The map $\mathbb H\text{P}^{\infty} \to K(\mathbb Z,4)$ representing a generator of $H^4(\mathbb H\text{P}^{\infty};\mathbb Z) = \mathbb Z$ is a rational equivalence.

But is there any honest map in the other direction (before rationalizing, of course) that is not null-homotopic?

And if so, how can we understand the set of all homotopy classes $[K(\mathbb Z,4),\mathbb H\text{P}^{\infty}]$?

  • 2
    $\begingroup$ Note that $\Omega$ of such a map would be a map $K(\mathbb{Z},3)\rightarrow S^3$, which I think is nullhomotopic, so if there is a nontrivial map like you want it'd act trivially on $\pi_4$, $H^4$ and so all cohomology. $\endgroup$ – Achim Krause Sep 29 '16 at 18:24
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    $\begingroup$ Yes, according to a theorem of Zabrodsky, using Miller's theorem resolving the Sullivan conjecture, every map $K(\mathbb{Z}, 3) \to S^3$ must be a phantom map; but since there is no pair of dimensions $n$ and $n+1$ for which the rational cohomology of the domain is nontrivial in dimension $n$ and the rational homotopy of the target is nontrivial in dimension $n+1$, the only phantom map available is the trivial map. $\endgroup$ – Jeff Strom Sep 29 '16 at 20:43
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    $\begingroup$ In "Connective Coverings, Phantom Maps and Genus Sets" by McGibbon and Roitberg they construct a pair of rational equivalences $S^{4n-1}\rightarrow S^{2n}\langle 2n\rangle \rightarrow K(\mathbb{Z},4n-1)$ and show there are no maps in the opposite direction which become rational inverses. Perhaps their methods may be of interest. $\endgroup$ – Tyrone Oct 2 '16 at 14:11
  • $\begingroup$ I don't know if this helps, but a pointed map $K(\mathbb{Z}, 4) \to \mathbb{HP}^\infty$ should be equivalent to an $E_4$-map $\mathbb{Z} \to \Omega^3S^3$. $\endgroup$ – Lennart Meier Oct 5 '16 at 9:16
  • $\begingroup$ An honest map from K(Z,4) must induce 0 in integral cohomology. This is because there are mod p primary cohomology operations of odd degree acting nontrivially on the mod p reductions of the fundamental class in K(Z,4), for all primes p. Thus an honest map is zero in dimension 4, hence 0 in all positive dimensions of the polynomial algebra. John R. Harper. $\endgroup$ – John R. Harper Feb 13 '17 at 18:39

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