I have been searching for a proof of the following fact. Let $A$ and $B$ be C$^\ast$-algebras such that $A$ is a subalgebra of $B$ (in the C$^\ast$-algebraic sense of course) and Let $C = \overline{ABA}$ be the hereditary C$^\ast$-algebra generated by $A$. Then there exists an inclusion \begin{equation}\tag{$1$} \mathcal{M}(C) \subseteq \mathcal{M}(B) \end{equation}

of multiplier algebras. So far, the sole criterion that I have found, which yields an embedding as above comes from Brown, Ozawa's book, stating that if $\lbrace e_\alpha \rbrace$is an approximate unit of $A$ such that it acts as an approximate unit for $B$ as well, then $$ \mathcal{M}(A) \subseteq \mathcal{M}(B). $$

Therefore, I was wondering whether an approximate unit of $A$, which necessarily must be an approximate unit of $C$, defines an approximate unit of $B$ as well? Any help or reference will be appreciated, greatly.


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