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

## Statement

For any group , there is a natural surjective homomorphism from its tensor square to its exterior square:

defined as follows: it sends a generator of the form to the corresponding generator .

The claim is that if 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

- Exact sequence giving kernel of mapping from tensor square to exterior square: The exact sequence is:

## Proof

The proof follows from Fact (1), and the observation that if is perfect, then the abelianization is the trivial group, hence the universal quadratic functor applied to that also gives the trivial group. Thus, the exact sequence becomes:

Because the mapping is sandwiched by zero groups on both sides, it must be an isomorphism.