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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 $g \in F(X)$?

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  • $\begingroup$ How are you defining $\varphi$ here? $\endgroup$ – Tim Raczkowski Dec 22 '14 at 16:43
  • $\begingroup$ As far as I can see you have not defined $\varphi$. $\endgroup$ – bof Dec 22 '14 at 16:44
  • $\begingroup$ Suppose $F=H=\mathbb Z$ and $G=\mathbb Z\times\mathbb R$. There is a surjective homomorphism $\phi:G\to H$ and there is a homomorphism $\psi:F\to H$ but there is no surjective homomorphism $\varphi:F\to G$. $\endgroup$ – bof Dec 22 '14 at 16:47
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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 $F$.)

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  • $\begingroup$ Is it necessary to prove that the homomorphism $\varphi$ is indeed a homomorphism? Or is it automatically given since we have extended the map $x \mapsto y_x$ in a way that gives a homomorphism? $\endgroup$ – FlyingDutchman Dec 22 '14 at 17:36
  • $\begingroup$ @FlyingDutchman: The latter. "$F$ a free group with basis $X$" means that any map $\varphi_0:X\to G$ (where $G$ is a group) can be extended (uniquely of course) to a homomorphism $\varphi:F\to G$. $\endgroup$ – bof Dec 22 '14 at 17:45
  • $\begingroup$ So the problem is very different from what I thought. Thank you for your answer! $\endgroup$ – FlyingDutchman Dec 22 '14 at 17:56

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