Homotopy Theory

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5:15 AM

@TimCampion: It does seem to at least be the case if X is an H-group, via multiplication (g,a) |--> g.a in one direction and

b|--> (b(0), b(0)^{-1} . b)

in the other. Do you know where you saw the more general result?

In my defense, it hadn't been my intention to ask only you, specifically, but I suppose if that's functionally what I did, it would go to support your claim this chat room has not been terribly active of late :)

3 hours later…

Maxime Ramzi

8:42 AM

@jdc by "more general result", do you mean the fiber sequence ? If so that's just the general fiber sequence $map_*(X,Y)\to map(X,Y) \to Y$ for pointed spaces : the fiber of "evaluation at the basepoint" over the basepoint is exactly the space of pointed maps

(apply this to $X=S^1$ and $Y=X$)

8 hours later…

Bryan Shih

4:21 PM

I'm trying to understand why $K_0(R)->K_0(R/I)$ is injective given a henselian pair (R,I). I want to understand the proof given in Lemma 4.20: https://arxiv.org/pdf/1803.10897.pdf

Stated: 1. Let M,N be finite projective modules over R. Any iso M/IM-> N/IN lifts to a map M->N. I agree.
2. this map is necessary iso by looking at kernel and Nakayama's llemma.

Ok: So the kernel should lie in IM. but its not clear to me if this IM is fg. - or even what type of Nakayama's lemma is applied here. (Should we prove Jac(R)IM=IM? This is not clear to me either.)

Dustin Clausen

5:06 PM

Let K be the kernel and C the cokernel. Then C is finitely generated and vanishes mod I. As I is contained in the Jacobson radical, by Nakayama's lemma this implies C=0, so the map is surjective. As N is projective this implies that the kernel K splits off as a summand. Thus we deduce both that K also vanishes mod I and that K is also finitely generated. Thus by Nakayama again we get K=0, so the map is injective as well, hence an iso.

1 hour later…

jdc

6:18 PM

@MaximeRamzi I was OK with that part, but thank you! What I was wondering was how you get the splitting of the free loop space with only the grouplike hypothesis, which I thought was less demanding than being an H-group. I couldn't see how the proof generalizes

Maxime Ramzi

6:39 PM

No no, grouplike is an assumption on an H-space, namely that the induced multiplication on $\pi_0$ be a group multiplication (probably what you called "H-group")

4 hours later…