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Is there a pointed space $(X, p)$ such that for infinitely many integers $n\geq 1$ there is a map $(X, p)\to (X,p)$ inducing an automorphism other than $\mathrm{id}$ on $\pi_n(X, p)$?

In particular $\pi_n(X, p)$ must be non-trivial for infinitely many $n$.

What if require in addition $X$ to be a finite-dimensional CW complex?

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    $\begingroup$ What do you mean by non-trivial? Is the identity homomorphism on a non-trivial group non-trivial? $\endgroup$
    – Mark Grant
    Commented Mar 11, 2021 at 14:30
  • $\begingroup$ @MarkGrant I mean a map other than identity $\endgroup$
    – mitsur
    Commented Mar 11, 2021 at 14:32
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    $\begingroup$ @mitsur So the zero map (which is induced by the constant map) works? $\endgroup$
    – Najib Idrissi
    Commented Mar 11, 2021 at 15:02
  • $\begingroup$ @NajibIdrissi I messed up the formulation. Now edited. $\endgroup$
    – mitsur
    Commented Mar 11, 2021 at 19:20

2 Answers 2

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Yes: take the product $\def\K{{\rm K}} \def\Z{{\bf Z}} A=∏_{k≥0}\K(\Z,n)$ of Eilenberg–MacLane spaces.

Then for each $n≥0$ there is a map $f_n\colon A→A$ given by identities on all factors with index $k≠n$ and by the map $\K(\Z,n)→\K(\Z,n)$ induced by the homomorphism $\Z→\Z$ that multiplies by 2 if $k=n$.

The induced map $π_n(f_n)\colon\Z\to\Z$ multiplies by 2 for all $n≥0$

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  • $\begingroup$ What is the topology on this product? I feel it is important, but don’t have any argument against your example. $\endgroup$
    – user51223
    Commented Mar 11, 2021 at 15:48
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    $\begingroup$ @user51223 The product topology, as always. That this map exists and is continuous follows from the universal property of the product topology. Exercise in universal property: if $f_i: X_i \to X_i$ is a family of continuous maps, then $\prod f_i: \prod_{i \in I} X_i \to \prod_{i \in I} X_i$ is a continuous map. All of Dmitri's statements about the induced maps etc similarly follow from the universal property. $\endgroup$
    – mme
    Commented Mar 11, 2021 at 16:17
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A degree 2 map $f: S^3 \rightarrow S^3$ will induce an isomorphism on the odd primary part of $\pi_*(S^3)$ which is nonzero infinitely often. And I would bet that it is also nonzero infinitely often on the 2 primary part as well. (If you really want to explore that, I would start with Neisendorfer's book Algebraic methods in unstable homotopy theory.)

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  • $\begingroup$ I would even conjecture that, if a simply connected finite complex $X$ has nontrival mod $p$ homology, then any self map of $X$ that is not null after localization at $p$ will induce a nontrivial homomorphism infinitely often on the p primary part of $\pi_*(X)$. $\endgroup$
    – Nicholas Kuhn
    Commented Mar 11, 2021 at 15:50
  • $\begingroup$ I think the conjecture already fails for $S^3$ which has 2-primary exponent $4$ (this is a result of James). $\endgroup$
    – Tyrone
    Commented Mar 11, 2021 at 15:58
  • $\begingroup$ I should add that $S^3$ also has $p$-primary exponent $p^2$ for any odd prime $p$ (this is due to Toda). In general, for $p$ an odd prime, the $p$-primary exponent of $S^{2n+1}$ is $p^{2n}$. Since $S^{2n+1}$ is an H-space when localised at an odd prime, its degree $p^{2n}$ self-map annihilates all $p$-torsion in $\pi_*S^{2n+1}$. $\endgroup$
    – Tyrone
    Commented Mar 11, 2021 at 16:16
  • $\begingroup$ @Tyrone You have the Cohen-Neisendorfer-Moore odd prime exponent theorem stated incorrectly and it is stronger than you say: $S^{2n+1}$ has exponent $p^n$. (The $n=1$ theorem is due to Selick, and grounds an inductive proof.) You are right that I forgot that $S^3$ is an H-space, so my conjecture was a bit too quick. $\endgroup$
    – Nicholas Kuhn
    Commented Mar 12, 2021 at 2:41
  • $\begingroup$ The result I quoted predates Cohen-Moore-Neisendorfer and is due to Toda. I didn't think Neisendorff could prove the CMN stuff in his text (I see now he did but left out the fun parts). Amusingly you also have an easy non-H-space counterexample a dimension lower in $S^2$. $\endgroup$
    – Tyrone
    Commented Mar 13, 2021 at 0:48

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