An application of parallel transport is to prove the existence of smooth frame along a curve on the manifold. But it seems that the existence has nothing to do with the metric, so I would like to find a proof without using the Riemannian metric and Levi-Civita connection. Formally speaking,
Let $M^n$ be a smooth manifold and $\gamma: I\to M$ be a smooth curve. Prove that there exists vector fields $E_1, E_2,\cdots, E_n\in\chi(\gamma)$ along $\gamma$ such that $E_1(t), E_2(t),\cdots, E_n(t)$ form a basis of $T_{\gamma(t)}M$ for all $t\in I$.
My attempt is as follows: we can always get local frame by using the coordinate vector fields. By compactness of $I$,we can partition it into a finite collection of subintervals on which $\gamma$ admits a frame. Then I try to use a $GL(n,\mathbb{R})$-valued function to adjust these frames so they match up.
If $J=[0,1]$ is a interval and there exist two frame {$X_1(t), \cdots, X_n(t)$} and {$Y_1(t), \cdots, Y_n(t)$}. If we can find a frame beginning with {$X_1(a), \cdots, X_n(a)$} and ending with {$Y_1(b), \cdots, Y_n(b)$} then we are done. We have $$ (X_1(t), \cdots, X_n(t))A(t)=(Y_1(t), \cdots, Y_n(t)), $$ where $A(t)\in GL(n,\mathbb{R})$. But if we let $$ Z_i=((1-t)I+tA(t))X_i, $$ $\det((1-t)I+tA(t))$ is not necessarily nonzero. I think there may be some perturbing trick here but I cannot figure it out.
Appreciate any help!