as $p_n$ and $q_n$ are distinct and $E$ is a local diffeomorphism in a neighbourhood of $(p, \epsilon_0)$. We set $m = E(p, \epsilon_0) = E(q, \epsilon_0)$.

We now show that $\gamma_p$ and $\gamma_q$ meet ‘head on’, i.e. that $\gamma_p'(\epsilon_0) = -\gamma_q'(\epsilon_0)$. If this is not the case, choose a vector $X \in T_{m} ({M \backslash K})$ such that $$\label{neginnerprod}
 g( X, \gamma_p'(\epsilon_0) ) < 0 \; \text{and} \; g( X, \gamma_q'(\epsilon_0) ) < 0.$$

Consider any path $\sigma\colon (-\delta, \delta) \rightarrow {M \backslash K}$ for some $\delta > 0$ with $\sigma(0) = m$, $\sigma'(0) = X$. Because $E$ is a local diffeomorphism, near $p$ we have a smooth curve $\sigma_Y^p\colon (-\delta, \delta) \rightarrow Y$ and a positive smooth function $r^p$ such that $\sigma(t) = E( \sigma_Y^p(t), r^p(t))$. We may obtain a variation through $Y$-normal geodesics by considering the map $\Gamma(s,t) = E( \sigma_Y^p(s), t r^p(s) )$. The first variation formula and equation above imply that $r^p(s)$ is a decreasing function of $s$; in particular for $s$ sufficiently small and positive, $r^p(s) < \epsilon_0$. The same argument may be applied near $q$, and this implies for small enough $s$ that $E( \sigma_Y^p(s), r^p(s) ) = E( \sigma_Y^q(s), r^q(s) )$ which 