Let $A$ be a C\*-algebra. We identify $A$ with its canonical image in the bidual $A^{\*\*}$. Consider the following conditions:


(1) $A$ is a von Neumann algebra.


(2) There is a multiplicative conditional expectation from $A^{\*\*}$ onto $A$, that is, a map $\pi\colon A^{\*\*}\to A$ that is a \*-homomorphism and such that $\pi(a)=a$ for all $a\in A$.


Then (1) implies (2): Consider the predual $A\_\*$ of $A$, and the canonical embedding $\kappa\colon A\_\*\to (A\_\*)^{\*\*}$. Then the dual of $\kappa$ has the desired properties.


Note that $A^{\*\*}$ is always a von Neumann algebra.
One can show that (2) implies that $A$ is a monotone complete AW\*-algebra.
However, is it also a von Neumann algebra?



> 
> Q1: Does (2) imply (1) ?
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Assuming (2), let $J$ denote the kernel of $\pi$. Then $A$ is (\*-isomorphic to) the quotient $A^{\*\*}/J$. 
However, a quotient of a von Neumann algebra by a closed, two-sided ideal need not be a von Neumann algebra. (For example, the Calkin algebra is such a quotient and not a von Neumann algebra.)


In the commutative case, we might also consider the following more general question:



> 
> Q2: Is the quotient of a commutative von Neumann algebra by a closed, two-sided ideal again a von Neumann algebra?
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Equivalently, is every closed subset of a hyperstonean space again hyperstonean?



von Neumann algebras as C*-algebras with multiplicative conditional expectation $A^{**}\to A$
This is a seven-year-old question, so I'm not sure if an answer is still meaningful at this point, but no, $A$ needs not be a vNa. There is even a commutative counterexample. In fact, fix any monotone complete commutative $C^\*$-algebra $A$ which is not a vNa. Then $A$ is injective in the category of unital commutative $C^\*$-algebras and $\*$-homomorphisms. (See, for example, <https://arxiv.org/abs/0706.2995> .) Hence, the identity map $A \rightarrow A$ can be extended to a multiplicative conditional expectation $\pi: A^{\*\*} \rightarrow A$.


