Surjective *-homs between multiplier algebras 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 $M(B)$.
 A: This is true if $A$ is $\sigma$-unital, and is sometimes called the "noncommutative Tietze extension theorem".  A good reference is Proposition 6.8 in Lance's Hilbert $C^*$-modules.  Proposition 3.12.10 in Pedersen's $C^*$-algebras and their automorphism groups covers the separable case, which was first proved by Akemann, Pedersen, and Tomiyama in a 1973 paper called "Multipliers of C*-algebras".
Pedersen points out in Section 3.12.11 that you can get counterexamples in the commutative case by considering non-normal locally compact Hausdorff spaces, so that Tietze's extension theorem doesn't apply.

Akemann, Pedersen, and Tomiyama are more explicit:

In fact let $X$ be a locally compact Hausdorff
  space which is not normal, and consider two disjoint closed sets $Y_1$
  and $Y_2$ such that the function $b$ which is one on $Y_1$ and zero on $Y_2$ has
  no continuous extension to $X$. The restriction map of $C_0(X)$ to
  $C_0(Y_2\cup Y_2)$ is surjective, and $b\in M(C_0(Y_1\cup Y_2))$, but $b$ is not the
  image of a multiplier of $C_0(X)$.

