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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?

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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 $\Sigma^2 \eta$.

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