# Wrong proof: Hopf fibration is 2-torsion (solved)

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:

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 $$\pi_3(S^2)$$.

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

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.
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.