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6:02 AM
@Dedalus It doesn't, because as a pro system, {R/(a_1^m,..., a_n^m)} is independent of whether you take the derived or underived quotient. (If the sequence is regular, then the derived quotient is already underived even without passing to the pro systems.)
@AaronMazel-Gee It's brave of you to share this. I had similar feelings when I used to be more in the homotopy theory community. My theory is that the very small size of the community leads to this heightened sense of competition and hence this "zero sum" attitude. Positive sum thinking is perhaps directly in conflict with this, and I think people feel little to no incentive to practice it, which is also probably why your comments were largely ignored by this chat room unfortunately.
 
 
5 hours later…
10:39 AM
@AaronMazel-Gee I am very hesitant to respond, as I don't feel as qualified to try and shape the homotopy theory community, being far from a professional homotopy theorist. But ill try anyway. On one hand, the idea of people telling their cool thoughts and results for free and me being able to just listen sounds great, and im for it completely, if just for this very selfish reason of being able to learn something new from you.
On the other hand, im slightly concerned about the side effects of deleting this norm. Some people come here as total beginners with not yet accomplishments in the field, and I am concerned about the pressure it might put on people to present results. There's also something cool about a place dedicated to people asking questions without fearing about their quality too much and people trying to answer the best they can.
 
10:53 AM
But this is just a slight concern, not an actual objection, of course. As I said, it would be cool if people actively tell cool stuff they do without any shame!
 
 
5 hours later…
3:33 PM
@RizaHawkeye oof when R is noetherian and commutative (Michel André is the reference I think) - just because I thought of getarounds of this limitation for an abandoned, hopeless project I once pursued
@AaronMazel-Gee @RizaHawkeye @S.carmeli And just wanted to say that though I don't frequent this forum often, I appreciate the thoughtfulness of this community :)
 
I also really appreciate how kind and thoughtful people seem to be here. Kudos to the regulars
 
3:54 PM
Anyone knows of a "modern" reference to the proof of the Segal conjecture (Carlsson's theorem I guess)? I tried to read the original paper and it is very nice but it would still be fun if there's a more recent reference explaining it.
 
4:11 PM
@S.carmeli The proof of Dylan's and Jeremy for the Lin version is awesome: arxiv.org/pdf/1911.05687.pdf
if I can also self-advert there's an interpretation in terms of real motives in arxiv.org/pdf/1912.11557.pdf - in particular Question 6.9
 
4:25 PM
thanks Elden :)
 
4:39 PM
@EldenElmanto thanks yah I know this one. I hoped for a reference also for the reduction from general p group to elementary abelian. The strategy in this paper can be helpful also for this task?
 
As far as I know - not really, but sounds like a question for @DylanWilson
 
5:44 PM
@S.carmeli nah, the methods in our paper are designed for the 'computational input' part, as opposed to the 'reduction' steps
it would be nice to modernize the discussion of those reduction steps
 
6:27 PM
Is there a nice proof out there that H_1 is the abelianization of pi_1? (Hatcher's needs more geometric intuition than I can supply on short notice...)
Teaching sure has a way of making you realize you don't actually understand things you've taken for granted forever...
4
 
6:54 PM
(On second thought Hatcher's proof here is fine, I suppose any proof of this with singular chains will be kind of messy. I still stand by my second comment, though.)
 
Yeah that proof is harder than you'd think. When I taught it this spring, I handled it by putting a lot of it into a lemma, which states that for a connected based X, you can compute homology using a subcomplex $C'\subset C$ of the singular chains, spanned by simplices whose vertices all go to the basepoint.
 
@RuneHaugseng I don't know whether this argument can be presented in a non-circular manner that is actually useful in something like a first course but I tend to think of it as follows: First reduce to the case of connected 2 dimensional CW complexes using cellular approximation. Any such complex is a cofiber of a map between wedges of circles. Using Van Kampen and that abelinization is a left adjoint reduce to the case of a circle. For the circle you compute.
The reason I'm suspicios of it being circular is that maybe the statement about connected 2-dimensional CW complexes requires some form of hurewitz which is proved using that statement about abelinezation. Also in general I would say it is very easy to give circular proofs of basic algebraic topology facts when you're not careful.
 
 
2 hours later…
9:04 PM
So just out of curiosity: If we impose no technological restrictions (such as those given when teaching a first class on algebraic topology) is there some simple/elegant way of proving Hurewicz? It would seem to be a statement about the lowest non-zero homotopy group of any spectrum (over the sphere).
 
9:31 PM
Seeing how textbook proofs rely on the fact that $H_n$ and $\pi_n$ agree on $S^n$, it should probably follow somehow from the fact that $\pi_0$ of the sphere spectrum is $\mathbb{Z}$.
 
@AdrianClough I once wondered if you manage build the infinity category of spectra as in Higher Algebra then it might be possible to show that the heart is the symmetric monoidal category of abelian groups by just showing that it is the unit of the category of presentable symmetric monoidal additive 1-categories.
Then the unit is Z which gives pi_0(S)=Z.
But you'd still have to deal with freudenthal to relate this to spaces...
 
I'm fine just assuming that that I know about $\Sigma^\infty \dashv \Omega^\infty$, and that all (co)homological information is contained in $\Sigma^\infty X$. Oh, but I would still need that $\pi_n X = \pi_n \Sigma^\infty X$ for $n$-connected spaces.
 
And some proofs of freudenthal actually use hurewitz so it's a little bit tricky
 
Does every map $K(\mathbb{Z}, n) \to K(\mathbb{Z}_m, n + k)$ factor through $K(\mathbb{Z}_m, n)$?
 
Interesting, I didn't know that. (I mostly learnt from Concise)
 
9:49 PM
@DylanWilson thanks.
 

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