How to show diffeomorphisms map boundaries to boundaries? Suppose I have a diffeomorphism $f: U \rightarrow V$ where $U$ is a simply connected open subset of $\mathbb{R}^n$. Consider any simply connected open or closed set $B$ such that the closure of $B$ is contained in $U$. 
Let $f(B) = D$. Does it then always follows that $f(\partial B) = \partial D$?
How can I prove this? It seems like it should be true... Any comments and suggestions are appreciated.   
 A: You actually just need a homeomorphism for this, and $B$ and $D$ can be any two closed subsets of some topological spaces (I assume that they are closed, since $f(\overline{B}) = \overline{D}$). You can prove the converse, i.e., that elements of the interior of $B$ are mapped to the interior of $D$ and viceversa. Indeed, an element $x$ of the interior of $D$ is contained in an open subset $W$ of $D$. Just map this neighborhood back: $f^{-1}(W)$ is an open neighborhood of the preimage $f^{-1}(x)$. The same works in the other direction.
To prove that in general $f(\overline{B}) = \overline{D}$ if $f$ is a homeomorphism and $f(B)=D$: the inclusion $\subseteq$ holds because $f$ is continuous, and the other inclusion because $f^{-1}$ is continuous.
A: Well, 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 $$\{p\in \mathbb R^n: \exists \,(x_n) \to x \text{ such that } F(x_n)\to p\}$$
