Let \(f:X\to Y\) be a continuous, bijective, open map. Show that \(f\) is a homeomorphism. [Open map can be replaced with closed map and the same result holds.]
Solution
Since \(f\) is bijective, it has an inverse \(f^{-1}:Y\to X\). To prove that \(f\) is a homeomorphism it remains only to show that \(f^{-1}\) is continuous.
If \(U\subseteq X\) is open, then \[ (f^{-1})^{-1}(U)=f(U), \] which is open in \(Y\) because \(f\) is an open map. Hence \(f^{-1}\) is continuous.
If instead \(f\) is assumed closed, then for every closed \(F\subseteq X\), \[ (f^{-1})^{-1}(F)=f(F) \] is closed in \(Y\). Thus \(f^{-1}\) is continuous by the closed-set criterion. In either case \(f\) is a homeomorphism.