Proposition

Let $q: X \rightarrow Y$ be any map. For a subset $U \subset X$, show that the following are equivalent:

$U$ is saturated.
$U = q^{-1}(q(U))$.
$U$ is a union of fibers.
If $x \in U$, then every point $x’ \in X$ such that $q(x) = q(x’)$ is also in $U$.

Solution

$1 \implies 2$

$U \subset q^{-1}(q(U))$?
- For any $x \in U$, $q(x) \in q(U)$, so $x \in q^{-1}(\{ q(x) \}) \subset q^{-1}(q(U))$.
$q^{-1}(q(U)) \subset U$?
- Since $U$ is saturated, there exists a subset $V \subset Y$ such that $U = q^{-1}(V)$. Then $q(U) = q(q^{-1}(V)) \subset V$. Therefore, $q^{-1}(q(U)) \subset q^{-1}(V) = U$.

Therefore, $U = q^{-1}(q(U))$.

$2 \implies 3$

\[\begin{align*} U &= q^{-1}(q(U)) \\ &= q^{-1}(\bigcup_{y \in q(U)} \{ y \}) \\ &= \bigcup_{y \in q(U)} q^{-1}(\{ y \}). \end{align*}\]

Thus $U$ is indeed a union of fibers.

$3 \implies 4$

Suppose $U$ is a union of fibers. Then there exists $V \subset Y$ such that $U = \bigcup_{y \in V} q^{-1}(y)$.

Let $x \in U$. Then $x \in q^{-1}(y)$ for some $y \in V$. Let $x’ \in X$ such that $q(x) = q(x’)$. Then $q(x’) = y$, so $x’ \in q^{-1}(u) \subset U$.

$4 \implies 1$

We will show that $U = q^{-1}(q(U))$. By the property of the inverse image, we know that $U \subset q^{-1}(q(U))$. Let $x \in q^{-1}(q(U))$. Then $q(x) \in q(U)$. Therefore, there exists a point $x’ \in U$ such that $q(x) = q(x’)$, so $x \in U$. Hence, $U = q^{-1}(q(U))$, so $U$ is saturated.