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If $\varphi: A\to B$ is a (norm-continuous, unital, involutive) homomorphism of $C^*$-algebras, then the image $\varphi(A)$ is closed in $B$ and therefore is a $C^*$-algebra with the $C^*$-norm induced from $B$ (G.Murphy, 3.1.6).

If in addition $A$ is a von Neumann algebra, will $\varphi(A)$ be a von Neumann algebra as well?

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No. The "why" depends on your definition of von Neumann algebra.

If you define von Neumann algebra the way it is usually used, it is not an intrinsic notion: both the double commutant and the ultraweak operator topology depend on the environment (i.e. $B(H)$). In that case, take a non-type I factor, say a II$_1$, and take an irreducible representation. Then the image will be dense in some $B(H)$, but it cannot be everything (because in a II$_1$ factor the identity is finite, for instance). So the image is not a von Neumann algebra.

The above shows that the usual definition of von Neumann algebra has its caveats. An intrinsic definition can be given: a C$^*$-algebra that admits a Banach pre-dual. More concretely this means that a von Neumann algebra is a C$^*$-algebra that admits an isometric surjective embedding onto the dual of a Banach space.

So in this second case, if $\varphi$ is injective, its image will be a von Neumann algebra. When $\varphi$ is not injective, it is easy to come up with a counterexample: you can take $A=B(H)$ and $\varphi$ the quotient map onto the Calkin algebra, which is not a von Neumann algebra (in an intrinsic way: it is not C$^*$-isomorphic to any von Neumann algebra).

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  • $\begingroup$ I thought, this question becomes senseless, if the definition depends on the environtment... I didn't understand, why is the answer "yes" in the second case? $\endgroup$ – Sergei Akbarov Nov 19 '14 at 21:03
  • $\begingroup$ Because if two C$^*$-algebras are isomorphic and one embeds isometrically in a dual, you can use the isomorphism to get an embedding of the other algebra in the dual. So with the intrinsic definition of von Neumann algebra, $\varphi(A)$ is a von Neumann algebra. $\endgroup$ – Martin Argerami Nov 19 '14 at 21:07
  • $\begingroup$ Excuse me, perhaps you are speaking about the situation where $\varphi$ is injective? I meant that it is not necessarily injective. Or I didn't understand something? $\endgroup$ – Sergei Akbarov Nov 19 '14 at 21:13
  • $\begingroup$ Yes, I though that you were saying injective. But it doesn't change things too much. I have edited the answer. $\endgroup$ – Martin Argerami Nov 19 '14 at 21:24
  • $\begingroup$ Ah, OK... Could you, please, give a reference with the explanation of why Calkin is not von Neumann? $\endgroup$ – Sergei Akbarov Nov 19 '14 at 21:36

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