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?

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. 
