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. 



Von Neumann Algebra isomorphism extension
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.


