Perfect implies natural mapping from tensor square to exterior square is isomorphism

Statement

For any group $G$, there is a natural surjective homomorphism from its tensor square to its exterior square: $G \otimes G \to G \wedge G$

defined as follows: it sends a generator of the form $g \otimes h$ to the corresponding generator $g \wedge h$.

The claim is that if $G$ is a perfect group, then this surjective homomorphism is also injective, i.e., its kernel is trivial, and hence, it is an isomorphism.

Facts used

1. Exact sequence giving kernel of mapping from tensor square to exterior square: The exact sequence is: $H_3(G;\mathbb{Z}) \to \Gamma(G^{\operatorname{ab}}) \to G \otimes G \to G \wedge G \to 0$

Proof

The proof follows from Fact (1), and the observation that if $G$ is perfect, then the abelianization $G^{\operatorname{ab}}$ is the trivial group, hence the universal quadratic functor applied to that also gives the trivial group. Thus, the exact sequence becomes: $H_3(G;\mathbb{Z}) \to 0 \to G \otimes G \to G \wedge G \to 0$

Because the mapping $G \otimes G \to G \wedge G$ is sandwiched by zero groups on both sides, it must be an isomorphism.