Suppose that I have a quotient map $f: X \rightarrow Y$ where $X$ has some topology and I say that a set of $Y$ is open if and only if its preimage under $f$ is.
I have a claim I think is valid, but wanted to double check.
I claim that a set $E \subseteq Y$ is open if and only if $E = f(U)$ for a $f$-saturated, open set $U$ in $X$.
First, say that $E \subseteq Y$ is open. Then, $f^{-1}(E)$ is open in $X$ and furthermore is $f$-saturated so $E = f(f^{-1}(E))$ .
Suppose that $E = f(U)$ for $U$ $f$-saturated, open in $X$. See that $E$ is open in $Y$ if and only if $f^{-1}(E)$ is open in $X$, so it suffices to show that $f^{-1}(E)$ is open in $X$. Since $U$ is saturated, $p^{-1}(p(U)) = U = p^{-1}(E)$. Therefore $p^{-1}(E)$ is open in $X$, which means $E$ is open in $Y$ as desired.
Corollary from this: open sets of $Y$ are precisely the images of saturated open sets of $X$.
Yes, this is a very standard characterisation of the quotient topology. The proof is correct, though there is a weird notation change from $f$ to $p$ in the second part.
Condensed version:
If $O \subseteq Y$ is quotient-open, $U=f^{-1}[O]$ is open and is $f$-saturated by definition. And $f[U]=O$ by ontoness of $f$.
If $O=f[U]$ where $U$ is $f$-saturated, then $f^{-1}[O] = f^{-1}[f[U]] = U$ (as $U$ is $f$-saturated) and so $O$ is quotient-open.