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?

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. 
