 A sphere cannot fibre over a lower-dimensional odd sphere

There is no fiber bundle $F\to S^m\to S^q$ when $m>q\ge 3$ 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 $d_{q}a=u$, the base generator. Its degree is even. If $a^s=0$ is its first vanishing power, the derivation law gives $0=d_q(a^s)=su{a}^{s-1}\ne 0$ 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 $S^3$.