Let $M^3$ be a closed orientable $3$manifold. If $H_2(M,\mathbb Z)=0$ and $H_2(M, \mathbb Z_2)\ne 0$, can we show that $M$ contains a 1sided incompressible surface?

$\begingroup$ What is the definition of 1sided incompressible surface? Is $P^2\times \{x\}$ incompressible inside $P^2 \times S^1$? $\endgroup$– Bruno MartelliAug 24 at 9:19

$\begingroup$ @BrunoMartelli A surface is 1sided if its normal bundle in the ambient 3manifold is not trivial. $P^2 \times {x}$ is incompressible in $P^2 \times S^1$. $\endgroup$– ZhiqiangAug 24 at 10:14
This follows from two exercises.
Exercise 1. Every class in $H_2(M, \mathbb{Z}_2)$ is represented by a smooth embedded surface. (See Exercise 4.5.12(b) in "4Manifolds and Kirby Calculus" by Gompf and Stipsicz.)
Suppose that the surface is $S \subset M$. If $S$ is orientable, then it bounds modulo $\mathbb{Z}$ and thus modulo $\mathbb{Z}_2$, a contradiction. Thus $S$ is not orientable. Since $M$ is orientable, any such surface is onesided. We now "compress".
Exercise 2. Compressing a surface does not change the homology class represented. If the compression separates, then at least one component will be nontrivial in $H_2(M, \mathbb{Z}_2)$ and so will again be onesided. (See the proof of Lemma 6.3 of "3manifolds" by Hempel for a very similar technique.)
Thus we eventually arrive at the desired onesided incompressible surface.

2$\begingroup$ Nonorientable does not imply odd Euler characteristic, but you don't need that to deduce that it is 1sided. On the other hand, I am a bit confused on the definition of incompressible surface in the 1sided setting. $\endgroup$ Aug 24 at 9:17

1$\begingroup$ I see. The surface $S$ obtained in Ex.1 is nonorientable and 1sided since $M$ is orientable. If $S=P^2$, then we are done by Ex.2. If $S$ is compressible, one can do surgery along the compressing disk to obtain a new surface $S'$ with $\chi(S') \ge \chi(S)+1$. Eventually one obtains an incompressible surface. $\endgroup$– ZhiqiangAug 24 at 10:23

3$\begingroup$ A compression can turn a 1sided surface into a 2sided surface, eg. think about the Klein bottle in $S^2\times S^1$. $\endgroup$ Aug 24 at 16:57

3$\begingroup$ Here $S$ is nonseparating. After compressing as much as possible, some component will be incompressible and nonseparating. That component cannot be 2sided since $H_2(M;\mathbb{Z})=0$. $\endgroup$ Aug 24 at 16:59

1$\begingroup$ @JoshHowie  Ok, this time I was not wrong, exactly... I have clarified. $\endgroup$– Sam NeadAug 24 at 17:29