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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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    $\begingroup$ Take $B$ to be the bidual of $A$ and take your morphism to be the canonical inclusion of $A$ into $A^{**}$. $\endgroup$
    – Yemon Choi
    Commented Sep 20, 2011 at 23:08
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    $\begingroup$ 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. $\endgroup$
    – Dmitri Pavlov
    Commented Sep 21, 2011 at 0:24
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    $\begingroup$ It might be worth adding that any C*-isomorphism between $A$ and $B$, if there is one, is also a W*-isomorphism. $\endgroup$
    – Nik Weaver
    Commented Nov 20, 2012 at 23:39

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