# Does a diffeomorphism always map boundaries to boundaries?

If $F:\overline{M} \to \overline{M'}$ is a diffeomorphism between two open bounded domains $M$ and $M'$ in $\mathbb{R}^n$, what conditions do I need on $F$ or $M$ or $M'$ to make sure that $F(\partial M) = \partial M'$?

Or is this simply something that I must specify? Thanks

• open domaine don't have boundries. Jan 1, 2014 at 15:35
• ?? $(0,1)$ has boundary $\{0,1\}.$ Jan 1, 2014 at 15:36
• your diffeomorphism are not defined on the boundary. Do you want to find an extension with such a property? Jan 1, 2014 at 15:40
• @EmanuelePaolini Oops. I meant it's defined on closure of $M$. Let me edit. Jan 1, 2014 at 15:42
• I think that a diffeomorphism between manifolds with boundary has the requested property by definition. Jan 1, 2014 at 15:46

Every homeomorphism between two bounded open sets $M$ and $M'$ in $\mathbb R^n$ induces a boundary correspondence in the following sense:

• $x\in \partial M$ corresponds to the cluster set $$\{y\in \mathbb R^n: \exists (x_n) \to x \text{ such that } F(x_n)\to y\}$$

The following properties hold:

1. Each cluster set is a nonempty closed subset of $\partial M'$
2. The union of cluster sets over $x\in\partial M$ is equal to $\partial M'$.

Proofs:

1. "Nonempty and closed" are easy to see. Suppose $y$ is an element of a cluster set and $y\notin \partial M'$. Then $y\in M'$. Let $U$ be a neighborhood of $y$ such that $\overline{U}\subset M'$. The set $F^{-1}(\overline{U})$ is a compact subset of $M$. Any sequence $x_n$ in $M$ that approaches the boundary of $M$ will eventually leave $F^{-1}(\overline{U})$; thus, $F(x_n)$ will eventually be outside of $U$. Contradiction.

2. Take $y\in\partial M'$, pick a sequence $y_n\to y$ in $M'$, then choose a convergent subsequence of $F^{-1}(y_n)$.

So far, no assumptions on $F$ were needed besides it being a homeomorphism. If you want to have boundary correspondence as a map, point-goes-to-a-point, then add the assumption that for every $x\in\partial M$ the limit $\lim_{z\to x}F(x)$ exists. This limit, which can also be denoted $F(x)$, satisfies $F(\partial M ) = \partial M'$ by the above.