First let's recall some definitions. Let $G$ be a perfect group, so that
$$H^2(G, A) \cong \text{Hom}(H_2(G), A)$$
for all abelian groups $A$ by universal coefficients. This means that when $A = H_2(G, \mathbb{Z})$ there is a distinguished class in $H^2(G, A)$ corresponding to the identity $H_2(G) \to H_2(G)$. The universal central extension of $G$ is the central extension classified by this map; it fits into a short exact sequence
$$1 \to H_2(G) \to \widetilde{G} \to G \to 1.$$
Now, you can find it claimed in many places that the braid group
$$B_3 \cong \langle a, b \mid a^2 = b^3 \rangle$$
is the universal central extension of the modular group
$$\Gamma \cong PSL_2(\mathbb{Z}) \cong \langle a, b \mid a^2 = b^3 = e \rangle.$$
But there's something fishy about this claim: $\Gamma$ isn't a perfect group! In fact, since $\Gamma \cong \mathbb{Z}_2 \ast \mathbb{Z}_3$, it's clear that $H_1(\Gamma) \cong \mathbb{Z}_6$ (and that $H_2(\Gamma) \cong 0$). So:
What is meant by the claim that $B_3$ is the universal central extension of $\Gamma$?
We have that
$$H^2(\Gamma, \mathbb{Z}) \cong \text{Ext}^1(\mathbb{Z}_6, \mathbb{Z}) \cong \mathbb{Z}_6$$
so presumably $B_3$ is the central extension classified by a generator of this group. But I don't understand in what sense this central extension is universal.
