Show that if there were fiber bundles \(S^{n-1} \to S^{2n-1} \to S^n\) for all \(n\), then the groups \(\pi_i(S^n)\) would be finitely generated free abelian groups computable by induction, and nonzero for \(i \geq n \geq 2\).
Solution
Assume bundles \[ S^{n-1}\longrightarrow S^{2n-1}\longrightarrow S^n \] exist for every \(n\ge2\). The fiber inclusion is nullhomotopic since \[ \pi_{n-1}(S^{2n-1})=0. \] Exercise 31 therefore gives, for all \(i\), \[ \pi_i(S^n)\cong \pi_i(S^{2n-1})\oplus\pi_{i-1}(S^{n-1}). \tag{*} \]
For fixed \(i\), repeated use of () terminates: following the first summand increases the sphere dimension from \(n\) to \(2n-1\), so after finitely many such steps the sphere dimension exceeds \(i\) and that summand is zero; following the second summand decreases both indices by one. Hence every \(\pi_i(S^n)\) is obtained by finitely many direct sums from groups \(\pi_j(S^j)\cong\mathbb Z\) and zero groups. Consequently all groups are finitely generated free abelian, and () provides an effective recursive computation.
It remains to prove nonvanishing for \(i\ge n\ge2\). First consider \(S^2\). The \(n=2\) bundle gives \[ \pi_i(S^2)\cong\pi_i(S^3) \qquad(i>2). \] For \(n=3\), () gives \[ \pi_i(S^3)\cong\pi_i(S^5)\oplus\pi_{i-1}(S^2). \] Since \(\pi_2(S^2)\cong\mathbb Z\), induction on \(i\) shows \[ \pi_i(S^2)\ne0\qquad(i\ge2). \] Now for arbitrary \(i\ge n\), repeatedly take the second direct summand in () to obtain an injection \[ \pi_{i-n+2}(S^2)\hookrightarrow\pi_i(S^n). \] The source is nonzero by the preceding paragraph. Thus \[ \boxed{\pi_i(S^n)\text{ would be finitely generated free abelian and nonzero for every }i\ge n\ge2.} \]