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Let $M$ be a 1-dimensional manifold with boundary, $x_0\in M$, and $\varphi:U\rightarrow V$ be a local coordinate chart such that $x_0\in U$. Take $x_0$ such that $\varphi(x_0)=0\in \partial\mathbb{H}$. Show that for any other local coordinate chart $\psi:U'\rightarrow V'$ such that $x_0\in U'$, $\psi(x_0)\in\partial\mathbb{H}$.

I understand that this makes sense in theory, but I'm not sure how to prove it. I'm not sure if it's correct to say that since $M$ is 1-dimensional, $x_0$ is the only boundary point contained in an open neighborhood of $x_0$.

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Too long for a comment.

I guess you are asking about a following (or a similar) claim, which should be easy to prove.

For possible future applications, I consider $n$-dimensional case which can be treated similarly to one-dimensional that.

Given a natural number $n$ put $\mathbb R^n_+=\{(t_1,\dots,t_n)\in\mathbb R^n: t_n\ge 0\}$. Then the boundary $\partial \mathbb R^n_+$ in $\mathbb R^n$ is a set $\{(t_1,\dots,t_n)\in\mathbb R^n: t_n=0\}$.

Claim. For a point $x_0$ of $n$-dimensional manifold $M$ the following conditions are equivalent

1) The point $x_0$ belongs to the boundary of $M$.

2) The point $x_0$ has no neighborhood homeomorphic to $\Bbb R^n$.

3) There exists a neighborhood $U$ of the point $x_0$ and a local coordinate chart $\varphi:U\rightarrow V\subset \mathbb R^n_+$ such that $\varphi(x_0)\in\partial \mathbb R^n$.

4) For each neighborhood $U$ of the point $x_0$ and each local coordinate chart $\varphi:U\rightarrow V\subset \mathbb R^n_+$ we have $\varphi(x_0)\in\partial \mathbb R^n$.

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