Proving bicontinuity of $\phi : \mathbb{R^+}\times S^1 \rightarrow \mathbb{C^+}$.

Question

The function is given by $\phi(a,e^{i\theta}) = re^{i\theta}$, with $0\leq \theta<2\pi$.

I'm trying to prove that is an isomorphism of groups and also and homeomorphism, my teacher said we need to use the definition of continuity that use open balls, and while I don't have problem knowing what are the open balls in $\mathbb{C}$.

I don't know how the open balls in $\mathbb{R^+}\times S^1$ look. Could you guys help me with that part.

Answer 1

I think that it is more natural to express $\phi$ as: $\phi(r,z)=rz$ ($t\in\mathbb{R}^+$ and $z\in S^1$). But then it is clear that $\phi$ is continuous. And $\phi^{-1}(z)=\left(\lvert z\rvert,\frac z{\lvert z\rvert}\right)$. So, it is also clear that $\phi^{-1}$ is continuous.