Three quick responses to Stasheff question............DMD
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Subject: To Jim
From: Peter May
Date: Thu, 16 Nov 2006 09:54:49 -0600
Let c_g: G >--> G be conjugation by g, c_g(h) = g^{-1}hg.
Let f: EG >--> EG be right multiplication by g, f(x) = xg.
Then f(xh) = f(x)c_g(h), which means that f induces Bc_g
on passage to classifying spaces. The induced map on BG
is obtained by passage to orbits and is thus the identity.
Peter
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Subject: Re: two postings
From: Rainer Vogt
Date: Thu, 16 Nov 2006 17:10:50 +0100
>
> Subject: query
> From: jim stasheff
> Date: Sat, 11 Nov 2006 14:19:29 -0500
>
> Consider the classifying space functor B:TopGrp --> Top
>
> I sem to recall that B maps conjugate maps to homotopic maps
>
> But my aging neurons come up with a proof only for path connected groups
- the path from the conjugating element to the
> identity goes over to the homotopy
>
> Can anyone do better?
> thanks
>
> jim
>
Dear Jim,
let f:G \to G be conjugation by x. Then B(f) is homotopic to the identity:
Consider G as a topologically enriched category with one
object. Then f is a functor and multiplication with x defines a
natural transformation of f to the identity. The natural transformation
becomes a homotopy after application of B.
Regards
Rainer
Prof. Dr. Rainer Vogt
Studiendekan
Direktor des Instituts fuer Mathematik
Fachbereich Mathematik/Informatik
Albrechtstrasse 28
Osnabrueck
49076
Germany
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Subject: Response to Stasheff
From: Kari Ragnarsson
Date: Thu, 16 Nov 2006 10:27:01 -0600 (CST)
Here's a suggestion:
For groups G and H, Consider the fibre sequence
Map_*(BG,BH)-> Map(BG,BH) -> BH.
At the end of the LES in homotopy you get
\pi_1(BH) -> \pi_0(Map_*(BG,BH)) -> \pi_0(Map(BG,BH)) -> \pi_0(BH) = *
This should be read as "\pi_1(BH) = \pi_0(H) acts on \pi_0(Map_*(BG,BH))
with quotient \pi_0(Map(BG,BH))." This action of \pi_0(H) corresponds to
conjugation. Now add your proof for actions within the connected component
of the identity, and you should have all the ingredients.
Kari