I am studying the proof that if $A$ is a $C^\*$-algebra such that $A^{\*\*}$ is a semidiscrete vN algebra, then $A$ has the completely positive approximation property (CPAP). I was able to handle the unital case, but I am stuck in the non-unital setting: The authors (N. Brown and N. Ozawa) suggest that one should prove that if $A^{\*\*}$ is semidiscrete then so is $(\tilde{A})^{\*\*}$ and then conclude by proving that if $\tilde{A}$ has the CPAP then so does $A$.


My problem is this: I can't prove that the double dual of the unitization will be semidiscrete. I cannot understand the double dual of the unitization in relevance to the double dual of $A$ at all. The authors state that $(\tilde{A})^{\*\*}\cong A^{\*\*}\oplus\mathbb{C}$ and mention that it is furthermore true that if $B$ is any $C^\*$-algebra with a (closed, two-sided) ideal $J$, then $B^{\*\*}\cong J^{\*\*}\oplus(B/J)^{\*\*}$. First of all, does $\cong$ mean as vector spaces or as $C^\*$-algebras? How can one prove this isomorphism? Extra bonus question: If all double duals involved are endowed with their ultraweak topologies, is $\cong$ a homeomorphism?



The double dual of the unitization of a $C^*$-algebra
Believe it or not, these are $\*$-isomorphisms as C${}^\*$-algebras. If $J$ is a closed two-sided ideal of $B$ then $J^{\*\*}$ is a weak\* closed two-sided ideal of $B^{\*\*}$, and every weak\*-closed two-sided ideal of a von Neumann algebra is a direct summand. I suppose these are good exercises. The supremum in $B^{\*\*}$ of an approximate unit for $J$ will be a central projection $p$ such that $pB^{\*\*} = J^{\*\*}$.


Bonus, any $\*$-isomorphism of von Neumann algebras is automatically a weak\* homeomorphism. That is because it preserves order and hence must be normal.


