Let $q:X\rightarrow Y$ be a continuous surjective map
$q:X \rightarrow Y$ is a quotient map $\iff$ $q$ takes saturated open subsets to open subsets.
My attempt:
Let $q:X \rightarrow Y$ be a continuous surjective quotient map. This means $Y$ is equipped with the quotient topology. So, if $U\subseteq$ $X$ is a saturated open subset of $X$, $U=q^{-1}(q(U))$. Since $q(U)$ is a subset of $Y$, it is open in Y, by definition of the quotient topology.
For the converse, suppose $q$ takes saturated open subsets to open subsets. Let $\tau_Y$ be the topology on $Y$ and $\tau'_Y$ the quotient topology induced by $q$, I must show that the two topologies are equal. If $A\in \tau_Y$ then since $q$ is continuous, $q^{-1}(A)$ is open in $X$, hence $A\in \tau'_Y$. If $V\in \tau'_Y$ then since $q^{-1}(V)$ is open in $X$. By assumption, $V$ is open ($\tau_Y)$ in $Y$.
Is my attempt correct? (Please answer this question)
Best Answer
No, the last paragraph is wrong, as you do not even use the assumption on saturated open sets.
Better: let $O$ be a subset of $Y$ such that $O'=q^{-1}[O]$ is open in $X$. We need to show that $O$ is open in $Y$ to have that $q$ is a quotient map. (we already have the continuity).
Note that $O'$ is saturated (the inverse image of any set under $q$ always is, exercise!) and so by assumption $q[O']$ is open. By surjectiveness however, $q[O']=O$ and we are thus done.