Impossibility of realizing codimension 1 homology classes by embedded non-orientable hypersurfaces

Question

Suppose we have an $n+1$-dimensional compact closed oriented manifold $M$ and an $n$-dimensional integral homology class $[\Sigma]\in H_n(M,\mathbb{Z})$ on $M.$ Then is it true that $[\Sigma]$ mod $2$ in $H_n(M,\mathbb{Z}/2\mathbb{Z})$ does not admit an embedded unorientable representative? Here by $[\Sigma]$ mod $2$, I mean the natural map of tensoring with $\mathbb{Z}/2\mathbb{Z}$ in the universal coefficient theorem. I do not require the connectedness of the representative.

If one drops the conditions of either codimension 1 or orientability of $M,$ then counterexamples can be found. Immersed counterexamples can also be found. And of course, classes that do not arise from reduction mod 2 of integral classes, evidently have non-orientable representatives only. However, to my best efforts, I could not find a counterexample with all the conditions imposed. I'm aware of Thom's results on representing homology classes with embedded submanifolds. However, to my limited knowledge, they do not address the above question.

The motivation for this question comes from geometric measure theory. Roughly speaking, the question is related to the Lavrentiev gap between minimal area in integral and mod 2 homology. I'm sorry if this question is trivial, as my field is very far from topology. Many thanks!

Answer 1

If $[\Sigma]\in H_{n}(M,\mathbb Z)$ is primitive, its image $[\Sigma] \in H_{n}(M,\mathbb Z/2\mathbb Z)$ can always be represented by a non-orientable manifold! Since it is primitive, we may suppose that $\Sigma$ is connected, and since $[\Sigma]\neq 0$ the manifold $M\setminus \Sigma$ is connected and hence there is always an arc in $M$ connecting two points of $\Sigma$ from opposite sides. If you self-connect sum $\Sigma$ along this arc you get a non-orientable representative of the same homology class in $H_{n}(M,\mathbb Z/2\mathbb Z)$.