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The integral Poincare duality theorem says that a closed connected oriented $n$-manifold has a fundamental class $[M]\in H_n(M;\mathbb{Z})$ for which cap product gives $H^q(M;\mathbb{Z})\cong H_{n-q}(M;\mathbb{Z})$. The general Poincare duality with the orientation local system uses $[M]\in H_n(M;{\mathcal{O}}_M)$ and givesfor a local coefficient system $\mathcal{A}$. Over ${\mathbb{F}}_2$ it holds without orientability. For constant integral coefficients on a nonorientable connected closed manifold, $H_n=H^0(M;{\mathcal{O}}_M)=0$, while $H^n=H_0(M;{\mathcal{O}}_M)=\mathbb{Z}/2$: the invariants of sign reversal in $\mathbb{Z}$ are zero and its coinvariants are $\mathbb{Z}/(a\sim -a)$.

If the three-manifold is orientable, duality and the universal coefficient theorem for cohomology giveFor a nonorientable three-manifold, $H_3=0$ and $H^3=\mathbb{Z}/2$. The top-degree universal coefficient sequence therefore givesSince the homology groups are finitely generated, their cyclic decomposition shows that this Ext group is isomorphic to the torsion subgroup of $H_2$. Thus the torsion is exactly $\mathbb{Z}/2$, not the possibly larger torsion subgroup of $H_1$. Also odd-dimensional closed manifolds have zero Euler characteristic, by pairing complementary mod-two Betti numbers. If $r$ is the free rank of $H_2$, it follows that $0=\chi (M)=1-b+r$, giving the second homology of a closed nonorientable three-manifoldThe inequality is forced by the nonnegative free rank $b-1$.

For the final graded-group construction, one always has $K^0=K^3=K^6=\mathbb{Z}$, since each of these is copied from $H^0$ of a connected three-manifold. Suppose it were the cohomology of a closed six-manifold $N$. The group $K^0$ makes $N$ connected, and $K^6=\mathbb{Z}$ makes it orientable: the nonorientable alternative has top integral cohomology $\mathbb{Z}/2$, as above. Its third rational cohomology would then be one-dimensional. But the middle Poincare duality pairing on $H^3(N;\mathbb{Q})$ is nondegenerate and skew-symmetric by graded commutativity, hence must have even dimension. Equivalently, a generator $u$ would satisfy $u⌣u=-u⌣u$, so its square in the torsion-free top group is zero and the entire rank-one pairing vanishes. This contradicts duality and the even rank of middle cohomology in dimension four k plus two. ThereforeBoth existence questions have negative answers, independently of the other groups of the original three-manifold.