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Let $X$ be a smooth projective curve over $\mathbb{C}$ and puncture it at a point $x$. We get a map from $spec(\mathbb{C}((z)))$ to the punctured curve $X^{°}$ corresponding to this puncture. Is the induced map of étale fundamental groups non-zero if the genus of $X$ is greater than 0? This might be equivalent to finding an étale cover of the punctured curve not extending to an étale cover of the complete one?

Edit: I should really say corresponding to the puncture and the choice of a coordinate at $x$.

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Yes. Since you are over $\mathbb{C}$, we can work with topological fundamental groups and then take the profinite completion. Suppose the genus is $g>0$, then we have a standard model for $X$ as a $2g$-gon $P$ with the sides identified appropriately. Assume that $x$ is a point of the interior of $P$. Let $\alpha_1,\ldots, \alpha_{2g}\in \pi_1(X)$ be generators corresponding to adjacent sides of $P$, then $\partial P=[\alpha_1,\alpha_2]\ldots[\alpha_{2g-1},\alpha_{2g}]=1$ in $\pi_1(X)$. However, $\partial P$ in nontrivial in $\pi_1(X-x)$ because this group is freely generated by the $\alpha_i$. Let $D \subset P$ be a small disk centered at $x$. Then $\partial D$ generates $\pi_1(D-x)$. This curve is homotopic to $\partial P$ which as I said is nontrivial in $\pi_1(X-x)$. Since free groups are residually finite, this remains nontrivial in the profinite completion.

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  • $\begingroup$ Am I to understand that the answer is "yes" to my claim that it's equiv to the existence of an étale cover not pulled back from the compactification, i guess Beauville-Laszlo? Your answer then provides a construction of such. (Which in hindsight was kind of clear given Riemann existence!) Thanks a lot :) $\endgroup$
    – user108998
    Commented Nov 13, 2017 at 14:20
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    $\begingroup$ I actually had in mind "yes" for your first question. Although the second is also true as well. $\endgroup$ Commented Nov 13, 2017 at 14:33
  • $\begingroup$ OK thanks. I was confused since ofc a v small punctured disc is not the analytification of spec of Laurent power series. I've made sense of it now. $\endgroup$
    – user108998
    Commented Nov 13, 2017 at 16:12

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