A textbook (not available in english - and actually more of a collection of notes than a "textbook" textbook) works with the following definition:
Definition: Let $M$ be a smooth $n$-manifold. A $\mathcal{D}\subset M$ subset is a regular domain is every $p\in\text{cl}(\mathcal{D})$ point is either in the interior of $\mathcal{D}$, or there exists such a chart $(U,\psi)$ containing $p$, for which $U\cap\text{cl}(\mathcal{D})=\psi^{-1}(B^n\cap\{\mathbf{x}:x_1\le0\})$ and $p=\psi^{-1}(\mathbf{0})$.
The boundary of a regular domain is denoted as $\partial\mathcal{D}$ and is the set of points $p\in M$ for which exist such charts presented before.
Then we have a theorem:
Theorem: If $M$ is an orientable smooth manifold and $\mathcal{D}\subset M$ is a regular domain, then $\partial\mathcal{D}$ is an orientable smooth submanifold.
The proof first proves that $\partial\mathcal{D}$ is a smooth manifold. For $p\in\partial\mathcal{D}$ we let $V=U\cap\partial\mathcal{D}$, where $U$ is the coordinate domain of the "special" chart given in the definition, and we define a map $r=\psi^{-1}|_{x_1=0}$.
(Note that this text defines the chart maps as maps from $U$ to the unit ball $B^n$)
Then the text says that "by definition", $r(B^{n-1})=V$. This is a step I don't understand. It is clear that $B^n\cap\{\mathbf{x}:x_1=0\}=B^{n-1}$, so this is true if $\psi^{-1}(\{\mathbf{x}:x_1=0\})=\partial D$. But based on the definitions so far, this is absolutely not trivial to me.
From the above given definition of the boundary $\partial\mathcal{D}$, how can I see that $\psi^{-1}$ maps the $x_1=0$ hyperplane to the boundary?