# Show that ∂(M×N)=M × ∂N.

Let M smooth manifolds (without boundary) and N is a smooth manifold with boundary.

Could someone help me to show that $∂(M × N) = M × ∂N$?

I saw a suggestion here on the site how to do it, but I'm stalling.

ps.: sorry, I know that already asked something similar to this problem, but I'm not able to do it.

• Is $M\times{N}$ the Cartesian product of $M$ and $N$? – Michael McGovern Apr 14 '18 at 19:45
• Yes! With product topology @Michael – Mancala Apr 14 '18 at 19:46

## 2 Answers

Let be $\Bbb H^n$ the half-space $\{(x_1,\dots,x_n)\in\Bbb R^n:x_n\ge 0\}$. Each point of $M\times N$ will have a nhood basis of open sets homeomorphic to products $U\times V$ of open sets $U\subset\Bbb R^m$, $V\subset\Bbb H^n$. Now, you have two cases (what cases?)...

in general $\partial (M\times N)=(\partial M\times N)\cup (M\times\partial N)$

So what can be said?

• This is for set-theoretic boundary, not boundary of manifolds with boundary. Note that if $M$ and $N$ are both manifolds with boundary, $M\times N$ is not a manifold with boundary. – Ted Shifrin Apr 14 '18 at 21:26