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 $H_3(\mathbb{R}P^3) \otimes H_1(\mathbb{R}P^2)=\mathbb{Z}\otimes \mathbb{Z}/2\mathbb{Z} \simeq \mathbb{Z}/2\mathbb{Z}$?