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Oct 17, 2019 at 21:17 comment added Noah Snyder A fun consequence is that stably there's a commutative product which agrees with both of these compositions, so stably $2 \eta = \eta 4$ and hence $2 \eta$ is trivial stably.
Dec 9, 2018 at 16:10 vote accept X1921
Dec 3, 2018 at 12:53 comment added Bertram Arnold One can also directly see this from the description of the Hopf invariant in terms of the cohomology ring of the mapping cone: If $f:S^3\to S^2$ and $g:S^2\to S^2$ are any two maps, there is an induced map $C_f\to C_{g\circ f}$ which is the identity on $H^4$ and multiplication by deg$(g)$ on $H^2$ using the obvious generators of these groups. The statement then follows from naturality of the cup product.
Dec 3, 2018 at 12:36 history answered Neil Strickland CC BY-SA 4.0