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Recall that the cohomotopy set $\pi^k(\mathcal{M})$ is $[\mathcal{M},S^k]$, i.e., the set of pointed homotopy classes of continuous mappings $\mathcal{M}\to S^k$. Recall also the Whitehead theorem:

Theorem: Suppose $X,Y$ are connected CW complexes. Suppose then that $f:X\to Y$ is a continuous map which induces an isomorphism $f_{*}:\pi_k(X,x_0)\to\pi_K(Y,f(x_0))$ for any $x_0\in X$. Then $f$ is a homotopy equivalence.

Does the following conjecture hold?

Conjecture: Suppose $X,Y$ are connected CW complexes. Suppose then that $f:X\to Y$ is a continuous map which induces an isomorphism $f^{*}:\pi^k(Y)\to\pi^k(X)$. Then $f$ is a homotopy equivalence.

An obvious motivation comes from the Whitehead theorem. The only lead I have on finding a proof of this conjecture is a theorem of Hopf states that $\pi^k(X)$ is in bijection with $H^k(M)$.

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Should it be $f^*: \pi^k(Y) \rightarrow \pi^k(X)$? –  Mingcong Zeng May 13 at 0:34
    
@mingcongzeng Why do you believe so? Has this conjecture been proved previously? –  SDevalapurkar May 13 at 0:38
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I mean $\pi^k$ should be contravariant instead of covariant in the statement you want to prove? –  Mingcong Zeng May 13 at 0:41
    
Foolish me! Should have noticed that. I'll edit the question. –  SDevalapurkar May 13 at 0:43

3 Answers 3

up vote 17 down vote accepted

No, this is false. According to the Sullivan Conjecture (Miller's Theorem), $\mathrm{map}_*(B\mathbb{Z}/p, S^n) \sim *$ for all $n$, which means $$ [\Sigma^n B\mathbb{Z}/p, S^k] = * $$ for all $n$. So if we let $f: \Sigma^k \mathbb{Z}/ p \to *$, the induced map $$ f^*: \pi^k(*) \to \pi^k ( \Sigma^n B\mathbb{Z}/p ) $$ is the equivalence $* \to *$. Since $f$ is not a homotopy equivalence, this counterexamps the conjecture.

Perhaps it would be more interesting to restrict attention to maps $f:X\to Y$ between finite complexes.

EDIT (further thoughts): If $K$ and $L$ are finite complexes, then something like your co-Whitehead statement is true!

Theorem 1: If $f: K\to L$ is a map of finite complexes such that $\pi^k( \Sigma^n f)$ is an isomorphism for all $k\geq k_0$ and all $n \geq n_0$, then $\Sigma f$ is a homotopy equivalence.

The proof uses a theorem of mine:

Theorem M: If $X$ is simply-connected and of finite type and $\mathrm{map}_*(X,S^k) \sim *$ for all sufficiently large $k$, then $\mathrm{map}(X,Y)\sim *$ for all finite-dimensional CW complexes $Y$.

Proof of Theorem 1: The hypotheses imply that the cofiber $C_{\Sigma^{n_0} f} \simeq \Sigma^{n_0} C_f$ satisfies $\mathrm{map}_*(\Sigma^{n_0} C_f, S^k) \sim *$ for all $k \geq k_0$. Theorem M implies that $\mathrm{map}_*(\Sigma^{n_0} C_f, \Sigma^{n_0}C_f) \sim *$, which implies $\Sigma^{n_0} C_f \sim *$ and hence that $\Sigma C_f \sim *$. This suffices to show that $\Sigma f$ is a homotopy equivalence.

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No, certainly not. If $X$ is a space with trivial reduced integral homology (for example the Poincare homology sphere with a ball removed) then the cohomotopy sets are all trivial, like those of a point.

EDIT Here are some details. Suppose $X$ has trivial homology, therefore trivial cohomology for all constant coefficients. Let $f:X\to Y$ be a based map with $Y$ connected and $\pi_1Y$ abelian. $Y$ is homotopy equivalent to a space that fibers over a $K(G,1)$ with $1$-connected fiber $Y_2$. $f$ can be lifted to a map $f_2:X\to Y_2$. $Y_2$ is homotopy equivalent to a space that fibers over a $K(G,2)$ with $2$-connected fiber $Y_3$. $f_2$ can be lifted to a map $X\to Y_{3}$, and this process can be repeated indefinitely. In the limit $f$ is lifted to an $\infty$-connected space, showing that it is homotopic to a constant.

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I was thinking of using the Postnikov tower of the sphere. –  Tom Goodwillie May 13 at 22:52
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I have added some details. –  Tom Goodwillie May 14 at 0:39
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What's a space without a Postnikov tower? –  Tom Goodwillie May 14 at 14:50
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@Jeff Strom, I guess you mean a Postnikov tower of principal fibrations. By the way, if you want a more general result, I think the following holds: if $X$ is acyclic and $Y$ has hypoabelian fundamental group (i.e. the fundamental group has no non-trivial perfect subgroup), then any map $f:X\to Y$ is null. I think you can prove it by using a generalized Postnikov tower for $Y$ originally given by Robinson (projecteuclid.org/euclid.ijm/1256052280). (to be continued...) –  Ricardo Andrade May 14 at 23:18
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(continuation) Then observe that this generalized Postnikov tower provides obstruction groups for extending $f$ to a map $CX\to Y$ (where $CX$ is the cone on $X$). These obstruction groups are identified with relative cohomology groups of the pair $(CX,X)$ with local coefficients. Finally, since $X$ is acyclic, all these relative cohomology groups are zero (i.e. the map $X\to CX\simeq\ast$ is acyclic). The condition on the fundamental group of $Y$ is there just to deal with the very first lift to $K(\pi_1 Y,1)$, which does not correspond to any cohomology/obstruction group. –  Ricardo Andrade May 15 at 0:45

Let me just add that stably it is true that a map f: $X\xrightarrow{} Y$ between finite complexes that induces isomorphisms on cohomotopy (of all degrees, now, including negative) is a weak equivalence. This is because the collection of all Z such that $f^{*} : [Y,Z] \xrightarrow{} [X,Z]$ is an isomorphism is a thick subcategory (also closed under products) so if it contains the spheres it contains all finite complexes. It is not true for general Y and Z because there are many many spectra with no cohomotopy at all; the Morava K-theories for example.

This also points to what goes wrong unstably, because mapping out of things behaves well with respect to fibrations, not cofibrations. So you'd expect the co-Whitehead theorem to be true not for finite complexes, but for spaces that can be built from spheres by taking iterated total spaces of fibrations; Lie groups for example, maybe.

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Do you really need to assume the cohomotopy isomorphism for all degrees? Can you get away with "sufficiently large" values? –  Jeff Strom May 14 at 17:56
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I think you cannot get away with sufficiently large values. Suppose X and Y are finite complexes, with the top cell in degree n. Then the cohomotopy in large degrees is going to be 0 for both of them. So the question is sort of the wrong way around, it seems to me; one should instead ask if you know about low (negative) cohomotopy can you conclude something. –  Mark Hovey May 22 at 11:21
    
Of course! Thanks. –  Jeff Strom May 22 at 11:22

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