Holonomy depends only on leafwise homotopy relative to endpoints

The homotopy image lies in one leaf. Since $H(s,0)=x$ for every $s$ and every slice $t\mapsto H(s,t)$ is a leafwise path, the point $H(s,t)$ lies in the leaf $L$ through $x$ for every $(s,t)$. Consequently for every curve $c:[0,1]\to [0,1{]}^2$ the composite $H\circ c$ is a leafwise path in $L$, and if $c$ runs from $(0,0)$ to $(1,1)$ then $H\circ c$ runs from $x$ to $y$.

Staircase paths in the square. The sets $H^{-1}(U)$, over foliation charts of $F$, form an open cover of the square; by [F3] there are grids $0=s_0<\cdots <s_m=1$ and $0=t_0<\cdots <t_n=1$ such that $H$ maps every cell $R_{jk}=[s_{j-1},s_j]\times [t_{k-1},t_k]$ into a single foliation chart. Consider the monotone lattice paths from $(0,0)$ to $(1,1)$ built from the unit steps $E$ (increasing $s$) and $N$ (increasing $t$). Starting from the path ${\sigma }_0=E^m{N}^n$, the bottom edge followed by the right edge, bubble the $N$ steps to the left: each of the $n$ steps $N$ crosses each of the $m$ steps $E$ once, in $mn$ successive interchanges of an adjacent pair $EN$ into $NE$, until the path ${\sigma }_{mn}=N^n{E}^m$, the left edge followed by the top edge, is reached. Parametrise the paths ${\sigma }_0,\dots ,{\sigma }_{mn}$ so that ${\sigma }_{\mathrm{\ell }}$ and ${\sigma }_{\mathrm{\ell }+1}$ coincide outside a subinterval on which they run from the common start of the interchanged steps to their common end along the two L-routes (the two two-segment side paths) of the cell spanned by those steps. Then each $P_{\mathrm{\ell }}:=H\circ {\sigma }_{\mathrm{\ell }}$ is, by step 1.1, a leafwise path from $x$ to $y$, with $P_0$ a reparametrisation of the concatenation of the constant path at $x$ with $b$, and $P_{mn}$ a reparametrisation of the concatenation of $a$ with the constant path at $y$.

Adding one interchange changes nothing. Fix $\mathrm{\ell }$, let $R$ be the cell spanned by the interchanged steps, mapped by $H$ into a foliation chart $U$, and let $C_1,C_2$ be the common start and the common end of the two interchanged steps, so that ${\sigma }_{\mathrm{\ell }}$ and ${\sigma }_{\mathrm{\ell }+1}$ agree outside one parameter interval on which they run from $C_1$ to $C_2$ along the two L-routes of $R$. The image $H(R)$ is connected and lies in $U\cap L$ by step 1.1, and it lies in one plaque as follows. By [F5], give $L$ its intrinsic second-countable manifold structure. Each plaque of $U$ in $L$ is intrinsically open: its inclusion factors smoothly through $L$ with invertible differential, since both tangent images equal $TF$, and [F6] applies. Distinct plaques are disjoint, so assigning the least index of a nonempty basic open set contained in each plaque injects this family into an enumerated basis. Thus the transverse values of $L\cap U$ are countable. Every transverse coordinate of the connected continuous image $H(R)$ is constant, since two values would force a nondegenerate interval of values, contrary to [F6]. The image lies in one connected level-set component, hence one plaque. In particular $p:=H(C_1)$ and $q:=H(C_2)$ lie in a common plaque of $U$, and both routes have images in $U$. Choose local transversals $S_p$ at $p$ and $S_q$ at $q$, a subdivision of $[0,1]$ that contains the two parameter values belonging to $C_1$ and $C_2$ and has no further subdivision point between them, foliation charts equal to $U$ on the middle interval and covering the common outer parts of the two paths, and intermediate transversals accordingly: this subdivision, these charts and these transversals satisfy the admissibility condition of [F4] for $P_{\mathrm{\ell }}$ and for $P_{\mathrm{\ell }+1}$, because outside the middle interval the two paths coincide and inside it both routes have images in $U$ with endpoints in a common plaque. Every chart-wise transport of [F4] is determined by its chart and its two transversals alone, so $P_{\mathrm{\ell }}$ and $P_{\mathrm{\ell }+1}$ receive one and the same composite germ; by [F4] that germ is a germ of each of the two paths, and by [F2] it is the intrinsic holonomy germ of each. Hence $h_{P_{\mathrm{\ell }}}(T^{\mathrm{\prime }},T)=h_{P_{\mathrm{\ell }+1}}(T^{\mathrm{\prime }},T)$.

Conclusion. Chaining step 2.2 over $\mathrm{\ell }=0,\dots ,mn-1$ gives $h_{P_0}(T^{\mathrm{\prime }},T)=h_{P_{mn}}(T^{\mathrm{\prime }},T)$. By step 2.1 the paths $P_0$ and $P_{mn}$ are reparametrisations of $c_x\ast b$ and of $a\ast c_y$, where $c_x,c_y$ denote the constant paths at $x,y$; reparametrising a chart chain changes only its subdivision points, so by [F2] it suffices to compare the germs of the two concatenations. Apply [F4] to $c_x\ast b$ with a chart chain whose subdivision contains the junction, whose chart on the constant piece is a foliation chart around $x$, and whose intermediate transversal at the junction is $T$ itself: the transport along the constant piece matches equal transverse coordinates at the single point $x$, so it is the identity germ of $(T,x)$, while the composite along the remaining pieces is a chain composite of $b$ and therefore equals $h_b(T^{\mathrm{\prime }},T)$ by [F2]; hence $h_{P_0}(T^{\mathrm{\prime }},T)=h_b(T^{\mathrm{\prime }},T)$. The same argument applied to $a\ast c_y$ with intermediate transversal $T^{\mathrm{\prime }}$ at $y$ gives $h_{P_{mn}}(T^{\mathrm{\prime }},T)=h_a(T^{\mathrm{\prime }},T)$. Therefore $h_a(T^{\mathrm{\prime }},T)=h_b(T^{\mathrm{\prime }},T)$ for leafwise homotopic paths with the same endpoints, and the holonomy germ depends only on the leafwise homotopy class relative to endpoints and on the endpoint transversals.