Commutative von Neumann algebras and localizable measure spaces This is not my subject so I apologize if my question is too obvious or understood from other pages.
I read some pages such as
Reference for the Gelfand-Neumark theorem for commutative von Neumann algebras and
von neumann algebras and measurable spaces.
If I understand correctly, there is some correspondence between localizable measure spaces and commutative von Neumann algebras given by
$$(\Omega,\nu)\mapsto L^{\infty}(\Omega,\nu).$$ 
But I wanted to clarify:
(1) What is the correct notion of a morphism of commutative von Neumann algebras? Is it a normal *-homomorphism? What is the exact definition of normal? is it the same as being σ-weakly continuous?  
(2) Is it true that the opposite category of the category of commutative von Neumann algebras (with the appropriate class of morphisms) is equivalent to the category of localizable measure spaces and measurable maps? Or do we need to use another type of morphisms between localizable measure spaces?
Any good references on the above will be highly appreciated.
 A: The technical part of your problem is resolved in the sadly little-known article "On point realization of $L^\infty$-endomorphisms" by Vesterstrøm and Wils in Math. Scand. 25 (1970) (journal link). By the way, von Neumann algebras are considered from a categorical viewpoint
by Guichardet in a paper in Bull. Math. Soc. 90 (1966).
A: The split interval $I^{\parallel} = \{t^+ : t \in [0,1]\} \cup \{t^- : t \in [0,1]\}$ yields the standard counterexample to the second question (details can be found in Fremlin Vol 3I, section 343, especially 343J, the exercises and the notes and comments at the end of the section). There usually are many maps of the measure space that induce the same morphisms of the measure algebra.
The identity homomorphism of the measure algebra of $I^{\parallel}$ is induced both by the identity map $f$ and by the map $g$ exchanging $t^+$ with $t^-$. Since the measure algebra uniquely determines the associated von Neumann algebra, both these maps induce the identity map of the algebra. Obviously $f(x) \neq g(x)$ for all $x \in I^{\parallel}$, so no identification modulo zero helps here.
Similarly, the measure algebra of $I^{\parallel}$ is canonically isomorphic to the measure algebra of the unit interval, and the two maps $t^{\pm} \mapsto t^-$ and $t^{\pm} \mapsto t^+$ both induce this isomorphism, but they are nowhere equal.
