# Schur multiplier of free group is trivial

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.