[Math] A product of smooth manifolds together with one smooth manifold with boundary is a smooth manifold with boundary

Question

Suppose $M_1, \dots M_k$ are smooth manifolds and $N$ is a smooth manifold with boundary. Then how do I see that $M_1 \times \dots \times M_k \times N$ is a smooth manifold with boundary, and$$\partial(M_1 \times \dots \times M_k \times N) = M_1 \times \dots \times M_k \times \partial N?$$

Answer 1

Using the simpler result that finite products of smooth (boundaryless) manifolds are smooth manifolds, your problem can be reduced to the case where $k=1$.

Now, given any point $\left(x,y\right) \in M_{1} \times N$, there exists a chart $\left(U,\phi\right)$ at $x$ in the atlas of $M_{1}$ and a chart $\left(V,\psi\right)$ at $y$ in the atlas of $N$. Now, $U \times V$ is a neighborhood of $\left(x,y\right)$ in the product topology, and the map $\phi \times \psi$ is a homeomorphism from $U\times V$ to $\mathbb{R}^{m} \times \mathbb{H}^{n}$. Noting that $\mathbb{H}^{n} := \left\{\left(x_{1},\ldots,x_{n}\right)\in\mathbb{R}^{n} | x_{n} \geq 0\right\}$, we have $\mathbb{R}^{m} \times \mathbb{H}^{n} = \mathbb{H}^{m+n}$. Use this to show that $M_{1} \times N$ is a manifold with boundary.

For the second part, given any point $\left(x,y\right)$, either $y \in \partial N$ (that is, $\psi\left(y\right) \in \partial \mathbb{H}^{n}$) or $y \notin \partial N$ (that is, $\psi\left(y\right) \notin \partial \mathbb{H}^{n}$. Now, $\partial \mathbb{H}^{n} := \left\{\left(x_{1},\ldots,x_{n}\right)\in\mathbb{R}^{n} | x_{n} = 0\right\}$. Thus, $\partial \left(\mathbb{R}^{m} \times \mathbb{H}^{n}\right) = \mathbb{R}^{m} \times \partial \mathbb{H}^{n}$. Use this to show that $\partial M_{1} \times N = M_{1} \times \partial N$.