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.
(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.