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If a $*$-homomorphism $A\to B$ between $C^*$-algebras is quasi-unital1 then there is an induced $*$-homomorphism $M(A) \to M(B)$ between the multiplier algebras of $A$ and $B$. Is this latter map also quasi-unital? Here multiplier algebras are given their strict topology.

Edit Actually, I only care about the case that $B=M(C)$ for some $C^*$-algebra $C$, and then since $M(C)$ is unital, $M(M(C)) \simeq M(C)$, giving an extension of quasi-unital $A \to M(C)$ to strictly continuous $M(A) \to M(C)$. It is this latter map that I hope is quasi-unital.


1 Recall that a $*$-homomorphism $f\colon A\to B$ is said to be quasi-unital if $\overline{f(A)B} = pB$, for some projection $p\in M(B)$. An equivalent definition is given in (Higson, Definition 1.1.6), and the proof of extension. The extension is the unique continuous map in the strict topology, though Higson doesn't mention this.

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(The following argument was given to me by Aidan Sims)

Assume $f:A \to M(C)$ is quasi-unital, so that there is a projection $p\in M(C)$ so that $\overline{f(A)M(C)} = pM(C)$. It turns out that the projection $q\in M(C)$ that ensures the extension $\hat{f}: M(A) \to M(C)$ is quasi-unital (so that $\overline{f(M(A))M(C)} = qM(C)$) is just the given $p$.

Recall that $p=\hat{f}(1_{M(A)})$. It follows that $pM(B) = \hat{f}(1_{M(A)})M(B) \subseteq \hat{f}(M(A))M(B)$. On the other hand for $a \in M(A)$ and $b \in M(B)$ we have $$ \hat{f}(a)b = \hat{f}(1_{M(A)}a)b = \hat{f}(1_{M(A)})\hat{f}(a)b = p (\hat{f}(a)b) \in p M(B). $$ Linearity and continuity then show that $\hat{f}(M(A))M(B) \subseteq pM(B)$, so we have equality.

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