connecting $(L_1,L_2)$ and $(L'_1,L_2)$ inside . Also, as $\Lambda_0(L'_1)$ is path-connected, there is a path connecting $(L'_1,L_2)$ and $(L'_1,L_4)$ inside . Finally, there is a path from $(L'_1,L_4)$ to $(L_3,L_4)$ inside as $\Lambda_0(L_4)$ is path-connected.

This step of the proof is based on the following proposition in which we denote by $\Omega^2$ the set of all admissible pairs of paths in $\Lambda(n)$ (see ). Let us note that by Section \[section-gammanor\] and (v’) in Section \[section-Maslov\], $\mu_{Mas}(\gamma_{nor},L_1)=1$ and $\mu_{Mas}(L_0,\gamma'_{nor})=-1$, where $L_0=\mathbb{R}^n\times\{0\}$ and $L_1=\{0\}\times\mathbb{R}^n$.

\[prop-Maslov\] Let

$$\mu:\Omega^2\rightarrow\mathbb{Z}$$ be a map such that the same properties (i’)-(iii’) from Section \[section-Maslov\] are satisfied, as well as

-   $\mu(\gamma_{nor},L_1)=1$ and $\mu(L_0,\gamma'_{nor})=-1$, where $L_0=\mathbb{R}^n\times\{0\}$ and $L_1=\{0\}\times\mathbb{R}^n$.

Then $\mu=\mu_{Mas}$ on $\Omega^2$.

We note at first that we have by the properties (ii’) and (iii’) homomorphisms

$$\begin{aligned}
\label{homomorphisms}
\mu, \mu_{Mas}:\pi_1(\Lambda(n)\times\Lambda(n),(L_0,L_1))\rightarrow\mathbb{Z}\end{aligned}$$

and we now claim that they coincide.\
We first note that

$$\pi_1(\Lambda(n)\times\Lambda(n),(L_0,L_1))\cong\pi_1(\Lambda(n),L_0)\times\pi_1(\Lambda(n),L_1)\cong\mathbb{Z}\oplus\mathbb{Z},$$ where the first isomorphism is induced by the projections onto the components and the second one is given by the Maslov index. As $\gamma_{nor}(0)=L_0$, $\gamma'_{nor}(0)=L_1$ and $\mu_{Mas}(\gamma_{nor})=\mu_{Mas}(\gamma'_{nor})=1$, we see that the pairs of paths

$$\{(\gamma_{nor},L_1), (L_0,\gamma'_{nor})\}$$ define a basis of $\pi_1(\Lambda(n)\times\Lambda(n),(L_0,L_1))$. Since the homomorphisms in coincide on this basis by (N), it follows that $\mu$ 