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I'm trying to follow Lee's book Introduction to Smooth Manifold in details. In chapter 2, Exercise 2.19 asks the reader to prove that if $F:M \to N$ is a smooth diffeomorphism between manifolds with boundary, then $F (\partial M) = \partial N$.

It suffices to show that $F(\partial M) \subseteq \partial N$. I thought about mimicking the proof given in the book that, in a smooth manifold, a point cannot be a boundary point and be in the domain of an interior chart at the same time (Theorem 1.46).

However, I noticed that the argument would rely on knowing forehand that the two manifolds have the same dimension, that is to say their charts map to open subsets of the same $\mathbb{H}^n$, but it is not clear for me at all that I can assume that.

For completeness, a sketch of my proof would go like this:

Given $p \in \partial M$, by definition of $F$ smooth there exists $(U, \phi)$ and $(V, \psi)$ charts such that $F(U) \subseteq V$ and $\hat{F} = \psi \circ F \circ \phi^{-1}$ is smooth. Since $p$ is a boundary point, this means there exists an open set $U_x$ containing $x = \phi(p)$ and $F_x: U_x \to \mathbb{R}^n$ such that $F_x = F$ in $\phi (U) \subset \mathbb{H}^m$ (and not necessarily $m = n$).

Now let us suppose that $F(p) \in \mathrm{int} \, M$. Then, since $F$ is a diffeomorphism (and thus an open mapping), by restricting if necessary we may suppose that $F(U) = V$. If $G = F^{-1}: N \to M$, it is known that $\hat{G} = \phi \circ G \circ \psi^{-1}: \psi (V) \to \phi (U)$ is smooth (in the sense of $\mathbb{R}^n$, because $F(p)$ is an interior point).

Now, we may find a ball $B$ around $\psi (F(p)) = \hat{F} (x)$ such that $\hat{G}(B) \subseteq \phi(U) \cap U_x$. Thus, we could write $$\mathrm{Id}_B = F_x \circ \hat{G} $$ as a composition of smooth functions in the regular sense (defined on open subsets of $\mathbb{R}^n$ and $\mathbb{R}^m$, respectively).

The conclusion would follow by noticing that pointwise $DF_x \circ D\hat{G} = \mathrm{Id}$ and thus $D \hat{G}$ is non-singular, but this would only be true if $m = n$ - otherwise I can only conclude it has a left-inverse.

I wonder if:

  1. there's some independent more basic argument that allows me suppose that $m = n$ right from the start; or
  2. the actual proof of this fact goes in a completely different direction, and implies somewhere along the way that $m = n$.

An answer for 1. or a tip for 2. would be appreciated.

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  • $\begingroup$ Invariance of the domain is a difficult topic in topology, but in the differentiable context It follows from the inverse function theorem (or, equivalently, the implicit function theorem), that a diffeomorphism preserves the dimension. See the third remark on this page: en.wikipedia.org/wiki/Diffeomorphism $\endgroup$ Commented Sep 30, 2017 at 20:12
  • $\begingroup$ I know that a diffeomorphism preserves dimension when it exists between two manifolds without boundary. This is easy because all charts are interior charts, and thus the expression of $F$ in coordinates yields a diffeomorphism between open sets of $\mathbb{R}^m$ and $\mathbb{R}^n$, and thus by the inverse function theorem the dimensions are the same. Here the question is slightly more subtle. $\endgroup$ Commented Sep 30, 2017 at 20:17
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    $\begingroup$ No, it's not more subtle. You determine the dimension away from boundary points. (If you like, start with a point of $N-\partial N$.) Assuming connectedness, it stays constant. $\endgroup$ Commented Sep 30, 2017 at 22:39
  • $\begingroup$ Ok, I thought about that, but then what ensures that the inverse image of a neighborhood of such point, say $F^{-1} (U)$, will be contained in $\mathrm{int} \, M$? The argument relies in finding interior charts around both $p$ and $F(p)$ as well. $\endgroup$ Commented Sep 30, 2017 at 23:40
  • $\begingroup$ @TedShifrin Could you explain the connectedness part? $\endgroup$ Commented Oct 21, 2017 at 22:53

1 Answer 1

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I thought exactly the same and I think I've found a satisfactory solution. Basically your po\text{int } 1. is right, but the argument is indeed quite subtle. I'm gonna divide this proof in 4 parts:

Part 1: If $M$ is a nonempty manifold with boundary then $\text{int }(M)\neq \emptyset$.

Proof: I'll leave this one as homework :) because it's not subtle.

Part 2: If $M\to N$ are nonempty diffeomorphic smooth manifolds with boundary with dimensions $m$ and $n$ then $m=n$.

Proof: This part is subtle, so I'll write the details carefully. Let $f:M\to N$ be a diffeomorphism. Then $f|\text{int } M:\text{int } M\to f(\text{int } M)$ is a diffeomorphism where $\text{int } M$ is a nonempty smooth manifold (this is because of part 1.) and $f(\text{int } M)$ is a nonempty smooth manifold with boundary. Let's rename $f|\text{int } M:\text{int } M\to f(\text{int } M)$ as $f_1:M_1\to N_1$. Using the same trick $f_1|f_1^{-1}(\text{int } N_1):f_1^{-1}(\text{int } N_1)\to \text{int } N_1$ is a diffeomorphism between nonempty smooth manifolds. By Proposition 2.17 (Diffeomorphism Invariance of Dimension) we have that $f_1^{-1}(\text{int } N_1)$ and $\text{int } N_1$ have the same dimension, so the same is true for $M$ and $N$.

Part 3: If $M$ is a smooth manifold with boundary, $(U,\phi)$ is a smooth chart for $M$ and $f:\phi(U)\to V$ is a diffeomorphism between open subsets of $\mathbb{H}^n$ or $\mathbb{R}^n$ then $(U,f\circ \phi)$ is a smooth chart for $M$.

Proof: I'll also leave this one for homework because it's also not subtle.

Part 4: (Theorem 2.18, Diffeomorphism Invariance of the Boundary): Suppose $M$ and $N$ are smooth manifolds with boundary and $F:M\to N$ is a diffeomorphism. Then $F(\partial M)=\partial N$.

Proof: Let $p\in \partial M$. This means there is a smooth chart $(U,\phi)$ for $M$ such that $p\in \partial U$, $\phi(U)\subseteq \mathbb{H}^n$ and $\phi(p)\in \partial \mathbb{H}^n$. By restricting $U$ (this is essentially what Lee does in his proof of Theorem 2.17, Diffeomorphism Invariance of Dimension) we may find a chart $(F(U),\psi)$ for $N$. Then $\psi\circ F\circ \phi^{-1}:\phi(U)\to \psi(F(U))$ is a diffeomorphism between open subsets of the same $\mathbb{H}^n$ or $\mathbb{R}^n$ because of part 2. Define $\hat{F}=\psi\circ F\circ \phi^{-1}$, then by part 3. $(U,\hat{F}\circ \phi)$ is a smooth chart for $M$. By Theorem 1.46 (Smooth Invariance of the Boundary) $\hat{F}(\phi(U))\subseteq \mathbb{H}^n$ and $\hat{F}(\phi(p))\in \partial \mathbb{H}^n$, this is the same as saying $\psi(F(U))\subseteq \mathbb{H}^n$ and $\psi(F(p))\in \partial \mathbb{H}^n$, which implies $F(p)\in \partial N$. So $F(\partial M)\subseteq \partial N$ and we are done.

This is the kind of details that always drives me crazy, but hopefully now I can fill most of the gaps. When I ask people about this kind of details I sometimes get offtopic answers by people that think this is a trivial detail. For example, I don't see how connectedness plays a role here, care to explain @Ted Shifrin?

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  • $\begingroup$ A little comment. I think we don't have to do the restriction if we proceed in this way: Let $p\in \partial M$. By the smoothness of $F$, we can find a smooth chart $(U,\phi)$ of $M$ and a smooth chart $(V,\psi)$ of $N$ s.t. $p\in U, F(p)\in V$ and $F(U)\subseteq V$. Then by Theorem 1.46, we know that $\phi(U)\subseteq \mathbb{H}^n$ and $\phi(p)\in \partial \mathbb{H}^n$. And now we can define $\hat{F}$ on $\phi(U)$ without restriction. $\endgroup$ Commented Feb 20, 2021 at 19:37
  • $\begingroup$ And there might be a slight typo which doesn't really matter. It's in the second line of the proof of part 4. I think it might should be $\color{darkred}{\phi(p)\in \partial \mathbb{H}^n}$. Anyway, thanks for the detailed answer. $\endgroup$ Commented Feb 20, 2021 at 19:46
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    $\begingroup$ I have in mind something that doesn't use Theorem 1.46. let $ p \in \partial M $ then there is $(U,φ)$ smooth chart s.t. $p \in U$, wouldn't $(V,ψ)=(F(U) , φ \circ F^{-1})$ be a smooth chart of $F(p)$ s.t. $ψ(V) \subset \mathbb{H}^n$ and $ψ(F(p)) \in \partial \mathbb{H}^n $, hence boundary point? $\endgroup$ Commented Sep 24, 2021 at 15:43
  • $\begingroup$ @Nash-iOS yes but I wanted to get $m = n$ along the way. $\endgroup$ Commented Apr 14, 2023 at 1:45

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