Diffeomorphism to $\mathbb{S}^1 \times \mathbb{S}^{n-1}$ I'm into manifolds, and inside the proof of an example, the author says is is "clear to see" that $$(\mathbb{R}^n-\{0\})/G \text{ is diffeomorph to } \mathbb{S}^1 \times \mathbb{S}^{n-1},$$ where $G = \{f^k | f^k(x)=2^kx, k \in \mathbb{Z}\}$ and $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$.
It is clear to me that $G$ acts discontinously in $\mathbb{R}^n-\{0\}$, but I could not go further.
Could someone give me a hand in here?

Converting my comments into an answer for clarity: we have a diffeomorphism
$$\mathbb{R}^n \setminus \{ 0 \} \ni v \mapsto \left( \| v \|, \frac{v}{\| v \|} \right) \in \mathbb{R}_{+} \times S^{n-1}$$
generalizing the polar decomposition for $n = 2$. The multiplication-by-$2$ map acts only on the first factor $\mathbb{R}_{+}$, and the quotient is diffeomorphic to $S^1$ via the map
$$\mathbb{R}_{+} \ni r \mapsto \exp \left( 2 \pi i \log_2 r \right) \in S^1.$$
There is no issue with "missing coordinates." The point is that after taking $\log_2$, the multiplication-by-$2$ map on $\mathbb{R}_{+}$ corresponds to the addition-by-$1$ map on $\mathbb{R}$, which gives a circle $S^1 \cong \mathbb{R}/\mathbb{Z}$ as usual.
