# Diffeomorphism preserves interior, exterior and boundary

Let $g : A \to B$ be a diffeomorphism of open sets $A, B \subseteq \mathbb{R}^n$. That is, $g$ is a bijection and both $g,g^{-1}$ are of class $C^r$ for some $r \geq 1$.

Let $S$ be a subset of $A$, and let $g(S) = T$. Is it true that $g(\text{Int }S) = \text{Int }T$, $g(\text{Ext }S \cap A) = \text{Ext }T \cap B$ and $g(\text{Bd }S) = \text{Bd }T$?

[Here we define $\text{Int }S$ as the union of all open sets contained in $S$, $\text{Ext }S$ as the union of all open sets disjoint from $S$, $\text{Bd }S$ as the set of points whose every neighborhood contain points from $S$ and $\mathbb{R}^n-S$. These 3 sets are disjoint and their union is $\mathbb{R}^n$.]

I think it's true, and believe I have a proof:

Indeed $g^{-1}$ is continuous, and since $\text{Int }S$ is open, $g(\text{Int }S)$ is open, so $g(\text{Int }S) \subseteq \text{Int }T$. Similarly $g$ is continuous, so $g^{-1}(\text{Int }T) \subseteq \text{Int }S$. It follows that $g(\text{Int }S) = \text{Int }T$.

Similarly $g(\text{Ext }S \cap A)$ is open, and $\text{Ext }S \cap A \subseteq A - S$, so that $g(\text{Ext }S\cap A) \subseteq \text{Ext }T \cap B$. Then $g(\text{Ext }S\cap A) = \text{Ext }T \cap B$.

Finally $g(A) = B$ so that $g(\text{Int }S \cup \text{Bd }S \cup (\text{Ext }S\cap A)) = \text{Int }T \cup \text{Bd }T \cup (\text{Ext }T\cap B)$. Since $g$ is a bijection, it follows that $g(\text{Bd }S) = \text{Bd }T$.

But I'm having doubts because Munkres' Analysis on Manifolds requires $S$ to be compact (Theorem 18.2, page 154). Is the statement true, or is there a flaw with the above proof?

• I assume that by Bd you mean boundary rather than the set of limit points. – user87690 Aug 23 '16 at 14:56
• @user87690 Yes, sorry about the mistake! I have edited my question. – Wakaka Aug 23 '16 at 15:08

It matters whether you consider $\def\Int{\operatorname{Int}}\Int$, $\def\Ext{\operatorname{Ext}}\Ext$, and $\def\Bd{\operatorname{Bd}}\Bd$ with respect to $A$, $B$ or with respect to the ambient space $ℝ^n$.
In the second case you really need to translate it to the first case. In general it holds that $Ext_A(S) = \Ext_{ℝ^n}(S) ∩ A$, and since the sets $A$, $B$ are open, we have also $\Int_A(S) = \Int_{ℝ^n}(S)$. Finally, we have $\Bd_A(S) = A \setminus (\Int_A(S) ∪ \Ext_A(S))$ $= A \setminus (\Int_{ℝ^n}(S) ∪ \Ext_{ℝ^n}(S) ∩ A)$ $= (ℝ^n \setminus (\Int_{ℝ^n}(S) ∪ \Ext_{ℝ^n}(S))) ∩ A$ $= \Bd_{ℝ^n}(S) ∩ A$. And the same for $B$ so it seems that it holds.
• Thank you for this detailed answer! I was thinking of with respect to $A, B$ but it's great that it works in both cases! – Wakaka Aug 23 '16 at 15:10