How many epimorphisms are there from $F_2$ (the free group with $2$ generators) to $\mathbb{Z}_5$?
If $F_2$ is generated by $2$ elements, so it has a basis of rank$2$, meaning that it is isomorphic to$\mathbb{Z}_2$, and therefore cannot be any epimorphisms to $\mathbb{Z}_5$..
Is this correct?
On
The group $F_2 = \langle{x_1, x_2\rangle}$ is not isomorphic to $\mathbb{Z}_2$; it is free, infinite, and nonabelian. By definition, the map $\operatorname{Hom(F_2, \mathbb{Z}_5}) \to \mathbb{Z_5}^2$ given by $f \to (f(x_1), f(x_2))$ is an isomorphism. Since $\mathbb{Z}_5$ is cyclic of prime order, such a map $f:F_2 \to \mathbb{Z}_5$ fails to be surjective iff $f(x_1), f(x_2) = 0$.
No this is not correct, because a free group has no relations by definition and therefore it is not abelian, and cannot be isomorphic to $\Bbb Z_2$.
Hint: say $F_2=\langle a,b\mid \, \rangle $.
A homomorphism $F_2\to \Bbb Z_5$ is entirely determined by the image of $a$ and $b$, which can be anything and are independent.
How many homomorphisms are not surjective?