Q:

Saturated Sets that Don't Equal Some Preimage

Let $X$ and $Y$ be topological spaces, and let $p : X \to Y$ be some surjective map. The map $p$ is said to be a quotient map provided $U \subseteq Y$ is open if and only if $p^{-1}(U)$ is open in $X$. Here is the definition of a saturated set and a relevant passage from Munkres:

We say a that a subset $C$ of $X$ is saturated (with respect to the surjective map $p: X \to Y$) if $C$ contains every set $p^{-1}(\{y\})$ that it intersects. Thus $C$ is saturated if it equals the complete inverse image of a subset of $Y$.

Are there any saturated sets that aren't equal to some preimage under some surjective map? The answer is probably trivial, but I am temporarily suffering from boneheadedness.
PS: I just realized the definition of a quotient map isn't as germane to the question as I initially thought. Oh! Well! I'll leave it anyways. 
EDIT:
The preceding is related to what follows, especially in light of quasi's answer. I am trying to prove the following: 

$p : X \to Y$ is a quotient map if and only if $p$ is continuous and maps saturated open sets to open sets. 

I was able to prove the forward direction, and here is my proof of it. 

Suppose $p$ is a quotient map. It follows immediately from the definition that $p$ is continuous. Now suppose that $U$ is saturated and open in $X$. e want to show $p(U)$ is open. Since $p$ is a quotient, this occurs if and only if $p^{-1}(p(U)$ is open, but, according to what quasi proved, $p^{-1}(p(U)) = U$ which is, by hypothesis, open.

Unfortunately, I am having trouble with the other direction. Specifically, I am having trouble showing surjectivity. Suppose that $p : X \to Y$ is a continuous map sending saturated open sets to open sets. Suppose that $U$ is open in $Y$. Then $p^{-1}(U)$ is open since $p$ is continuous. Now suppose that $U \subseteq Y$ is such that $p^{-1} (U)$ is open. Since it is saturated with respect to $p$, we have that $p(p^{-1}(U) \subset u$ is open. If we had surjectivity, then we would be finished...Let's try to prove surjectivity. Note that $X = p^{-1}(Y)$ which means that $X$ is saturated and open. Therefore $p(X) = p(p^{-1}(Y)$ is open in $Y$...Hmm...I tried many other things, but I couldn't figure out how to  prove surjectivity. 

A:

A set $A \subseteq X$ is saturated w.r.t. $p: X \to Y$, iff it satisfies one of the following equivalent conditions:

$\forall y \in Y: p^{-1}[\{y\}] \cap A \neq \emptyset \implies p^{-1}[\{y\}] \subseteq A$.
$A = p^{-1}[p[A]]$.
$\exists B \subseteq Y: A = p^{-1}[ B]$. 

Proof of 1 implies 2: For all $A: A \subseteq p^{-1}[p[A]]$ (as $x \in A$ trivially implies $p(x) \in p[A]$ so by definition $x \in p^{-1}[p[A]]$), so let $x \in p^{-1}[p[A]]$. This means that $p(x) \in p[A]$ or equivalently that there is some $a \in A$ with $p(x) = p(a)$. But then $a$ witnesses that $p^{-1}[\{p(x\}] \cap A \neq \emptyset$ and so $x \in p^{-1}[\{p(x)\}] \subseteq A$, by 1. and so $x \in A$, as required, and the equality under 2 has been shown.
2 implies 3 is trivial: just take $B = p[A]$.
3 implies 1: Suppose that $A= p^{-1}[B]$ and suppose that $y \in Y$ is such that $p^{-1}[\{y\}] \cap A \neq \emptyset$. Let this be witnessed by $a \in A$ so that $a \in A$ and $a \in p^{-1}[\{y\}]$, i.e. $p(a) = y$. But then $y  =p(a) \in B $ so $p^{-1}[\{y\}] \subseteq p^{-1}[B] = A$ as required.
Note that this holds for any function $p$ between $X$ and $Y$.
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Now suppose that $p$ is quotient. This means by definition that $p$ is surjective and 

(q) $\forall O \subseteq Y$: $p^{-1}[O]$ is open iff $O$ is open.

The right to left part of (q) says that $p$ is continuous and if $A$ is saturated and open then we know by condition 2 above that $A = p^{-1}[p[A]]$ and as $A$ is open and $p$ is quotient (using the left to right implication for $O = p[A]$) this exactly means that $p[A]$ is open, as required.
Now suppose that $p$ is continuous and maps saturated open sets to open sets. We want to show (q): the right to left implication is just the continuity of $p$, so that’s OK. Now if $p^{-1}[O]$ is open, it is open and saturated (using condition 3) so $p[p^{-1}[O]] = O$ (by surjectivity of $p$) is open by the assumption. Hence (q) holds for $p$ and $p$ is quotient.

