The map $q: \mathbb{R}^3 / \{0\} \to S^2$ where $q(x) = \frac{x}{|x|}$ is a quotient map

Question

In exercise 3.64 in Lee's topology, he claims that the fibers of $q$ are open rays in $ \mathbb{R}^3 / \{0\}$, which makes sense. He then says that it is easy to check that $q$ takes open saturated sets to open sets, which seems intuitive, but I can't quite put into words why exactly that would be.

Intuitively, open saturated sets in the domain are open "pie slices" (and their unions) with the tip in the origin, and $q$ will just project those to the unit ball in a very straightforward sense. How do I formalize this reasoning?

Thank you.

Answer 1

The saturated sets of $f$ are those sets $S\subseteq \mathbb R^3\setminus \{0\}$ for which $S=\bigcup_{z\in \Lambda} f^{-1}(z)$ for some fixed $\Lambda\subseteq S^2.$ Suppose $S$ is open. We need to show that $f(S)$ is open, too.

Since open sets are invariant under rotations, without loss of generality, $z=(1,0,0)\in f(S).$ Then, $f^{-1}((1,0,0))$ is the ray $(0,\infty)\subseteq \mathbb R^3\setminus \{0\},\ (1,0,0)\in S$ and since $S$ is open, there is a ball $(1,0,0)\in B((1,0,0))\subseteq S.$

To show that $f$ maps $B$ to an open set in $S^2$ containing $(1,0,0),$ you can do some elementary analytical geometry or note that $h: \Bbb R^3\setminus\{0\}\to S^2\times(0,\infty):\ h(x)=(f(x),|x|)$ is a continuous bijection with continuous inverse $h^{-1}(f(x),x)=|x|f(x) $ and since the projection $\pi_1$ is open, it follows that $f=\pi_1\circ h$ is open.

In fact, the last paragraph shows that $f$ is an open map, and the fact that $S$ is saturated is superfluous.

Answer 2

Roughly, and with an elementary approach:

A saturated set is of the form $$\{\alpha \mathcal V \mid \alpha>0\}$$ where $\mathcal V \subseteq S^2$ is some set of unit vectors.

Let $A\subseteq \Bbb R^3\setminus \{0\}$ be open and saturated, and pick a point $x\in q(A)$. Note that in particular $x\in A$ (since $q(A)\subseteq A$), which means that there exists a $\delta>0$ such that $B_\delta(x)\subseteq A$, since $A$ is open. But then $$ B_\delta(x)\cap S^2 \subseteq q(A) $$ is an open neighborhood in $q(A)$ containing $x$.

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Even simpler: $$ q(A) = A \cap S^2, $$ which is open by definition of the subspace topology.

Answer 3

Say we have $p\colon X \to Y$ a continuous map of topological spaces, and moreover, there exists $i \colon Y \to X$ continuous map such that $p\circ i = \mathbb{1}_Y$ ($i$ is a section of $p$). Then $p$ is surjective (set theory) and a quotient map.

Proof: we need to check that if $f \colon Y \to Z$ is a map such that $f\circ p \colon X \to Z$ is continuous, then $f$ is also continuous. But we have $f = f \circ \mathbb{1}_Y= f \circ (p \circ i) = (f\circ p) \circ i$, a composition of continuous maps, so continuous.

In particular, a retraction $p \colon X \to Y \subset X$ is a quotient map.

More generally we can show that a continuous map is a quotient map if there are local sections around all points of the target space.

In our case $p \colon \mathbb{R}^n\backslash\{0\}\to S^{n-1}$ is a retraction, so a quotient map. Now, we have an action of the group $\mathbb{R}_+^{\times}$ on $\mathbb{R}^n\backslash\{0\}$ and the orbits are the fibres of the map $p$. We conclude that there exists a homeomorphism of quotients spaces $\mathbb{R}^n\backslash\{0\}/ \mathbb{R}_{+}^{\times} \simeq S^{n-1}$. Now in general, the quotient map $X \to X/G$ is open, because the saturation of an open subset $U\subset X$ is $\cup_{g\in G} g\cdot U$, again open. We conclude that the map $p$ is open ( it is not closed).

Since we are in fact dealing with manifolds, the existence of a (smooth) section implies that the map is a submersion ( another way to see it is an open map).