We've been looking at a proof that the suspension of the Hopf map $\eta: S^3 \to S^2$ is an order 2 element in $\pi_4(S^3)$. In the course of that we were looking at that diagram: enter image description here

One can see that $\sigma$ has degree 1 while $\tau$ has degree $-1$. So $\sigma_* = id_{\pi_3(S^3)} $ and $\tau_* = - id_{\pi_2(S^2)}$ as maps on $\pi_3(S^3)$ and $\pi_2(S^2)$ respectively. That implies that $[\eta] = -[\eta]$ in $\pi_3(S^2)$ which would imply that $\eta$ is 2-torsion. That we know to be wrong. Where's our mistake?

Edit/Solution: The diagram only proves that $[\tau \circ \eta] = [\eta]$. Hence we have shown that a degree $-1$ map on a $S^2$ induces the identity on $\pi_3(S^2)$.

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    $\begingroup$ I do not think it is a good idea to edit a question by incorporating the solution after you received the answer. $\endgroup$
    – Paul Frost
    Nov 21 at 10:04
  • $\begingroup$ Indeed, you should accept my answer so that the question does not stay on the unanswered list. $\endgroup$ Nov 21 at 18:39

1 Answer 1


Let $H(f)$ denote the Hopf invariant of $f\colon S^{2n-1}\to S^n$. Then $H(\tau\circ\eta)=(\text{deg}\,\tau)^2 H(\eta)$ and $H(\eta\circ\sigma) = (\text{deg}\,\sigma)H(\eta)$. Since $(-1)^2=+1$, there is no contradiction.

See Problem 14 in Milnor's Topology from a Differentiable Viewpoint or Exercise 1 on p. 428 of Hatcher.

Regarding your argument, why should $[\tau\circ\eta]=[\eta]\in \pi_3(S^2)$? What you say up to "That implies that ..." is all correct.


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