Finding a counterexample; quotient maps and subspaces Let $X$ and $Y$ be two topological spaces and $p: X\to Y$ be a quotient map. 
If $A$ is a subspace of $X$, then the map $q:A\to p(A)$ obtained by restricting $p$ need not be a quotient map. Could you give me an example when $q$ is not a quotient map?
 A: M.B. has given an example that shows that the restriction of a quotient map to an open subspace need not be a quotient map. Here is an example showing that the restriction to a saturated set also need not. Restrictions to an open saturated set however always works.
Let $X=[0,2],\ \ B=(0,1],\ \ A=\{0\}\cup(1,2]$ with the euclidean topology. Then the identification $q:X\to X/A$ is a quotient map, but $q:B\to q(B)$ is not.
Here is why: As Ronald Brown wrote, $q:B\to q(B)$ is a quotient map iff each subset of $B$ which is saturated and open in $B$ is the intersection of $B$ with a saturated open set in $X$. Now $U=\left(\frac12,1\right]$ is open and saturated in $B=(0,1]$, but if it were the intersection of $B$ and an open saturated $V$, then this $V$ would intersect $A$ and, since it is saturated, it would contain $0$, and then again by openness it had to contain $[0,\epsilon)$ for some $\epsilon>0$. So the intersection of $V$ and $B$ can never be $U$.
A: Standard counter-example:
Let $X=[0, 2\pi]$ and $f: X\to \mathbb{S}^1$ where $f(x) = (\cos(x), \sin(x))$
Then $f$ is surjective and closed (why?) and thus a quotient map. But the restriction of $f$ to $[0, 2\pi)$ cannot be a quotient map as $[0,2\pi)$ is not homeomorphic to the circle. 
A: It may be helpful to have the general result. The following is 4.3.1 of Topology and Groupoids in the above notation: 
Suppose $p: X \to Y$ is an identification map, and $A$ is a subspace of $X$.  
The following conditions are equivalent.
(a)
the restriction $q: A \to  p(A)$ of $p$ is an identification map;
(b)
each $q$-saturated set which is open in $A$ is
the intersection of $A$ with a $p$-saturated set open in $X$;
(c)
as for (b) but with 'open' replaced by 'closed'.
This result has the usual corollaries, for example as stated above. 
