Proof: First of all, the condition on $g$ from Lemma 6 is the condition on a loop $g_{|T_{z}B}$ of scalar products on 2-dimensional spaces, here $z\in\partial U_{1}\simeq S^{1}$. We can write $g=\exp(\psi)g_{0}$ where $\det(g_{0})=1$ and $\psi$ is a smooth function. Then we have

$$\nabla_{e_{z}}(\exp{\psi}g_{0})=\exp({\psi})\nabla_{e_{z}}g_{0}+\exp(\psi)\partial_{e_{z}}(\psi)\,g_{0}\,\,.$$

The equation of Lemma 6 gives $\partial_{e_{z}}\psi=-(\nabla_{e_{z}}g_{0})(n_{z},n_{z})/g_{0}(n_{z},n_{z})$. The RHS of this expression is known as long as we know $g_{0}$. Hence we can say that $d\psi=\beta_{g_{0}}$, where $\beta_{g_{0}}$ is a 1-form depending on the restriction $(g_{0})_{|\partial U_{1}}$. We see that it suffices to find such $g_{0}$ that $\int_{\partial U_{1}}\beta_{g_{0}}=0$ (then $\psi$ and hence $g$ does exist).

Let us consider the functional $I(g_{0})=\int_{S^{1}}\beta_{g_{0}}$. We can interpret a metric $g_{0}$ as a point in the Lobachevsky plane ${\cal H}=SL(2,{\bf R})/SO(2)$. More precisely, let us consider the space $S$ of pairs $(g_{0},P)$ where $g_{0}$ is a positive quadratic form on ${\bf R}^{2}$ such that $\det(g_{0})=1$ and $P$ is a half-plane in ${\bf R}^{2}$ (the meaning of $P$ is the inward oriented tangent half-plane to $\partial U_{1}$ at point $z\in\partial U_{1}$). This space is naturally diffeomorphic to $S^{\ast}({\bf R}^{2})\times{\cal H}$. The latter manifold can be identified in $SL(2,{\bf R})$-equivariant way with the manifold consisting of pairs $(x,y)$, where $x\in{\cal H}$ and $y$ belongs to the absolute. Hence $(g_{0})_{|\partial U_{1}}$ is (locally) a non-parametrized path in $S$ (it would be a global path, if the bundle over $S^{1}$ given by the all metrics on $S^{1}$ with the determinant $1$ was trivial).

Next we observe that the variation $\delta I(g_{0})=\int_{N}\omega$, where $N$ is a $2$-dimensional surface bounded by the paths defined by $g_{0}$ and $g_{0}+\delta g_{0}$, and $\omega$ is a canonical $SL(2,{\bf R})$-invariant $2$-form on $S$. One can show that even by a small variation of the path defined by $g_{0}$ we can make $I(g_{0})$ an arbitrary real number. In particular, we can find $g_{0}$ such that $I(g_{0})=0$. This concludes the proof of Lemma 6. $\blacksquare$