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I've given a proof of the exactness of the Birman exact sequence of groups: $$1\to\pi_1(S_{g,r}^s)\to MCG(S_{g,r}^{s+1})\overset{\lambda}{\to} MCG(S_{g,r}^s)\to 1$$ making use of classifying spaces solely. Shortly, I looked at the induced map $\lambda_\ast$ between moduli spaces, observing that the fiber should be a surface of type $S_{g,r}^s$, as "naively" one can see that it corresponds to the surface where "the removed cusp could have been". This bundle of classifying spaces induces again the starting sequence in homotopy, and the proof is complete.

Unfortunately, I've used that the moduli spaces are $K(Mod(S_{g,r}^s),1)$, whereas that's not true, in general. Do you think I can correct the proof somehow? Of course a sufficient condition would be having a free $MCG(S)$-action, but it is false for general values of $s$, $g$ and $r$.

Could I say anything, for example, assuming that $s$ is positive? What are the conditions needed to guarantee that the mapping class group is torsion-free?

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1 Answer 1

See Andy Putman's answer to this math overflow question. This at least gives you good references.

To answer your question about torsion, almost all values of $s,g$, and $r$ will have torsion in the mapping class group, and moduli space will only be a rational classifying space. Just build a symmetric looking surface with those values and isometries of that surface will form a torsion subgroup of the mapping class group.

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His name is Andy PutMaN. XD –  Patrick Da Silva Jul 11 '11 at 19:36
    
Thanks, will correct. –  Grumpy Parsnip Jul 11 '11 at 23:20
    
It was tempting to read Putnam though. –  Patrick Da Silva Jul 12 '11 at 7:06

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