Universal central extension groups are superperfect

For an abelian group $A$ and a homomorphism $\varphi :U\to A$, the maps $x\mapsto (u(x),0)$ and $x\mapsto (u(x),\varphi (x))$ from $U$ to the split central extension $G\times A\to G$ are both over $G$. Universality makes them equal, so every such $\varphi$ vanishes. Taking $A=U_{\mathrm{a}\mathrm{b}}$ shows that $U$ is perfect.

Let $e:E\twoheadrightarrow U$ be any central extension. The composite $[E,E]\to U\to G$ is surjective because $e([E,E])=[U,U]=U$. If $x\in [E,E]$ maps to $1$ in $G$, then $e(x)\in \ker u\subseteq Z(U)$, whence $[x,E]\subseteq \ker e\subseteq Z(E)$ and therefore $[x,[E,E]]=1$. Thus $[E,E]\to G$ is a central extension.

Universality of $u$ gives a map $s:U\to [E,E]$ over $G$. Both $e\circ s$ and ${\mathrm{id}}_U$ are maps from $U$ to the central extension $u:U\to G$ over $G$, so uniqueness gives $e\circ s={\mathrm{id}}_U$. Consequently every central extension of $U$ splits.

Since $U$ is perfect, the free-presentation theorem under Choice gives a universal central extension $v:V\to U$. Under the resolution hypotheses, its kernel is $M(U)$ by Kernel of the universal central extension. Step 3.1 gives a homomorphic section, so centrality yields $V\cong M(U)\times U$. The argument of step 1.1 applied to $v$ makes $V$ perfect. Abelianizing this product gives $M(U)=0$. Also $U_{\mathrm{a}\mathrm{b}}=0$ by step 1.1, so $H_1(U;\mathbb{Z})=H_2(U;\mathbb{Z})=0$ and $U$ is superperfect.