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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 $M(B)$.

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up vote 5 down vote accepted

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 $C_0(X)$.

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Nice work, Jonas. In the commutative case, do you know if $\sigma$-unitality is necessary? That is, if $A=C)0(X)$ where $X$ is locally compact but not $\sigma$-compact, does there exist a quotient HM from $A$ onto some $B$ such that the induced map of multiplier algebras is not onto? – Yemon Choi Jul 8 '10 at 20:58
(grr, typo, but hopefully you know what I mean) – Yemon Choi Jul 8 '10 at 20:58
One of the few books I have at home is Lance's book... Must read more carefully in future! Many thanks... – Matthew Daws Jul 8 '10 at 21:00
Yemon: My immediate answer is no, I do not know. Matthew: You're welcome. – Jonas Meyer Jul 8 '10 at 21:02
Yemon: It is not necessary, because there are normal locally compact Hausdorff spaces that are not $\sigma$-compact, like the discrete topology on an uncountable set. According to the reference table in the back of Counterexamples in topology, examples 42,45, and 109 also give examples of this. – Jonas Meyer Jul 9 '10 at 1:43

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