Let $f:F_n\to F_m$ be an epimorphism ($n\geq m$). Then it is true that there is a basis $X=X_1\sqcup X_2$ in $F_n$ such that $f$ maps $\langle X_1\rangle$ isomorphically onto $F_m$, and maps $X_2$ to identity. One can probably deduce this statement from the Grushko-Neumann theorem, but somehow I cannot get an elegant proof of it.
I can see that by the Grushko-Neumann we can assume that $m=1$, and hence our epimorphism factors through the abelianization. And then an ugly linear algebra appears... Any suggestions how to make the proof slick?
I don't think you need Grushko-Neumann for this, you just need to know about the Nielsen reduction process.
Start with any (ordered) free generating set $Y = \{ y_1, y_2, \ldots, y_n \}$ of $F_n$ and then apply the Nielsen reduction process to the generating sequence $\phi(Y) = (\phi(y_1),\ldots,\phi(y_n))$ of the free group $F_m$. But carry out the reduction operations on $Y$ as well as $\phi(Y)$.
A t the end of the process you will have what you are looking for. The set $X_1$ will be the inverse image under $\phi$ of the resulting free generating set of $\phi(F)$, whereas $X_2$ will map onto the generators that get eliminated during the reduction process as being the identity element of $F_m$.