homeomorphisms mapping interiors to interiors and boundaries to boundaries

Question

Why do homeomorphisms map interiors to interiors and boundaries to boundaries? I cannot find a good proof for it that does not involve algebraic topology. I only need it for spaces in $\mathbb{R}^n.$

Answer 1

Here are elementary proofs (given in mathonline.wikidot): Let $X,Y$ be topological spaces, $f:X\to Y$ an homeomorphism and $A\subset X$ a subspace.

(1) $f(A^\circ)=f(A)^\circ$.

$\subset)$ Let $x\in f(A^\circ)$. Then $f^{-1}(x)\in A^\circ$ so there exists an open neighbourhood $U\subset X$ of $f^{-1}(x)$ such that $f^{-1}(x)\in U\subset A$. Hence we have $x\in f(U)\subset f(A)$. Since $f$ is a homeomorphism $f(U)$ is an open neighbourhood of $x$ (and it is contained in $f(A)$). Hence $x\in f(A)^\circ$.

$\supset)$ Now let $x\in f(A)^\circ$. Then there exists an open neighbourhood $V\subset Y$ of $x$ such that $x\in V\subset f(A)$ and so $f^{-1}(x)\in f^{-1}(V)\subset A$. Since $f$ is a homeomorphism we have that $f^{-1}(V)$ is open in $X$. Therefore $f^{-1}(V)$ is an open neighbourhood of $f^{-1}(x)$ contained in $A$ and therefore $f^{-1}(x)\in A^\circ$ so $x\in f(A^\circ)$.

(2) $f(\partial A)=\partial f(A)$ (recall that, in general, $\partial B=\overline{B}\cap\overline{X\setminus B}=\overline{B}\cap \overline{B^c}$).

$\subset)$ Let $x\in\partial A$. Then $f(x)\in f(\partial A)$. Let $V$ be any open neighbourhood of $f(x)$ in $Y$ so $f^{-1}(V)$ is open in $X$ and contains $x$. So there exists $a,b\in X$ with $a\in A\cap f^{-1}(V)$ and $b\in A^c \cap f^{-1}(V)$ where $a,b\neq x$ since $f$ is bijective. Therefore $f(a)\in f(A)\cap U$ and $f(b)\in (f(A))^c\cap U$ where $f(a),f(b)\neq f(x)$. So $f(x)\in \partial(f(A))$ which shows that $f(\partial A)\subset \partial(f(A))$.

$\supset)$ Now let $x\in \partial f(A)$ and let $V\subset Y$ be an open neighbourhood of $x$. Then there exists $a,b\in Y$ with $a\in f(A)\cap V$ and $b\in f(A)^\circ \cap V$ where $a,b \neq x$. Then $f^{-1}(a)\in A\cap f^{-1}(V)$ and $f^{-1}(b)\in A^\circ \cap f^{-1}(V)$ and since $f$ is continuous, $f^{-1}(V)$ is open in $X$ which shows that $f^{-1}(x)\in\partial A$. So $x\in f(\partial A)$.

An observation: From (2) you can conclude that homeomorphisms take closures to closures too.

Answer 2

I think it's worth providing a shorter proof:

If $(E_i,\tau_i)$ is a topological space, $f:E_1\to E_2$ is an open map, $B_1\subseteq E_1$ and $B_2:=f(B_1)$, then $B_1^\circ\in\tau_1$ and hence $$f(B_1^\circ)\in\tau_2\tag1.$$ By definition, $B_2^\circ$ is the largest set in $\tau_2$ contained in $B_2$ and hence $$f(B_1^\circ)\subseteq B_2^\circ\tag2.$$ If $f$ is a homeomorphism, then $f^{-1}$ is an open map as well and we may replace $(f,B_1,B_2)$ by $(f^{-1},B_2,B_1)$ to obtain $$f^{-1}(B_2^\circ)\subseteq B_1^\circ.\tag3$$

Answer 3

The answers given here are valid if you start with $Y_1\subset X_1$, $Y_2\subset X_2$ and a homeomorphism$X_1\rightarrow X_2$. But say you begin with $Y, Z\subset X$ and a homeomorphism $f:Y \rightarrow Z$. Then $f$ in general need not preserve interiors! It might help to recall:

Definition: If $Y\subset X$ is a subspace then $Y°$ is the largest open set of $X$ contained in $Y$.

Note: $f$ is an open map, but that only means $f(Y°)$ is open in $Z$, not necessarily in $X$, as the following counter-example will illustrate. This is where the general statement fails.

Counter-example: Let $X=\{a,b\}$ be a two point set with topology such that $a$ is open and $b$ is not. Let $Y=\{a\}$, $Z=\{b\}$ and $f$ be the map sending $a$ to $b$. As every one point set has the same topology, $f$ is a homeomorphism, but note $Y°=\{a\}$ while $f(Y°)=Z\neq Z°$ as $Z°$ is empty.

The good news is that we can prove your statement when $X=\mathbb{R}^n$ but the bad news is that we can't avoid a bit of algebraic topology. Allow me to just use Invariance of Domain (IOD), which says that if the domain $U$ of a continuous bijection $f:U\rightarrow \mathbb{R}^n$, is an open subset of $\mathbb{R}^n$, then $f(U)$ is open in $\mathbb{R}^n$ and $f$ is a homeomorphism onto its image. In fact we only need the former.

Theorem If $Y, Z\subset \mathbb{R}^n$ and $f:Y\rightarrow Z$ is a homeomorphism then $f(Y°)=Z°$

Proof: $Y°$ is an open subset of $\mathbb{R}^n$ and $f_{|Y°}$ is a continuous bijection, so by IOD $f(Y°)$ is open in $\mathbb{R}^n$. But $f(Y°)\subset Z$, thus $f(Y°)\subset Z°$. Now if $V\subset Z$ is open in $\mathbb{R}^n$ then $f^{-1}_{|V}$ is also a continuous bijection onto $f^{-1}(V)$, so by IOD, $f^{-1}(V)$ is open in $\mathbb{R}^n$ and thus $f^{-1}(V)\subset Y°$. Therefore $$f(f^{-1}(V))=V\subset f(Y°)$$ This proves $Z°\subset f(Y°)$ and the desired equality follows.