Let $F_n$ be the free group generated by $x_1,\cdots,x_n$ and let $K_n$ be the reduced free group, that is, $F_n$ modulo the relation that $[x_i,x_i^g]=1$ for all $i\in\{1,\cdots,n\}$,$g\in F_n$, where $x_i^g=g^{-1}x_ig$ and $[\cdot,\cdot]$ is the commutator. There is a natural quotient map $q_n:F_n\to K_n$.
My question is, if a group homomorphism $f:F_m\to F_n$ induces two homomorphisms $f'$ and $f'':K_m\to K_n$; that is, the following two diagrams are commutative: $$\begin{array} A & F_m &{\stackrel{f}{\longrightarrow}} & F_n & \\ & \downarrow{q_m} & &\downarrow{q_n}& \\ &K_m & \stackrel{f'}{\longrightarrow} & K_n & & \end{array}$$ $$\begin{array} A & F_m &{\stackrel{f}{\longrightarrow}} & F_n & \\ & \downarrow{q_m} & &\downarrow{q_n}& \\ &K_m & \stackrel{f''}{\longrightarrow} & K_n & & \end{array}$$ Is it true that $f'=f''$?
Edit: Actually the existence of such a function $f'$ is probably worthy asking as well; but at the moment I am mainly interested in the uniqueness.