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Let $MCG(S_{g,b}^s)$ be the mapping class group of a surface $S_{g,b}^s$. Assume that it is not trivial.

Is it true that $MCG(S_{g,b}^s)$ is isomorphic to $MCG(S_{g',b'}^{s'})$ if and only if $ S_{g,b}^s$ is homeomorphic to $S_{g',b'}^{s'}$ ? Or what is the complete list of counterexamples?

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  • $\begingroup$ Aren't the mapping class group of the $2$-sphere and the $2$-disk both trivial? $\endgroup$
    – Vidit Nanda
    Oct 18, 2013 at 3:04
  • $\begingroup$ Sorry, I will edit. $\endgroup$
    – Anonymous
    Oct 18, 2013 at 3:08
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    $\begingroup$ ams.org/mathscinet-getitem?mr=1666775 $\endgroup$
    – Ian Agol
    Oct 18, 2013 at 19:47

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