at 0:46 | comment | added | Jeff Strom | @DylanWilson I think you'd want to check your constructions carefully before you claim anything involving spectra is done without some kind of appeal to Freudenthal. | |
| Oct 12, 2017 at 21:04 | comment | added | Dylan Wilson | @QiaochuYuan why do you need more to see it represents cohomology? Are we not allowed to use the Eilenberg-Steenrod axioms? I don't think those use the Hurewicz theorem... you have a spectrum, so it satisfies all the axioms except maybe the dimension one, and, as you say, that follows from looping down till you get to $\mathbb{Z}$. I think with this you forego the need for Freudenthal and such in the "even more" part of the answer. | |
| Oct 12, 2017 at 19:37 | history | edited | Jeff Strom | CC BY-SA 3.0 |
deleted 4 characters in body
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| Oct 12, 2017 at 12:51 | history | edited | Jeff Strom | CC BY-SA 3.0 |
clarity
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| Oct 12, 2017 at 12:50 | comment | added | Jeff Strom | More detail on Eilenberg-Mac Lane spaces added. | |
| Oct 11, 2017 at 19:47 | history | edited | Jeff Strom | CC BY-SA 3.0 |
MORE detail
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| Oct 11, 2017 at 18:09 | comment | added | Qiaochu Yuan | @Oscar: I don't know what construction you have in mind that requires the Hurewicz theorem, but all that I think I need to construct $K(\mathbb{Z}, n)$ is a model for the classifying space functor $B$ which takes topological abelian groups to topological abelian groups, which IIRC was written down by Segal, applied $n$ times to $\mathbb{Z}$. Verifying that this space has the correct homotopy groups just requires knowing that $\Omega B G \cong G$. But then I guess verifying that it represents (co)homology requires more. | |
| Oct 11, 2017 at 18:05 | comment | added | Oscar Randal-Williams | Without the Hurewicz theorem how do you know $K(\mathbb{Z},n)$ exists? | |
| Oct 11, 2017 at 18:05 | comment | added | Neil Strickland | What is your definition of $K(\mathbb{Z},n)$? The various usual definitions are not obviously equivalent without the Hurewicz Theorem. Similarly, many of the standard facts about connectivity of spectra and maps implicitly use the Hurewicz Theorem, so it is far from clear that you have a non-circular argument. | |
| Oct 11, 2017 at 17:48 | history | edited | Jeff Strom | CC BY-SA 3.0 |
added detailed argment
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| Oct 11, 2017 at 5:08 | comment | added | David Roberts♦ | How so? Is it because $S^n$ is a Moore space? That the Eilenberg-Mac Lane space represents on cohomology? | |
| Oct 11, 2017 at 3:03 | review | Low quality posts | |||
| Oct 11, 2017 at 3:28 | |||||
| Oct 11, 2017 at 2:47 | history | answered | Jeff Strom | CC BY-SA 3.0 |