= A sphere cannot fibre over a lower-dimensional odd sphere
There is no <fiber bundle> $F\to S^m\to S^q$ when $m>q\geq3$ and $q$ is odd. A local trivialization makes $F$ a retract of an open subset of $S^m$, so $H^j(F;\mathbb Q)=0$ for $j>m$; the <long exact sequence of homotopy groups of a fibration> makes $F$ connected. In the multiplicative rational <Serre spectral sequence>, only base columns zero and $q$ occur. Since $H^q(S^m)=0$, a fibre class $a\in H^{q-1}(F;\mathbb Q)$ must satisfy $d_qa=u$, the base generator. Its degree is even. If $a^s=0$ is its first vanishing power, the derivation law gives $0=d_q(a^s)=sua^{s-1}\ne0$ on the $E_q$ page, a contradiction. Bounded fibre cohomology ensures such a vanishing power exists. In particular $S^4$ cannot be the total space of a bundle over $S^3$.
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