My question is the following: Let $G:=F_n$ If we look at the commutator subgroup $[G,G]$ of $G$, we get the canonical epimorphism
$\varphi: G\to G/[G,G]$
Since $[G,G]$ is characteristic in $G$, we know that $Aut(G)$ acts in a natural way on the factor group $G/[G,G]$ and we get a map:
$\Phi:\mathrm{Aut}(G)\to \mathrm{Aut} (G/[G,G]);\alpha(g) \mapsto \bar{\alpha}(g*[G,G]):=\alpha(g)*[G,G]$
But how can I show that $\Phi$ is an epimorphism?
Added.
When I was asking the question we were in the general case, where $G$ is an arbitrary group. Because of some answers, I edited the question into the case, where $G=F_n$, the free group of rank $n$.
Thanks to the last comment, I now know that there is a solution in the book "Combinatorial Group Theory" from Magnus. I don't have the book beside me. So does someone knows a proof for the existence of the epimorphism $\Phi$. I think, if we assume that $Aut(F_n)$ is generated by the right nielsen transformations, we only have to show that $Aut(F_n/[F_n,F_n])$ is generated by these trasnformations, since we know that every $\alpha\in Aut(F_n)$ induces an $\bar{\alpha}\in Aut(F_n/[F_n,F_n])$. Is this true? And how can we get this?