Assume I have a $C^{\infty}$ domain $\Omega$ in $\mathbb{R}^n$ with boundary $\partial \Omega$. Let $x \in \partial \Omega$. There exists a smooth map $\psi$ and a neighborhood $V_x \subset \mathbb{R}^n$ with $x \in V_x$ such that $$ \psi(V_x \cap \Omega) = B(0,1) \cap \{p \in \mathbb{R}^n, p_n > 0\}$$ and $$ \psi(V_x \cap \partial \Omega) = B(0,1) \cap \{p \in \mathbb{R}^n, p_n = 0\}.$$ My question is the following: can I choose a particular $\psi$ so that $\psi(tx + (1-t)y) = t \psi(x) + (1-t) \psi(y)$ for all $x, y \in V_x$, $t \in (0,1)$ ? If so, why ? If not, why not ?
I am trying to prove that if $(z-y) \cdot n_y$, where $n_y$ is the unit normal vector to $\partial \Omega$ at $y$ and $z \in \Omega$, for $t$ small enough, $\epsilon$ small enough $tz + (1-t)y' \in \Omega$ for all $y' \in B(y,\epsilon) \cap \partial \Omega$ and this choice of particular chart is the only piece I'm missing.