Suppose we have a principal bundle $p: P \to M$ with principal connection form $\omega: TP \to \mathfrak{g}$.

Given any arbitrary curve $\gamma: I \to P$ in the total space of the principal bundle, where $I=[0,1]$, we can use the nonabelian fundamental theorem of calculus [R. W. Sharpe Thm 7.14] to integrate $\gamma^*\omega: TI \to \mathfrak{g}$ to a map $g: I \to G$ such that $g(0) = e$ and $g^* \omega_G = \gamma^* \omega$, where $\omega_G: TG \to \mathfrak{g}$ is the usual Maurer-Cartan form on $G$.

Now we can define a sort of transport of the starting point $\gamma(0)$ along this curve by
\begin{equation}\gamma(0) \mapsto \gamma(t).g(t)^{-1}.\end{equation}
**Note:** I might need to replace $g(t)^{-1}$ by $g(t)$ in the formula above.

On the other hand, we can use the standard definition of parallel transport: take the projection $p \circ \gamma: I \to M$, and lift it again to a horizontal curve $\widetilde{\gamma}: I \to P$ with the same starting point $\widetilde{\gamma}(0)=\gamma(0)$. Then we can parallel transport the starting point $\widetilde{\gamma}(0)=\gamma(0)$ by traveling along the horizontal curve: \begin{equation}\mathrm{Pt}_{p \circ \gamma}(t)\gamma(0)= \widetilde{\gamma}(t). \end{equation}

Do these two notions of transport agree? In other words, do we have an equality $\widetilde{\gamma}(t) = \gamma(t).g(t)^{-1}?$

For anyone not familiar with the nonabelian FTC, I'll give the global version here:

**Theorem (nonabelian fundamental theorem of calculus):** Suppose we are given a connected manifold $M$ equipped with a $\mathfrak{g}$-valued $1$-form $\omega: TM \to \mathfrak{g}$. Then there exists a smooth map $f: M \to G$ such that $\omega =f^* \omega_G$ if and only if $\omega$ satisfies the Maurer-Cartan equation $d\omega + \frac{1}{2}[w \wedge \omega]=0$ and the monodromy representation $\pi_1(M, x) \to G$ is trivial. Moreover, if these conditions are satisfied then the map $f$, called the *integral* of $\omega$, is unique up to left translation by an element of $G$.

**Corollary:** On the interval $I=[0,1]$ any 1-form $\omega: TI \to \mathfrak{g}$ has an integral $f: I \to G$ unique up to choice of $f(0) \in G$.