Parallel transport depends only on the endpoints of a curve

Take $E={\mathbb{R}}^2\times \mathbb{R}$ and $\omega =xdy$ in its unit frame. This is a smooth connection by [F1]. Traverse the unit square through $(0,0),(1,0),(1,1),(0,1),(0,0)$, using each side's affine parameter $t\in [0,1]$. The four values of $x{y}^{\mathrm{\prime }}$ are respectively $0,1,0,0$. Thus the four equations in [F2] are $v^{\mathrm{\prime }}=0,v^{\mathrm{\prime }}=-v,v^{\mathrm{\prime }}=0,v^{\mathrm{\prime }}=0$, with solution multipliers $1,e^{-1},1,1$ respectively. The multiplier on the second side follows directly by differentiating $v(t)=e^{-t}v(0)$.

By [F3] the square's transport multiplies by $e^{-1}$, whereas the constant loop at $(0,0)$ multiplies by $1$. These differ on the unit vector of the endpoint fibre, since $e^{-1}<1$. The zero vector is fixed by both and is not a witness. Both paths are continuous finite piecewise smooth loops with exactly the same endpoints; corners cause no additional derivative condition. No curvature or homotopy theorem is used.