Let $\phi: G \to H$ be a surjective homomorphism and let $\psi: F \to H$ be a homomorphism with $F$ a free group with basis X. Prove that there exists a homomorphism $\varphi: F \to G$ such that $\phi \circ \varphi = \psi$.
This is what I did so far:
Let $g \in F(X)$ with $g = x_1^{\epsilon_1} x_2^{\epsilon_2} ... x_n^{\epsilon_n}$ with $\epsilon_n \in {\pm 1}$ then every element $g' \in G$ can be expressed as $\varphi(g) = g'$, because F(X) is a free group. Thus $\varphi$ is surjective.
$\varphi$ is a homomorphism because let $g, h \in F(X)$ with $g= x_1 x_2 .. x_n y_1 y_2 ... y_m$ and $h = y_m^{-1} y_{m-1}^{-1}... y_1^{-1} x_{n+1} x_{n+2}... x_{k}$. Then $\varphi(gh) = \varphi(x_1 x_2 .. x_n x_{n+1} ... x_{k}) = \varphi(x_1 .. x_n y_1 ... y_m y_m^{-1} .. y_1^{-1} x_{n+1} ... x_k) = \varphi(g) \varphi(h)$.
We have now two consecutive surjective homomorphisms so that $\phi \circ \varphi = \psi$.
Now my questions to this problem:
Is it the right idea to show that we have to consecutive surjective homomorphisms in order to prove $\phi \circ \varphi = \psi$?
Is it enough to say that $F(X)$ is a free group and because of this every element in $G$ can be expressed as $\varphi(g)$ with $g \in F(X)$?
For each $x\in X$, choose $y_x\in G$ so that $\phi(y_x)=\psi(x)$. (I guess you need the Axiom of Choice for this if $X$ is infinite.) Extend the map $x\mapsto y_x$ to a homomorphism $\varphi:F\to G$. Since the homomorphisms $\psi$ and $\phi\circ\varphi$ agree on $X$, they are identical.
(There is no reason to expect $\varphi$ to be surjective; $G$ could even have higher cardinality than $F$.)