I am trying to prove the following:
Let $p: X \mapsto Y$ be a quotient map. Let $A$ be an open set of $X$ such that $A$ is saturated with respect to $p$. Then $q: A \mapsto p (A)$ defined as the restriction of $p$ to $A$ is a quotient map.
Here is how I have gone about it. To show that $q$ is a quotient map, we show that (1) $q$ is continuous, and (2) $q$ maps saturated open sets of $A$ to open sets of $p (A)$. We are of course taking $A$ as a subspace of $X$ and $p (A)$ as a subspace of $Y$.
Proving (1) is straightforward and derives from the fact that $p$ itself is continuous. Regardless of whether $A$ is open or not, restricting the domain of a continuous function still yields a continuous function (taking the new domain in the subspace topology). That is, if we take $g: A \mapsto Y$ defined by $g (x) = p (x)$, then $g$ is continuous. Following this, restricting the range of a continuous function to the image of the domain still yields a continuous function. Thus, $q: A \mapsto g (A) = p (A)$ defined by $q (x) = g (x) = p (x)$ is continuous also.
To prove (2), suppose we have a saturated open set $S$ in $A$ (saturated, that is, with respect to $q$). Then $S = q^{-1} (q (S))$. Since $S \subseteq A$, then $q (S) = p (S) \subseteq p (A)$. For $T \subseteq p (A)$, it is also the case that $q^{-1} (T) = p^{-1} (T)$. Taking $T = q (S) = p (S)$, these facts show that $S = p^{-1} (p (S))$; i.e., $S$ is saturated with respect to $p$ as well. And since $S$ is open $A$, and $A$ is open in $X$, then $S$ is open in $X$ as well. By the fact that $p$ is a quotient map, $p (S) = q (S)$ must be open in $Y$. But $q (S) \subseteq p (A) \implies q (S) = p (A) \cap q (S)$ directly shows that $q (S)$ must be open in $p (A)$ under the subspace topology.
Is there anything incorrect or missing in my proof?