$\mathbf R^2-\{\mathbf 0\}$ is homeomorphic to $S^1\times \mathbf R$. I am trying to to prove the following:

$\mathbf R^2-\{\mathbf 0\}$ is homeomorphic to $S^1\times \mathbf R$.

Since $\mathbf R^+=\{x\in \mathbf R:x>0\}$ is homeomorphic to $\mathbf R$, it suffices to show that $S^1\times \mathbf R^+$ is homeomorphic to $\mathbf R^2-\{\mathbf 0\}$.
There is a natural thing to try.
Define $f:S^1\times \mathbf R^+\to \mathbf R^2-\{\mathbf 0\}$ as 
$$g(\mathbf p,t)=t\mathbf p$$
It is clear that $g$ is continuous and bijective.
So we need to show that $g^{-1}$ is continuous.
It is intuitively clear to me that $g^{-1}$ is continuous but I cannot see how I can prove this in a clean way.
Perhaps there is another approach?
Thanks.
 A: $g^{-1}$ sends $re^{i\theta}$ to $(e^{i\theta},r)$. It is enough to show $re^{i\theta}\mapsto e^{i\theta}$ and $re^{i\theta}\mapsto r$ are continuous. The latter is the norm. Hence it is continuous. The first is a quotient of the identity by the norm. It remains only to prove that the quotient (reciprocal + multiplication) are continuous.
A: A standard strategy is to simply to compute $g^{-1}$ explicitly:
Since $$g({\bf p}, t) := {\bf p} t,$$ computing (and using that ${\bf p} \in S^1$) gives
$$|g({\bf p}, t)| = |{\bf p} t| = t.$$
Then, we can recover $\bf p$ by
$$\frac{g({\bf p}, t)}{| g({\bf p}, t)|} = \frac{{\bf p} t}{t} = {\bf p}.$$
Hence, the inverse map is just:
$$g^{-1}({\bf x}) = \left( \frac{\bf x}{|{\bf x}|}, |\bf{x}| \right),$$ which is manifestly continuous on $\mathbb{R}^2 - \{{\bf 0}\}$.
Your function $g$ has a nice conceptual definition, and you could just as well use it to construct this inverse: $g$ takes a direction and magnitude and gives the resulting vector, so $g^{-1}$ must take a vector and give its direction and magnitude.
