Conjugate instants are isolated unless the geodesic is constant

If $n=0$, every tangent fiber is zero by [F8], so [F2] gives $C_a(\gamma )=\mathrm{\varnothing }$. Assume $n>0$ and choose a basis $e_1,\dots ,e_n$ of $T_{\gamma (a)}M$. For each $v$ in this fiber, let $J_v$ be the unique Jacobi field with $J_v(a)=0$ and $D_t{J}_v(a)=v$, supplied by [F3]. The local frame formula [F6] and curvature-slot linearity [F16] show that the Jacobi equation [F1] is linear in the field; therefore linear combinations of these fields have the corresponding linear-combination initial data. Uniqueness in [F3] makes $v\mapsto J_v$ linear by [F9]. For $t>a$, define the endpoint map $E_t(v)=J_v(t)$. By [F2], $t\in C_a(\gamma )$ exactly when $\ker E_t\mathrm{\ne }\{0\}$: a nonzero endpoint-vanishing Jacobi field has nonzero initial derivative by [F3], and conversely a nonzero $v\in \ker E_t$ gives such a field. In the fixed basis at $a$ and any local frame at $\gamma (t)$, $E_t$ is an $n\times n$ matrix $A(t)$. Its columns are smooth by [F3] and [F5], so $A$ is smooth locally and conjugate instants are exactly its singular times.

Near the initial time $a$, use a local frame along $\gamma$ whose value at $a$ is the chosen basis. The endpoint matrix satisfies $A(a)=0$. By [F6], the derivative of the coefficient column of $J_v$ at $a$ equals the coefficients of $D_t{J}_v(a)$ because the connection-matrix term is multiplied by $J_v(a)=0$. Thus $A^{\mathrm{\prime }}(a)=I_n$ and $A(t)\mathrm{/}(t-a)\to I_n$ as $t\downarrow a$. Since the determinant is a polynomial in the matrix entries and $\det I_n=1$, $A(t)$ is invertible for all sufficiently close $t>a$ in $I$. This also rules out accumulation at $a$ when $a$ is an included endpoint.

Fix $t_0\in C_a(\gamma )$ and use one local frame near $\gamma (t_0)$ to write $A_0=A(t_0)$. Put $K=\ker A_0$, $Y=\mathrm{im}{A}_0$, and $C=Y^{\perp }$ in the target fiber at $t_0$. The inner product and orthogonal decomposition [F7, F11, F13] give the projections $P_Y,P_C$; [F10] and [F12] give $\dim K=\dim C$. For $u\in K$ and any $v\in T_{\gamma (a)}M$, both $J_u(a)$ and $J_v(a)$ vanish, so their Wronskian is zero at $a$ and hence at $t_0$ by [F4]. Since $J_u(t_0)=0$, this says $g(D_t{J}_u(t_0),J_v(t_0))=0$. As $J_v(t_0)$ ranges over $Y$, $D_t{J}_u(t_0)\in C$. Define $B:K\to C$ by $B(u)=D_t{J}_u(t_0)$. The map is linear by step 1.1, the real-linearity of $D_t$ in [F6], and [F9]. If $B(u)=0$, then $J_u(t_0)=0$ and $D_t{J}_u(t_0)=0$, so uniqueness [F3], applied with initial time $t_0$, gives $J_u=0$ and then $u=D_t{J}_u(a)=0$. Thus $B$ is injective; the equal finite dimensions make it an isomorphism.

Split the domain as $K\oplus K^{\perp }$ and the target as $Y\oplus C$. The map $B_0=P_Y{A}_0{\mathrm{\mid }}_{K^{\perp }}:K^{\perp }\to Y$ is an isomorphism: its kernel is zero, and every element of $Y$ is $A_0(u+v)=A_{0}v$ for the decomposition $u+v\in K\oplus K^{\perp }$. By continuity, for all sufficiently small $h$, $B_h=P_{Y}A(t_0+h){\mathrm{\mid }}_{K^{\perp }}$ is still invertible with uniformly bounded inverse; this follows in bases from continuity of its nonzero determinant and the adjugate formula. For $u\in K$, the derivative $A^{\mathrm{\prime }}(t_0)u$ is the coordinate vector of $D_t{J}_u(t_0)$, since $J_u(t_0)=0$ and [F6]; step 2.1 puts this vector in $C$. Hence, in operator norm, $P_{Y}A(t_0+h){\mathrm{\mid }}_K=o(\mathrm{\mid }h\mathrm{\mid })$, $P_{C}A(t_0+h){\mathrm{\mid }}_K=hB+o(\mathrm{\mid }h\mathrm{\mid })$, and $P_{C}A(t_0+h){\mathrm{\mid }}_{K^{\perp }}=O(\mathrm{\mid }h\mathrm{\mid })$. If $A(t_0+h)(u+v)=0$ with $u\in K$ and $v\in K^{\perp }$, the $Y$-component and the bounded inverse of $B_h$ give $\mathrm{\parallel }v\mathrm{\parallel }=o(\mathrm{\mid }h\mathrm{\mid })\mathrm{\parallel }u\mathrm{\parallel }$. The $C$-component then gives $0=hB(u)+o(\mathrm{\mid }h\mathrm{\mid })\mathrm{\parallel }u\mathrm{\parallel }+O(\mathrm{\mid }h\mathrm{\mid })\mathrm{\parallel }v\mathrm{\parallel }=hB(u)+o(\mathrm{\mid }h\mathrm{\mid })\mathrm{\parallel }u\mathrm{\parallel }$. Because $B$ is an isomorphism, this forces $u=0$ for all sufficiently small nonzero $h$, and then the $Y$-component forces $v=0$. Thus $A(t_0+h)$ is injective and hence invertible. Restricting to $h$ with $t_0+h\in I$ proves that $t_0$ is isolated, including at a one-sided endpoint. The argument uses only finite-dimensional matrix estimates, not a selected sequence of kernel vectors.

On a neighborhood of any time in $I\cap (a,\mathrm{\infty })$, the endpoint matrix has continuous entries by step 1.1. Its singular set is the zero set of the continuous determinant, so $C_a(\gamma )$ is closed relative to $I\cap (a,\mathrm{\infty })$. Step 3.1 shows this closed set is discrete. If $K_0$ is a compact subinterval of $I\cap (a,\mathrm{\infty })$, then $C_a(\gamma )\cap K_0$ is closed in $K_0$ and hence compact by [F14]. Its cover by its open singletons has a finite subcover by [F15], so it is finite. If the intersection is empty the conclusion is immediate.

If $\gamma$ is constant, [F2] gives no conjugate times, so the set is empty. The nondegenerate interval and one-sided endpoint conventions are those in [F1] and [F3]; the initial time itself is excluded by $t>a$, and the zero Jacobi field does not witness conjugacy by [F2]. The local matrix and compact-cover arguments after the Wronskian invocation make no selection. Exactly ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited and spent to use [F4], through its stated curvature-symmetry dependency; no full Axiom of Choice or further choice principle is used. The proposition is not an iff claim.