# Show that the restriction of a quotient map onto a saturated open subset is still a quotient map.

Def: Let $$X,Y$$ be sets and $$p:X \rightarrow Y$$ be a surjective map, then $$A\subset X$$ is called saturated with respect to p if it is the preimage of its image under $$p$$, i.e. $$A=p^{-1}(p(A)).$$

Def et $$X, Y$$ be topological spaces, then a map $$p:X \rightarrow Y$$ is called a quotient map if $$p$$ is surjective, continuous and mapping saturated open sets of $$X$$ to open sets of $$Y$$.

Problem: Let $$X, Y$$ be topological spaces and $$p:X \rightarrow Y$$ be a quotient map (i.e. p is a continuous surjection s.t. $$\forall$$ open subset $$A\subset X$$ that is saturated with respect to $$p$$, $$p(A)$$ is an open subset of $$Y$$). Then for a saturated open subset $$A\subset X$$, $$p|_A=q:A \rightarrow p(A)$$ is a quotient map, where $$A,p(A)$$ are endowed with subspace topologies.

Attempt: It suffices to prove that for any saturated open subset $$B\subset A$$, i.e. $$B=q^{-1}(q(B))$$, we have $$q(B)$$ open in $$p(A)$$. Because $$q$$ is obviously surjective, and $$q=p|_A\Rightarrow q$$ is continuous.

Since for any open subset $$B\subset A$$, $$B=A\cap U$$ where $$U$$ is an open set in $$X$$, we may assume that we are given a saturated open set $$A\cap U$$. Observa that $$A\cap U=q^{-1}\circ q(A\cap U)=p^{-1}\circ p(A\cap U)$$, since $$A$$ is saturated with respect to $$p$$. So we have $$A\cap U$$ is saturated with respect to $$p$$. Also note that $$A\cap U$$ is open in $$X$$ because $$A,U$$ are open in $$X$$. By the assumption that $$p$$ is a quotient map, we know that $$q(A\cap U)=p(A\cap U)$$ is open in Y. Then $$q(A\cap U)=p(A\cap U) \cap p(A)$$ is open in $$p(A)$$.

Since the choice of $$A\cap U$$ is arbitrary, we conclude that $$q$$ is a quotient map.

Note that a function $$p : X \to Y$$ is a quotient map in the sense of your definition if and only if the following are satisfied:
(1) $$p$$ is surjective.
(2) A subset $$U \subset Y$$ is open in $$Y$$ if and only if $$p^{-1}(U)$$ is open in $$X$$.
This is the usual definition of a quotient map. The continuity of $$p$$ is nothing else than the "only if" part in (2). To verify the equivalence of both definitions observe that the saturated [open] subsets of $$X$$ are precisely the sets having the form $$p^{-1}(B)$$ [with $$B \subset Y$$ open]. This is true because $$p(p^{-1}(B)) = B$$ by surjectivity of $$p$$.
If you start with the above definition, then your problem is equivalent to showing that if $$V \subset Y$$ is open , then $$q : p^{-1}(V) \to V$$ satisfies the "if" part of (2). So let $$U \subset V$$ be a set such that $$W = q^{-1}(U)$$ is open in $$p^{-1}(V)$$. Since $$p^{-1}(V)$$ is open in $$X$$, also $$W$$ is open in $$X$$. But $$W = p^{-1}(U)$$, hence $$U$$ is open in $$Y$$. This implies that $$U$$ is open in $$V$$.