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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 $\mathbb{R}^n.$

Kashif
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  • (related: http://math.stackexchange.com/q/46353/) – Grigory M Jun 29 '11 at 06:41
  • 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. – Kashif Jun 29 '11 at 07:58
  • 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? – Henno Brandsma Jun 29 '11 at 08:18
  • Some proofs of the change of variables theorem in analysis such as the one in Buck's advanced calculus text. – Kashif Jun 29 '11 at 08:29
  • They might use it, but then you can just assume it's true, right? – Henno Brandsma Jun 29 '11 at 09:13

2 Answers2

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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.

Alejandro Tolcachier
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    It'd be nice if someone posted the full answer in case that site gets taken down. – Kashif Jan 09 '20 at 03:00
  • I posted the full answer, based on http://mathonline.wikidot.com/the-interior-of-a-set-under-homeomorphisms-on-topological-sp and http://mathonline.wikidot.com/the-boundary-of-a-set-under-homeomorphisms-on-topological-sp – Alejandro Tolcachier Jan 14 '20 at 00:30
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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 $$f^{-1}(B_2^\circ)\subseteq B_1^\circ.\tag3$$

0xbadf00d
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  • 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 $\tau_2$. – Bruno Krams Jan 21 '21 at 07:58
  • @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. – 14159 May 06 '21 at 09:04
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    @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 $E_2$ – Bruno Krams May 08 '21 at 12:22