A conjugate antipode breaks global Bolker injectivity

Put $F(n,x)=n\cdot x$ on $S^2\times S^2$. At a point with $F=0$, the vector $x$ lies in $T_n{S}^2$ by [F5], and the tangent variation $(\dot{n},\dot{x})=(x,0)$ satisfies $dF(\dot{n},\dot{x})=x\cdot x=1$. Thus $dF\mathrm{\ne }0$; [F6] makes $Z=F^{-1}(0)$ an analytic hypersurface of dimension $3$ in the $4$-dimensional product.

A tangent vector to $Z$ satisfies $x\cdot \dot{n}+n\cdot \dot{x}=0$. Given any $\dot{n}\in T_n{S}^2$, set $\dot{x}=-(x\cdot \dot{n})n$; then $\dot{x}\in T_x{S}^2$ by [F5], and the tangent equation holds. Hence ${\pi }_G$ is a submersion. Given any $\dot{x}\in T_x{S}^2$, set $\dot{n}=-(n\cdot \dot{x})x$; the same calculation proves ${\pi }_X$ is a submersion. Both maps are analytic restrictions of product projections. Since $Z$ is compact, ${\pi }_G$ is proper on ${\pi }_X^{-1}(K)$ for every compact $K\subset X$. The density induced by the positive-definite analytic product round metric has strictly positive analytic coefficient in every local incidence chart. The dimensions satisfy $4>3>2\ge 2$, so [F1] applies.

The defining differential on tangent vectors to the product is $dF=(x^{\mathrm{♭}},n^{\mathrm{♭}})$ by the metric identification [F4]. Therefore by [F3] the relation at $(n,x)$ consists of $(n,\eta ;x,\xi )=(n,\tau x^{\mathrm{♭}};x,-\tau n^{\mathrm{♭}}),\tau \mathrm{\ne }0.$ At the antipodal point $-x$, choose multiplier $-\tau$. Its data covector is $(-\tau )(-x{)}^{\mathrm{♭}}=\tau x^{\mathrm{♭}}$, the same as at $x$, while the object point changes from $x$ to $-x$. Both are nonzero relation points. Hence the left projection is not injective, and [F2] implies failure of global Bolker.

Let $v=n\times x$, a unit tangent vector at $x$. For $s$ near zero put $n_s=\cos (s)n+\sin (s)v$, $v_s=n_s\times x$, and ${\gamma }_s(t)=\cos (t)x+\sin (t)v_s,0\le t\le \pi \mathrm{.}$ Each $n_s$ is a unit vector orthogonal to $x$, and $v_s$ is a unit vector orthogonal to $x$; hence ${\gamma }_s$ parametrizes a great circle at unit speed and is a geodesic by [F10]. Thus [F7] makes this a geodesic variation. Its endpoints are fixed: ${\gamma }_s(0)=x$ and ${\gamma }_s(\pi )=-x$. Differentiating in $s$ at zero gives $J(t)={{\mathrm{\partial }}_s{\gamma }_s(t)\mid }_{s=0}=\sin (t)(v\times x)=-\sin (t)n,$ because $(n\times x)\times x=-n$ when $n\cdot x=0$ and $\mathrm{\mid }x\mathrm{\mid }=1$. By [F11] this field is Jacobi and hence satisfies the Jacobi equation in [F8]. It is nonzero for $0<t<\pi$ and vanishes at both endpoints, so [F9] makes the endpoints conjugate. The proof uses no choice axiom; the sphere has no boundary, so this example lies outside the simple-surface hypotheses.