I’m having trouble verifying what ought to be a relatively simple detail in a proof of Milnor’s book on algebraic K-theory, in the section on universal central extensions.
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 $h[R,F] = 1$?