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Let $M_g$ be the moduli space of Riemann surfaces, as described for example in the book of Harris and Morrison - Moduli of curves. As a topological space, or better as orbifold, it has smooth points and singular points. Let $S\subseteq M_g$ be the subset of smooth points. In the article by Cornalba ("On the locus of curves with automorphisms") this subset is identified, in the case $g>3$, with the set of those curves which have just the trivial automorphism.

Is there any idea (or any reference) of how to compute the homotopy groups $$\pi_2(M_g,S) \qquad \mbox{and} \qquad \pi_1(S)?$$

Edit: I should have included a base point above, but I intentionally didn't because I wouldn't know which one to choose.

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I'm not sure why you're talking about codimension $1$ components of $S$ since the smooth locus $S$ is actually dense in $\mathcal{M}_g$.

Anyway, the fundamental group of the locus $S$ of smooth points is the mapping class group $Mod_g$ of the surface $\Sigma_g$. To see this, recall that $\mathcal{M}_g$ is the quotient of Teichmuller space $\mathcal{T}_g$ by the action of $Mod_g$. Let $\widetilde{S}$ be the preimage of the smooth locus in $\mathcal{M}_g$ under the projection $\mathcal{T}_g \rightarrow \mathcal{M}_g$. The action of $Mod_g$ on $\mathcal{T}_g$ preserves $\widetilde{S}$, and moreover the restriction of the action of $Mod_g$ to $\widetilde{S}$ is free (the only fixed points of the action of $Mod_g$ on $\mathcal{T}_g$ come from curves with automorphisms). It follows that $\widetilde{S} \rightarrow S$ is a regular cover with deck group $Mod_g$. To prove that the fundamental group of $S$ is $Mod_g$, it is enough to prove that $\widetilde{S}$ is $1$-connected. For this, recall that the locus of curves with automorphisms has complex codimension $2$ in $\mathcal{M}_g$ as long as $g > 3$. This implies that the complement in $\mathcal{T}_g$ of $\widetilde{S}$ also has complex codimension $2$ and thus real codimension $4$, which implies in particular that the inclusion $\widetilde{S} \hookrightarrow \mathcal{T}_g$ induces an isomorphism on fundamental groups. The desired result now follows from the fact that $\mathcal{T}_g$ is contractible.

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  • $\begingroup$ yes you are right about $S$ being dense. I mixed up things because in the article I quoted the author was talking about the complement of $S$. Great argument for $\pi_1(S), very clear and clean! Thank you a lot! $\endgroup$ Commented Nov 3, 2017 at 9:51
  • $\begingroup$ from the long exact sequence of the pair $(M_g,S)$ it is clear at least that $\pi_2(M_g,S)$ cannot be zero. $\endgroup$ Commented Nov 3, 2017 at 11:33

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