Manifolds with Boundary - I think I misunderstood the definition of boundary points, because I found a contradiction

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The following definitions are from the 2nd edition of John Lee's "Introduction to Smooth Manifolds", page 25.

Definition 1

We will call a chart $(U,\phi)$ an interior chart if $\phi(U)$ is an open subset of $\mathbf{R}^n$ and a boundary chart if $\phi(U)$ is an open subset of $\mathbf{H}^n$ such that $\phi(U)\cap\partial\mathbf{H}^n$ is not empty.

Definition 2

A point $p\in M$ is called an interior point of $M$ if it is in the domain of some interior chart. It is a boundary point of $M$ if it is in the domain of a boundary chart that sends $p$ to $\partial\mathbf{H}^n$.

Consider the manifold $S^1$. The inverses of \begin{align} [0,\pi)&\to S^1\\ x&\mapsto\mathrm{e}^{ix} \end{align} and \begin{align} [0,\pi)&\to S^1\\ x&\mapsto-\mathrm{e}^{ix} \end{align} are boundary charts and according to the definition above, $(-1,0)$ and $(1,0)$ are boundary points of $S^1$. But the inverse of \begin{align} (-\pi,\pi)&\to S^1\\ x&\mapsto\mathrm{e}^{ix} \end{align} is an interior chart (smoothly compatible with the other two charts) and $(1,0)$ turns out to be an interior point.

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Let $p : [0, \pi) \to S^1,~~ x \mapsto e^{ix}$ be one of the parametrizations you chose, and let $U := p([0, \pi)) \subset S^1$.

Then $p^{-1} : U \to [0, \pi)$ is not a chart because $U$ is not an open subset of $S^1$.