let $ \vartheta $ be the tautological soldering form on $ \TBundle Q $. Note that $ (\phi^* \vartheta)_{t, \varepsilon} (\difp_t) = \dot{q}_\varepsilon (t) $ and $ (\phi^* \vartheta)_{t, \varepsilon} (\difp_\varepsilon) = \diF q_\varepsilon (t) $. Since the connection $ K $ on $ \TBundle Q $ is torsion-free with respect to $ \vartheta $, using  and , we obtain $$\begin{split}
        0 &= \phi^* T (\difp_t, \difp_\varepsilon)
            \\
            &= \dif_{\phi^* K} (\phi^* \vartheta) (\difp_t, \difp_\varepsilon)
            \\
            &= \nabla^\phi_{\difp_t} \bigl(\phi^* \vartheta (\difp_\varepsilon)\bigr) - \nabla^\phi_{\difp_\varepsilon} \bigl(\phi^* \vartheta (\difp_t)\bigr) - \phi^* \vartheta \bigl(\commutator{\difp_t}{\difp_\varepsilon}\bigr)
            \\
            &= \DifFrac{}{t} (\diF q) - \diF \dot{q}(t).
    \end{split}$$ That is, $ \diF \dot{q}(t) = \DifFrac{}{t} (\diF q) $. Moreover, by , we have $$K \left(\difFracAt{}{\varepsilon}{0} \xi \ldot q_\varepsilon\right)
            = K \bigl(\tangent \xi_* (\diF q)\bigr)
            = \nabla_{\diF q} \xi_*,$$ where $ \xi_* $ is viewed as a section $ Q \to \TBundle Q $ and where we suppressed the explicit time dependence. In passing, we note that $ \tangent \xi_* (\diF q) $ is the fundamental vector field at $ \diF q $ on $ \TBundle Q $ generated by $ \xi $. Thus, under the isomorphism , we obtain $$\begin{split}
        \difFracAt{}{\varepsilon}{0}\bigl(q_\varepsilon(t), \dot{q}_\varepsilon(t) + \xi(t) \ldot q_\varepsilon(t)\bigr)
            &= \Bigl(\diF q(t), 