Free-presentation construction is universal

The chosen basis lifts extend uniquely to a homomorphism $h:F\to E$ over $G$. Every $r\in R$ maps into the central kernel of $E\to G$, so $h([F,R])=1$. Restricting $h$ to $[F,F]$ therefore gives a homomorphism $[F,F]\mathrm{/}[F,R]\to E$ over $G$. Since $G$ is perfect, the displayed source maps onto $G$, and the preceding central-extension lemma gives its central kernel.

If $v,w:[F,F]\mathrm{/}[F,R]\to E$ are two maps over $G$, their pointwise ratio $v(x)w(x{)}^{-1}$ lies in the central kernel. Centrality makes this ratio a homomorphism to an abelian group. The source is perfect: write elements $a,b\in F$ as $a=a^{\mathrm{\prime }}r$, $b=b^{\mathrm{\prime }}s$ with $a^{\mathrm{\prime }},b^{\mathrm{\prime }}\in [F,F]$ and $r,s\in R$, which is possible because $G$ is perfect; modulo $[F,R]$ one has $[a,b]=[a^{\mathrm{\prime }},b^{\mathrm{\prime }}]$. Its generators are thus commutators inside the source. Every homomorphism from it to an abelian group is zero, so $v=w$. Existence from step 1.1 and this uniqueness prove universality.