It is well-known that the braid group $B\_{n}$ injects into the group of automorphisms of the free group $F\_{n}$. However, there is certainly a kernel when mapping to the outer automorphism group $Out(F\_{n})$. Namely, the kernel contains the generator of the center of $B\_{n}$. Could someone please explain or give a reference to the fact (?) that the whole kernel of $B\_{n} \rightarrow Out(F\_{n})$ is the center of $B\_{n}$?



Center of the braid group and outer automorphisms of the free group
Assuming we're talking about the same map and there's just a $n \longmapsto n+1$ mix-up, this question reduces to studying $Inn(F\_n)$ intersected with the image of $B\_n \to Aut(F\_n)$. 


An inner automorphism of $F\_n$ is a conjugation automorphism, so we're looking at braids that act on $F\_n$ by conjugation. Let the generators of $F\_n$ be denoted $x\_1, \cdots, x\_n$ and consider the product $x\_1x\_2\cdots x\_n$. The image of the braid group fixes this element (the image of $B\_n \to Aut(F\_n)$ is precisely the subgroup of $Aut(F\_n)$ that fixes this element). But if $c x\_1x\_2 \cdots x\_n = x\_1x\_2 \cdots x\_n c$, $c$ must be a power of $x\_1x\_2\cdots x\_n$ since otherwise $F\_n$ wouldn't be a free group. 


