TITLE: $H_p(\mathbb{R}P^3 \times \mathbb{R}P^2)$
QUESTION [5 upvotes]: 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 $H_3(\mathbb{R}P^3) \otimes H_1(\mathbb{R}P^2)=\mathbb{Z}\otimes \mathbb{Z}/2\mathbb{Z} \simeq \mathbb{Z}/2\mathbb{Z}$?

REPLY [4 votes]: 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.