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I am reading Lee's Introduction to Topological Manifolds and attempting to prove the following proposition:

Proposition 2.58. If $M$ is an n-dimensional manifold with a boundary, then $\textrm{Int} M$ is an open subset of $M,$ which is itself an n-dimensional manifold without boundary.

Where manifolds with a boundary are defined in terms of charts mapped to open sets in $\mathbb{H}^n = \mathbb{R}^{n-1} \times [0, \infty).$ I need to prove this without using the invariance of the boundary (i.e. that the manifold boundary and interior are disjoint.

My attempt so far involved the construction of charts $(U_i, \varphi_i)$ that cover $M$ and the identification of points mapped to $\partial\mathbb{H}^n$ with $\partial M;$ however, I can't use $\textrm{Int} M = M \setminus \partial M$ without invoking the invariance of boundary. I would prefer hints or partial answers suggesting how I should proceed to full proofs.

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  • $\begingroup$ Why was the tag changed to differential geometry? I'm somewhat new to the study of manifolds but I believe differential geometry focuses on the study of smooth or Riemannian manifolds. My question is about topological manifolds so isn't algebraic topology more fitting? $\endgroup$ – Fady Nakhla Jul 16 at 20:19
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It's hard to give a hint to this problem without giving the whole thing away. By definition, $\text{Int} M$ is the subset of all $x \in M$ for which there exists a chart $(U_i,\phi_i)$ such that $x \in U_i$ and such that $\phi_i(U_i) \subset \mathbb R^{n-1} \times (0,\infty)$. This existence property is clearly also true for every $y \in U_i$, using the exact same chart $(U_i,\phi_i)$, and therefore $U_i \subset \text{Int}(M)$.

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  • $\begingroup$ Oh. Thereby showing that Int M is open since every point has an open neighborhood, and locally Euclidean since each of those neighborhoods has an image contained in the open upper half space. Correct? $\endgroup$ – Fady Nakhla Jul 12 at 20:17
  • $\begingroup$ That's correct. $\endgroup$ – Lee Mosher Jul 12 at 20:18
  • $\begingroup$ Great. Thank you for your help Lee $\endgroup$ – Fady Nakhla Jul 12 at 20:22

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