I am trying to prove a topology statement.
Let $X,Y$ be topological spaces, and let $f: X \to Y$ be a bijection. Prove that $f$ is a homeomorphism if and only if $f$ is continuous and open.
My Attempt:
Suppose that $f$ is a homeomorphism.
Then by definition, $f$ is continuous.
And by definition of continuity, for every open $U \subset Y$, $f^{-1}(U)$ is open in X. Let $f^{-1}(U) =V$. Then, since $f$ is bijective, we have that $f(V)=f(f^{-1}(U))$. Since $f$ is continuous, all the open subsets of $X$ can be obtained as $f^{-1}(U)$ and since $U$ is open, $f(V)$ is open for all $V \subset X$.
(Other direction) Assume that $f$ is continuous and open. Then Since $f$ is bijective, it has an inverse $f^{-1}$. Since $f$ is open, for any open set $U$ of $X$, $f(U)$ is open in $Y$. We need to prove that $f^{-1}$ is continuous. Since $f$ is open, for all open sets $U \in X$, $(f^{-1})^{-1}(U)=f(U)$ is open in $Y$. Thus $f^{-1}$ is continuous and $f$ is a homeomorphism
Any help would be appreciated.
Best Answer
Continuing where you stopped: Since $f$ is bijective, we have that $f(f^{-1}(U)) = U$ and since $U$ is open, we have that $f(V)$ is open (in $Y$)
To prove that $f^{-1}$ is continuous, we should prove that for all open sets $U \in X$, $(f^{-1})^{-1}(U)$ is open in $Y$, since $(f^{-1})^{-1}(U) = f(U)$. And since $f$ is open, this follows directly.