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.


I guess your question is whether your proof is correct although you didn't say this. Yes, it is.

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 $V$.


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