Commutator map is surjective homomorphism from exterior square to derived subgroup

Statement with left action convention
Suppose $$G$$ is a group. Consider the commutator map which is a set map $$G \times G \to G$$:

$$(x,y) \mapsto [x,y] = xyx^{-1}y^{-1}$$

This map induces a surjective homomorphism of groups from the exterior square $$G \wedge G$$ of $$G$$ to the derived subgroup of $$G$$ as follows: it is the unique homomorphism $$G \wedge G \to G$$, that, when composed with the set map $$G \times G \to G$$ given by $$(x,y) \mapsto x \wedge y$$ gives the commutator map $$(x,y) \mapsto [x,y]$$.

The kernel of this homomorphism is defined as the Schur multiplier $$M(G)$$ of $$G$$ (see Schur multiplier is kernel of commutator map homomorphism from exterior square to derived subgroup of central extension). Explicitly, we have the following short exact sequence of groups with the middle group being a central extension:

$$0 \to M(G) \to G \wedge G \to [G,G] \to 1$$

Related facts

 * Commutator map is homomorphism from exterior square to derived subgroup of central extension
 * Hopf's formula for Schur multiplier
 * Commutator map on free group is isomorphism between exterior square and derived subgroup
 * Schur multiplier is kernel of commutator map homomorphism from exterior square to derived subgroup

Facts used

 * 1) uses::Commutator map is exterior pairing

Proof
The proof follows essentially from Fact (1) and the universal property of exterior product with respect to exterior pairings.