Let $M_i(i=1,2)$ and $N_i(i=1,2)$ be von Neumann algebras. If $M_i\cong N_i(i=1,2)$, then we have $\mathcal{M}_1\overline\otimes\mathcal{M}_2\cong \mathcal{N}_1\overline\otimes\mathcal{N}_2$.
If $\mathcal{M}_1\overline\otimes\mathcal{M}_2\cong \mathcal{N}_1\overline\otimes\mathcal{N}_2$, can we conclude that $M_i\cong N_i(i=1,2)$? If the conclusion is not correct, can anyone show me a counterexample, thanks!
Obviously this is false. Surely, if the analogue fails for finite-dimensional vector spaces, you wouldn't expect to hold it in another setting right?
Take $M \ne N$. Then $M \overline{\otimes} N \cong N \overline{\otimes} M$, the isomorphism being implemented by a unitary on the underlying Hilbert spaces.
If you don't like this example, think finite-dimensional (where the von Neumann-tensor product coincides with the usual tensor product):
$$\mathbb{C}^2\overline{\otimes}\mathbb{C}^2 \cong \mathbb{C}^4 \overline{\otimes}\mathbb{C} \cong \mathbb{C}^4.$$