Homeomorphisms of the unit ball in $\mathbb{R}^n$

Question

Let $\mathbb{D}^{n}= \{ x \in \mathbb{R}^n : |x| \leq 1 \}$ be the unit ball in $\mathbb{R}^n$, and let $\phi : \mathbb{D}^{n} \to \mathbb{D}^{n}$ be a homeomorphism. Suppose $\mathbb{D}^{n} = E_1 \cup E_2 \cup \Gamma$, where $E_1$ and $E_2$ are two open sets, and $\Gamma = \partial E_1 = \partial E_2$. It seems that $\phi (\partial \mathbb{D}^{n})= \partial \mathbb{D}^{n}$ by (among other things) using Brouwer fixed point theorem (as Henno Brandsma mentioned).

I want to know whether it is true that $$ \mathbb{D}^{n} = \phi (E_1) \cup \phi (E_2),\text{ and } \phi (\Gamma)= \partial (\phi (E_1)) = \partial (\phi (E_2)) ? $$

Is it necessary to impose some additional conditions on $E_i$s or $\phi$ (for example assume it is also a diffeomorphism?) What about the case $\mathbb{D}^{n} = \phi (E_1) \cdots \cup \phi (E_m)?$. Roughly speaking, I would like to know if the ball is divided into $m$ partitions, is its homeomorphic image is also divided into $m$ parts in a way that preserve boundaries of the partitions?

I am very beginner in the subject and this may be very elementary question. Any reference for study is warmly welcomed.

Answer 1

Any continuous homeomorphism of $\Bbb D^{n-1}$ obeys $f[\partial \Bbb D^{n-1}] = \partial \Bbb D^{n-1}$. This is a standard fact from dimension theory, e.g. You need Brouwer's fixed point theorem to show it.

Answer 2

Regarding $\partial E$, I'll assume you mean the topological frontier $\partial E = \text{closure}(E) \cap \text{closure}(E^c)$, where $E^c$ denotes the complement of a subset; no other notion of boundary makes sense in this general context.

The properties you ask about are true, and can be proved by direct application of basic definitions of set theory and topology.

First, the set theoretic union and intersection operators are invariant under any bijection. For example if $\phi$ is any bijection of a set and if $A,B$ are subsets then $\phi(A \cup B) = \phi(A) \cup \phi(B)$. Since every homeomorphism is a bijection, we have $\phi(E_1) \cup \phi(E_2) = \phi(E_1 \cup E_2) = \phi(\mathbb D^n) = \mathbb D^n$, so this answers your first question affirmatively.

Your second question can also be answered affirmatively, in a similar fashion but tying together a few more steps: the complement operator $E^c$ is invariant under any bijection so $f(E^c) = f(E)^c$; and the closure operator is invariant under any homeomorphism so $\text{closure}(f(E)) = f(\text{closure}(E))$. I'm sure you tie these together (with invariance of intersection under bijection) to conclude that $f(\partial E) = \partial (f(E))$ for any subset $E$.