Is the interior of a smooth manifold with boundary connected?

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).

Solution 1:

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 $M_2$.

Solution 2:

This is true because a manifold is homotopically equivalent to its interior (this follows from the collar neighborhood theorem)