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) uses::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.