[step:Shrink $\varepsilon$ to enforce global injectivity]We claim that for some $0 < \varepsilon \le \varepsilon_1$, the map $\Phi|_{\partial M \times [0, \varepsilon)}$ is injective.
Suppose, for contradiction, that no such $\varepsilon$ exists. Then for every $k \in \mathbb{N}$ there exist distinct points
\begin{align*}
(p_k, s_k),\, (q_k, t_k) \in \partial M \times [0, \tfrac{1}{k}) \quad \text{with} \quad \Phi(p_k, s_k) = \Phi(q_k, t_k).
\end{align*}
By compactness of $\partial M$, pass to a subsequence (still denoted $k$) such that $p_k \to p^* \in \partial M$ and $q_k \to q^* \in \partial M$. Since $s_k, t_k \in [0, 1/k)$, we have $s_k \to 0$ and $t_k \to 0$. Continuity of $\Phi$ gives
\begin{align*}
\Phi(p^*, 0) = \lim_{k \to \infty} \Phi(p_k, s_k) = \lim_{k \to \infty} \Phi(q_k, t_k) = \Phi(q^*, 0),
\end{align*}
i.e. $p^* = q^*$ (since $\Phi(\cdot, 0) = \mathrm{id}_{\partial M}$). By the previous step, $\Phi$ is a diffeomorphism from some open neighbourhood $W_{p^*}$ of $(p^*, 0)$ onto its image. For all sufficiently large $k$, both $(p_k, s_k)$ and $(q_k, t_k)$ lie in $W_{p^*}$ (they converge to $(p^*, 0)$), yet $\Phi$ identifies them — contradicting injectivity of $\Phi|_{W_{p^*}}$.
Fix such an $\varepsilon \in (0, \varepsilon_1]$ from now on.[/step]