# Quotient map restricted to a saturated (but not open) subset

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?

Note that it also true for saturated closed subsets of $$X$$.
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.