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

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

• The open subset homeomorphic to $S^{n-1}$ is the image of $\mathbb{R}^n-0$ under the quotient map. And the point which belongs to every closed nonempty set is the image of $0$. – feynhat Jan 1 at 11:48
• You write that "the action of $\mathbb R^+$ is given by the following lemma". But that lemma does not give an action; it instead requires an action to be already given. – Lee Mosher Jan 4 at 5:33

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.

• Thanks; for the first hint I found a homeomorphism between $R^n - \{0\}$ and $S^{n-1} \times \mathbb{R}^+$ which is defined as below : $f:\mathbb{R}^n\backslash\{0\}\rightarrow\mathbb{S}^{n-1}\times\mathbb{R}^+$ given by $f(x)=(\dfrac{x}{||x||},||x||)$ . Is it true? – A maths freak Jan 2 at 15:23
• Forget the $\mathbb{R}^+$ factor in the range. Show that your map $\mathbb{R}^n - \{0\} \to S^{n-1}$ descends to the quotient $(\mathbb{R}^n -\{0\})/\mathbb{R}^+ \to S^{n-1}$ (basically, you need to show that your map doesn't change after scaling by a positive real number). Next, show that this map has an inverse. Let me know if something is still unclear, I can write a more detailed answer if you want. – feynhat Jan 3 at 17:14
• I could just find that homeomorphism between them. Would you mind writing your answer for the first hint in detail ? – A maths freak Jan 4 at 6:49
• @Reza I have added more details for the first hint. – feynhat Jan 10 at 9:41

Take the unit sphere $$S^{(n-1)}$$ in $$\mathbb{R}^{(n)}$$; the quotient map $$\pi$$ restricted to the unit sphere is a topological embedding since it is injective and continuous. Also, we have $$\pi(S^{n-1})=\pi(\mathbb{R}^n\setminus\{0\})$$ so $$\pi(S^{n-1})$$ is open. For the second claim, consider $$\pi(0)$$. If $$V$$ is a neighborhood of this point in the quotient space, then $$\pi^{-1}(V)$$ is a neighborhood of the origin in $$\mathbb{R}^n$$. Take an open ball $$B$$ around the origin in this neighborhood. You can see that $$\pi(B)=\pi(\mathbb{R}^n)$$ because you can make every non-zero vector small enough by multiplying a small positive factor. Hence, the only open set containing $$\pi(0)$$ is the quotient space itself and every closed non-empty subset must contain this point.