Exact sequence giving kernel of mapping from tensor square to exterior square

Statement
Let $$G$$ be any group. There is a natural exact sequence as follows:

$$H_3(G;\mathbb{Z}) \to \Gamma(G^{\operatorname{ab}}) \to G \otimes G \to G \wedge G \to 0$$

Here:


 * $$H_3(G;\mathbb{Z})$$ denotes the third homology group for trivial group action of $$G$$ on $$\mathbb{Z}$$.
 * $$\Gamma$$ denotes the universal quadratic functor and $$G^{\operatorname{ab}}$$ denotes the abelianization of $$G$$.
 * $$G \otimes G$$ denotes the tensor square of $$G$$, i.e., the tensor product of $$G$$ with itself where both parts of the compatible pair of actions are the action of $$G$$ on itself by conjugation.
 * $$G \wedge G$$ is the exterior square of $$G$$.

The maps are as follows:


 * The first map is mysterious (some kind of connecting homomorphism?)
 * The second map, $$\Gamma(G^{\operatorname{ab}}) \to G \otimes G$$, sends
 * The third map, $$G \otimes G \to G \wedge G$$, is the natural quotient map that sends a generator of the form $$x \otimes y$$ to the corresponding element $$x \wedge y$$.

Related facts

 * Perfect implies natural mapping from tensor square to exterior square is isomorphism
 * Kernel of natural homomorphism from tensor square to group equals third homotopy group of suspension of classifying space
 * Schur multiplier is kernel of commutator map homomorphism from exterior square to derived subgroup