I have a curious question about a natural sequence, which I haven't seen answered in the literature.

Let $\Sigma$ be an oriented surface of genus $g$ without boundary with a set $\mathcal P_n$ of $n$ punctures on it. The mapping class group $\mathcal M(\Sigma,\mathcal P_n)$ is the group of orientation-preserving mapping classes which fix $\mathcal P_n$ as a set (where mapping classes are considered up to isotopies of $\Sigma$ which preserve $\mathcal P_n$ pointwise). Similarly, the pure mapping class group $\mathcal{PM}(\Sigma,\mathcal P_n)$ is the subgroup of $\mathcal M(\Sigma,\mathcal P_n)$ consisting of mapping classes which fix each puncture from $\mathcal P_n$. We have a short exact sequence: $$ 1\to \mathcal{PM}(\Sigma,\mathcal P_n)\to \mathcal M(\Sigma, \mathcal P_n)\to \mathbb S_n\to 1, $$ where $\mathbb S_n$ is the symmetric group on $n$ letters.

**Question:** for which values $(g,n)$ does this sequence split? I.e. when is there a subgroup of $\mathcal M(\Sigma,\mathcal P_n)$ isomorphic to $\mathbb S_n$ inducing the same action on punctures?

I suspect that the answer is: $$ \text{it splits only for }(g,n)=(g,1), (g,2) \text{ for arbitrary $g$, and } (g,3) \text{ for }g\not\equiv 2 \bmod 3. $$

The values of $n\ge4$ can be excluded by the following consideration: any finite subgroup of the stabilizer of a puncture in $\mathcal M(\Sigma,\mathcal P_n)$ must be cyclic (see e.g. Lemma A.3.11 in F.Castel “Geometric representations of the braid groups”, Asterisque, 378, 2016.).

That the values $n=1,2$ for arbitrary $g$ and $n=3$ for $g\equiv 0,1 \bmod 3$ give us split sequences can be seen by drawing $\mathbb S_n$-symmetrical pictures of the surface $\Sigma$. However, I don't know how to prove that for $g\equiv 2 \bmod 3$ and $n=3$ the sequence is not split. Any ideas?

**Update:** It turns out, that this sequence splits for $n=1,2,3$ regardless of genus, and does not split for $n\ge4$. See a beautiful construction in the answer of Dave Benson.