The free group on two generators $F_2=\langle x,y|\rangle$ is the fundamental group of $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$. Now, there are plenty of galois covers of this space whose galois group is not solvable. Thus the "maximal solvable cover" (i.e. the limit over all galois covers with solvable galois group) is not the universal cover, but rather a quotient thereof. In other words, the natural map: $$F_2\to\lim_{\begin{smallmatrix}\longleftarrow\cr H\unlhd F_2\cr F_2/H\text{ finite solvable}\end{smallmatrix}}F_2/H$$ is not injective.

Can someone exhibit an explicit element of the kernel? What about the shortest element (by word length in $F_2$) in the kernel? In other words, the question is: what universal word in $x,y$ always vanishes when $x,y$ are specialized to elements of some solvable group $G$? (note that since $G$ is solvable, so is the subgroup generated by $x$ and $y$).

Such an element now has the following seemingly impossible property. Consider it as a closed path $\gamma$ in $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$. Now try to lift $\gamma$ to the cover of $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$ corresponding to the linking number with $0$ (i.e. we take the universal cover of $\mathbb P^1(\mathbb C)\setminus\{0,\infty\}$ and take the inverse image of $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$). Of course $\gamma$ lifts to a *closed* curve, since otherwise we just exhibited an abelian quotient of $F_2$ in which $\gamma$ is not sent to zero. The cover we just considered is $\mathbb P^1(\mathbb C)\setminus\mathbb Z$, and we can try to lift $\gamma$ to some abelian cover of this, etc. Of course, $\gamma$ always lifts to a *closed* curve, since all these covers are solvable! I'm having a hard time visualizing what such a curve $\gamma$ would look like geometrically in $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$ (I once thought all elements of $F_2$ could be "broken" by a sequence of such covers!)