Existence criterion for universal central extensions

For every abelian group $A$ and homomorphism $\varphi :G\to A$, the two maps $x\mapsto (u(x),0)$ and $x\mapsto (u(x),\varphi (u(x)))$ from $U$ to the split central extension $G\times A\to G$ must agree by universality. Since $u$ is surjective, $\varphi =0$. Taking $A=G_{\mathrm{a}\mathrm{b}}$ and $\varphi$ the quotient map gives $G_{\mathrm{a}\mathrm{b}}=0$, so $G$ is perfect.

Conversely, let $G=F\mathrm{/}R$ be perfect. Then $F=[F,F]R$, so $[F,F]\mathrm{/}[F,R]\to G$ is a central surjection. Given any central extension $E\to G$, lift the free generators of $F$ to $E$. The resulting map $F\to E$ kills $[F,R]$ on $[F,F]$ because the kernel of $E\to G$ is central. Different choices of lifts differ by central kernel elements and hence agree on $[F,F]$, giving a canonical map $[F,F]\mathrm{/}[F,R]\to E$ over $G$. Writing $F^{\mathrm{\prime }}=[F,F]$, the identity $F=F^{\mathrm{\prime }}R$ implies $F^{\mathrm{\prime }}=[F^{\mathrm{\prime }},F^{\mathrm{\prime }}][F,R]$, so $[F,F]\mathrm{/}[F,R]$ is perfect. The pointwise difference of any two maps from this group to $E$ over $G$ is therefore a homomorphism to the central abelian kernel of $E\to G$, and must vanish. Thus the canonical map is unique, proving universality.

Steps 1.1 and 1.2 prove the two implications.