$\text{GL}(1,\mathbb{R})$ is homeomorphic to $\text{O}(1)\times \mathbb{R}$

Question

This statement is from my Geometry book, "Notes on Geometry" by Rees.

The book says it's clear to see this is true, but I can't think of a map that is a homeomorphism between them.

Here, $\text{GL}(1,\mathbb{R})=\mathbb{R}\setminus \{0\}$

And, we have $\text{O}(1)=\{1,-1\}$, so $\text{O}(1)\times \mathbb{R}=\left\{(1,a),(-1,b)|a,b\in \mathbb{R} \right\}$.

(I tried $x\mapsto \left ( \frac{x}{|x|},x \right )$, but this doesn't seem to be continuous)

Is there any simple example?

Answer 1

$\Bbb R\setminus \{0\}$ is a disjoint union of two topological copies of $\Bbb R$, namely $(0,+\infty)$ and $(-\infty,0)$. $\{-1,1\} \times \Bbb R$ is essentially the same thing, written differently.

Note that $\ln(x)$ is a homeomorphism between $(0,\infty)$ to $\Bbb R$ (inverse is $x \to e^x$). So we use that map twice e.g.

$$h: O(1) \times \Bbb R \to \Bbb R\setminus \{0\}: h(i, x)= i e^x ; x \in \Bbb R; i \in \{-1,1\}=O(1)$$

The continuity is easy to check.