Do we have $ f(\mathbb{C}^{*}) = \mathbb{C}\setminus\{0,1\} $ for $ f $ continuous? It is known that the exponential function continuously sends  $ \mathbb{C} $ onto  $  \mathbb{C}^{*} $ .
There is no "hole" in  $ \mathbb{C} $ and there is one in $  \mathbb{C}^{*}  $.
But what if we try to continuously send a one-hole set onto a two-hole one, for example $  \mathbb{C}^{*}   $ 
onto  $  \mathbb{C}\setminus\{0,1\}  $ ? 
 A: The following function continuously sends $\mathbb{C} \backslash \{0\}$ to $\mathbb{C} \backslash \{0,1\}$ : $$f : a+ib \longmapsto \exp\Big(a+\frac{2 i \pi}{1+(a-b)^2}\Big).$$
Indeed, $f(\mathbb{C}) \subset \mathbb{C}\backslash \{0\}$, and as $\frac{2\pi}{1+(a-b)^2} \in ]0,2\pi]$, if $f(a+ib)=1$, then $a=0$ and $\frac{2i\pi}{1+(a-b)^2}=2i\pi$, so $b=a$, and thus $a+ib=0$... Hence $f(\mathbb{C} \backslash \{0\}) \subset \mathbb{C} \backslash \{0,1\}$.
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For $y \in \mathbb{C} \backslash \{0,1\}$, there exists $r>0$ and $\theta \in ]0,2\pi]$ such that $y=re^{i\theta}$, and $r \neq 1$ or $\theta \neq 2\pi$.
Let us denote $a=\mathrm{ln}(r)$ and $b=a+\sqrt{\frac{2\pi}{\theta}-1}$. Then $f(a+ib)=\exp \Big( \mathrm{ln}(r)+i \theta \Big) = y$. We can't have $a+ib=0$ as it would imply $r=1$, $\theta=2\pi$, and so we have $\mathbb{C} \backslash \{0,1\} \subset f(\mathbb{C}\backslash \{0\})$.
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As $f$ is continuous, $f$ continuously sends $\mathbb{C} \backslash \{0\}$ to $\mathbb{C} \backslash \{0,1\}$.
