Subject: Re: question about topological groups and naturality
From: "Ronald Brown"
Date: Tue, 14 Feb 2006 18:41:40 -0000
Hi Johannes
An old paper of mine
``Cohomology with chains as coefficients'', {\em Proc. London Math. Soc.}
(3) 14 (1964), 545-565.
discussed various related naturality questions and gave counterexamples.
Here is a way of seeing what is going on.
If C is a chain complex over the integers, there is a free chain complex F
and morphism f: F \to C inducing an iso in homology, and then there is a
morphism g: F \to H(F) inducing an iso in homology. There are lots of
choices involved, so naturality is unlikely. Now move from chain complexes
to simplicial abelian groups (Dold-Kan) and then to spaces.
This paper was part of a project to calculate k-invariants of function
spaces X^Y, of which the first step was to *calculate* k^Y when k is a
cohomology operation, using Moore's theorem. The techniques of the paper
help to solve this and do calculations.
Hope that helps.
Ronnie Brown
www.bangor.ac.uk/r.brown
www.popmath.org.uk
>>
>> Subject: Question about topological groups
>> From: Johannes Ebert
>> Date: Mon, 13 Feb 2006 19:41:06 +0100 (CET)
>>
>> Hello,
>>
>> I have an innocent question on topological groups. A theorem of Morre
>> states that if G is a commutative topological group, then it is
homotopy
>> equivalent to a product of Eilenberg-Maclane spaces, namely
>> G \simeq \prod_{i=0}^{\infty} K(\pi_i G,i).
>> Question: Is it true that this splitting is functorial? More precisely:
If
>> f:G \to H is a group homomorphism (H also abelian), then f is homotopic
to
>> \prod_{i=0}^{\ifty} \pi_{i}(f) under the decomposition above?
>>
>> For my purpose, it would suffice to assume that f is an automorphism of
G.
>>
>> Best regards,
>> Johannes Ebert