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Apr 13, 2017 at 12:58 history edited CommunityBot
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S Dec 17, 2013 at 12:31 history bounty ended CommunityBot
S Dec 17, 2013 at 12:31 history notice removed CommunityBot
Dec 17, 2013 at 4:21 comment added David White The notion of a left adequate set might be helpful. This was introduced by Heller in "On the Representability of Homotopy Functors." Such a set $K$ has the property that if $Hom(A,f)$ is a bijection for all $A\in K$ then $f$ is an isomorphism. So maybe what you're looking for is a localization of the homotopy category in which isomorphisms are abelian homology isomorphisms? The notion of left adequate is also discussed by Raptis in "On the cofibrant generation of model categories." So getting the abelian homology isomorphism homotopy category as a Bousfield localization might work.
Dec 10, 2013 at 8:46 comment added Johannes Ebert @Misha: how do you apply Yoenda's lemma here?
Dec 10, 2013 at 7:42 comment added Misha Verbitsky I think what you say is almost true (and follows from Yoneda's lemma) if you replace "abelian" by "finitely presented". Should not be too hard to prove, in fact. And for abelian there should be an easy counterexample. Sorry for being not very specific.
S Dec 9, 2013 at 11:03 history bounty started Johannes Ebert
S Dec 9, 2013 at 11:03 history notice added Johannes Ebert Authoritative reference needed
Dec 6, 2013 at 17:14 comment added Martin Palmer Incidentally, they also prove (Theorem 1.4) the sufficient condition for being a weak equivalence that Tyler Lawson gave in his answer to your other question.
Dec 6, 2013 at 17:13 comment added Martin Palmer Hi Johannes. I just came across this paper of Casacuberta and Rodríguez in which they mention that the answer to your last question is "no" if you make the weaker assumption that $[K,X]\to [K,Y]$ is a bijection for all spheres $K$. (This would not be a weaker assumption if we were talking about based homotopy classes of course, but for unbased homotopy classes I don't know.) They construct (Example 1.2) an inclusion of discrete groups $N\to G$ such that $BN\to BG$ has this property but is not an integral homology equivalence.
Dec 6, 2013 at 16:32 history edited Johannes Ebert CC BY-SA 3.0
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Dec 6, 2013 at 13:52 history edited Johannes Ebert CC BY-SA 3.0
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Dec 6, 2013 at 13:40 history asked Johannes Ebert CC BY-SA 3.0