What is the Relationship Between Fiber Bundles of Spheres?

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Prove that if there is a fiber bundle $S^k \to S^m \to S^n$, then $k = n-1$ and $m = 2n-1$.

 

[Infrared]

Spoiler

By checking dimensions, $k+n=m$ so it's enough to check just one of those equalities.

We may assume that $n>0$ since for $S^m\to S^n$ to be surjective, the codomain $S^n$ needs to be connected, and it's not possible to have $m=0.$

Case 1: $n>1.$ We examine the following part of the associated long exact sequence:

$\pi_{k+1}(S^m)\to\pi_{k+1}(S^n)\to\pi_k(S^k)\to\pi_k(S^m).$ The outer groups are zero since $\pi_i(S^j)=0$ when $0<i<j.$ So, $\pi_{k+1}(S^n)\cong\mathbb{Z}$ and hence $k+1\geq n.$

We now consider the following part of the same exact sequence:
$\pi_n(S^m)\to\pi_n(S^n)\to\pi_{n-1}(S^k)\to\pi_{n-1}(S^m).$ The case $n=m$ is impossible because that would make $k=0,$ in which case $S^m\to S^m$ is a double cover, but $n=m>1$ so $S^n$ is simply connected. So, again the two outer groups vanish and $\pi_{n-1}(S^k)\cong\mathbb{Z},$ giving $n-1\geq k.$ Together with the above, this means $k=n-1.$

Case 2: $n=1.$ Examine the tail end of the exact sequence: $\pi_1(S^m)\to\pi_1(S^1)\to\pi_0(S^k)$ (where $\pi_0$ isn't given a group structure but exactness still makes sense and is valid.) If $m>1$ then the image of $\pi_1(S^m)\to\pi_1(S^1)$ is trivial and cannot match the kernel of $\pi_1(S^1)\to\pi_0(S^k)$, which is the whole $\pi_1(S^1).$ So $m=1,$ which forces $k=0$ and the triple $(k,m,n)=(0,1,1)$ satisfies the right equations.

It's been a while since I took algebraic topology but hopefully this is mostly right.