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I want to prove the following:

Let $f:X\rightarrow Y$ be a bijection. $f$ is an homeomorphism iff ($f(U)$ is open iff $U$ is open in $X$)

Starting from the usual definition for an homeomorphism: bijective map which is continuous and has continuous inverse.

The "only if" part of the proof is fairly easy: I let $f$ be an homeomorphism and then prove the "iff" in the parentheses.

The reciprocal is what boggles me: I let the proposition "$f(U)$ is open iff $U$ is open in $X$" be true. I do not know how to use it to prove that $f$ is an homeomorphism.

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  • $\begingroup$ Is your definition of continuous "the preimage of an open set is again open"? $\endgroup$ – Andres Mejia Jan 16 '17 at 19:59
  • $\begingroup$ @AndresMejia Yes, that's right. $\endgroup$ – Soap Jan 16 '17 at 21:47
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Well, let $U \subseteq Y$ be open. Then there exists some $V$ so that $f(V)=U$ (because $f$ is bijective), and in particular, the preimage of $U$ is exactly $V$, which is open by our hypothesis that

$f(V)=U$ open $\implies$ $V$ is open.

Hence the preimage of an open set is again open, showing $f$ continuous.

The same argument (more or less) goes for showing that $f^{-1}$ is continuous.

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If $f^{-1}(U)$ is open $\forall U $ open set follows f is continuous. Now, f is a bijection, then there exists its inverse function. If $f (U)=(f^{-1}) ^{-1 }(U)$ is open $\forall U $ open, then $f^{-1}$ is continuous for the same theorem.

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  • $\begingroup$ I know that it is standard, but I must admit that your notation for preimage and inverse are somewhat confusing in your last sentence, unless you already know what to look for. $\endgroup$ – Andres Mejia Jan 16 '17 at 20:01

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