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Let us assume that $f:X \to Y$ is a map of connected CW complexes, having the following property: if $K$ is a finite CW complex, then the induced map $f_{\ast}:[K,X] \to [K,Y]$ on \emph{free} homotopy classes of maps is a bijection.

Question: what do I know about the map $f$?

Remarks: it is clear that if $Y$ is simply connected, then so is $X$ and thus $f$ is a weak homotopy equivalence. Tyler Lawson's answer to my old question Counterexamples in algebraic topology? proves some other cases that guarantee that $f$ is a weak equivalence. Namely, if $\pi_1 (f)$ is surjective, then $f$ has to be a weak equivalence, and surjectivity is guaranteed by $\pi_1 (Y)$ abelian or finitely generated.

I am interested in an application where the fundamental group $\pi_1 (Y)$ is nonabelian and not finitely generated, and $\pi_1 (f)$ is not surjective.

A concretely counterexample that shows that $f$ does not have to be a weak equivalence is given by the stabilization map $B \Sigma_{\infty}\to B \Sigma_{1+\infty}$.

What I hope for is the answer: $f$ will be an ''abelian homology equivalence''. This means that if $L$ is a system of local coefficients on $Y$, thought of as an $\mathbb{Z}\pi_1 (Y)$-module, such that $\pi_1 (Y)$ acts through an abelian group, then it follows that $f_{\ast}:H_{\ast}(X,f^{*}L) \to H_{\ast}(Y,L)$ is an isomorphism.

Is this true??

Improve this question
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    $\begingroup$ 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. $\endgroup$
    – Martin Palmer
    Commented Dec 6, 2013 at 17:13
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    $\begingroup$ 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. $\endgroup$
    – Martin Palmer
    Commented Dec 6, 2013 at 17:14
  • $\begingroup$ 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. $\endgroup$
    – Misha Verbitsky
    Commented Dec 10, 2013 at 7:42
  • $\begingroup$ @Misha: how do you apply Yoenda's lemma here? $\endgroup$
    – Johannes Ebert
    Commented Dec 10, 2013 at 8:46
  • $\begingroup$ 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. $\endgroup$
    – David White
    Commented Dec 17, 2013 at 4:21

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