Let $\pi : X \to Y$ be any map, and $U$ be a subset of $X$.
The problem is:
"$\forall x\in U$, $ \pi (x) = \pi(x') $, then $x' \in U$" then $U$ is saturated.
(U is saturated $\iff$ $\exists V \subset Y $ s.t. $U = \pi ^{-1}(V)$ )
My understanding is:
To show $U$ is saturated, it is enough to show $U = \pi ^{-1}(\pi(U))$.
$U \subset \pi ^{-1}(\pi(U))$ is satisfied for any map $\pi$.
We will show $U \supset \pi ^{-1}(\pi(U))$.
Suppose $ x\in \pi ^{-1}(\pi(U))$.
$\iff \pi(x) \in \pi(U)$
By definition of $\pi(U)$, there is some $a \in U$ s.t. $\pi(x) = \pi(a)$,
so hypothesis says $x\in U$.
My proof is fine? Thanks.