Given the following definition of manifold with boundary :
$S \subset \mathbb{R}^{N}, N \in \mathbb{N}$ is a $n-$differential manifold with boudary if : $n \in \mathbb{N}, \forall \hspace{0.1cm} p \in S, \hspace{0.1cm}\exists \hspace{0.1cm} V $open neighborhood of $p$ in $\mathbb{R}^{N}$ and $U$ open in $\mathbb{R}^n$, with $\varphi : U \longmapsto V \hspace{0.1cm} C^\infty$ with differential of rnk $n$ such that $\varphi(U \cap \{x \in \mathbb{R}^n : x_n \geq 0\}) = V \cap S$ is a homeomorphism. In this setting we say that $\partial S \ni p \iff \varphi(U \cap \{x_n = 0\}) \ni p$
With definition I'd like to prove the well-definess of the boundary of a manifold, without using arguments which involve algebraic topology.
To be more accurate, it's not clear to me how the following follows from the definition :
If $\varphi_1,\varphi_2$ are parametrization with $ p \in \varphi_1(U \cap \{x \in \mathbb{R}^n : x_n \geq 0\}), p \in \varphi_2(U \cap \{x \in \mathbb{R}^n : x_n \geq 0\})$ we have that $$p \in \varphi_1(U_1 \cap \{x_n = 0\}) \iff p \in \varphi_2(U_2 \cap \{x_n = 0\})$$
Any help hint or reference of the proof will be appreciated.
To prove the consistency of the definition for a point $p$ to be in the boundary, it is equivalent to prove the consistency of the converse, i.e the definition for a point to be in the interior of $M$.
Let $p \in M$ and suppose there exists a chart $\varphi : U \to \mathbb{R}^n_+=\{(x_1,\ldots,x_n)\in \mathbb{R}^n~|~x_n \geqslant 0\}$ such that $\varphi(p)_n >0$. Suppose $\psi : V \to \mathbb{R}^n_+$ is another chart with $p \in V$. Let us show that $\psi(p)_n>0$.
By the very definition of charts, $\psi\circ \varphi^{-1} : \varphi(U\cap V) \to \psi (U\cap V)$ is a diffeomorphism. Now, as $\varphi(p)_n>0$, there exists an open subset around $\varphi(p)$ is $\varphi(U\cap V)$ is an open subset $W\subset \varphi(U\cap V)$ such that for all $x \in W$, $x_n>0$ (take an open ball with small enough radius). $W$ is then an open subset of $\mathbb{R}^n_+$ such that every $x$ in it has $x_n >0$: it is an open subset of $\mathbb{R}^n$.
Now, extand the codomain of $\psi\circ \varphi^{-1}$ to be all of $\mathbb{R}^n$. We have a map $\psi \circ \varphi^{-1} : W \to \mathbb{R}^n$ which defined on an open subset of $\mathbb{R}^n$ and such that it is a diffeomorphism onto its image: its image is thus open because of the inverse function theorem. Therefore, it is an open subset of $\mathbb{R}^n$, contained in $\mathbb{R}^n_+$. Consequently, all $x$ in $W$ satisfy $\psi\circ\varphi^{-1}(x)_n >0$. That is, all $q$ in $\varphi^{-1}(W)$ satisfies $\psi(q)_n>0$. Thus, $\psi(p)_n >0$. We have shown that if there exists a chart such that $\varphi(p)_n >0$, then for all chart $\psi$ for which $\psi(p)$ is defined, then $\psi(p)_n>0$.
Conversly, if there exsits a chart for which $\varphi(p)_n = 0$, then for all chart $\psi$ such that $\psi(p)$ is defined, $\psi(p)_n =0$.
Comment: this is an exact reformulation of this but we do not use here the invariance domain theorem because in this context, the fact that a diffeomorphism is open is a straightforward consequence of the inverse function theorem, that is an easier result.
Edit It has been asked several times in comments to explain why transition maps of a smooth manifold are diffeomorphisms. This is directly following from the definitions of:
Here is the reason. A diffeomorphism $f : U \to V$ between open subset of $\mathbb{R}^n$ is a smooth homeomorphism with smooth inverse.
The transitions maps are $\varphi_{\alpha\beta} = \varphi_{\beta}\circ {\varphi_{\alpha}}^{-1} : \varphi_{\alpha}(U_{\alpha}\cap U_{\beta}) \to \varphi_{\beta}\left(U_{\alpha}\cap U_{\beta} \right)$ and by definition, are smooth homeomorphisms. Now, it is straightforward to show that ${\varphi_{\alpha\beta}}^{-1} = \varphi_{\beta\alpha}$ is also a transition function and thus, is a smooth homeomorphism by definition.
Hence, $\varphi_{\alpha\beta}$ is a smooth homeomorphism with smooth inverse: it is a diffeomorphism.