Von Neumann Algebra isomorphism extension I have Von Neumann algebra $\mathfrak{U}$ and a weakly dense *-subalgebra $A$. I have another Von Neumann Algebra $\mathfrak{V}$ and an injective *-homorphism $$\phi: A\longrightarrow \mathfrak{V}$$ such that $\phi(A)$ generates $\mathfrak{V}$. Is it possible to say that one can extend $\phi$ to an *-isomorphism from $\mathfrak{U}$ onto $\mathfrak{V}$? 
P.D. In fact the case I am interested in is when A is a Hilbert algebra and $\mathfrak{V}$ is the natural Von Neumann Algebra of A as defined by Dixmier in its books C *-Algebras or Von Neumann Algebras. I however believe the question as posted has all the required ingredients. 
 A: You can give a characterisation of when it holds.  Without loss of generality, we may suppose that $A$ is closed, i.e. is a C$^*$-algebra.  Consider the universal enveloping von Neumann algebra, which I will consider as being the bidual $A^{**}$.  If $\phi:A\rightarrow\mathfrak{M}$ is a $*$-homomorphism with $\phi(A)$ weak$^*$-dense in $\mathfrak{M}$ then there is a unique extension $\tilde\phi:A^{**}\rightarrow\mathfrak{M}$ which is a surjective, weak$^*$-continuous $*$-homomorphism.
As $A$ generates $A^{**}$, your question has a positive answer if and only if $\tilde\phi$ is always an isomorphism.  This is equivalent (consider the kernel of $\tilde\phi$) to $A^{**}$ having no proper weak$^*$-closed ideals.  In turn, this is equivalent to $A^*$ having no proper $A$-invariant closed subspaces; and is equivalent to $A^{**}$ having no non-trivial central projections.
See Volume 1 of Takesaki, Section III, Chapter 2.
(I presume that $\mathfrak{U}'$, in your question, is not the commutant of $\mathfrak{U}$.  This is slightly unfortunate notation...)
A: No.  This will almost never be true (subalgebras of the compacts are the only cases I can think of where it could work). The easiest example is probably $C[0,1].$  It has an injective homomorphism to $\ell^\infty(\mathbb{N})$--by point evaluation at the rationals--that generates $\ell^\infty(\mathbb{N})$.  It also sits inside $L^\infty[0,1].$  But $\ell^\infty(\mathbb{N})$ has minimal projections and $L^\infty[0,1]$ doesn't.
