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.



The splitting for the cohomology version of Kunneth formula
The classifying space $B(G\_1\times G\_2)$ is homotopy equivalent to $BG\_1\times BG\_2$. Since the Leray spectral sequence of $B(G\_1\times G\_2)\to BG\_1$ degenerates, there exists a filtration on $H^2(B(G\_1\times G\_2),A)=H^2(G\_1\times G\_2,A)$ where the consecutive quotients are $H^2(G\_1,A), H^1(G\_1,H^1(G\_2,A)), H^0(G\_1,H^2(G\_2,A))$. The first group is just the image of $p^2\_1$. The third group is isomorphic to $H^2(G\_2,A)$ and it lifts to $H^2(G\_1\times G\_2)$ as the image of $p\_2^2$. Here $p\_k^l$ is the map of $H^l$'s induced by the $k$-th projection.


So the kernel of the map $i$ of the posting is isomorphic to $H^1(G\_1,H^1(G\_2,A))$ (and for similar reasons to $H^1(G\_2,H^1(G\_1,A))$). This is isomorphic to $Hom(G\_1,Hom(G\_2,A))$. If $G\_1$ and $G\_2$ are abelian, we get $Hom(G\_1\otimes G\_2,A)$ as per Tom's comment. However this is not necessarily the same as $Hom(G\_1,A)\otimes Hom(G\_2,A)$: take e.g. $G\_1=G\_2=\mathbf{Z},A=\mathbf{Q}/\mathbf{Z}$, then one group is $\mathbf{Q}/\mathbf{Z}$ and the other is 0. If $A$ is a ring, we can take the tensor product of the Hom's over $A$ and we have a natural map $F:Hom(G\_1,A)\otimes\_A Hom(G\_2,A)\to Hom(G\_1\otimes G\_2,A)$ given by $F(f\_1\otimes f\_2)(g\_1\otimes g\_2)=f\_1(g\_1)f\_2(g\_2)$. However, this again is not necessarily an isomorphism: morally, the tensor product of the duals can be smaller that the dual of the tensor product.


upd: the first version contained an incorrect example.


