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Let $X$ be a simply connected algebraic curve over $\mathbb{C}$ and $f:X\rightarrow \mathbb{A}^1_{\mathbb{C}}$ is a surjective étale map. Then is it true $f$ is finite?

All the domains of non finite surjective étale maps I have ever seen are not simply connected. So I conjectured this.

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$X$ is smooth since $f$ is étale and simply connected, so $X$ is $\mathbb P^1$ or $\mathbb A^1$. Every map $\mathbb P^1 \to \mathbb A^1$ is constant hence not étale. Every map $\mathbb A^1 \to \mathbb A^1$ is given by a polynomial, hence finite unless it's constant, and again constant maps are not étale.

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  • $\begingroup$ Thank you for answering. If $X$ is smooth, why $X$ is $\mathbb{P}^1$ or $\mathbb{A}^1$? $\endgroup$
    – George
    Commented Apr 3 at 20:12
  • $\begingroup$ @George Those are the only simply-connected smooth algebraic curves. The fundamental group of a curve of genus $g$ with $n$ missing points is free on $2g-2+n$ generators (or a surface group if $n=0$) hence simply-connected implies $g=0$ and $n\leq 1$. $\endgroup$
    – Will Sawin
    Commented Apr 3 at 20:21
  • $\begingroup$ I'd like to know more about these fact. I have little knowledge about fundamental group of algebraic curve. Could you recommend a reference that has a proof that those are the only simply connected smooth algebraic curves. $\endgroup$
    – George
    Commented Apr 3 at 20:34
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    $\begingroup$ @George In the complex-analytic category, one can use the en.wikipedia.org/wiki/Uniformization_theorem together with the observation that the unit disc is not an algebraic curve $\endgroup$
    – Will Sawin
    Commented Apr 3 at 20:53
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    $\begingroup$ @George One has to know enough of the classification of algebraic curves to know there is a unique one biholomorphic to $\mathbb A^1$. The key thing is that every smooth algebraic curve is a smooth projective algebraic curve with some points deleted. One checks the projective curve has to have genus zero and then that only one point can be deleted and concludes it is $\mathbb A^1$. $\endgroup$
    – Will Sawin
    Commented Apr 4 at 0:24

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