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S Apr 21 at 12:22 history suggested Ken CC BY-SA 4.0
fixed some typos
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Dec 17, 2019 at 15:38 history edited მამუკა ჯიბლაძე CC BY-SA 4.0
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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jan 23, 2015 at 16:43 comment added archipelago The corrections to the McDuff-Segal paper Peter May mentioned can be found as Lemma 3.1 in D.McDuff - "The homology of some groups of diffeomorphisms.".
Mar 21, 2013 at 22:15 history edited Johannes Ebert CC BY-SA 3.0
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Nov 21, 2012 at 10:06 comment added Oscar Randal-Williams I have written up what I know about this problem, and it is available at dpmms.cam.ac.uk/~or257/GCrem.pdf.
Oct 26, 2012 at 21:04 vote accept Johannes Ebert
Oct 16, 2012 at 16:57 answer added Oscar Randal-Williams timeline score: 18
Oct 15, 2012 at 15:18 comment added Johannes Ebert "In that generality, M∞ as you define it using just one element can be misleading." I agree, and taking another element could a priori lead to a different answer to the question.
Oct 14, 2012 at 23:06 comment added Peter May Sorry, I was just explaining the term ``group completion'' for those readers who might not know. Your question is all about isomorphisms, so perhaps doesn't make that clear. From the point of view of group completion as I defined it, introducing $M_{\infty}$ is irrelevant: if $M$ is a topogical monoid, $\pi_0(M)$ is homotopically central in $M$, and $\pi_0(\Omega BM)$ is homotopically central in $\Omega BM$, then the natural map $M\to \Omega BM$ is a group completion. In that generality, $M_{\infty}$ as you define it using just one element can be misleading. I doubt this is helpful to you.
Oct 14, 2012 at 16:59 comment added Johannes Ebert @Peter May: Am I overlooking something here, or is it just a reformulation of the problem? I guess the important case is where $R$ is the group ring of the fundamental group of $Y$. But verifying the hypothesis that $\pi_0$ is central in this ring needs (say for $Out (F_{\infty})$) homological stability with abelian coefficients, just in the same way the argument works for constant coefficients. With the ''standard notion of group completion'', the question becomes: how do I see that the above maps are group completions?
Oct 14, 2012 at 16:15 comment added Peter May I don't have time for a long answer, but the standard notion of group completion in infinite loop space theory is an H-map X >--> Y, where \pi_0(Y) is a group and (to avoid an unconvincing morasse in the literature) X and Y are homotopy associative and commutative such that \pi_0(X) >--> \pi_0(Y) is a group completion in the obvious sense and for every commutative ring (not just Z) the map H_*(X;R) >--> H_*(Y;R) is a localization of graded rings obtained by inverting the elements of \pi_0(X) [these elements being an R-basis for H_0(X;R)]. McDuff somewhere published corrections to M-Segal.
Oct 14, 2012 at 16:10 history edited Johannes Ebert CC BY-SA 3.0
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Oct 14, 2012 at 12:21 history edited Dmitri Pavlov CC BY-SA 3.0
fixed a formula that did not display properly
Oct 14, 2012 at 11:51 history asked Johannes Ebert CC BY-SA 3.0