$\newcommand{\al}{\alpha}$ Let $M$ be a smooth Riemannian manifold. Fix $p \in M, v \in T_pM$. Define $$ \mathcal{A}_v:=\{ w \in T_pM\,|\,\exists\alpha:(-\epsilon,\epsilon) \to M, \, \, \alpha(0)=p, \dot \alpha(0)=v, (\nabla_{\frac{\partial}{\partial_t}}\dot \alpha)(0)=w\}, $$ where $\nabla_{\frac{\partial}{\partial_t}}\dot \alpha$ is the covariant derivative along the path $\alpha$ w.r.t the Levi-Civita connection.
Conjecture: For every $v \in T_pM$, we have $\mathcal{A}_v=T_pM$.
Question: Is there a coordinate-free proof?
A possible solution:
Set $\gamma(t) = \exp_p(t(v + \frac{t}{2}w))$. Then $(\nabla_{\frac{\partial}{\partial_t}}\dot \alpha)(0)=w$. Here is an attempt to find a coordinate-free proof of that.
A coordinate-based approach:
If $\dot \alpha(t)=a^i(t)\partial_i( \alpha(t))$, then $$ \nabla_{\frac{\partial}{\partial_t}}\dot \alpha(0)=\dot a^i(0)\partial_i+a^i(0)a^j(0)\Gamma_{ij}^k\partial_k. $$ Now the $a^i(0)$ are uniquely determined by requiring $v=\dot \alpha(0)=a^i(0)\partial_i$.
Since we can freely choose $\dot a^i(0)$, we can always arrange for $\nabla_{\frac{\partial}{\partial_t}}\dot \alpha(0)$ to equal any given vector $w$, regardless of the value of $a^i(0)a^j(0)\Gamma_{ij}^k\partial_k$.
