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Hello,

Is there a counterexample to the following statement: let $A,B$ two von Neumann algebras, every morphism $A \rightarrow B$ of $C^* $-algebras is a $W^*$-homomorphism ?

( a $W^* $-homomorphism is a continuous morphism for the weak topologies $\sigma(A,A_* )$ and $\sigma(B,B_* )$, where $A_* $ and $B_* $ are the preduals)

Thanks in advance.

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5  
Take $B$ to be the bidual of $A$ and take your morphism to be the canonical inclusion of $A$ into $A^{**}$. –  Yemon Choi Sep 20 '11 at 23:08
12  
Take any commutative von Neumann algebra A without atomic projections. There are no W*-morphisms from A to complex numbers, yet there are many C*-morphisms of this type. In fact, the Gelfand-Neumark theorem tells us that C*-morphisms distinguish different elements of A. –  Dmitri Pavlov Sep 21 '11 at 0:24
    
Thanks for your answer. –  user12806 Sep 21 '11 at 6:15
3  
It might be worth adding that any C*-isomorphism between $A$ and $B$, if there is one, is also a W*-isomorphism. –  Nik Weaver Nov 20 '12 at 23:39
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