Is the image of a von Neumann algebra under a C*-homomorphism a von Neumann algebra as well? 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?
 A: 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). 
