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Wikipedia states that "a bijective continuous map is a homeomorphism if and only if it is open, or equivalently, if and only if it is closed.".
How do we prove this fact?
I can prove the obvious direction, but im unsure how to proceed the other ways

Martin Sleziak
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terry
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  • Welcome to math.SE. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Michael Albanese Jun 25 '15 at 21:11
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2 Answers2

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A bijection $f: X \to Y$ that is closed is also open (and vice versa).

The reason is that if $f$ is a closed bijection and $V$ is an open set in $X$, then the complement $V^c$ of $V$ is closed in $X$, and therefore the image $f(V^c)$ is closed in $Y$ since $f$ is closed. But bijectivity means that $f(V^c) = (f(V))^c$. This means that $f(V)$ is open in $Y$, being the complement of a closed set. Thus, $f$ is an open map as well.

The proof that an open bijection is closed is analogous.

As for why this implies that the map is a homeomorphism: $f$, as given, is continuous by assumption. Further, $f$ has an inverse $f^{-1}$ by bijectivity. Why would this be continuous? Well, we need to know whether $(f^{-1})^{-1}(U)$ is open when $U$ is. But this is the same as $f(U)$, which is open since $f$ is an open map.

user35443
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Arthur
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1

If it's open, then the inverse image by its inverse of any open set is open. This gives the continuity of the inverse and hence the required result.

  • All that remains is to show that a bijective continuous closed map is also open. – Noah Schweber Jun 25 '15 at 21:30
  • The hard part remains – terry Jun 25 '15 at 21:35
  • No, it remains to show that a homeomorphism is open (or closed). Which is pretty straightforward. The "inverse image" theorem is a two-way result. –  Jun 25 '15 at 21:40
  • It's not hard - suppose $f$ is closed and $U$ is open. Then what can we say about $f(\overline{U})$? What does this have to do with $\overline{ f(U)}$? – Noah Schweber Jun 25 '15 at 21:41