Parallel transport under reparametrization reversal and concatenation

On a smooth frame segment the chain rule gives $D_u(V\circ h)=h^{\mathrm{\prime }}(u)(D_{t}V)(h(u))$ from the two terms in [F1]. Thus a reparametrized parallel section is parallel even where $h^{\mathrm{\prime }}=0$. Continuity past corners and a common finite refinement give the same result piecewise. By uniqueness, the transported endpoint vector is the value of this section, so increasing endpoint-preserving time changes leave $P$ unchanged. A pause has constant section and makes no change.

A decreasing time change exchanges endpoints; the same calculation gives backward transport, which equals the inverse by [F3]. To concatenate, first take the parallel section with input $v$ along ${\gamma }_1$, then the one with input $P_{{\gamma }_1}v$ along ${\gamma }_2$. They agree at the joining point, hence give a continuous piecewise parallel section on the concatenation. Uniqueness identifies its endpoint with $P_{{\gamma }_2\ast {\gamma }_1}v$, proving the composition order. A singleton or constant piece has identity transport; zero-rank fibres have the unique identity map. All refinements and concatenations here have finitely many pieces.