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