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Is it possible to get an holomorphic immersion form $\mathbb{C}$ to $\hat{\mathbb{C}}$ which is surjective?

Here immersion means that f'(z)\neq 0 when $f(z)$ is finite and {\left(\frac 1f\right)}'(z)\not=0 when $f(z)=\infty$. Of course if you show that $f$ is a covering you get a contradiction, but don't succeed in proving that.

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In fact it is impossible, since such an immersion would be injective and so an homeomorphism, which is clearly impossible. To prove the injectivity, let's consider $A=\cup_{z\in \hat {\mathbb{C}}} \{x \hbox { such that } f(x)=z \hbox{ and with the smallest module} \}$ if there is two $x$ with the same module and same image under $f$ we choose the one with the smalllest real part. Then we easily show that this set is open and closed and hence equal to $\mathbb{C}$ which achieves the proof.