Essential maps of spectra which are null when localized at any prime

Question

There are maps of spaces which are not null-homotopic, but when localized at any prime become null. I don't know explicit constructions of any, but an example is given in Section 6 of Chapter 25 of the Handbook of Algebraic Topology (this chapter was written by C.A. McGibbon), of a phantom map $\Omega^2S^5\to \mathbb{H}P^\infty$ which has this property.

My question is really simple to state: are there maps of spectra which are globally non-trivial, but become null when localized at any prime? Or, can anyone give a proof that this is impossible?

It's well known that there's an "arithmetic square" for reconstructing spectra from their rationalizations and their $p$-completions (cf. these notes), and I believe this might have some bearing on this situation?

Answer 1

Such maps exist, even when the target is the sphere $S^{0}$.

A map $X \rightarrow S^{0}$ is $p$-locally trivial for each $p$ if it vanishes when composed with each of the $p$-localizations $S^{0} \rightarrow S^{0}_{(p)}$. This is the same as being in the kernel of the map $[X, S^{0}] \rightarrow [X, \prod S^{0}_{(p)}]$ induced by the diagonal $S^{0} \rightarrow \prod S^{0}_{(p)}$.

Assume, by contradiction, that all such maps are zero. This would mean that $S^{0} \rightarrow \prod S^{0}_{(p)}$ is a monomorphism in the stable homotopy category. Since the latter is triangulated, a monic necessarily splits, but $S^{0} \rightarrow S^{0}_{(p)}$ is not split, as that would in particular mean that we have a splitting on $\pi_{0}$, even though $\mathbb{Z} \hookrightarrow \prod \mathbb{Z}_{(p)}$ is not split. We deduce that $S^{0} \rightarrow \prod S^{0}_{(p)}$ is not split and hence not a monomorphism.

The proof of the fact that in a triangulated category any monic splits (see the answer to this question) allows one to construct an explicit example of a map that vanishes $p$-locally at each prime. Namely, the inclusion of the fibre $F \rightarrow S^{0}$ of the map $S^{0} \rightarrow \prod S^{0}_{(p)}$ is non-zero, as otherwise the latter would be split, although the composition $F \rightarrow S^{0} \rightarrow \prod S^{0}_{(p)}$ does necessarily vanish.