bijections ${\overline{E}}_i\colon Y_i \times (0,1] {\rightarrow}{M^{*}}\backslash K_i$. Endow each subset ${M^{*}}\backslash K_i$ with the topology $\tau_i$ that makes ${\overline{E}}_i$ a homeomorphism. We now show that these topologies are equivalent. Let $K$ be a compact set such that $K \supset K_1 \cup K_2$. Consider the identity map from $({M^{*}}\backslash K, \tau_1) {\rightarrow}({M^{*}}\backslash K, \tau_2)$, i.e. the composition $\operatorname{\overline{\psi}}= {\overline{E}}_2^{-1} \circ {\overline{E}}_1$. To show that the topology on ${M^{*}}\backslash K$ is independent of $K_i$, it suffices to show that $\operatorname{\overline{\psi}}$ is a homeomorphism. By the symmetric roles of the $K_i$, it suffices to prove that $\operatorname{\overline{\psi}}$ is an open map.

We already have that $\psi = {E}_2^{-1} \circ {E}_1$ is a diffeomorphism. We need only check that open neighbourhoods in $\tau_1$ of points in ${M(\infty)}$ are taken to open neighbourhoods in $\tau_2$. Choose a basis element of the form ${\overline{E}}_1(U \times (c,1])$ where $U$ is open in $Y_1$. For every $[\gamma_p] \in {M(\infty)}\cap {\overline{E}}_1(U \times (c,1])$ where $\gamma_p$ is asymptotic to $\sigma_q$ we must find a neighbourhood $V_q \subset Y_2$ of $q$ and $d > 0$ so that ${\overline{E}}_2( V_q \times (d, 1] ) \subset {\overline{E}}_1(U \times (c,1])$. It is sufficient to show that ${\overline{E}}_2( V_q \times (d, 1) ) \subset {\overline{E}}_1(U 