Consider the surface group $S_g=langle a_1,b_1,a_2,b_2,dots,a_g,b_g mid (a_1,b_1)(a_2,b_2)cdots(a_{g},b_{g})=1rangle$, which is the fundamental group of the closed orientable genus-$g$ surface.

Suppose $2leq m<n$ and let $p:S_nto S_m$ be the canonical projection, which is the identity on generators $a_i,b_i$, $1leq ileq m$ and which trivializes $a_i,b_i$, $m+1leq ileq n$. Let $F_r$ denote the free group with rank $r$.

**Question:** Is it possible to have finite rank free group $F_r$ and epimorphisms $f:S_nto F_r$ and $g:F_rto S_m$ such that $gcirc f=p$?

There is apparently a kind of classification of epimorphisms of surface groups onto free groups that seems relevant (in the following reference). However, I have not been able to see exactly how it helps answer this question.

R. I. Grigorchuk, P. F. Kurchanov, and H. Zieschang, “Equivalence of homomorphisms of surface groups to free groups and some properties of 3-dimensional handlebodies,” Preprint, Ruhr-Universität Bochum, 1990.

The classification: If $r>n$, then no epimorphism $S_nto F_r$ can exist. Apparently, when $rleq n$, then there is only one epimorphism $S_nto F_r$ “up to automorphism.” The representative is this: choose $F:S_nto F_r$ to send $a_i$, $1leq ileq r$ to the free generators and trivialize all other generators. This is unique in the sense that for any other epimorphism $f:S_nto F_r$, we must have $F=fcirc sigma$ for some $sigmain Aut(S_n)$.

Considering this classification and the structure of $p$, my feeling is that the question has a negative answer but that I am perhaps overlooking something simple.