# Diffeomorphisms between smooth manifolds with boundary

In the professor Lee's introduction to smooth manifolds 2nd edition, the notion of diffeomorphism is defined for smooth manifolds with or without boundary. However, I saw some propositions that seems also to be true for smooth manifolds with boundary where professor Lee made the hypothesis only for smooth manifolds.

Example 2.14(b)on page 38: If $M$ is any smooth manifold and $(U,\varphi)$ is a smooth coordinate chart on $M$; then $\varphi:U\to \varphi(U)$ is a diffeomorphism.

Theorem 2.17 (Diffeomorphism Invariance of Dimension) on page 39: A nonempty smooth manifold of dimension $m$ cannot be diffeomorphic to an $n$-dimensional smooth manifold unless $m=n$.

Usually if some proposition is also true for the smooth manifolds with boundary, he will state clearly in the hypothesis. But this time he doesn't, so I wonder if there is any subtle reasons making the two propositions above no-longer true for smooth manifolds with boundary(I feel like a simple modification of the proof works also for the smooth manifolds with boundary), or professor Lee simply forgot to say "smooth manifolds with or without boundary"?

You are right, it is equally true. Look at the chapter of tangent space, he defines the space tangente for points on boundary and it has the same dimension as $M$. So using that it is a diffeo it would implie that $df$ is a isomorphism, what is possible if, and only if, $m=n$.

• Do you think that $\varphi:U\to \varphi(U)$ is also a diffeomorphism when $M$ has boundary? Why? Commented Mar 26, 2016 at 3:25
• take a chart for $\partial M$ as follows: $(U,\phi)$ chart for $M$. Then $(\partial M \cap U, \phi_{\partial M})$ is a chart for $\partial M.$ Take $p \in \partial M.$ Then $\phi_{\partial M}(U\cap \partial M) = \partial \mathbb{H}^n = \mathbb{R}^{n-1}.$ Using that $\phi : U \to \phi(U)$ is a diffeomorphism we get the result. Commented Mar 26, 2016 at 3:34
• Notice that $U\cap \partial M$ is open on $\partial M$. Then $\phi_{\partial M}$ is a homeomorphism. But it is also a diffeomorfism for the same reason of the Lee example... Commented Mar 26, 2016 at 3:39
• Thank you and +1, I need to think maybe for a couple of days before I take you solution~~~ Commented Mar 26, 2016 at 3:41
• there is no reason to accept my solution now or in any moment. I suggest you to look at the chapters 3 and 5.... Commented Mar 26, 2016 at 3:43