[proofplan] Both parts use the explicit formula $J(t) = d(\exp_p)_{ta}(tw)$ for a Jacobi field along $\gamma(t) = \exp_p(ta)$ that vanishes at the base point with $J'(0) = w$. For Part 1, reparametrise so the conjugate point is at $t = 1$; the existence of a non-trivial $J$ with $J(0) = J(1) = 0$ produces a non-zero $w \in T_p M$ in the kernel of $d(\exp_p)_a$, so the differential is not injective, and on a finite-dimensional space not injective is equivalent to not surjective. For Part 2, define $f(t) := g(J(t), \dot\gamma(t))$. Show $f'' = 0$ via metric compatibility of $\nabla$ and the Jacobi equation (using $R(\cdot, \cdot, \dot\gamma, \dot\gamma) = 0$ from the curvature symmetries), with boundary values $f(0) = f(\beta) = 0$. The boundary-value problem $f'' = 0$, $f(0) = f(\beta) = 0$ then forces $f \equiv 0$, i.e., $J \perp \dot\gamma$ along $\gamma$. [/proofplan] [step:Set up Jacobi field uniqueness via a second-order linear ODE] Let $\gamma : I \to M$ be a smooth geodesic with $\gamma(0) = p$, $\dot\gamma(0) = a$. A Jacobi field along $\gamma$ is a section $J$ of $\gamma^* TM$ satisfying the Jacobi equation \begin{align*} \nabla_t \nabla_t J + R(J, \dot\gamma) \dot\gamma = 0, \end{align*} where $R$ is the Riemann curvature (chapter convention $R = -\nabla\circ\nabla$). Choose a parallel orthonormal frame $\{e_1(t), \dots, e_n(t)\}$ along $\gamma$, so each $e_i$ is a smooth section of $\gamma^* TM$ with $\nabla_t e_i = 0$ and $g(e_i, e_j) = \delta_{ij}$. Writing $J(t) = \sum_i J_i(t) e_i(t)$ with smooth scalar coefficients $J_i : I \to \mathbb{R}$, the Jacobi equation becomes the second-order linear ODE system \begin{align*} \ddot J_i(t) + \sum_j A_{ij}(t) J_j(t) = 0, \quad A_{ij}(t) := g(R(e_j(t), \dot\gamma(t)) \dot\gamma(t), e_i(t)), \end{align*} where the matrix $A(t) = (A_{ij}(t))$ is smooth in $t$. By the Picard-Lindelöf existence and uniqueness theorem for linear ODE systems with smooth coefficients on an interval, for any $t_0 \in I$ and any prescribed initial data $J(t_0), \nabla_t J(t_0) \in T_{\gamma(t_0)} M$, there is a unique Jacobi field $J$ along $\gamma$ with these initial values. In particular: \begin{align*} J(t_0) = 0 \text{ and } \nabla_t J(t_0) = 0 \implies J \equiv 0 \text{ on } I. \end{align*} We will use this uniqueness statement repeatedly. By [Jacobi Fields via the Exponential Map](/theorems/2717), for any $w \in T_p M \cong T_a(T_p M)$, the unique Jacobi field $J$ along $\gamma(t) = \exp_p(ta)$ with $J(0) = 0$ and $\nabla_t J(0) = w$ is given by \begin{align*} J(t) = d(\exp_p)_{ta}(t w), \quad t \in I. \end{align*} We use this formula throughout. [/step] [step:Reduce Part 1 to the case $\beta = 1$ by reparametrisation] For Part 1, by hypothesis $q = \exp_p(\beta a)$ is conjugate to $p$ along $\gamma(t) = \exp_p(ta)$, with $\beta \ne 0$. Set $\tilde a := \beta a \in T_p M$ and $\tilde\gamma(t) := \exp_p(t \tilde a) = \exp_p(\beta t a) = \gamma(\beta t)$. Then $\tilde\gamma$ is a smooth geodesic with $\tilde\gamma(0) = p$, $\tilde\gamma(1) = q$, and the conjugate point along $\tilde\gamma$ at $\tilde a$ is $\tilde\gamma(1) = q$. By [Geodesic Rescaling](/theorems/2710), the rescaling $t \mapsto \beta t$ takes Jacobi fields along $\gamma$ to Jacobi fields along $\tilde\gamma$ via $\tilde J(t) := J(\beta t)$, and $\tilde J(t) = 0 \iff J(\beta t) = 0$, so the conjugate-point relation transfers. We may therefore assume without loss of generality $\beta = 1$ in what follows: $q = \exp_p(a)$ and we seek to show $d(\exp_p)_a$ is not surjective. [/step] [step:From a non-trivial Jacobi field with $J(0) = J(1) = 0$, extract a non-zero kernel element of $d(\exp_p)_a$] By the definition of conjugate point, there is a non-trivial Jacobi field $J$ along $\gamma$ with $J(0) = 0$ and $J(1) = 0$. Set $w := \nabla_t J(0) \in T_p M$. Suppose for contradiction $w = 0$. Then $J$ is a Jacobi field with $J(0) = 0$ and $\nabla_t J(0) = 0$, so by the ODE uniqueness statement of Step 1, $J \equiv 0$. This contradicts non-triviality of $J$. Hence $w \ne 0$. By Step 1, $J(t) = d(\exp_p)_{ta}(tw)$ for $t \in [0, 1]$. Evaluating at $t = 1$: \begin{align*} 0 = J(1) = d(\exp_p)_a(w). \end{align*} So $w \in \ker d(\exp_p)_a$ with $w \ne 0$, i.e., $d(\exp_p)_a$ is not injective. The differential $d(\exp_p)_a : T_a(T_p M) \to T_q M$ is a linear map between finite-dimensional vector spaces of equal dimension $\dim M$ (using the canonical identification $T_a(T_p M) \cong T_p M$). For such linear maps, the rank-nullity theorem gives "injective $\iff$ surjective", so $d(\exp_p)_a$ is not surjective either. Equivalently, $q$ is a singular value of $\exp_p$ — by the regular-value theorem, if $d(\exp_p)_a$ were surjective, $q$ would be a regular value at $a$. This proves Part 1. [guided] We are given a non-trivial Jacobi field $J \not\equiv 0$ along $\gamma(t) = \exp_p(ta)$ with $J(0) = J(\beta) = 0$, and we want $(d\exp_p)_{\beta a}$ not surjective. Choose a parallel orthonormal frame $\{e_1, \dots, e_n\}$ along $\gamma$ and write $J(t) = \sum_i J_i(t) e_i(t)$. Since $\nabla_t e_i = 0$, the Jacobi equation becomes the linear second-order system \begin{align*} \ddot J_i(t) + \sum_j A_{ij}(t) J_j(t) = 0, \quad A_{ij}(t) := g(R(e_j, \dot\gamma)\dot\gamma, e_i), \end{align*} with smooth coefficient matrix $A(t)$. By Picard–Lindelöf for linear ODE systems with smooth coefficients on $[0, \beta]$, the solution is uniquely determined by initial position and initial velocity. In particular, the linear map \begin{align*} T : T_pM \to T_{\gamma(\beta)}M, \quad w \mapsto J_w(\beta), \end{align*} where $J_w$ is the Jacobi field with $J_w(0) = 0$ and $\nabla_t J_w(0) = w$, is well-defined and linear between vector spaces of equal dimension $\dim M$. Set $w := \nabla_t J(0)$. If $w = 0$, then $J$ has $J(0) = 0$ and $\nabla_t J(0) = 0$, forcing $J \equiv 0$ by ODE uniqueness — contradicting non-triviality. So $w \ne 0$. By uniqueness, $J = J_w$, so $T(w) = J_w(\beta) = J(\beta) = 0$. Thus $T$ has non-trivial kernel; by rank-nullity (equal dimensions), $T$ is non-surjective. Finally, identify $T$ with the differential of the exponential map. By [Jacobi Fields via the Exponential Map](/theorems/2717), $J_w(\beta) = (d\exp_p)_{\beta a}(\beta w)$. Since $\beta \ne 0$, $w \mapsto \beta w$ is a linear isomorphism, so non-surjectivity of $T$ is equivalent to non-surjectivity of $(d\exp_p)_{\beta a}$. [/guided] [/step] [step:Set up the orthogonality argument for Part 2 using a parallel decomposition] For Part 2, again parametrise the geodesic so $\gamma(0) = p$ and $\gamma(\beta) = q$ with $J(0) = J(\beta) = 0$ for the Jacobi field $J$ in question. Define \begin{align*} f : [0, \beta] &\to \mathbb{R} \\ t &\mapsto g(J(t), \dot\gamma(t)). \end{align*} We aim to show $f \equiv 0$. Since $\nabla_t \dot\gamma = 0$ (geodesic equation), differentiating $f$ via metric compatibility of $\nabla$ gives \begin{align*} f'(t) = \frac{d}{dt} g(J, \dot\gamma) = g(\nabla_t J, \dot\gamma) + g(J, \nabla_t \dot\gamma) = g(\nabla_t J, \dot\gamma). \end{align*} Differentiating again, with $\nabla_t \dot\gamma = 0$ once more: \begin{align*} f''(t) = g(\nabla_t \nabla_t J, \dot\gamma). \end{align*} The Jacobi equation $\nabla_t \nabla_t J = -R(J, \dot\gamma) \dot\gamma$ inserted into $f''$: \begin{align*} f''(t) = -g(R(J, \dot\gamma) \dot\gamma, \dot\gamma) = -R(J, \dot\gamma, \dot\gamma, \dot\gamma). \end{align*} By the skew-symmetry of $R$ in its second pair of indices — see [Symmetries of the Riemann Curvature Tensor](/theorems/2704) — $R(\cdot, \cdot, \dot\gamma, \dot\gamma) = 0$, so \begin{align*} f''(t) = 0 \quad \text{for all } t \in [0, \beta]. \end{align*} [/step] [step:Solve the boundary-value problem $f'' = 0$, $f(0) = f(\beta) = 0$] The function $f$ satisfies $f'' \equiv 0$ on $[0, \beta]$, so $f$ is affine: $f(t) = \alpha t + \beta_0$ for constants $\alpha, \beta_0 \in \mathbb{R}$. The boundary values are \begin{align*} f(0) = g(J(0), \dot\gamma(0)) = g(0, \dot\gamma(0)) = 0, \quad f(\beta) = g(J(\beta), \dot\gamma(\beta)) = g(0, \dot\gamma(\beta)) = 0. \end{align*} From $f(0) = 0$, $\beta_0 = 0$; from $f(\beta) = 0$ and $\beta \ne 0$, $\alpha = 0$. So $f \equiv 0$, i.e., \begin{align*} g(J(t), \dot\gamma(t)) = 0 \quad \text{for all } t \in [0, \beta]. \end{align*} This proves Part 2. [guided] We have a Jacobi field $J$ along $\gamma$ with $J(0) = J(\beta) = 0$ (and $\beta \ne 0$). Study the tangential coefficient $f(t) := g(J(t), \dot\gamma(t))$. **Computing $f''$.** Two structural facts: (i) $\nabla_t \dot\gamma = 0$ (geodesic equation), and (ii) $R(\cdot, \cdot, \dot\gamma, \dot\gamma) = 0$ by skew-symmetry of $R$ in its second pair (see [Symmetries of the Riemann Curvature Tensor](/theorems/2704)). By metric compatibility of $\nabla$: \begin{align*} f'(t) = g(\nabla_t J, \dot\gamma) + g(J, \nabla_t \dot\gamma) = g(\nabla_t J, \dot\gamma). \end{align*} Differentiating again, \begin{align*} f''(t) = g(\nabla_t \nabla_t J, \dot\gamma) = -g(R(J, \dot\gamma)\dot\gamma, \dot\gamma) = -R(J, \dot\gamma, \dot\gamma, \dot\gamma) = 0, \end{align*} using the Jacobi equation $\nabla_t\nabla_t J = -R(J,\dot\gamma)\dot\gamma$ and (ii). **Solving $f'' = 0$ with $f(0) = f(\beta) = 0$.** Since $f''\equiv 0$, $f$ is affine: $f(t) = \alpha t + \beta_0$. The boundary values give $\beta_0 = 0$ and $\alpha\beta = 0$, hence $\alpha = 0$ (using $\beta \ne 0$). So $f \equiv 0$, i.e., $J \perp \dot\gamma$ along $\gamma$. **Why $\beta \ne 0$ is essential.** Without the second boundary condition we get $\beta_0 = 0$ but $\alpha$ unconstrained. The condition $\beta \ne 0$ is what lets the second equation force $\alpha = 0$. [/guided] [/step]