Radial jacobi tensor is invertible before the first conjugate point

The initial-derivative map and normality of twice-vanishing fields. [F1, F3, given] By [F1] the assignment $w\mapsto J_w$ from $N_0$ to the Jacobi fields along $\gamma$ is linear, and uniqueness in [F1] makes it injective: if $J_w=0$ then $w=D_t{J}_w(0)=0$. Its image is exactly the space of normal Jacobi fields vanishing at $0$: each $J_w$ is normal by [F1], and conversely a normal Jacobi field $J$ with $J(0)=0$ equals $J_{D_{t}J(0)}$ with $D_{t}J(0)\in N_0$ by normality and uniqueness. Now let $J$ be any Jacobi field with $J(0)=J(t)=0$. The function $u(s):=g(J(s),T(s))$ satisfies $u^{\mathrm{\prime }\mathrm{\prime }}=g(D_s^{2}J,T)+2g(D_{s}J,D_{s}T)+g(J,D_s^{2}T)=g(D_s^{2}J,T)=-\mathrm{Rm}(J,T,T,T)=0$ by [F1] and the skewness of [F3], so $u$ is affine on the interval from $0$ to $t$; since $u(0)=u(t)=0$, the affine function $u$ vanishes identically, and therefore $0=u^{\mathrm{\prime }}(0)=g(D_{t}J(0),T(0))+g(J(0),D_{t}T(0))=g(D_{t}J(0),T(0))\mathrm{.}$ Hence $D_{t}J(0)\in N_0$ and $J=J_{D_{t}J(0)}$: every field in ${\mathcal{K}}_{\gamma }(0,t)$ is a normal field vanishing at $0$, necessarily of the form $J_w$.

The kernel of $A(t)$. [F1, F2, step 1.1] For $w\in N_0$ we have $A(t)w=J_w(t)$ by definition, so $w\in \ker A(t)⟺J_w(t)=0⟺J_w\in {\mathcal{K}}_{\gamma }(0,t)\mathrm{.}$ By step 1.1 the map $w\mapsto J_w$ is a linear bijection from $N_0$ onto the normal Jacobi fields vanishing at $0$, and it carries $\ker A(t)$ onto exactly the elements of that space which also vanish at $t$, namely onto ${\mathcal{K}}_{\gamma }(0,t)$; the inverse is $J\mapsto D_{t}J(0)$, which step 1.1 shows to be normal for every $J\in {\mathcal{K}}_{\gamma }(0,t)$. Therefore $\ker A(t)$ is linearly isomorphic to ${\mathcal{K}}_{\gamma }(0,t)$, so $\dim \ker A(t)=\dim {\mathcal{K}}_{\gamma }(0,t)$; by [F2] the latter is the multiplicity of the pair $(0,t)$ when that pair is conjugate and is $0$ when it is not.

Before the first conjugate instant the kernel is zero. [F2, step 2.1, given] Let $0<t<\tau$. By the definition of $\tau$ as an infimum, no pair $(0,s)$ with $0<s<\tau$ is conjugate: if some $0<s<\tau$ were conjugate, then $\tau \le s<\tau$, a contradiction. Since $t<\tau$, the pair $(0,t)$ is not conjugate, so by [F2] the space ${\mathcal{K}}_{\gamma }(0,t)$ is $\{0\}$; step 2.1 gives $\ker A(t)=\{0\}$, and $A(t)$ is injective.

Injectives between equidimensional spaces are isomorphisms. [F1, F4, step 3.1] By [F1] the map $A(t):N_0\to N_t$ is linear and both spaces have dimension $n-1<\mathrm{\infty }$. Injective means $\dim \ker A(t)=0$, so rank-nullity [F4] gives $\dim \mathrm{im}A(t)=n-1=\dim N_t$; hence $A(t)$ is surjective as well and therefore an isomorphism. In the parallel identification of Radial Jacobi tensor the matrix $\bar{A}(t)$ represents this isomorphism, so it is invertible at this $t$. The argument used only the given geodesic; no completeness, compactness or choice principle enters.