(Munkres 2ed, Theorem 22.1)
Let $p:X\to Y$ be a quotient map; let $A$ be a subspace of $X$ that is
saturated with respect to $p$ [i.e. $A=p^{-1}(p(A))$]. Let $q:A\to
> p(A)$ be the map obtained by restricting $p$. If $A$ is either open or
closed in $X$, then $q$ is a quotient map.
My professor said that the saturation condition
$$A=p^{-1}(p(A))$$
is necessary because of the following counterexample.
Let $\pi_1:\mathbb R^2\to\mathbb R$ be the projection to the first coordinate.
$\pi_1$ is a continuous surjective map.
Moreover, it is an open map, thus it is a quotient map.
Consider a proper subset
$$A=\{x\times y:xy=1\}\cup\{0\times 0\}$$
of $\mathbb R^2$.
$A$ is closed in $\mathbb R^2$ and it doesn't satisfy the saturation condition since ${\pi_1}^{-1}({\pi_1}(A))={\pi_1}^{-1}(\mathbb R)=\mathbb R^2$.
Then, $q=\pi_1|A:A\to\mathbb R$ is not a quotient map.
For, although $(0,\infty)$ is an open subset of $\mathbb R$, its inverse image
$$A^+=(\pi_1|A)^{-1}((0,\infty))=\{x\times y:xy=1,x>0\}$$
is not open in $A$.
Thus the conclusion of the theorem fails to hold.
But why is $A^+$ not open in $A$?
How can I figure out this?
I think it is open.
Consider the first quadrant $F=\{x\times y:x>0\quad y>0\}$.
$F$ is open in $\mathbb R^2$.
Then $A^+=A\cap F$ is open in $A$.
Am I wrong?
Best Answer
You are correct: $A^+$ is open in $A$. It is the preimage of $(0,\infty)$, which is open in $\mathbb{R}$, via $\pi_{|A}$ after all. So what exactly is that argument supposed to show? That $\pi_{|A}$ is not continuous? It definitely is, restriction of a continuous function is continuous.
This has to be tweaked. Consider
$$A'=\pi_{|A}^{-1}([0,\infty))=A^+\cup\{(0,0)\}$$
and note that $A'$ is again open in $A$, but this time $[0,\infty)$ is not open in $\mathbb{R}$. Thus $\pi_{|A}$ is not a quotient map.