Product manifolds

Question

I have a question on the product of two manifolds. I have $M, N$ two real manifolds (with a smooth differentiable structure), with $\partial M=0$. I have showed that $M\times N$ has a natural induced real smooth differentiable structure, by costructing an atlas for $M\times N$ using an atlas of $M$ and an atlas of $N$. But I have the following questions.

• Why I have to assume that $\partial M=0$?
• What goes wrong if both manifolds have a boundary? For example, let's take $[0,1) \times [0,1)$.

My professor says that there no exist a local homeomorphism and so this is not a manifolds with boundary. But how I can show it?

Thank you so much.

Answer 1

If you understand why $M \times N$ is naturally a smooth manifold when $M$ and $N$ are manifolds without boundary, then the only thing you need to worry about is the specific boundary issue.

Can you properly show that if $x \in M$ where $M$ has no boundary and $y$ is a boundary point of $N$, then $(x,y)$ is a boundary point of $M \times N$ (in other words has a neighborhood diffeomorphic to a Euclidean half-space) ?

If so, you should see why $(x,y)$ does not have a "suitable" neighborhood (diffeomorphic to a Euclidean space or half-space) when $x$ and $y$ are both boundary points. In your example, see what a neighborhood of the point $(0,0)$ looks like (it's a "corner").

Edit: Okay, I'll edit to give some more details (see comments). Say $x\in \partial M$ and $y \in \partial N$. Using charts, $(x,y)$ has a neighborhood diffeomorphic to a Euclidean quarter-space, i.e. to $Q = \{x = (x_1, \dots, x_p) \in \mathbb{R}^p,~x_{p-1}\geqslant 0 \textrm{ and } x_{p}\geqslant 0\}$. The question is: why is this not acceptable?

• It's pretty clear that $Q$ is not diffeomorphic to a Euclidean space.
• Why is $Q$ not diffeomorphic to a Euclidean half-space? I think it's also pretty clear intuitively. Let me suggest a proof: Suppose there is a diffeomorphism $\varphi : Q \to H$ where $H = \{x\in\mathbb{R}^p, ~ x_p \geqslant 0\}$. By definition of the differentiability of such a map, $\varphi$ extends to a diffenrentiable map $\tilde{\varphi} : U \to V$ where $U$ and $V$ are neighborhood of the origin in $\mathbb{R}^p$. It should be clear that $\varphi$ sends $\partial Q$ to $\partial H$. In particular it is not hard to derive that the linear map $D\tilde{\varphi}(0) : \mathbb{R}^p \to \mathbb{R}^p$ must send both hyperplanes $\{x_{p-1}=0\}$ and $\{x_p = 0\}$ to the hyperplane $\{x_p = 0\}$. Since $D\tilde{\varphi}(0)$ is invertible, this is not possible.