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Let $f:X\to Y$ be a pointed map of pointed connected $n$-dimensional CW complexes. Whitehead's theorem says that if $f_*:\pi_qX\to \pi_qY$ is an isomorphism for $q\le n$ and a surjection for $q=n+1$, then $f$ is a homotopy equivalence (e.g. Theorem (Whitehead) on p.75 of May's "Concise Course in Algebraic Topology").

I am interested in counterexamples to this when you drop the surjectivity condition for $q=n+1$. That is,

Question: What examples are there of a map $f:X\to Y$ of pointed connected $n$-dimensional CW complexes that induces isomorphism on $\pi_q$ for $q\le n$, but is not a homotopy equivalence?

I would also like to know what are the "minimal" examples of this. For example, it seems impossible for $n\le 2$ (the induced map on universal covers is a homology isomorphism by the Hurewicz theorem, and hence is a weak equivalence and thus induces isomorphism on all higher homotopy groups). I also wonder if there is an example with finite complexes.

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  • $\begingroup$ I think for $n=+\infty$ what you look for is related to Jet maps which are discussed in Gray's book. Restricted to finite dimensional complexes, they give isomorphism, but not a homotopy equiv. in general. I also suggest to compare the above version of Whitehead's theorem to the version related to $n$-types. Look at Mosher and Tangora, Theorem 3 on page 131. They are not the same notions, but very much related. $\endgroup$
    – user51223
    Jul 5, 2015 at 21:42
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    $\begingroup$ @user51223: I don't follow. For $n = \infty$ you can just apply the usual Whitehead's theorem. $\endgroup$ Jul 6, 2015 at 4:36
  • $\begingroup$ Sorry, I made a mistake. I meant Phantom maps. Look at `SPACES OF THE SAME n-TYPE, FOR ALL n' by Brayton Gray,TOPOLOGY Vol.5,pp.241. I think it has sort of examples you are after. $\endgroup$
    – user51223
    Jul 6, 2015 at 8:26

2 Answers 2

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It seems to me that the stronger statement is true. If $f$ is only an isomorphism on homotopy in degrees $* \leq n$ then the homotopy fibre $F$ is $(n-1)$-connected and its Hurewicz map is an isomorphism in degree $n$. Considering the Serre spectral sequence of the fibration seqeuence $F \to \widetilde{X} \to \widetilde{Y}$, and using the fact that it vanishes above the $n$th column (by the dimension restriction on $Y$) it follows that there is a short exact sequence $$0 \to H_n(F) \to H_n(\widetilde{X}) \to H_n(\widetilde{Y}) \to 0$$ and so the composition $\pi_n(F) \to H_n(F) \to H_n(\widetilde{X})$ is injective. But this factors through the map $\pi_n(F) \to \pi_n(\widetilde{X})$, which is trivial, and hence $\pi_n(F)=0$ too.

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I think isos up to dimension $n$ is enough to deduce $f_*:[A,X]\to[A,Y]$ onto when $dim(A)\leq n$. This gives a right inverse to $f$ which also induces isos up to $n$ and hence has a right inverse. So $f$ has a homotopy inverse.

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    $\begingroup$ This is wonderful! Too bad I can't accept this too... $\endgroup$
    – KotelKanim
    Jul 6, 2015 at 5:26
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    $\begingroup$ I just add that the fact you refer to in the first sentence is also stated as a theorem (and proved) in May's book right before the theorem I quoted. $\endgroup$
    – KotelKanim
    Jul 6, 2015 at 15:03

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