Does there exist a smooth, closed, non-orientable $6$-manifold $M$ such that $H\_4(M;\mathbb{Z})=\mathbb{Z}/2$? 



Non-orientable $6$-manifold with $H_4(M)=\mathbb{Z}/2$?
$M=S^2\times \mathbb{RP}^2\times\mathbb{RP}^2$. 


Since $Tor(\mathbb{Z}/2,\mathbb{Z}/2)=\mathbb{Z}/2$, the Künneth formula tells you that the homology is:


* $H\_0(M,\mathbb{Z})=\mathbb{Z}$
* $H\_1(M,\mathbb{Z})=\mathbb{Z}/2 \oplus \mathbb{Z}/2$
* $H\_2(M,\mathbb{Z})=\mathbb{Z}/2\oplus \mathbb{Z}$
* $H\_3(M,\mathbb{Z})=\mathbb{Z}/2\oplus\mathbb{Z}/2\oplus \mathbb{Z}/2$
* $H\_4(M,\mathbb{Z})=\mathbb{Z}/2$
* $H\_5(M,\mathbb{Z})=\mathbb{Z}/2$
* $H\_6(M,\mathbb{Z})=0$.


