From an example in Munkres Topology:
Consider the projection map $\pi_{1}: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ onto the first coordinate; it is continuous and surjective. It is also open which tells us it is a quotient map. Given the subset: $$C = \{(x, y) \text{ | } xy=1 \}$$ We construct: $$A = C \cup \text{{(0,0)}}$$
$A$ is a subspace of $\mathbb{R} \times \mathbb{R}$ and the map $\pi_{1}$ restricted to $A$ is not a quotient map because even though $\text{{(0,0)}}$ is saturated with respect to the restriction, it's image is closed in $\mathbb{R}$.
Is $A$ saturated with respect to $\pi_{1}$? From what I have understood, saturation means that for all $r \in \mathbb{R}$, if $\pi_1^{-1}{({r})}$ intersects $A$, then $\pi_1^{-1}{({r})} \subset A$. What exactly is $\pi_1^{-1}{({r})}$ here? Is it $(\{r\} \times \mathbb{R})$?
Best Answer
Another equivalent way to define saturated sets: $f:X \to Y$ and $A \subseteq X$ is saturated wrt $f$ iff $f^{-1}[f[A]]=A$ or $\forall x \in X: f(x) \in f[A] \implies x \in A$.
We consider the map $\pi'_1: A \to \mathbb{R}$, the restricted projection (which is of course still continuous).
The only point that maps to $\{0\} = \pi'_1[\{(0,0)\}]$ is $(0,0)$ (because the domain is $A$!) and that already lies in $\{(0,0)\}$. So $\{(0,0)\}$ is saturated wrt $\pi'_1$. If you think about it, $\pi'_1$ is in fact 1-1 so all subsets are saturated.