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 $\mathbb{R}^n.$

  • $\begingroup$ (related: math.stackexchange.com/q/46353) $\endgroup$
    – Grigory M
    Jun 29, 2011 at 6:41
  • $\begingroup$ 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. $\endgroup$
    – Kashif
    Jun 29, 2011 at 7:58
  • $\begingroup$ 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? $\endgroup$ Jun 29, 2011 at 8:18
  • $\begingroup$ Some proofs of the change of variables theorem in analysis such as the one in Buck's advanced calculus text. $\endgroup$
    – Kashif
    Jun 29, 2011 at 8:29
  • $\begingroup$ They might use it, but then you can just assume it's true, right? $\endgroup$ Jun 29, 2011 at 9:13

2 Answers 2


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.

  • 1
    $\begingroup$ It'd be nice if someone posted the full answer in case that site gets taken down. $\endgroup$
    – Kashif
    Jan 9, 2020 at 3:00
  • $\begingroup$ I posted the full answer, based on mathonline.wikidot.com/… and mathonline.wikidot.com/… $\endgroup$ Jan 14, 2020 at 0:30
  • $\begingroup$ 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? $\endgroup$ 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 $$f^{-1}(B_2^\circ)\subseteq B_1^\circ.\tag3$$

  • $\begingroup$ 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 $\tau_2$. $\endgroup$ Jan 21, 2021 at 7:58
  • $\begingroup$ @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. $\endgroup$
    – 14159
    May 6, 2021 at 9:04
  • 3
    $\begingroup$ @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 $E_2$ $\endgroup$ May 8, 2021 at 12:22

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