Hello! Given a surjective group homomorphism $\displaystyle \phi: A \rightarrow B$. Why is the restriction to the commutator subgroup $\displaystyle \phi': [A,A] \rightarrow [B,B]$ also surjective?
Greetings
Banach
you only need to show that the generators of [B,B], the commutators, have preimage in [A,A]. so let $\displaystyle b=b_1b_2b_1^{-1}b_2^{-1},$ where $\displaystyle b_1,b_2 \in B.$ since $\displaystyle \phi$ is surjective, $\displaystyle b_1=\phi(a_1), \ b_2=\phi(a_2),$ for some
$\displaystyle a_1,a_2 \in A.$ let $\displaystyle a=a_1a_2a_1^{-1}a_2^{-1} \in[A,A].$ then it's clear that $\displaystyle \phi(a)=b.$
Quite simply, because it's a homomorphism.
Let $\displaystyle [g,h] \in [B,B]$. Then, $\displaystyle g$ and $\displaystyle h$ are homomorphic images of elements from $\displaystyle A$, because the mapping is onto. So substitute in these elements, $\displaystyle g=a \phi$, $\displaystyle h=b \phi$ and then use the fact that $\displaystyle \phi$ is a homomorphism to see that it is in the image of $\displaystyle [A,A]$. Thus the generators of $\displaystyle [B,B]$ are contained in the image, and so the whole subgroup is.
$\displaystyle [g,h] = [a \phi, b \phi] = [a,b] \phi \in [A,A]\phi$.
the induced map $\displaystyle \phi^{ab}$ is defined by $\displaystyle \phi^{ab}([A,A]a)=[B,B]\phi(a).$ for any $\displaystyle [B,B]b \in B^{ab}$ choose $\displaystyle a \in A$ such that $\displaystyle \phi(a)=b.$ then $\displaystyle \phi^{ab}([A,A]a)=[B,B]\phi(a)=[B,B]b.$
Well, it is clear: Let $\displaystyle [b] \in B^{ab}$, then there is a $\displaystyle b\in B: p_B(b)=[b]$. But there is an $\displaystyle a \in A: \phi(a)=b. $. Since $\displaystyle \phi^{ab}(p_A(a))=p_B(\phi(a))=[b]$, we are there.
Greetings
Banach