Let A and B be C*-algebras, and let $\phi:A\rightarrow B$ be a *surjective* *-homomorphism. Then $\phi$ is non-degenerate, and so we can extend it to *-homomorphism between the multiplier algebras: $\tilde\phi: M(A)\rightarrow M(B)$. It's rather tempting to believe that then, surely, $\tilde\phi$ is also surjective. But I cannot for the life of me think of a proof. Any ideas...?

**Background:** The multiplier algebra $M(A)$ is the largest C*-algebra containing A as an essential ideal. Concretely, pick some "large enough" representation of A (either $A\rightarrow B(H)$ a non-degenerate *-representation, or $A\rightarrow A^{**}$ say) and then $M(A) = \{ x : xa,ax\in A \ (a\in A)\}$ the idealiser of $A$ in our large ambient algebra. As $\phi$ surjects, it's very easy to define $\tilde\phi$: we simply have that $$\tilde\phi(x) \phi(a) = \phi(xa), \quad \phi(a) \tilde\phi(x) = \phi(ax).$$
This is well-defined, for if $\phi(a)=0$, then given an approximate identity $(e_i)$ for A, we have that $\phi(xa) = \lim_i \phi(xe_i a) = \lim_i \phi(xe_i) \phi(a) = 0$, and so forth. Indeed, if $B\subseteq B(K)$ say, then $\tilde\phi(x)$ is the limit (strong operator topology say) of the net $\phi(xe_i)$ in $B(K)$; then clearly this is in the idealiser of $B$, and so does define a member of $M(B)$.