# 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 $\mathbb{R}^n.$

• (related: math.stackexchange.com/q/46353) Jun 29, 2011 at 6:41
• It's fine. I don't understand the proof in that link either. I just need a basic proof that doesn't involve knowledge of homotopy or fundamental topological groups. Jun 29, 2011 at 7:58
• There are proofs without algebraic topology techniques, but they need dimension theory and Brouwer's fixed point theorem (which can be proved elementarily). It's non-trivial, and you won't find a really short proof. What's the purpose of having such a proof? Can't you just refer to it, using a reference? Jun 29, 2011 at 8:18
• Some proofs of the change of variables theorem in analysis such as the one in Buck's advanced calculus text. Jun 29, 2011 at 8:29
• They might use it, but then you can just assume it's true, right? Jun 29, 2011 at 9:13

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.

• It'd be nice if someone posted the full answer in case that site gets taken down. Jan 9, 2020 at 3:00
• I posted the full answer, based on mathonline.wikidot.com/… and mathonline.wikidot.com/… Jan 14, 2020 at 0:30
• The proof of (2) is ok? I mean.. why we can find $a, b \in X$ with $a \in A \cap f^{-1}(V)$ and $b \in A^c \cap f^{-1}(V)$ ? And why is $a,b \ne x$ if $f$ is bijective? 21 hours ago

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 $$f^{-1}(B_2^\circ)\subseteq B_1^\circ.\tag3$$

• It should be noted, that this proof does not work if we merely assume $f$ to be a homeomorphism between $B_1$ and $B_2$ for in this case $f(B^\circ_1)$ need not belong to $\tau_2$. Jan 21, 2021 at 7:58
• @BrunoKrams Isn't any homeomorphism an open map? If $U$ is open in $E_1$, then $U=f^{-1}(f(U))$ since $f$ is a bijection and so $f(U)$ is open in $E_1$ since $f^{-1}$ is continuous. May 6, 2021 at 9:04
• @14159 Of course you are right that any homeomorphism is an open map. However if $f$ is a homeomorphism between $B_1$ and $B_2$ we can only conclude that $f(B_1^\circ)$ is open in $B_2$ endowed with the relative topology and that does not imply that $f(B_1^\circ)$ is an open subset of $E_2$ May 8, 2021 at 12:22