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Let $S_{g,b}$ an orientable surface with genus $g$ and $b$ boundary components and $S_g^b$ be an orientable surface with $b$ punctures.

Denote by $PMCG(S_g^b)$ and $PMCG(S_{g,b}) $ the pure mapping class groups, that is, the group of orientation preserving homeomorphisms of the surface fixing the punctures or the boundary pointwise modulo isotopy.

Is $$0 \to \mathbb Z^b \to PMCG(S_{g,b}) \to PMCG(S_g^b) \to 0 $$ a split short exact sequence ?

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No, it is not split (except in a few degenerate cases like $(g,n) = (0,1)$ or $(g,n)=(0,2)$; let's assume that $g \geq 2$ for the moment just to be careful). It is a central extension, so if it was split then the abelianization of the pure mapping class group of $S_{g,b}$ would contain a copy of $\mathbb{Z}^b$; however, this abelianization is trivial.

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  • $\begingroup$ Thank you for your answer. Could you clarify your point about abelianizations and precise why the abelianization of $PMCG(S_{g,b})$ is trivial? $\endgroup$
    – Anonymous
    Commented Oct 15, 2013 at 13:18
  • $\begingroup$ I am sorry, but there a still a couple of things I still don't get. I'll write here, there may be naive mistakes, please correct me if I am wrong. $PMCG(S_g^b)$ is generated by Dehn twists around non-trivial simple closed curves, right? Every non-trivial simple closed of $S_g^b$ can be seen as a non-trivial simple closed curve in $S_{g,b}$, so it seems that $PMCG(S_g^b)$ is "naturally" a subgroup of $PMCG(S_{g,b})$. So why does the inclusion $PMCG(S_{g}^b) \to PMCG(S_{g,b})$ not split the sequence above? Thank you in advance. $\endgroup$
    – Anonymous
    Commented Oct 15, 2013 at 13:19
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    $\begingroup$ @yanglee : For your first question, see any standard source on the mapping class group (eg Farb-Margalit). For your second question, try to write down a careful construction of your inclusion and you will see that it is not well-defined even though it is well-defined on Dehn twists. $\endgroup$
    – Andy Putman
    Commented Oct 15, 2013 at 16:21
  • $\begingroup$ I see the problem now, thank you very much! $\endgroup$
    – Anonymous
    Commented Oct 15, 2013 at 17:09

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