# A function between topological spaces is an open, continuous bijection if and only if it is a homeomorphism.

Definition 1: A function $$f:X\to Y$$ for topological spaces $$X$$ and $$Y$$ is continuous if the preimage $$f^{-1}(U)\subset X$$ is open in $$X$$ for each $$U\subset Y$$ open in $$Y$$.

Definition 2: A function $$f:X\to Y$$ for topological spaces $$X$$ and $$Y$$ is open if a set $$Z\subset X$$ being open in $$X$$ implies that $$f(Z)\subset Y$$ is open in $$Y$$.

Definition 3: A function between topological spaces is a homeomorphism $$\iff$$ it is a continuous bijection and the inverse function $$f^{-1}$$ is also continuous.

Definition 4: Let $$f:X\to Y$$ be a function between sets. The preimage of a subset $$U\subset Y$$ is the collection of $$x\in X$$ such that $$f(x)\in U$$, notated $$f^{-1}(U)$$. Concretely, $$f^{-1}(U)=\{x\in X\mid f(x)\in U\}.$$

Theorem: The function $$f:X\to Y$$ for topological spaces $$X$$ and $$Y$$ is an open, continuous bijection $$\iff$$ $$f$$ is a homeomorphism.

Proof Attempt: Let $$f:X\to Y$$ be an open, continuous bijection. Our goal is to show that $$f$$ is a homeomorphism; in particular, we need to show that the inverse function $$f^{-1}:Y\to X$$ is continuous. To do this, we need to show that for each set $$Q\subset X$$ open in $$X$$ the preimage $$(f^{-1})^{-1}(Q)\subset Y$$ is open in $$Y$$. Let $$Z\subset X$$ be open in $$X$$. Consider the following manipulation of the definition of preimage: $$(f^{-1})^{-1}(Z)=\{y\in Y \mid f^{-1}(y)\in Z\}=\{y\in Y \mid y\in f(Z)\}=f(Z)\subset Y.$$ Since $$f$$ is open and $$Z\subset X$$ is open in $$X$$, $$(f^{-1})^{-1}(Z)=f(Z)\subset Y$$ is open in $$Y$$. Thus, $$f^{-1}$$ is continuous, and in particular $$f$$ is a homeomorphism.

Now let $$f$$ be a homeomorphism. Then $$f$$ is a continuous bijection with a continuous inverse $$f^{-1}$$. We aim to show that $$f$$ is open. Let $$Z\subset X$$ be open in $$X$$. Since $$f^{-1}$$ is continuous, the preimage $$(f^{-1})^{-1}(Z)=f(Z)\subset Y$$ is open in $$Y$$. Thus $$f$$ is open.$$\;\;\;\boxed{}$$

Notes: Does this proof make sense? Is it valid given the definitions? The equality $$\{y\in Y \mid y\in f(Z)\}=f(Z)$$ was very tricky for me to come to understand, and I'm only partly sure it is true.

Response to Answerers: Thank you for your time and expertise! I think I have a more economical grasp of the concept now. My newfound conception is that for a bijection $$f$$, $$f$$ being open is equivalent to $$f^{-1}$$ being continuous. This should be illustrated through the following diagram, where for notational ease I've relabeled $$f^{-1}=g$$. The key for me has been the understanding that the implications in the definitions of openness (of $$f$$) and continuity (of $$g$$) are the same:

To keep notation clean use $$g: Y \to X$$ for the unique inverse of $$f$$ (so that $$f(x)=y$$ iff $$g(y)=x$$ etc.) and note that indeed for any $$O \subseteq X$$ we have $$y \in g^{-1}[O] \iff g(y) \in O \iff \exists x \in O: g(y)=x \iff \exists x \in O: f(x)=y \iff y \in f[O]$$ hence
$$g^{-1}[O]=f[O]\tag{1}$$
from which it immediately follows that $$f$$ open implies $$g$$ continuous (taking $$O$$ to be any open set) or $$f$$ closed implies $$g$$ continuous (taking $$O$$ to be closed etc.) too, and vice versa.
Your proof is correct, but you've used more notation than is necessary, I think. The equality that you've expressed doubt about is a vacuous tautology - it's true, but in such a trivial manner that it's not really worth writing down (in particular, for any set $$S \subseteq Y$$, we have $$S = \{y \in Y \mid y \in S\}$$).