So you have a free group $F\_n$, freely generated by $\alpha\_1 \cdots \alpha\_n$. Pick any $n$ elements $g\_1 \cdots g\_n$ and define an endomorphism $\psi$ of $F\_n$ by $\psi(\alpha\_i) = g\_i^{-1}\alpha\_ig\_i$ and extend as usual.  

It looks very much to me that $\psi$ will always be injective, but I'm having a hard time proving this. Any ideas? 


Cheers in advance.


Hair



An endomorphism of free groups
It's injective. Indeed, the image is free of rank $k\le n$, and is injective if and only if $k=n$ (as $F\_n$ is Hopfian: is not a proper quotient of itself). Since the image surjects onto the abelianization $\mathbf{Z}^n$, we have $k=n$. More generally, any endomorphism of a free group that maps onto a finite index subgroup of the abelianization has to be injective.


As Aurel mentions, it can fail to be an automorphism: writing $F\_2=\langle x,y\rangle$, the subgroup generated by $x$ and $(yx)y(yx)^{-1}$ is a proper subgroup (of infinite index).


