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Let $M$ be a smooth connected manifold with boundary. Is it true that the interior of $M$ is also connected? (As usual I am assuming $M$ is second-countable and Hausdorff).

I am trying to rule out something like two "tangent disks" (or circles) where the (topological) interior is obviously not connected. This case is not a counter-example since the union of two tangent disks (or circles) is not a manifold with boundary (the point of tangency is pathological).

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Assume you have two connected components $M_1$ and $M_2$ of $int M$. The boundary of each is a subset of the boundary of $M$. Since $M$ is connected, the closures of $M_1$ and $M_2$ intersect at some boundary point $p$. Let $X:\mathbb{H}^n\to U\subset M$ be a coordinate chart around $p$, where $\mathbb{H}^n$ is the upper half of $\mathbb{R}^n$ (including boundary). Then $X^{-1}(M_i)$ is an open set in $\mathbb{R}^n$, and whose boundary is a subset of the boundary of $\mathbb{H}^n$. But then $X^{-1}(M_i)$ must be the whole $\mathbb{H}^n\setminus \partial\mathbb{H}^n$, which contradict the disjointness of $M_1$ and $M_2$.

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    $\begingroup$ To be clear, when you refer to the boundary of $M_1$ and $M_2$ you mean their topological boundary as subsets of $M$, rather than their manifold boundary. $\endgroup$ – Eric Wofsey Jan 21 '17 at 21:58
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This is true because a manifold is homotopically equivalent to its interior (this follows from the collar neighborhood theorem)

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  • $\begingroup$ Thanks! I wonder if there is a more elementary approach to this question ,without invoking this corollary, which is a little non-trivial. (Although, we do not really need the full corollary, but only a part of the proof, that there is a continuous surjective map $M \to \text{int} M$ which is easier). I will still wait a while before accepting this answer, to see if anything simpler comes. Thanks again. $\endgroup$ – Asaf Shachar Jan 21 '17 at 14:30

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