Q:

(Munkres) Clarification with respect to what a saturated subset means (quotient maps)

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 $Y$?

A:

The definition of saturated is correct. It should not be the complete inverse image of $Y$, because the complete inverse image of $Y$ would simply be $X$ itself, and why go to all that trouble with a new definition that produces nothing other than $X$ itself?
Here's an example. For $p : \mathbb R^2 \to \mathbb R$ defined by $p(x,y)=x$, a saturated subset is any subset of $\mathbb R^2$ which can be expressed as a union of vertical lines. So, for example, the union of the two vertical lines $x=0$ and $x=1$ is saturated, and can be expressed either as $p^{-1}\{0,1\} = \{0,1\} \times \mathbb R = (\{0\} \times \mathbb R) \cup (\{1\} \times \mathbb R)$. Also, the whole vertical strip $p^{-1}[0,1] = [0,1] \times \mathbb R$ is saturated.
Regarding your very final question, the expression $p^{-1}(b)$ represents a different subset of $X$ for each different value of $b \in Y$. In my example, for $b=0$ we have $p^{-1}(b) = p^{-1}(0) = \{0\} \times \mathbb R$, whereas for $b=1$ we have $p^{-1}(b) = p^{-1}(1) = \{1\} \times \mathbb R$.

A:

$A \subseteq X$ 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.

