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Nov 28, 2012 at 16:22 comment added Robert Kucharczyk In fact the "naive" fundamental group of $\mathscr{M}_g(\mathbb{C})$ is trivial.
Nov 28, 2012 at 16:19 history edited Robert Kucharczyk CC BY-SA 3.0
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Nov 28, 2012 at 16:18 comment added Robert Kucharczyk @ Qiaochu: yes.
Nov 28, 2012 at 9:02 comment added S. Carnahan I've been told that $\mathcal{M}_g^{an}$ is a "more correct" notation than $\mathcal{M}_g(\mathbf{C})$, since the latter is apparently just a set.
Nov 28, 2012 at 1:12 comment added ThiKu @Qiaochu: I guess you meant $g>1$?
Nov 27, 2012 at 18:31 comment added Qiaochu Yuan In the last claim doesn't "fundamental group" need to be replaced by "orbifold fundamental group" (at least for $g = 1$)?
Nov 27, 2012 at 17:47 history answered Robert Kucharczyk CC BY-SA 3.0