Schur multiplier of free group is trivial

Statement
The statement has many equivalent forms:


 * 1) For a free group $$F$$, the commutator map homomorphism from the exterior square $$F \wedge F$$ to the derived subgroup $$[F,F]$$ is an isomorphism.
 * 2) For a free group $$F$$, the Schur multiplier $$M(F) = H_2(F;\mathbb{Z})$$ is the trivial group.