$E$ is bijective. Surjectivity of $E$ onto ${M \backslash K}$ is easy to see: for any point $q \in {M \backslash K}$ there is a closest point $p \in K$ to $q$, and it is straightforward to argue that $\gamma_p(t_0) = q$ for some $t_0$. In order to argue injectivity of $E$, let $$\begin{split}
\epsilon_0 = \sup \{ \epsilon > 0 \; | \; & E\colon Y \times [0, \epsilon) {\rightarrow}\overline{{M \backslash K}} \; \text{is a diffeomorphism}  \\ 
 & \text{onto its image} \}. 
\end{split}$$ If $E$ is not injective then for $n > 0$ we have distinct points $p_n, q_n \in Y$ and $r$-values $t_n, s_n \in [0, \epsilon_0 + 1/n)$ such that $E(p_n, t_n) = E(q_n, s_n)$. By the choice of $\epsilon_0$ it is impossible that both $t_n < \epsilon_0$ and $s_n < \epsilon_0$, so we may assume $ \epsilon_0 < s_n < \epsilon_0 + 1/n$. It is straightforward to argue that a bound of the form $\epsilon_0 - 2/n < t_n < \epsilon_0 + 1/n$ holds as well. By compactness of $Y$ we may pass to convergent subsequences and obtain points $p$ and $q$ such that $E(p, \epsilon_0) = E(q, \epsilon_0)$. The points $p$ and $q$ are distinct 