Example of quotient restriction of quotient mapping which is not quotient Here's the problem that I'm trying to solve:

Let $f: X \to Y$ be a quotient mapping.

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*Find an example of $f, X, Y$ where for some $A \subseteq X$, $f|_{A}:A \to f(A)$ is not a quotient mapping.


*If $B \subseteq Y$ is open or closed, prove that $f|_{f^{-1}(B)}:f^{-1}(B) \to B$ is a quotient mapping.


*Find an example of $f, X, Y$ and $B \subseteq Y$ which is neither open nor closed, such that $f|_{f^{-1}(B)}:f^{-1}(B) \to B$ is not a quotient map.

For 1, I noticed that $f:[0,1] \to \mathbb{S}^{1}$, $f(x)=e^{2\pi i x}$ is a quotient map, but $f|_{[0,1)}:[0,1) \to \mathbb{S}^{1}$ is not: $f^{-1}(f([0, \frac{1}{8})))=[0,\frac{1}{8})$, which is open in $[0,1)$, but $f([0,\frac{1}{8}))$ is not open in $\mathbb{S}^{1}$.
For 2, things are straightforward: if $U \subseteq B$ is open in $B$, then $U = B \cap V$, where $V$ is open in $Y$.Now, $f|_{f^{-1}(B)}(U)=f^{-1}(B) \cap f^{-1}(V) \in \tau_{f^{-1}(B)}$. The other direction follows because the restriction of a conitnuous function is continuous.
However, for $3$, the counterexample for $1$ doesn't work, since $[0,1)$ is not the inverse image of any subset of $\mathbb{S}^{1}$. What would be a good example of illustrating $3$?
 A: A quotient map $f:X \to Y$ for which $f$ simultaneously restricted to $f^{-1}[S]$ and $S \subseteq Y$ for all such $S$ is also quotient, is called a "hereditarily quotient map".
A theorem says that if $f$ is a quotient map onto a Hausdorff Fréchet-Urysohn space (so any first countable Hausdorff space as codomain e.g.) then $f$ is hereditarily quotient.
This means that your circle-interval example cannot work, as the image is even metrisable.
A quotient map which is not hereditarily quotient (example 2.4.17 in Engelking's "General Topology"):
$X= (0,\frac{1}{2}] \cup\{1\} \cup \{1+\frac{1}{n}: n\ge 2\}$ as a subspace of $\mathbb{R}$. Define $R$ to be the equivalence relation $R$ on $X$ where $xRy$ iff ($x=y$ or $|x-y|=1$), so we identify $\frac{1}{n}$ with $1+\frac{1}{n}$ for all $n\ge 2$. $Y$ is the resulting quotient space $X/R$ (in the quotient topology, obviously) and $q: X \to Y$ mapping $x$ to its class $[x]$ under $R$ is the required quotient map.
Then for $A = \{1\} \cup (0,\frac{1}{2}]\setminus \{\frac{1}{n}: n \ge 2\} \subseteq X$ note that $A = q^{-1}[q[A]]$. Consider $q| A \to q[A]$, which is a restriction of the
required form (for $S = q[A]$), and $q$ is 1-1 and continuous. If the restriction were quotient is would thus be a homeomorphism, which is is not as $1$ is isolated in $A$ while $q(1)$ is not isolated in $q[A]$.
A: The above answer is indeed valid, and has a lot of background knowledge.
But I think I found a simple counterexample in Munkres' book.
It is contained in Section 22, example 1.
I'll just quote it.

Example 1. Let $X$ be the subspace $[0,1]\cup[2,3]$ of $\mathbb R$, and let $Y$ be the subspace $[0,2]$ of $\mathbb R$.
  The map $p:X\to Y$ defined by
  $$p(x)=\begin{cases}x&\text{for }x\in[0,1]\\x-1&\text{for }x\in[2,3]\end{cases}$$
  is readily seen to be surjective, continuous, and closed.
  Therefore it is a quotient map.
Note that if $A$ is the subspace $[0,1)\cup[2,3]$ of $X$, then the map $q:A\to Y$ obtained by restricting $p$ is continuous and surjective, but it is not a quotient map.
  For the set $[2,3]$ is open in $A$ and is saturated with respect to $q$, but its image is not open in $Y$.

where he used the definition of a saturated set as

A subset $C$ of $X$ is said saturated (with respect to the surjective map $p:X\to Y$) if it contains every set $p^{-1}(\{y\})$ that it intersects.

and used the following acknowledgment ;

To say that $p$ is a quotient map is equivalent to saying that $p$ is continuous and $p$ maps saturated open sets of $X$ to open sets of $Y$.

