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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 $\pi_1^{-1}\text{({r})}$ intersects $A$, then $\pi_1^{-1}\text{({r})} \subset A$. What exactly is $\pi_1^{-1}\text{({r})}$ here? Is it $(\{r\} \times \mathbb{R})$?

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  • $\begingroup$ $(\{r\} \times \mathbb{R}) \cap A$, because we work with the restricted map. $\endgroup$ – Henno Brandsma Mar 3 at 22:54
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    $\begingroup$ Well my confusion arose because of the following theorem: Let $p : X \to Y$ be a quotient map; let $A$ be a subspace of $X$ that is saturated with respect to $p$; let $q : A \to p(A)$ be the map obtained by restricting $p$. Then if $p$ is either an open or closed map $\implies$ $q$ is a quotient map. $\endgroup$ – Praful Shankar Mar 3 at 23:19
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    $\begingroup$ But I see now that the theorem specifies saturation of $A$ with respect to $p$ ($\pi_{1}$ in our case), while in the example I chose, $A$ is saturated only with respect to the restriction $q$. Thanks! $\endgroup$ – Praful Shankar Mar 3 at 23:22
  • $\begingroup$ Yes, restrictions of quotient maps are tricky (there are so-called hereditarily quotient maps where we can restrict to subsets and keep a quotient map)... $\endgroup$ – Henno Brandsma Mar 3 at 23:35
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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.

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