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Kunneth formula for cohomology: The cross product $H^*(X;\mathbb Z)\otimes H^*(Y;\mathbb Z)\to H^*(X\times Y;\mathbb Z)$ is an isomoprhism of rings if $X$ and $Y$ are CW complexes and $H^k(Y,R)$ is finitely generated free $R$-module for all $k$.

To see the necessity of finitely generated condition, Hatcher gives an counterexample in the exercise of his Algebraic Topology section 3.2, is to take $X,Y$ to be infinite discrete sets, the cross product map $H^*(X;\mathbb Z)\otimes H^*(Y;\mathbb Z)\to H^*(X\times Y;\mathbb Z)$ is not an isomorphism.

In particular, the cross product is not isomorphic in the $0$-th cohomology. To prove this, since $X$ has infinite components, $H_0(X)\cong \bigoplus_{i\in I} \mathbb Z$, and by Universal coefficient theorem, $H^0$ is dual of $H_0$, so $H^0(X)\cong (H_0(X))^*\cong \prod_{i\in I}Z$. It is same for $Y$. Now it reduces to show the inclusion is not a surjective $$\left(\prod_{i\in I}\mathbb Z\right) \otimes_{\mathbb Z}\left(\prod_{j\in J}\mathbb Z\right)\subsetneqq \prod_{(i,j)\in I\times J}\mathbb Z$$

for $I, J$ infinite sets, and the inclusion map is given by $$(n_i)_{i\in I}\otimes (m_j)_{j\in J}\mapsto (n_im_j)_{(i,j)\in I\times J}$$

In particular, since $$ \prod_{i\in I}\left(\mathbb Z\otimes_{\mathbb Z}\left(\prod_{j\in J}\mathbb Z\right)\right)=\prod_{i\in I}\prod_{j\in J}\mathbb Z=\prod_{(i,j)\in I\times J}\mathbb Z$$

we just need to show $$\left(\prod_{i\in I}\mathbb Z\right) \otimes_{\mathbb Z}\left(\prod_{j\in J}\mathbb Z\right)\subsetneqq \prod_{i\in I}\left(\mathbb Z\otimes_{\mathbb Z}\left(\prod_{j\in J}\mathbb Z\right)\right)$$

Or in other words, we just need in this case, commuting direct product and tensor product does not give you a surjective map. But I have no idea how to go on here.

I know there is a post :Examples proving why the tensor product does not distribute over direct products shows in general tensor product does not commutes with direct product. But, it does not help the special case I want to prove. Thanks in advance someone can help.

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AG learner
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    There is no need to invoke the Universal Coefficient Theorem to compute $H^0$ of a discrete space! It follows from a trivial computation using the singular complex, for example/ – Mariano Suárez-Álvarez Mar 06 '16 at 23:48

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There is a natural isomorphism $$\prod_{I\times J}\mathbb{Z}\cong\operatorname{Hom}\left(\bigoplus_I\mathbb{Z},\prod_J\mathbb{Z}\right),$$ since Hom turns direct sums in the domain coordinate into direct products. Every element of $\prod_I\mathbb{Z}\otimes \prod_J\mathbb{Z}$ then corresponds to a homomorphism $\bigoplus_I\mathbb{Z}\to\prod_J\mathbb{Z}$ whose image is finitely generated. Indeed, an element $a\otimes b\in\prod_I\mathbb{Z}\otimes \prod_J\mathbb{Z}$ corresponds to the homomorphism that takes $c\in\bigoplus_I\mathbb{Z}$ to $(a\cdot c)b$ (where $a\cdot c=\sum_{i\in I} a_ic_i\in\mathbb{Z}$), and the image of this homomorphism is contained entirely in the subgroup generated by $b$. So if you have a general element $$x=\sum_{k=1}^n a_k\otimes b_k\in \prod_I\mathbb{Z}\otimes \prod_J\mathbb{Z},$$ the image of the homomorphism corresponding to $x$ is contained in the subgroup generated by $b_1,\dots,b_n$. (Conversely, with a bit more work you can show that every homomorphism whose image is finitely generated comes from an element of the tensor product.)

Since $I$ and $J$ are both infinite, there exists a homomorphism $\bigoplus_I\mathbb{Z}\to\prod_J\mathbb{Z}$ whose image is not finitely generated (for instance, corresponding to any map $I\to J$ with infinite image). Any such homomorphism gives an element of $\prod_{I\times J}\mathbb{Z}$ that is not in the image of $\prod_I\mathbb{Z}\otimes \prod_J\mathbb{Z}$.

Eric Wofsey
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