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Let $S_{g,1}$ be a orientable compact surface of genus $g$ with one boundary component and $\Gamma_{g,1}$ the mapping class group. By $F_n$ I denote the free group on $n$ generators.

One obtains a representation $\rho: \Gamma_{g,1} \rightarrow Aut(F_{2g})$.

What is the kernel of $\rho$?

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To be clear, you are placing a base point on the boundary of $S_{g,1}$. Otherwise, you only get a representation into $\mathrm{Out}(F)$. –  HJRW Jan 30 '12 at 13:30
    
@HW and lsw: Could you clarify what is $\rho$ and why it depends on where the basepoint is (as long as it is fixed)? –  Mark Sapir Jan 30 '12 at 15:07
    
@HW: Thank you for making this precise. @Mark Sapir: You have to consider the induced action on the fundamental group of $S_{g,1}$. By fixing a base-point there is no $Inn(\pi_1)$-action. –  lsw Jan 30 '12 at 15:16
    
@Isw: Why is the kernel non-trivial? –  Mark Sapir Jan 30 '12 at 16:24
    
@Mark Sapir: I don't know. Why is it trivial? –  lsw Jan 30 '12 at 16:39
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1 Answer

up vote 5 down vote accepted

The representation is faithful, since a mapping class is determined by its action on the fundamental group of the surface. A surface is a $K(\pi,1)$, so given any element $Aut(S_{g,1})$, one obtains a (pointed) map $\varphi:S_{g,1}\to S_{g,1}$ which is unique up to homotopy. Now one needs to know that two homotopic homeomorphisms of a surface are isotopic, which is classic (at least one may find this in a paper of Waldhausen). In fact, one may identify the image in $Aut(F_{2g})$ as the subgroup preserving the peripheral element. Also, note that everything should be fixing a basepoint in the boundary, as in HW's comment.

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Thank you, this is a very nice answer. –  lsw Jan 30 '12 at 19:50
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