# Saturated subsets of quotient map.

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 $$(\{r\} \times \mathbb{R})$$?

• $(\{r\} \times \mathbb{R}) \cap A$, because we work with the restricted map. – Henno Brandsma Mar 3 at 22:54
• 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. – Praful Shankar Mar 3 at 23:19
• 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! – Praful Shankar Mar 3 at 23:22
• 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)... – Henno Brandsma Mar 3 at 23:35

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.