Proof. Condition C1 implies that all transformations $\varphi_{L}$ admit an analytic continuation to $U$. Let us introduce a decreasing filtration by positive real numbers $Symp^{\geq r}(U),\,\,r\in{\bf R},r\geq 0$ on group $Symp(U)$ by the formula

$$\left\{g\in Symp(U)\,|\,\,\log|\xi^{\prime}/\xi-1|,\log|\eta^{\prime}/\eta-1|<-r\,\mbox{ where }(\xi^{\prime},\eta^{\prime})=g((\xi,\eta))\,\right\}$$

This is a complete filtration, and condition C2 implies that in any quotient $Symp(U)/Symp^{\geq r}(U)$ only a finite number of elements $\varphi_{L}$ are non-trivial. Therefore we can define the product in the quotient group.

In order to prove independence of $\gamma$, we consider the quotient group $Symp(U)/Symp^{\geq r}(U)$, and the finite $1$-dimensional CW-complex (graph) consisting of finitely many pieces $L$, such that $\varphi_{L}\neq 1$ in the quotient. For each vertex $v$ of the graph there is a natural cyclic order on the edges incident to $v$. The product $\varphi_{v}=\prod_{L}\varphi_{L}^{\pm 1}$ taken in the cyclic order over the set of edges incident to $v$ is equal to $id$ (this follows from the construction of $\varphi_{l}$ via factorizations). Since $U$ is simply-connected, we conclude that the image of $i_{x,y}^{\gamma}$ in $Symp(U)/Symp^{\geq r}(U)$ does not depend on $\gamma$. Using completeness of the filtration we see that $i_{x,y}:=i_{x,y}^{\gamma}$ does not depend on $\gamma$. Proof of the identity $i_{x,y}i_{y,z}=i_{x,z}$ is similar. $\blacksquare$