Suppose $G_1,G_2$ and *A* are abelian groups. Consider the cohomology groups for trivial action:

$$H^2(G_1 \times G_2,A), H^2(G_1,A), H^2(G_2,A)$$

We have projection maps $G_1 \times G_2 \to G_1$ and $G_1 \times G_2 \to G_2$, and these induce maps in the opposite direction on the cohomology groups. Combining these, we get a map:

$$p:H^2(G_1,A) \oplus H^2(G_2,A) \to H^2(G_1 \times G_2,A)$$

We can further see that this homomorphism is injective.

We also have natural inclusions $G_1 \to G_1 \times G_2$ and $G_2 \to G_1 \times G_2$ and so we get a map:

$$i:H^2(G_1 \times G_2,A) \to H^2(G_1,A) \oplus H^2(G_2,A)$$

Moreover, the maps *p* and *i* are one-sided inverses of each other. In particular, the kernel of *i* can be identified with the quotient $H^2(G_1 \times G_2,A)/(H^2(G_1,A) \oplus H^2(G_2,A))$. This latter quotient can also be computed using the Kunneth formula, according to which (if I'm reading it correctly) the quotient should be $\operatorname{Hom}(G_1,A) \otimes \operatorname{Hom}(G_2,A)$.

My question: is there a natural bijection between the kernel of the map *i* described above and what the quotient "should" be by the Kunneth formula. And can this bijection be proved directly?

Note: Concretely, the kernel of i is those cohomology classes that, if you restrict to either the embedded $G_1$ or the embedded $G_2$, you get the trivial cohomology class on that piece.