$F_n$ is the free group of rank $n$. I know that it is possible to construct an Epimorphism $f: F_n \to F_{n-1}$. How is this done? (I also want to consider the case $F_2 \to \mathbb{Z}$ but maybe there is no difference in the argument.)
My thoughts: Maybe one can use the fact that $F_2$ contains the free group of rank $k$ for every $k \in \mathbb{N}$. This implies that $F_n$ can be viewed as a subgroup of $F_{n-1}$ for $n$ at least $3$.
Free groups have a very special property. Each $F_n$ has a set of generators $\{a_1,\ldots,a_n\}$, called free generators. These generators have the following property:
If $G$ is any group and $h:\{a_1,\ldots, a_n\}\to G$ is any function, then there is a unique group homomorphism $H:F_n\to G$ such that $H(a_i)=h(a_i)$.
With this property we can easily show that there always is an epimorphism $F_n\to G$ if $G$ has a set of at most $n$ generators (regardless of whether $G$ is free or not). And this in fact generalizes to free groups over arbitrary alphabets, not only finite.