Problem is:
Continuos, surjective map $\pi$ is a quotient map $\iff$ $\pi$ sends saturated open to open or saturated closed to closed.
($U$ is saturated $\iff$ $\exists V \in Y$ s.t. $U = \pi^{-1} (V)$)
My understanding is:
First, $U$ is saturated $\iff$ $U= \pi^{-1}(\pi(U))$,
(since 1]: if $U$ is saturated and $x \in \pi^{-1}(\pi(U))$,
then $\pi(x) \in \pi(U)=\pi(\pi^{-1}(V)) \subset V$ (for some $V \subset Y$),
so that $x\in \pi^{-1}(V)=U$.
The other inclusion is satisfied for all maps.
2]: Converse is immediate by definition of saturated sets.)
Now, Suppose $\pi$ is a quotient map. Let $U$ be saturated and open in $X$.
We will show $\pi(U)$ is open in $Y$
$U = \pi^{-1}(\pi(U))$ is open in $X$ $\iff$ $\pi(U)$ is open in $Y$,
so $\pi(U)$ is open in $Y$.
Conversely, suppose $\pi$ is surjective and continuous map which sends saturated open sets in $X$ to open set in $Y$.
We will show for any set $V \subset Y$, $V$ is open in $Y \iff \pi^{-1}(V)$ is open in $X$.
If $V$ is open in $Y$, then $\pi^{-1}(V)$ is open since $\pi$ is continuous.
Conversely, suppose $\pi^{-1}(V)$ is open in X.
We want to say $V$ is open.
Surjectivity implies $V=\pi(U)$ for some $U \subset X$.
$\pi^{-1}(V) = \pi^{-1}(\pi (U))$,
so if $U$ is saturated, $\pi^{-1}(\pi (U)) = U$, and $\pi(U)=V$ is open in $Y$ so we are done.
I think the problem is $U$ cannot be always a saturated set.
Any hint is welcomed. Thanks.
Edit:
Thanks to the comment,
This part
can be changed as:
Since $\pi^{-1}(V)$ is saturated by definition, and $\pi$ sends a saturated open set to an open set,
$\pi(\pi^{-1}(V))$ is open in $Y$.
Since $\pi$ is surjective, $V = \pi(\pi^{-1}(V))$, and we are done.