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May 12, 2020 at 18:52 comment added Gustavo Granja Yes, should be $S^4$.
May 12, 2020 at 15:42 comment added skd Yep! Edited my answer. (Also, I think you meant to write S^4 where you wrote S^5.)
May 12, 2020 at 15:42 history edited skd CC BY-SA 4.0
added 573 characters in body
May 12, 2020 at 12:58 comment added Gustavo Granja More generally, localized at an odd prime $p$, $\mathbb{H}P^{\frac{p-1}{2}}$ is equivalent to the $(p-1)/2$-th stage of the James construction $J_{\frac{p-1}{2}} S^5$ as there are no obstructions to extending the identity on the bottom cell. This map can't be extended to $J^{\frac{p+1}{2}} S^5$ because $P^1$ acts non-trivially on the generator of $H^4(\mathbb{H}P^\infty;\mathbb Z/p)$.
May 12, 2020 at 4:27 vote accept skd
May 12, 2020 at 4:26 history answered skd CC BY-SA 4.0