TITLE: Non-orientable $6$-manifold with $H_4(M)=\mathbb{Z}/2$?
QUESTION [8 upvotes]: Does there exist a smooth, closed, non-orientable $6$-manifold $M$ such that $H_4(M;\mathbb{Z})=\mathbb{Z}/2$?

REPLY [6 votes]: $M=S^2\times \mathbb{RP}^2\times\mathbb{RP}^2$. 
Since $Tor(\mathbb{Z}/2,\mathbb{Z}/2)=\mathbb{Z}/2$, the Künneth formula tells you that the homology is:

$H_0(M,\mathbb{Z})=\mathbb{Z}$
$H_1(M,\mathbb{Z})=\mathbb{Z}/2 \oplus \mathbb{Z}/2$
$H_2(M,\mathbb{Z})=\mathbb{Z}/2\oplus \mathbb{Z}$
$H_3(M,\mathbb{Z})=\mathbb{Z}/2\oplus\mathbb{Z}/2\oplus \mathbb{Z}/2$
$H_4(M,\mathbb{Z})=\mathbb{Z}/2$
$H_5(M,\mathbb{Z})=\mathbb{Z}/2$
$H_6(M,\mathbb{Z})=0$.