# Covariant derivative in Ehresmann connection is independent of choice of curve

I'm taking a graduate differential geometry course and now I'm studying about Ehresmann connection. I've learned that there is a correspondence between connections and covariant derivatives. For a given Ehresmann connection of vector bundle $E\to M$, the covariant derivative of section $s$ at $x$ along tangent vector $v$ is given by $$\nabla_{v}s := \frac{d}{dt}\big|_{t=0} (P_{\gamma}^{t})^{-1}(s(\gamma(t)))$$

where $\gamma$ is a curve with $\gamma(0)=x, \gamma'(0)=v$ and $P_{\gamma}^{t}:E_{\gamma(0)}\to E_{\gamma(t)}$ is a parallel transport along $\gamma$. Our homework is showing this definition is independent of choice of curve $\gamma$. Here is my trial :

In class, we showed that the lift of curve $\gamma$, $\tilde{\gamma}$ with condition $\tilde{\gamma}'(t)\in H_{\tilde{\gamma}(t)}$ is unique by translating it into ODE. Let $\tilde{\gamma}$ be a lift of curve $\gamma$ to $E$ with $\tilde{\gamma}(0)=e\in E$.Suppose $\gamma\in U$ for some open $U$ with local trivialization $\Phi_{U}:\phi^{-1}(U)\to U\times \mathbb{R}^{k}, \Phi_{U}(e)=(\pi(e), \phi_{U}(e))$ for $e\in E$. Let $\Phi_{U}^{\gamma}:\gamma^{*}E\to [0, 1]\times \mathbb{R}^{k}$ be a induced trivialization of the pullback bundle $\gamma^{*}E\to [0, 1]$ and define $\vec{a}(t, e)$ as $\Phi_{U}^{\gamma}\circ \tilde{\gamma}(t)=(t, \vec{a}(t, e))\in [0, 1]\times\mathbb{R}^{k}$. Since horizontal tangent bundle $H_{e}T(\gamma^{*}E)$ has dimension 1, there exists unique horizontal lifting of $\frac{\partial}{\partial t}$ to $\gamma^{*}E$. We denote the image of this unique lifting under $\Phi_{U}^{\gamma}$ by $(1, \vec{b}^{\Phi}(t,e))$ which is a basis of $H_{(t, e)}(\gamma^{*}E)\subset T_{(t, e)}(\gamma^{*}E)$. Then the condition $\tilde{\gamma}'(t)\in H_{\tilde{\gamma}(t)}$ is translated into the following ODE

$$\frac{\partial \vec{a}}{\partial t}(t, e) = \vec{b}^{\Phi}(t, \Phi_{U}^{-1}(\gamma(t), \vec{a}(t, e)))$$ with initial condition $\vec{a}(0, e)=\phi_{U}(e)$. With this expression, we can rewrite covariant derivative as

$$\frac{d}{dt}\big|_{t=0}\vec{a}(-t, s(\gamma(t)))$$ But to differentiate this w.r.t $t$. when we use chain rule, there was a problem with $e$-component. How can we think about $\frac{\partial \vec{a}}{\partial e}$? Maybe I'm confusing about something.

There is another problem which I can't approach well : For a given (affine) connection $\nabla:TM\times \Gamma(E)\to \Gamma(E)$, define

$$H_{e}TE=\{ds(v)\in T_{e}E|s:\text{any local section near } x, s(x)=e, \nabla_{v}(s)=0\}$$ Then the assignment $e\mapsto H_{e}TE$ satisfies the properties of Ehresmann connection.