Last updated Unknown

Let $X, Y$ be Topological Spaces and let $F: X \to Y$ be a Function. Then $F$ is a Homeomorphism If and Only If $F$ is a Bijection, contintuous, and an Open Map.

$(\Rightarrow)$ Suppose $F$ is a Homeomorphism. Then $F$ is a Bijection and a Continuous Function by definition. Futhermore $F^{-1}$ is a Continuous Function by definition. Thus, for any Open $U \subset X$, $F(U) = (F^{-1})^{-1}(U)$ is Open. $\checkmark$

$(\Leftarrow)$: Suppose $F$ is a Bijection, a Continuous Function, and an Open Map. Then, let $U \subset X$ be Open. We want to show $F^{-1}$ is continuous. Note $(F^{-1})^{-1}(U) = F(U)$ is Open because $F$ is an Open Map. Thus $F^{-1}$ is a Continuous Function and $F$ is a Homeomorphism. $\checkmark$ $\blacksquare$