Heegaard Splittings of Non-orientable 3 manifolds A well known and oft-utilized fact from 3-manifold topology is that all closed, orientable 3-manifolds admit Heegaard splittings. 
I am trying to understand what the appropriate notion of Heegaard splitting for a closed, nonorientable 3-manifold should be, assuming I want lots of familiar facts to carry over to this setting. I'm also curious about the interaction with the orientable case.
In particular, some things  I am pondering include:

Given a closed, nonorientable 3-manifold $Y$,
1) Can one decompose $Y = H_{1} \cup_{\Sigma}H_{2}$, for some surface $\Sigma \hookrightarrow Y$, and some (possibly nonorientable) handlebodies $H_{1},H_{2}$ ?
2) Can one decompose $Y = \Sigma \coprod \tilde{H}$, for a one-sided surface $\Sigma \hookrightarrow Y$ and an open handlebody $\tilde{H}$?

And in the same vein as this question:

3) Can one realize $Y$ as a quotient $M/h$ of a free, involutive, orientation reversing homeomorphism $h:M \rightarrow M$ of an orientable 3-manifold $M$, where $h$ exchanges the two sides $U,V$ of some Heegaaard splitting $M= U \cup V$?

 A: Check the proof of Theorem 1 in J. H. Rubinstein. One-sided Heegaard splittings of 3-manifolds: While he proves a slightly different theorem than in your question, I think, his argument will yield what you want. The nontrivial mod 2 cohomology class that he is using will be replaced by the canonical element of $H_2(M,Z_2)$ Poincare-dual to the element of $H^1(M,Z_2)$ which sends each orientation-reversing 1-cycle to $1\in Z_2$.  
A: In answer to part 1, you can definitely realize a non-orientable $3$-manifold as a union of two possibly non-orientable handlebodies. Just triangulate the manifold and take a regular neighborhood of the $1$-skeleton. This will be your first handlebody. The second handlebody is the closure of the complement. The fact that the second space is a handlebody is not as easy to see, but follows in the same way as the orientable case. 
For number 3, you can definitely find an $M$ such that your manifold is a quotient by a free involutive homeomorphism. This is called the orientation double cover. Moreover, you can simply lift your two handlebodies in the quotient to handlebodies in $M$. You know that at least one of the handlebodies has a connected cover, otherwise $M$ would be disconnected. But then its boundary is connected, implying the other handlebody is connected as well.
I'm not sure about number 2. Seems unlikely in general.
A: Considering question 2) let me exemplificate a 3-manifold which can be splitted along an orientable surface:
Let $N_3$ be the nonorientable genus three surface.
Take $E=N_3\times S^1$. Since $N_3=T_0\cup_C M\ddot{o}$, where $T_0$ is a puntured 2-torus, $M\ddot{o}$ is a Möbiusband and 
$C=\partial T_0=\partial M\ddot{o}=T_0\cap M\ddot{o}$ then
$E=(T_0\times S^1)\cup_U (M\ddot{o}\times S^1)$, where $U$ is the 2-torus 
$C\times S^1$. 
But one can consider $M\ddot{o}\times S^1=\overline{{\cal N}(C\ddot{o}\times S^1)}$ (a regular neighbourhood and has, as aconnected boundary, another 2-torus) where $C\ddot{o}$ is the core of $M\ddot{o}$. 
Then $E=(T_0\times S^1)\cup \overline{{\cal N}(C\ddot{o}\times S^1)}$. That is 
$$E\smallsetminus{\rm int}{\cal N}(C\ddot{o}\times S^1) =T_0\times S^1.$$
That is, $C\ddot{o}\times S^1$ (an orientable surface) is splitting $E$ within $C\ddot{o}\times S^1$ and $T_0\times S^1$ both orientable.
