Let $X,Y$ metric spaces and $U \subset X , V\subset Y$ open sets.
Let $f:U\rightarrow V$ be a homeomorphism. Prove that $f$ is an open map. I need to show that for every open subset of $U′⊂U$, $f(U′)$ is open.
This is a question from a multi-variable calculus class. I have tried first to assume $U$ as a open ball then playing with the inverse of $f$, but I'm clueless.
Best Answer
Since $f:U\to V$ is a homeomorphism, $\space f$ must have a continuous inverse $\space f^{-1}: V \to U$. Now we use the fact that $f^{-1}$ is continuous, which means any open subset $W \subset U$ is such that $(f^{-1})^{-1}(W) = f(W) \subset V$ is open, completing the proof.
The notation isn't great, but basically use the "inverse image of an open set is open" definition of continuity to see that the inverse of $f^{-1}$ (which is $f$) produces open images in $V$