# Mapping cone of double suspension of Hopf map (Hatcher proposition 4L.11)

Let $$\eta\colon S^3\to S^2$$ denote the Hopf map. I don't understand a step in the proof of proposition 4L.11 of Hatcher, which states that $$\eta\circ(\Sigma\eta)$$ is not nullhomotopic.

Let $$C\eta=S^2\cup_\eta D^4$$ the mapping cone of $$\eta$$. If we assume that it is nullhomotopic, then we can define a map $$f\colon S^5\to C\eta$$ in the following way: Denote $$S^5$$ as the union of two cones $$CS^4$$. On one of these cones, you have the nullhomotopy of $$\eta\circ\Sigma\eta$$, composed with $$i\colon S^2\to C\eta$$. On the other cone, you have the nullhomotopy of $$i\circ\eta\colon S^3\to C\eta$$ (because $$C\eta\simeq\mathbb C P^2$$, it has trivial $$\pi_3$$), precomposed with $$\Sigma\eta$$. On the intersection of these cones, $$S^4$$, both are $$i\circ\eta\circ\Sigma\eta$$, so this gives a map $$f\colon S^5\to C\eta$$.

The space $$X=C\eta\cup_f D^6$$ is the mapping cone of $$f$$. Now, Hatcher states that $$X/S^2$$ is homotopy equivalent to the mapping cone of $$\Sigma^2\eta$$. Why is this true?

The $$2$$-cell you're quotienting out of $$X$$ belongs to $$C\eta$$, and the remaining piece is a $$S^4$$. Applying $$f$$ and then pinching out to this top cell of $$C\eta$$ should give you $$\Sigma^2 \eta: S^5 \to S^4$$, so $$X/S^2 \simeq S^4 \cup_{\Sigma^2 \eta} D^6$$, i.e. the mapping cone of $$\Sigma^2 \eta$$.
Let's try to make this more precise. On the first cone, $$f$$ postcomposed with $$C\eta \to S^4$$ is nullhomotopic rel the equatorial $$S^4$$, since $$S^2 \xrightarrow{i} C\eta \to S^4$$ is null. On the second cone, we have $$(CS^4, S^4) \xrightarrow{C\Sigma\eta} (CS^3, S^3) \to (S^4, *),$$ so collapsing $$S^4$$ gives $$\Sigma^2 \eta$$.