Extending a $*$-homomorphism between $C^*$-algebras to $*$-homomorphism between generated von Neumann algebras Let $A \subseteq \cal B(H)$, $B \subseteq \cal B(H')$ be $C^*$-algebras, where $\cal H$, $\cal H'$ are Hilbert spaces and let $\psi: A \rightarrow B$ be a $*$-homomorphism.
My question: When does this $*$-homomorphism extend to a $*$-homomorphism $\overline{\psi}: A''\rightarrow B''$ between the generated von Neumann algebras $A''$, $B''$ of $A$ and $B$? Is there a criterion which allows this?
 A: There are easy examples of non-extending $\ast$-homomorphisms. For instance recall that a groups homomorphism $\Phi: G \to H$ extends to the reduced $C^\ast$-algebras iff $\mathrm{ker}(\Phi)$ is amenable, but it extends to the reduced von Neumann algebras iff $\mathrm{ker}(\Phi)$ is compact. Take the trivial character of an amenable group $G$,
$$
  \chi \Big( \sum_{g\in G} a_g \lambda_g \Big) = \sum_{g\in G} a_g
$$
always extends to a $\ast$-homomorphism $\chi:C_{\mathrm{red}}^\ast G \to \mathbb{C}$ but never pass to the von Neumann algebra unless $G$ is finite.
In the normal case: Any $\ast$-homomorphism between von Neumann algebras is spatially implemented and of the form $a \mapsto p ( a \otimes 1 ) p$, where $p$ is a projection commuting with $A \otimes \mathbb{C} 1$. Then your criterion will be something like asking the $\ast$-homomorphism to be spatially implemented by a map $V: \mathcal{H} \to \mathcal{H'}$ and take $p$ to be the range projection.
In the non-normal case: I do not have an answer. Take $A = c_0(\mathbb{N}) +  \mathbb{C} 1 \subset B(\ell^2(\mathbb N))$ the $C^\ast$-algebra of sequences with a limit. There is a $\ast$-homomorphism $\psi(a_n)_n = \lim_n a_n$. It does not extends to $A^{''} = \ell^\infty(\mathbb{N})$ in a normal way but there are uncountably many non-normal extensions, one for every limit with respect to a proper ultrafilter
$$
  \bar\psi((a_n)_n) = \lim_{n, \, \omega} a_n
$$
A: The answer in general is no. Even when $A\simeq B$ and $\psi$ is an isomorphism. For instance you can take $A=B=UHF(2^\infty)$, and represent $B$ via GNS by the trace, and represent $A$ via a Powers' state. Then $B''=R$, the hyperfinite II$_1$-factor, and $A''$ is a type III factor. Not only $\psi$ does not admit an extension: there is no nonzero homomorphism $\pi:A\to B$ (proof: take a projection that halves the identity on $A$, then $\operatorname{tr}\pi(I)=\operatorname{tr}\pi(p)+\operatorname{tr}\pi(I-P)=2\operatorname{tr}\pi(I)$, so $\pi(I)=0$). 
Note that in the example above $\psi$ is normal, so it is as good as it can be as a $*$-homomorphism; and still it is not enough to guarantee an extension. 
In summary, you may find some concrete example where it works, but you cannot expect a general characterization. 
