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I was reading a proof and it concludes, from the fact that the rational Pontryagin classes are topological invariants, that the map $\pi_{4n}(BO)\rightarrow\pi_{4n}(BTOP)$ induced by the inclusion $BO\rightarrow BTOP$ is injective. I don't see how this works, can anyone explain it to me. Thanks!

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  • $\begingroup$ Topological invariance of rational Potnryagin classes is equivalent to the map $H^\ast(BTop;\mathbb{Q})\to H^\ast(BO;\mathbb{Q})$ being surjective (see page 1 of arxiv.org/abs/0901.0819). The conclusion should follow from this, a little rational homotopy theory, and the fact that $\pi_{4n}(BO)$ is torsion-free. But I haven't worked out the details of the argument. $\endgroup$
    – Mark Grant
    Feb 8, 2014 at 12:31
  • $\begingroup$ @Mark Grant we don't need rational homotopy theory, $\pi _{4n}(BO)$ injects to $H_{4n}(BO,\mathbb{Q})$, so your reference implies the inclusion. $\endgroup$
    – user43326
    Feb 8, 2014 at 15:17

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