Timeline for Integral versus real (universal) characteristic classes

7 events

when toggle format	what		by	license	comment
Dec 17, 2013 at 17:10	comment	added	Ricardo Andrade		Dear @Achim Krause: Yes, any free chain complex whose homology has finite type (i.e. is finitely generated in each degree) is chain equivalent to a finite type, free chain complex. See, for example, lemma 5.5.9 of Spanier's book "Algebraic topology". In fact, we can directly apply theorem 5.5.10 of Spanier's book which gives a universal coefficient theorem for cohomology in terms of cohomology (as opposed to the usual universal coefficient theorem where cohomology is expressed in terms of homology), as long as the homology has finite type.
Dec 17, 2013 at 16:47	comment	added	Achim Krause		Oh, you're right. So this argument won't apply to singular cohomology, but for the spaces in question the cochain complex is quasi-isomorphic to a degreewise finitely-generated complex, so everything should work out.
Dec 17, 2013 at 14:42	comment	added	Ricardo Andrade		It is essential to restrict $C$ to also be finitely generated, otherwise the first map $\operatorname{Hom}(C,\mathbb{Z}) \otimes \mathbb{R} \to \operatorname{Hom}(C,\mathbb{R})$ will not be an isomorphism in general.
Dec 17, 2013 at 10:10	review	First posts
Dec 17, 2013 at 10:16
Dec 17, 2013 at 10:04	history	undeleted	Achim Krause
Dec 17, 2013 at 10:04	history	deleted	Achim Krause		via Vote
Dec 17, 2013 at 9:52	history	answered	Achim Krause	CC BY-SA 3.0