Section 22 in Munkres' TOPOLOGY, 2nd edition: How to establish this equivalence? Let $X$ and $Y$ be topological spaces; let $p \colon X \to Y$ be a surjective map. 
Then $p$ is said to be a quotient map provided a subset $U$ of $Y$ is open in $Y$ if and only if $p^{-1}(U)$ is open in $X$ (or equivalently a subset $B$ of $Y$ is closed in $Y$ if and only if the set $p^{-1}(B)$ is closed in $X$). 
And, a subset $C$ of $X$ is said to be saturated with respect to the surjective map $p \colon X \to Y$ if $C$ contains every set $p^{-1}(\{ y \})$ that it intersects. 
Then how to establish the following? 
The map $p$ is a quotient map if and only if $p$ is continuous and $p$ maps saturated open sets of $X$ to open sets of $Y$ (or saturated closed sets of $X$ to closed sets of $Y$). 
An afterthought: 
If $C$ is a saturated subset of $X$, then, for every $x \in C$, we have 
$$ p^{-1}(\{ p(x) \} ) \subset C.$$
So 
$$C = \bigcup_{x \in C} \{ x \} \subset \bigcup_{x \in C} p^{-1}( \{ p(x) \} ) \subset C;$$
that is, 
$$C = \bigcup_{x \in C} p^{-1}( \{ p(x) \} ) = p^{-1} \left( \bigcup_{x \in C} \{ p(x) \} \right) = p^{-1} \left( p(C) \right).$$
Am I right? 
 A: Assume $p$ is a quotient map:


*

*As any open set $U \subset Y$ is open in $Y \iff p^{-1}(U)$ is open in $X$, we have that $p$ is continuous.

*For any saturated open subset, $V \subset X$, consider $p(V)$. As $V$ is saturated, we can write $$V = \cup_{y \in p(V)}p^{-1}(y)$$  That is: $$V = p^{-1}(p(V))$$ Thus as $p^{-1}(p(V))$ is open, $p(V)$ must be too, and so $p$ maps saturated open sets to open sets.


Lets go the other way, assume $p$ is continuous and takes saturated open sets of X to open sets of Y:


*

*If $U \subset Y$ is open in $Y$, $p^{-1}(U)$ is open in $X$ by the continuity of $p$. 
$$U \text{ open } \Rightarrow p^{-1}(U) \text{ open}$$
Now to show that if $p^{-1}(U)$ is open in $X$, then $U$ is open in $Y$

*$p^{-1}(U)$ is open in $X$ by assumption, and further is saturated - it is the union of the $p^{-1}(y)$ s.t. $y \in U$. I claim that $$U = p(p^{-1}(U))$$
This is true because $p$ is surjective. Thus as $p^{-1}(U)$ is a saturated open set, $U$ is open by the assumption that saturated open sets are mapped to open sets. That is $$p^{-1}(U) \text{ open } \Rightarrow U \text{ open}$$
Thus we are done.
