TITLE: Check whether a given map is a complex chart
QUESTION [0 upvotes]: An example of complex chart on $\mathbb{R^2}$ is given by $f_U$ defined below:
$$
f_U(x,y)=\frac{x}{1+\sqrt{x^2+y^2}}+i\frac{y}{1+\sqrt{x^2+y^2}}
$$
where $U$ is an open subset of $\mathbb{R}^2$. How do I show that $f_U$ is an homeomorphism? It is clearly continuous. An inverse could be $f^{-1}(z)=(\frac{Re(z)}{1-|z|},\frac{Im(z)}{1-|z|})$. However, the last one is not in general continuous. How to solve this problem?

REPLY [1 votes]: We have $|f_U(x,y)| <1$, hence $f^{-1}(z)=(\frac{Re(z)}{1-|z|},\frac{Im(z)}{1-|z|})$ is only defined for $z \in \mathbb C$ with $|z|<1$.
Conclusion: $f^{-1}$ is continuous.
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