homeomorphisms mapping interiors to interiors and boundaries to boundaries 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 $\mathbb{R}^n.$
 A: 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.
A: 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 $$f^{-1}(B_2^\circ)\subseteq B_1^\circ.\tag3$$
