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If $X$ and $Y$ are spaces then the Kunneth theorem implies that $$H_n(X \times Y;\mathbb{Z}) = \left(\bigoplus_{p+q = n}H_p(X;\mathbb{Z}) \otimes H_q(Y;\mathbb{Z})\right) \oplus \left(\bigoplus_{p+q=n-1} \operatorname{Tor}(H_p(X;\mathbb{Z}) ,H_q(Y;\mathbb{Z}))\right).$$

Now if $X$ and $Y$ are CW-complexes, then (correct me if I'm wrong) we know that $H_i(X;\mathbb{Z})$ and $H_i(Y;\mathbb{Z})$ are free abelian, which should imply that $\operatorname{Tor}(H_i(X;\mathbb{Z}) ,H_j(Y;\mathbb{Z})) = 0$ for any $i, j \geq 0$ and hence that $$H_n(X \times Y;\mathbb{Z}) = \bigoplus_{p+q = n}H_p(X;\mathbb{Z}) \otimes H_q(Y;\mathbb{Z}).$$

But now consider $K \times \mathbb{RP}^2$ (which is a CW-complex) where we let $K$ denote the Klein bottle, then in one calculation of $H_3(K \times \mathbb{RP}^2)$ I've seen, it was stated that $$H_3(K \times \mathbb{RP}^2) = \operatorname{Tor}(H_1(K;\mathbb{Z}), H_1(\mathbb{RP}^2;\mathbb{Z})) = \mathbb{Z}/2.$$

This seems to contradict what I've said above, is there some subtlety that I am missing?

Perturbative
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    Homologies of CW-complexes in no way need to be free. Every space has a CW-approximation, so if they're all free then all spaces have free abelian homology in all degrees.... – Randall Sep 08 '21 at 16:48
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    For a concrete space, $\mathbb{R}P^n$ has a CW structure with one cell in each dimension, but has torsion homology. – Randall Sep 08 '21 at 16:52
  • Good discussion of how to compute the homology of $\mathbb{R}P^2$ using the cell structure: https://math.stackexchange.com/questions/768804/homology-group-of-real-projective-plane – Randall Sep 08 '21 at 17:28
  • @Randall Thanks for your comments, indeed it seems that the homology groups of CW complexes need not be free, my confusion stems from Proposition 3.1 here: https://ncatlab.org/nlab/show/cellular+homology, however now that I look at it, I think it is the group of cellular chains that is free abelian on the basis of $n$-cells rather than the homology group itself. – Perturbative Sep 08 '21 at 20:47

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