# Second cohomology group with finite coefficients of the product of two varieties

1. This is surely well known. Let $X$ and $Y$ be smooth, projective, connected complex varieties. Then $$H^2(X\times Y,Z/n)=H^2(X,Z/n)\oplus H^2(Y,Z/n) \oplus (H^1(X,Z/n)\otimes H^1(Y,Z/n))$$ for any $n>1$. I think I can prove this using a counting argument and the K\"unneth formula with coefficients in $Z$, but it seems a ridiculous way of doing it. Is there a more natural way or a good reference?

2. The same question for the \'etale cohomology of varieties over a separably closed field of characteristic not dividing $n$. (Same comment)

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The identity map of $H^2(X,\mathbb{Z}/n)\oplus H^2(Y,\mathbb{Z}/n)$ can be decomposed as $$H^2(X,\mathbb{Z}/n)\oplus H^2(Y,\mathbb{Z}/n)\to H^2(X\times Y,\mathbb{Z}/n)\to H^2(X,\mathbb{Z}/n)\oplus H^2(Y,\mathbb{Z}/n)$$ where the first arrow is the sum of the pullbacks and the second arrow is the sum of the maps induced by inclusions of $X,Y$ into the product. So $H^2(X,\mathbb{Z}/n)\oplus H^2(Y,\mathbb{Z}/n)$ splits off. The remaining part can be identified with $H^1(X,H^1(Y,\mathbb{Z}/n))$ via the Leray spectral sequence.
Now let me consider the topological case. In this case we have $$H^1(Z,A)=Hom(H_1(Y,A),A))$$ for any non-pathological space $Z$ and any ring $A$ since $H_0(Z,A)$ is free. So we get
$$H^1(X,H^1(Y,\mathbb{Z}/n))\cong Hom(H_1(X,\mathbb{Z}/n)\otimes H_1(Y,\mathbb{Z}/n),\mathbb{Z}/n)$$
$$\cong Hom (H_1(X,\mathbb{Z}),\mathbb{Z}/n)\otimes Hom (H_1(Y,\mathbb{Z}/n),\mathbb{Z}/n)\cong H^1(X,\mathbb{Z}/n)\otimes H^1(Y,\mathbb{Z}/n).$$
By the way, one thing is not completely clear to me: one probably needs to prove the commutativity of a certain diagram, namely, that the injective map from $H^1(X,H^1(Y,Z/n))$ to $H^2(X\times Y,Z/n)$ is the same as the cup-product map from $H^1(X,Z/n)\otimes H^1(Y,Z/n)$. You show that the two groups are isomorphic, but maybe an abstract isomorphism is not enough. – Alexei Skorobogatov Oct 21 '11 at 19:45