# The interior of a manifold with a boundary is a manifold

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.

• 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? – Fady Nakhla Jul 16 at 20:19

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 $$U_i \subset \text{Int}(M)$$.