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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    $\begingroup$ To be clear, you are placing a base point on the boundary of $S_{g,1}$. Otherwise, you only get a representation into $\mathrm{Out}(F)$. $\endgroup$ – HJRW Jan 30 '12 at 13:30
  • $\begingroup$ @HW and lsw: Could you clarify what is $\rho$ and why it depends on where the basepoint is (as long as it is fixed)? $\endgroup$ – user6976 Jan 30 '12 at 15:07
  • $\begingroup$ @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. $\endgroup$ – lsw Jan 30 '12 at 15:16
  • $\begingroup$ @Isw: Why is the kernel non-trivial? $\endgroup$ – user6976 Jan 30 '12 at 16:24
  • $\begingroup$ @Mark Sapir: I don't know. Why is it trivial? $\endgroup$ – lsw Jan 30 '12 at 16:39

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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