In Munkres, Section 22 (The Quotient Topology) he says the following:
Another way of describing the quotient map is as follows: We say that a subset $C$ of $X$ is saturated (with respect to the surjective map $p:X \rightarrow Y)$ if $C$ contains every set $p^{-1}(\{y\})$ that it intersects.
He further says: Thus $C$ is saturated if it equals the complete inverse image of a subset of $Y$.
Question: Shouldn't it be the complete inverse image of $Y$? Since $p^{-1}({y})$ implies it takes all the points of $Y$?
Best Answer
$A \subseteq X$ is saturated for an equivalence relation $R$ on $X$, if $x \in A$ and $xRy$ implies $y \in A$ too. So if $A$ contains a member of an equivalence class it contains all of the class. This sort of explains the name: it "absorbs" with every point all its equivalent points as well: such a set is a union of classes.
If $q:X \to X/{R}$ is the map associated with that relation, so $q(x)$ is the class that $x$ belongs to, this translates to: if $A \cap q^{-1}(y)$ is non-empty (so $A$ intersects the class $y$) it contains that whole class $q^{-1}(y)$..
A quotient map is just the canonical map for its induces equivalence relation $xRx'$ iff $q(x)=q(x')$. We can also say that $A$ is saturated iff $A=q^{-1}[q[A]]$, which is in formulaic form what I said at the beginning.