A surjective map $\pi: X \rightarrow Y$ is a quotient map when a subset $V$ is open in $Y$ if and only if $\pi^{-1}(V)$ is open in $X$. It's well-known that the restriction of $\pi$ to a saturated open subset $S$ in $X$ is also a quotient map. By a saturated subset we mean $S=\pi^{-1}(\pi(S))$.
The proof can go as follows. First, let $V$ be a subset of $\pi(S)$ and $\pi|_S^{-1}(V)$ is open in $S$. Since $S$ is open, $\pi|_S^{-1}(V)$ is also open in $X$. Because $S$ is a saturated subset, we have $\pi|_S^{-1}(V) = \pi^{-1}(V)$. $\pi^{-1}(V)$ is open in $X$ implies $V$ is open in $Y$, thus open in $\pi(S)$. The other direction, that $V$ is open implies $\pi|_S^{-1}(V)$ is open, comes from the fact that $\pi|_S$ is continuous as a restriction of a continuous map.
We have used the condition that $S$ is open in the proof. My question is very simple, what if we don't assume $S$ is open? Can anyone kindly give an example that the restriction of a quotient map to a saturated subset is not a quotient map?
Best Answer
Note that it also true for saturated closed subsets of $X$.
Here is an example as requested.
Let $Y = \mathbb Q \cup \{\infty\}$ with a point $\infty \notin \mathbb R$. Define $$\pi : \mathbb R \to Y, \pi(x) = \begin{cases} x & x \in \mathbb Q \\ \infty & x \in \mathbb I = \mathbb R \setminus \mathbb Q \end{cases} $$ and give $Y$ the quotient topology with respect to $\pi$. Let $Q = \pi(\mathbb Q)$ denote the subspace of $Y$ with underlying set $\mathbb Q$. Since $\pi^{-1}(Q) = \mathbb Q$, we see that $\mathbb Q$ is a saturated subset of $\mathbb R$. It is neither open nor closed in $\mathbb R$.
The restriction $p = \pi \mid_{\mathbb Q} : \mathbb Q \to Q$ is a continuous bijection. It is not a quotient map:
Let $\mathbb Q^+ = \{x \in \mathbb Q \mid x > 0 \}$ and $V= p(\mathbb Q^+)$. We have $p^{-1}(V) = \mathbb Q^+$ which is open in $\mathbb Q$. Assume that $V$ is open in $Q$. Then it has the form $V = V' \cap Q$ with an open $V' \subset Y$. The only candidates for $V'$ are $V$ and $V' = V \cup \{\infty\}$. The set $V$ is not open in $Y$ because $\pi^{-1}(V) = \mathbb Q^+$ which is not open in $\mathbb R$. But also $V'$ is not open in $Y$ because $\pi^{-1}(V') = \pi^{-1}(V) \cup \pi^{-1}(\infty) = \mathbb Q^+ \cup \mathbb I$ which is not open in $\mathbb R$. Thus our assumption leads to a contradiction.