Lifting surjective von Neumann algebra homomorphisms Is the following true?  What's a nice proof?

Let $M$ and $N$ be von Neumann algebras, and let $\phi:M\rightarrow N$ be a normal, surjective, *-homomorphism.  Is there a normal *-homomorphism $\theta:N\rightarrow M$ with $\phi\circ\theta$ being the identity?  If I cannot choose $\theta$ as a *-homomorphism, can I at least get a normal complete contraction?

This is, well, hinted at in the proof of Lemma 3.2 of http://pjm.math.berkeley.edu/pjm/2002/205-1/p09.xhtml but I don't follow the details (and they are proving it for weak* TROs: I think surely the von Neumann algebra case should be easier).  Normal *-homomorphisms between von Neumann algebras have a very nice structure theorem, and maybe if I stared at that long enough I'd see an answer, but I thought I'd ask on MO...
 A: Yes you can get a $\phi$ that is a homomorphism. Here is a quick sketch. 
First let $p=sup$ {$p_\alpha,$ projections in $Ker \theta$}. So $p\in Ker \theta$. Furthermore $p\in Z(M)$, the center of M. 
To see this note that if this were not true then we could find a unitary $u\in M$ with $p\neq upu^\star$. So then $p\wedge upu^\star$ would be a projection in $Ker \theta$ bigger than $p$. 
From here you can get that $Ker \theta=pMp$, and so we can decompose $M=Ker \theta \oplus M_1$ and $\theta|_{M_1}$ is injective and thus an isomorphism, thus $\phi$ can be chosen to just be the inverse of $\theta$ on $M_1$. 
Note that if we demand that $\phi$ be unital, this doesn't work and I don't think it can be done in general. I will have to give it more thought.
A: Morphisms of von Neumann algebra have very nice properties.
More precisely, the kernel of a morphism f: M→N of von Neumann algebras is a σ-weakly closed two-sided ideal.
Such ideals are in bijective correspondence with central projections of M.
Moreover, you have a direct sum decomposition M=N⊕ker f.
If you allow non-unital morphisms then the map n→(n,0) solves the problem.
Otherwise choose an arbitrary morphism f: N→ker f and the map n→(n,f(n)) solves the problem.
However, it might happen that there are no morphisms from N to ker f,
and hence you have an obstruction.  For example, there are no morphisms from a non-trivial factor to C.
