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 $\rho$?



Kernel of the representation of the mapping class group to $Aut(F_n)$
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](http://www.ams.org/mathscinet-getitem?mr=224099)). 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. 


