Flatness implies path-independence on simply connected domains

Let $E\to M$ be a vector bundle with a connection on a vector bundle $\nabla$ , and let $F_\nabla$ denote its curvature.

Proposition (endpoint dependence on simply connected sets). Let $U\subset M$ be a connected, simply connected open set such that $F_\nabla|_U=0$ . Then for any two points $x,y\in U$ , the parallel transport map

\mathrm{PT}_\gamma : E_x \to E_y

depends only on the endpoints $x,y$ , not on the choice of piecewise smooth path $\gamma$ in $U$ from $x$ to $y$ .

Equivalently, if $\gamma_0,\gamma_1$ are two such paths with the same endpoints, then $\mathrm{PT}_{\gamma_0}=\mathrm{PT}_{\gamma_1}$ .

Principal bundles

A parallel statement holds for a principal bundle with a flat principal connection: on a simply connected $U$ where curvature vanishes, the transport $P_x\to P_y$ is independent of the path in $U$ .

Examples

Euclidean space. On $U=\mathbb R^n$ with the trivial bundle $U\times V$ and $\nabla=d$ , curvature vanishes and parallel transport is the identity, hence depends only on endpoints.
Restriction of a flat but globally nontrivial situation. A flat connection on a bundle over $S^1$ can have nontrivial holonomy around the circle, but on any simply connected arc $U\subset S^1$ the proposition applies, so transport inside $U$ is path-independent.
Why simply connectedness matters. On $S^1$ , take a flat $U(1)$ -connection with holonomy $e^{i\theta}\neq 1$ . Curvature is still zero, but transport from a point to itself along the generator loop is multiplication by $e^{i\theta}$ , so dependence on the homotopy class of the path persists when the domain is not simply connected.