# Proving alternative definition of homeomorphism

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.

• Is your definition of continuous "the preimage of an open set is again open"? Jan 16 '17 at 19:59
• @AndresMejia Yes, that's right.
– Soap
Jan 16 '17 at 21:47

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.
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.