Here is the set up. Let $$G$$ be a perfect group ($$G = [G,G]$$), and let $$gamma:F to G$$ be a homomorphism from a free group $$F$$ onto $$G$$. Let $$R subset F$$ be the kernel. Then $$[R,F] subset F$$ is a normal subgroup, and we have a surjection
$$phi:F/[R,F] to F/R cong G$$
The kernel of $$phi$$ is central, so by a previous lemma of Milnor’s, the commutator subgroup
$$(F/[R,F])’ = [F,F]/[R,F]$$
is a perfect central extension of $$G$$. Now we want to show that $$[F,F]/[R,F]$$ is the universal central extension. Let $$(X,psi)$$ be a central extension of $$G$$. Then by the universal property of free groups, there is a homomorphism $$h:F to X$$ over $$G$$, that is, $$psi circ h = gamma$$.
All this I can follow. Next, Milnor assers that $$h[R,F] = 1$$, which I don’t see. I know that $$psi circ h[R,F] = gamma[R,F] = 1$$, so $$h[R,F] subset ker psi$$. We know that $$psi$$ is a central extension, so $$ker psi$$ is contained in the center of $$X$$. How do I get from here to see that $$h[R,F] = 1$$?