Problem: Let $(X, T_X)$ and $(Y, T_Y)$ be topological spaces with bijective function $h:X \to Y$ continuous and open. Then $h$ is a topological equivalence between $X$ and $Y$.
It says that, $h:X \to Y$ is a bijection, therefore it is a 1-1 correspondence and also onto (i.e. it is injective and surjective). So, there is a one-to-one correspondence between $X$ and $Y$ with respect to $h$. Now, we know that, a function is said to be continuous iff for every $V \in T_Y$, $f^{-1}(V) \in T_X$, and it is said to be open iff for every $U \in T_X$, we have $f(U) \in T_Y$. So, $h:X \to Y$ is continuous implies if we let $V$ be an open set of $Y$ so that $V \in T_Y$, then using the property of continuous functions, we can say that its inverse is open in X and therefore $f^{-1}(V) \in T_X$.
A function is called open if it maps an open set to another open set, therefore $h:X \to Y$ is open => if we let any $U \subset X$, $U$ is open in $X$, so $U \in T_X$. So, $h(U)$ is open in $Y$, so $h(U) \in T_Y$.
So, $h:X \to Y$ is 1-1 correspondence, such that $\forall U \in T_X$, $f(U) \in T_Y$, and $\forall V \in T_Y$, we have the pre-image $f^{-1}(V) \in T_X$ which basically proves topological equivalence b/w $X$ and $Y$.
This is how I proved this proposition. Can someone verify this and let me know if this correct or not. Do feel free to give a better notation and style. Appreciate your help and support.
Best Answer
There is almost no proof. By assumption $h$ is already a bijection.
$h$ is open is saying the same as $$\forall U \in \mathcal{T}_X: h[U] \in \mathcal{T}_Y$$
and $h$ is continuous says
$$\forall V \in \mathcal{T}_Y: h^{-1}[V] \in \mathcal{T}_X$$
so practically by your definition an open continuous function fulfills the essential demands on being an equivalence. We need a bijection for an equivalence but that is part of your assumption too (the previous facts hold for all $h$, bijective or not).