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 $M(B)$.



Surjective *-homs between multiplier algebras
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](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WJJ-4CRHY72-43&_user=440026&_coverDate=07%252F31%252F1973&_rdoc=1&_fmt=high&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1394971669&_rerunOrigin=google&_acct=C000020939&_version=1&_urlVersion=0&_userid=440026&md5=9ed3e6657608f669d2437593c461f3f6)".


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.




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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 $C\_0(X)$.
> 
> 
> 


