Schur multiplier of free group is trivial

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., free group) must also satisfy the second group property (i.e., Schur-trivial group)
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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.