# $U$ takes the same value on $\pi$ then $U$ is saturated

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.

• Yes, your proof is fine. Jun 21, 2015 at 20:01
• My pleasure. $\,$ Jun 22, 2015 at 21:29