Induced group homomorphism Task:
Assume $G$, $Q$ and $H$ are groups. Furthermore, let $\phi: G \rightarrow H$ be a group homomorphism and $\xi: G \rightarrow Q$ be a surjective group homomorphism. Besides that, assume that $kern(\xi) \subset kern(\phi)$. Then there is exactly one group homomorphism $f: Q \rightarrow H$ such that $\phi = f \circ \xi$.
Proof:
First, we show the definiteness. For every $u \in Q$ there is at least one $g \in G$ such that $\xi(g) = u$. Since the diagram commutates it must be $f(u) = \phi(g)$. This means that there can't exist more than one $f$. 
We have to show that this leads to a well defined function. So assume that $g, g'$ $\in G$ are two preimages of $u$. Then, we can conclude that
$g'g^{-1} \in kern(\xi) \subset kern(\phi)$ and therefore $\phi(g) = \phi(g')$ such that the function is indeed well defined. 
Assume $u, v \in Q$ and let $g, h \in G$ be preimages of $u$ and $v$. Then $gh$ is a preimage of $uv$ and therefore
$f(uv) = \phi(gh) = \phi(g)\phi(h) = f(u)f(v).$
This means that $f$ is a group homomorphism. 
Questions:


*

*Why does he conclude that $g'g^{-1} \in kern(\xi) \subset kern(\phi)$? Since he assumed that $g$ and $g'$ are two preimages of $u$, it should be clear that $\phi(g) = \phi(g')$.

*Furthermore, why does he need to prove $\phi(g) = \phi(g')$ anyway? Is he actually showing that $\phi$ is well-defined? If so, why is this necessary? 

*Why is $f(uv) = \phi(gh)$? 
 A: *

*You have $\xi(g)=\xi(g')$, hence $e_Q=\xi(g)^{-1}\xi(g')=\xi(g^{-1})\xi(g')=\xi(g^{-1}g')$ since $\xi$ is a group homomorphism. And so $\phi(g)=\phi(g')$. So it doesn't matter for your $f$ which preimage you choose and $f$ is hence well defnied (see 2).

*You could have several preimages and he wants to show it doesn't matter which one you choose for your lift. Hence he shows $f$ is well definied ($\phi$ is well defined by definition).

*Consider $uv=z$ as an element of $Q$. Then by definition you chose $f(z)=\phi(g_z)$ with $g_z=gh$ a preimage of $z$ (since $\phi$ is a group homomorphism you can choose a product of preimages of $u$ and $v$).

A: It's usually easier to work out uniqueness first.
Suppose $f\colon Q\to H$ exists such that $\phi=f\circ\xi$. We want to see where $f$ should map an element of $Q$ onto. Since $\xi$ is surjective, for any $q\in Q$ there is $g\in G$ with $q=\xi(g)$, so
$$
f(q)=f(\xi(g))=\phi(g)\tag{1}
$$
and therefore $f$ is unique, provided it exists. It is evident that $f(q)$ should not depend on the preimage we choose for $q$. Now, suppose $q=\xi(g)=\xi(g')$; then $g^{-1}g'\in\ker\xi\subset\ker\phi$. Hence
$$
1=\phi(g^{-1}g')=\phi(g)^{-1}\phi(g')
$$
which implies $\phi(g)=\phi(g')$. Thus $f$ can be defined by $(1)$, because $f(q)$ does not depend on the particular preimage.
Is $f$ so defined a homomorphism? Yes, because if $q=\xi(g)$ and $q'=\xi(g')$, we also have
$$
qq'=\xi(g)\xi(g')=\xi(gg')
$$
and therefore $gg'$ is a preimage for $qq'$. Then, by definition,
$$
f(qq')=\xi(gg')=\xi(g)\xi(g')=f(q)f(q')
$$
and $f$ is a homomorphism.
As to your questions:


*

*If $\alpha\colon G\to G'$ is a group homomorphism, then $\alpha(g_1)=\alpha(g_2)$ if and only if $g_1^{-1}g_2\in\ker\alpha$.
Proof If $\alpha(g_1)=\alpha(g_2)$, then $1=\alpha(g_1)^{-1}\alpha(g_2)=\alpha(g_1^{-1})\alpha(g_2)=\alpha(g_1^{-1}g_2)$ and $g_1^{-1}g_2\in\ker\alpha$. Work out the other direction.

*It is necessary to show that $\xi(g)=\xi(g')$ for defining $f$.

*See above.
