A nowhere-zero vector field on an odd sphere

The map $z\mapsto iz$ is $\mathbb{R}$-linear, hence smooth, with $\mathrm{\mid }iz\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }=1$ on the sphere, so $X$ is a smooth nowhere-zero map of the sphere to itself.

Put $d=2m$. The $2d$ hemispheres $U_k^s=\{x\in S^{d-1}:s{x}_k>0\}$, $1\le k\le d$, $s\in \{-1,1\}$, have charts ${\phi }_k^s$ deleting coordinate $k$, with image the open unit ball and inverse inserting $x_k=s\sqrt{1-\mathrm{\mid }u{\mathrm{\mid }}^2}$. They cover the sphere and have smooth transitions. Their induced bundle charts define a topology by pulling back Euclidean open sets; [F3] makes the definitions agree on overlaps. The bundle is Hausdorff: distinct base points are separated by inverse images of disjoint base neighbourhoods, and vectors over the same point are separated in one bundle chart. The inverse images of rational balls in these finitely many bundle charts give an explicitly countable basis. Thus [F3] supplies a smooth tangent-bundle structure without choice. Every other smooth base chart has compatible induced charts, so this is the canonical structure.

For each $z$, the smooth curve $t\mapsto (z+tiz)\mathrm{/}\mathrm{\mid }z+tiz\mathrm{\mid }$ lies in the sphere and has velocity $iz$ at zero by [F1], so [F4] makes $X(z)$ a tangent vector. In a hemisphere chart, its bundle coordinates are $(u,D{\phi }_k^s(X(({\phi }_k^s{)}^{-1}(u))))$: the fibre part simply deletes coordinate $k$ from $i({\phi }_k^s{)}^{-1}(u)$, hence is smooth. Therefore $z\mapsto (z,iz)$ is a smooth section of the canonical bundle constructed in step 1.2 and is nowhere zero by step 1.1. This proves the unconditional field claim, consistent with $\chi (S^{2m-1})=0$ by [F2] and, under AC, with the converse of Poincare-Hopf.