# The splitting for the cohomology version of Kunneth formula

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 answer is yes, except that the group you want is $Hom(H_1(G_1)\otimes H_1(G_2),A)$ (and I think of this as a combination of Kuenneth and Universal Coefficients). – Tom Goodwillie Aug 26 '10 at 14:18

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.