In my lecture notes $M$ is said to be a $n$-dimensional submanifold of $\mathbb R^N$ if for all $p\in M$ there is a homeomorphism $\psi$ from an open neighborhood $\Omega_1\subseteq\mathbb R^N$ of $p$ onto an open subset $\Omega_2$ of $\mathbb R^N$ such that with $U:=\Omega_1\cap M$ either $$\psi(U)=\Omega_2\cap(\mathbb R^n\times\{0\})\tag1$$ ($0\in\mathbb R^{N-n}$) or $$\psi(U)=\Omega_2\cap(\mathbb H^n\times\{0\})\text{ and }\psi(p)\in\partial\mathbb H^n\times\{0\}\tag2.$$ The corresponding chart is $\phi:=\pi\circ\left.\psi\right|_U$, where $\pi$ is the projection of $\mathbb R^N$ onto the first $n$-components. If $(1)$ holds, then $p$ is an interior point; if $(2)$ holds, it is a boundary point.
I've read that we can assume that if $p$ is an interior point, then $\phi_k(p)>0$. Why? By the definition above we should only know that if $(2)$ doesn't hold, then $\phi_k(p)\ne0$.
EDIT: I get if $p$ is an interior point and $\varepsilon>0$ is arbitrary, then $\tilde\Omega_2:=\Omega_2-(\psi(p)-\varepsilon)$ is open and $\tilde\psi:=\psi-(\psi(p)-\varepsilon)$ is again a diffeomorphism from $\Omega_1$ onto $\tilde\Omega_2$ with $\tilde\psi(x)=\varepsilon$. But if that's the argument, wouldn't we be able to modify $\psi$ such that it even satisfies $(2)$?