The orbit space $\dfrac{\mathbb{R}^n}{\mathbb{R}^+}$ has an open subset homeomorphic to $S^{n-1}$

differential-geometrymanifoldstopological-groups

Let $X$ be the orbit space $\dfrac{\mathbb{R}^n}{\mathbb{R}^+}$, where the action of $\mathbb{R}^+$ is given by the following lemma. I want to show that $X$ has an open subset homeomorphic to $S^{n-1}$, and a point that belongs to every nonempty closed subset. Could anyone help me show this?
It's problem $2$ of the chapter $21$ of Introduction of smooth manifolds john lee.

here is the lemma :
For any continuous action of a topological group $G$ on a topological
space $M$; the quotient map $\pi : M \to \dfrac{M}{G}$ is an open map.

Recall: Suppose we are given an action of a group $G$ on a topological space $M$; which we
write either as $\theta: G \times G \to M$ or as $(g,p) \to g.p $. (For definiteness, let us assume
that $G$ acts on the left; similar considerations apply to right actions.) Recall that the
orbit of a point $p \in M$ is the set of images of $p$ under all elements of the group:
\begin{align}
G.p=\{g.p : g \in G\}
\end{align}
Define a relation on$M$ by setting $p \sim q$ if there exists $g \in G$ such that $g.p=q$.
This is an equivalence relation, whose equivalence classes are exactly the orbits of
$G$ in $M$. The set of orbits is denoted by $\dfrac{M}{G}$; with the quotient topology it is called
the orbit space of the action

Best Answer

Hint 1: Show that $(\mathbb{R}^n - \{0\})/\mathbb{R}^+$ is homeomorphic to $S^{n-1}$.

Hint 2: Recall quotient topology. A set in the quotient space is closed iff. its inverse image under the quotient map is closed. Now, I claim that any nonempty subset of the quotient space which does not contain the image of the origin, cannot be closed. (This is because its inverse image under the quotient map will contain points arbitrarily close to the origin but not the origin. Hence the inverse image is not closed).

UPDATE: I will recall the following basic theorem from point-set topology (which is labeled as "Corollary 22.3" in Munkres's Topology).

Theorem. Let $X$, $Y$ be topological spaces, and $f : X \to Y$ be a continuous surjective map. Let $X^\ast$ denote the fiber space of $f$, ie., $X^\ast = \{f^{-1}(y) : y \in Y\}$. Let $q : X \to X^\ast$ be the quotient map and equip $X^\ast$ with the quotient topology. Then the induced map $\widetilde{f} : X^\ast \to Y$ is a homeomorphism iff. $f$ is a quotient map.

Put $X = \mathbb{R}^n-\{0\}$, $Y = S^{n-1}$ and $f : x \mapsto x/\|x\|$. It is easy to show that the fibers of this map are in one-to-one correspondence with the orbits of $\mathbb{R}^n -\{0\}$ under the $\mathbb{R}^+$ action. So $X^\ast$ is this case is $(\mathbb{R}^n -\{0\})/\mathbb{R}^+$.

Now, all that we need is to show that $f$ is a quotient map. I claim that $f$ is actually an open map. Here is a simple argument. In one of your comments you suggested that you have a homeomorphism $h : \mathbb{R^n}-\{0\} \to S^{n-1}\times\mathbb{R}^+$. Let $p_1 : S^{n-1}\times\mathbb{R}^+ \to S^{n-1}$ be the projection onto the first factor. Then it is easy to show that our map $f$ is the composition $p_1\circ h$. So, being a composition of open maps, $f$ itself is open. Hence $f$ is a quotient map (if a map is continuous, surjective and open then it is a quotient map). And so, by the theorem the induced map $\widetilde{f} : (\mathbb{R}^n -\{0\})/\mathbb{R}^+ \to S^{n-1}$ is a homeomorphism.

Related Question