The holonomy germ is independent of the foliation chart chain

Inside a single chart the transport is a coordinate matching. Let $U$ be a foliation chart with coordinates $(x,y)$ containing the image of a subinterval, and let $S,S^{\mathrm{\prime }}$ be local transversals at the two subinterval endpoints $p,p^{\mathrm{\prime }}$, which lie in a common plaque of $U$. Then the transport germ $S\to S^{\mathrm{\prime }}$ constructed in the single-chart case of A leafwise path determines a germ of a transverse diffeomorphism is the map $u\mapsto$ (the point of $S^{\mathrm{\prime }}$ with the same transverse coordinate as $u$) near $p$. Consequently it is unchanged if an intermediate transversal $S^{\mathrm{\prime }\mathrm{\prime }}$ at an interior point $q$ of the subinterval is inserted or replaced: the composite of the transports $S\to S^{\mathrm{\prime }\mathrm{\prime }}$ and $S^{\mathrm{\prime }\mathrm{\prime }}\to S^{\mathrm{\prime }}$ matches transverse coordinates in the same chart and hence agrees near $p$ with the direct transport, since all three maps send a point to the point with the same $y$-coordinate.

Comparison on small overlaps. Around each point of a path segment contained in $U\cap U^{\mathrm{\prime }}$, choose a smaller product foliation box whose closure need not be fixed, with domain contained in $U\cap U^{\mathrm{\prime }}$. There the transition has transverse part $y^{\mathrm{\prime }}=r(y)$ by [F1]. Its differential is invertible: the full transition differential is block triangular and invertible, so its transverse diagonal block is invertible. By [F6], after shrinking $r$ is injective near the transverse value, and therefore matching $y$ is equivalent to matching $y^{\mathrm{\prime }}$ for transversals with endpoints in this small box. The transports computed in $U$ and $U^{\mathrm{\prime }}$ consequently agree as germs on that piece. Compactness and [F5] give a finite subdivision of the common segment into these boxes; composing and using step 1.1 proves equality over the entire segment. This comparison uses the transverse transitions on neighborhoods, not only equality of the central plaque germs.

Two chains compute the same germ. Let two finite chart chains with subdivisions be given. By step 1.1 the computed germ changes neither when a subinterval is subdivided inside one of the given charts nor when intermediate transversals are inserted, so we may refine both chains. The images of sufficiently small subintervals of a common refinement lie in a single chart of the first chain and a single chart of the second chain simultaneously; by [F5] finitely many such subintervals suffice to cover $[0,1]$, and by step 2.1 the transport over each such subinterval is the same germ whichever of the two charts is used to compute it. Composing the germs over the common refinement, both chains give the same composite germ from $(T,x)$ to $(T^{\mathrm{\prime }},y)$.

Changing intermediate transversals. At a subdivision point $a(t_i)$ both adjacent charts $U_i,U_{i+1}$ contain $a(t_i)$ (each contains the closed subinterval adjacent to the point), so $U_i\cap U_{i+1}$ is a neighbourhood of $a(t_i)$; shrinking the subdivision around $t_i$ and applying steps 1.1 and 2.1 to the transport across the resulting small subinterval shows that the composite is unchanged when the auxiliary transversal $T_i$ at $a(t_i)$ is replaced by another local transversal.

Conclusion. By steps 3.1 and 3.2 the composed germ does not depend on the displayed chain, the subdivision or the auxiliary transversals; only the leafwise path and the endpoint transversals remain. Hence $h_a=h_a(T^{\mathrm{\prime }},T)$ is a well-defined germ of a local diffeomorphism $(T,x)\to (T^{\mathrm{\prime }},y)$, as claimed.