Corollary 21.1.
\(\pi_4(S^2) \cong \mathbb{Z}/2\text{.}\)
Section 21.3 Computing \(\pi_4(S^3)\)
Pick a map \(f \colon S^3 \to K(\mathbb{Z}, 3)\) generating \(\pi_3(K(\mathbb{Z}, 3)) \cong \mathbb{Z}\text{,}\) and convert it into a fibration \(p \colon E_f \to K(\mathbb{Z}, 3)\) with \(S^3 \xrightarrow{\simeq} E_f\) a weak equivalence. Let \(F\) be the fibre. The long exact sequence in homotopy of \(F \to E_f \to K(\mathbb{Z}, 3)\) shows:
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\(F\) is \(3\)-connected;
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\(\pi_4(F) \cong \pi_4(S^3)\text{.}\)
By the Hurewicz theorem applied to \(F\text{,}\) \(\pi_4(F) \cong H_4(F; \mathbb{Z})\text{;}\) so it suffices to compute \(H_4(F; \mathbb{Z})\text{.}\)
Run the homological Serre spectral sequence of \(F \to E_f \simeq S^3 \to K(\mathbb{Z}, 3)\text{:}\)
\begin{equation*}
E^2_{p,q} \;=\; H_p\bigl(K(\mathbb{Z}, 3);\, H_q(F; \mathbb{Z})\bigr) \;\Longrightarrow\; H_{p+q}(S^3) \;=\; \begin{cases} \mathbb{Z} & p+q = 0, 3, \\ 0 & \text{otherwise}. \end{cases}
\end{equation*}
Since \(F\) is \(3\)-connected, \(H_0(F) = \mathbb{Z}\) and \(H_1(F) = H_2(F) = H_3(F) = 0\text{.}\) So the bottom row of \(E^2\) is the table of \(H_*(K(\mathbb{Z}, 3))\) above, and rows \(q = 1, 2, 3\) vanish.
The diagonal \(p + q = 4\) at \(E^2\) contains only the entry \(E^2_{0,4} = H_4(F)\text{.}\) For the abutment \(H_4(S^3) = 0\text{,}\) every entry on this diagonal must die. Differentials out of \(E^r_{0,4}\) all vanish, since their targets \((-r,\, 3+r)\) lie outside the first quadrant; so the only possibly nonzero differential into \(E^r_{0,4}\) is \(d^5 \colon E^5_{5,0} \to E^5_{0,4}\text{.}\) From the homology table, \(E^2_{5,0} = H_5(K(\mathbb{Z}, 3)) = \mathbb{Z}/2\text{.}\) Tracking what survives to \(E^5\) in position \((5, 0)\) requires checking that earlier differentials \(d^r\) for \(r < 5\) do not change it: incoming \(d^r\) at \((5, 0)\) originates at \((5+r, 1-r)\) which is in a zero row for \(r \geq 2\text{;}\) outgoing \(d^r\) at \((5, 0)\) lands at \((5-r, r-1)\) which is in a zero row for \(r = 2, 3, 4\text{.}\) So \(E^5_{5,0} = \mathbb{Z}/2\) unchanged.
The diagram on the diagonal \(p+q=4\text{:}\)
For \(E^\infty_{0,4}\) and \(E^\infty_{5,0}\) to both be zero, the differential
\begin{equation*}
d^5 \colon E^5_{5,0} \;=\; \mathbb{Z}/2 \;\longrightarrow\; E^5_{0,4} \;=\; H_4(F)
\end{equation*}
must be an isomorphism. We conclude
\begin{equation*}
\pi_4(S^3) \;\cong\; \pi_4(F) \;\cong\; H_4(F) \;\cong\; \mathbb{Z}/2.
\end{equation*}
Proof.
The Hopf fibration \(S^1 \to S^3 \to S^2\) gives \(\pi_4(S^3) \cong \pi_4(S^2)\) since \(\pi_i(S^1) = 0\) for \(i \geq 2\text{.}\)
Corollary 21.2.
\(\pi_1^s \cong \mathbb{Z}/2\text{,}\) generated by (the suspensions of) the Hopf map \(\eta \colon S^3 \to S^2\text{.}\)
Proof.
By Freudenthal’s suspension theorem (Theorem 11.10), \(\pi_{n+1}(S^n) \to \pi_{n+2}(S^{n+1})\) is an isomorphism for \(n \geq 3\text{.}\) Hence \(\pi_1^s = \pi_{n+1}(S^n)\) for any \(n \geq 3\text{;}\) taking \(n = 3\) gives \(\pi_4(S^3) = \mathbb{Z}/2\) above.
