Does $Ab(\pi_1(\Sigma_2)/N)=1$ imply $\pi_1(\Sigma_2)/N = 1$? Let $\Sigma_2$ be a genus two surface. If $N$ is a normal subgroup of $\pi_1(\Sigma_2)$ such that the abelianization of the quotient $Ab(\pi_1(\Sigma_2)/N)$ is  trivial, does that also imply that the quotient $\pi_1(\Sigma_2)/N$ is also trivial?

We don't really need a sledgehammer here.  You can get an explicit homomorphism from $\pi_1(\Sigma_2)=\langle x,y,u,v\mid [x,y][u,v]\rangle$ onto the alternating group $A_6$ by sending the generators $x$ and $u$ to $a=(4,5,6)$, and sending $y$ and $v$ to $b=(1,2,3,4,5)$ in $A_6$. Since $(4,5,6)$ and $(1,2,3,4,5)$ generate $A_5$, this is surjective.
