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I'm working through an example of the Kunneth formula in my book. Without showing any working it states that for $X = \mathbb{R}P^3 \times \mathbb{R}P^2$

$H_p(X)=\begin{cases} \mathbb{Z} & \mbox{if } p=0\\ \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} & \mbox{if } p=1 \\ \mathbb{Z}/2\mathbb{Z} & \mbox{if } p=2 \\ \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} & \mbox{if } p=3 \\ 0 & \mbox{if } p\ge 4 \end{cases} $

I agree for $0 \le p \le 3$, but for $p=4$ do we not have some contribution from $H_3(\mathbb{R}P^3) \otimes H_1(\mathbb{R}P^2)=\mathbb{Z}\otimes \mathbb{Z}/2\mathbb{Z} \simeq \mathbb{Z}/2\mathbb{Z}$?

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    @Sean - An Introduction to Al$g$ebraic Topology by Rotman (highly recommended!)2011-04-13

1 Answers 1

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You are correct. At first I thought you were neglecting the Tor part of the Künneth Exact sequence, but in degree $4$ all of the $Tor(H_p(\mathbb{RP}^3),H_{3-p}(\mathbb{RP}^2))$ terms vanish.

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    thanks - I was just a bit lazy and didn't write out the (zero) Tor terms2011-04-14