Is the following true? What's a nice proof?



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> 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?
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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...



Lifting surjective von Neumann algebra homomorphisms
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.


