Conjugacy is a property of two points independent of the geodesic between them

Let $p=e_0=(1,0,0)$, $e_1=(0,1,0)$, $e_2=(0,0,1)$, and $u(s)=\cos (s)e_1+\sin (s)e_2$ for $-1<s<1$; then $p,u(s)$ are orthonormal for each $s$. [construct, algebra] 2.1 Define $F(s,t)=\cos (t)e_0+\sin (t)u(s)$ on $(-1,1)\times [0,2\pi ]$. Orthonormality and ${\cos }^{2}t+{\sin }^{2}t=1$ give $\mathrm{\mid }F(s,t)\mathrm{\mid }=1$, so $F$ is a smooth map into $S^2$; by [F1], each longitudinal curve is a great-circle affine geodesic. Thus $F$ is a geodesic variation. [F1, step 1.1, algebra] 3.1 The central geodesic $\gamma (t)=F(0,t)=\cos (t)e_0+\sin (t)e_1$ has $\gamma (0)=\gamma (2\pi )=p$, and differentiating gives $J(t)={\mathrm{\partial }}_{s}F(0,t)=\sin (t)e_2$, with $J(0)=J(2\pi )=0$ and $J(\pi \mathrm{/}2)=e_2\mathrm{\ne }0$. By [F2], $J$ is Jacobi; [F3] makes the endpoints conjugate along this full loop. [F2, F3, step 2.1, algebra] 4.1 On the same interval let $\delta (t)=p$. By [F1] it is a constant geodesic with the same endpoints as $\gamma$, but [F4] says those endpoints are not conjugate along $\delta$; hence this endpoint pair has different conjugacy outcomes along the two segments. [F1, F4, step 3.1] 5.1 Since $0<2\pi$, the witness interval is nondegenerate; $J$ is smooth up to both included endpoints and has the displayed endpoint zeros, while its nonzero value in step 3.1 witnesses conjugacy. The fixed $S^2$ example needs no dimension-zero or dimension-one case, and its standard basis and variation are explicit, so neither AC nor ${\mathrm{A}\mathrm{C}}_{\omega }$ is used; no iff is asserted.