Extending a $C^*$-algebras morphism to the multipliers

Suppose $f:A\to B$ is an onto $C^*$-algebras morphism. How can I extend $f$ to $\bar{f}:M(A)\to M(B)$? Where $M(A)$ and $M(B)$ denote the two-sided multipliers of $A$ and $B$, respectively. Thank you in advance for any comments or replies.

Solution 1:

Definition: A $*$-homomorphism $f: A \to M(B)$ is called non-degenerate if $f(A)B = \operatorname{span}\{f(a)b: a \in A, b \in B\}\subseteq B$ is norm-dense in $B$.

If the $*$-morphism $f: A \to B$ is onto, then the co-extended map $f: A \to M(B)$ is non-degenerate, because $f(A)B = BB$ is norm-dense in $B$ (this follows for instance by the existence of an approximate unit for $B$).

The following theorem is well-known, but a good reference seems to be hard to find. The best reference I can find is proposition 2.1 in Lance's book "Hilbert $C^*$-modules", where one can take $E= A$ (considered as a right Hilbert $A$-module in the obvious way) and where we identify $\mathcal{L}_A(A) \cong M(A).$ For convenience, I sketch a proof below that does not use any technical language about Hilbert-modules.

Theorem: A non-degenerate $*$-homomorphism $f: A \to M(B)$ extends uniquely to a $*$-homomorphism $\overline{f}: M(A) \to M(B)$.

Proof (sketch): View elements $M(A)$ as double centralizers $(L,R)$, i.e. pairs of linear maps $L,R: A \to A$ satisfying $$R(a)b = a L(b)$$ for all $a,b \in A$. Recall that $A\hookrightarrow M(A)$ via $a \mapsto (L_a, R_a)$ where $L_a$ is left multiplication with $a$ and $R_a$ is right multiplication with $a$.

Given $(L,R) \in M(A)$, define $$L': B \to B: \sum_i f(a_i)b_i \mapsto \sum_i f(L(a_i))b_i$$ $$R': B \to B: \sum_i b_if(a_i) \mapsto \sum_i b_if(R(a_i))$$ (check that these assignments define bounded linear maps on $f(A)B$ resp. $Bf(A)$, which extend to maps on $B$ by non-degeneracy). Then check that $(L',R') \in M(B)$. The map $$M(A) \to M(B): (L,R) \mapsto (L',R')$$ is then the desired extension. $\quad \square$

Careful, it need not be true that the extension $\overline{f}$ is surjective! However, this is true if we require $A$ to be $\sigma$-unital.

Solution 2:

Consider $M(B)$ the set of double centralizer of $B$ and define \begin{align*} \overline{f}: M(A)~ \longrightarrow &~ M(B) \\ T ~ \longmapsto &~ (L,R) \end{align*} such that $L(f(a))=f(Ta)$ and $R(f(a))=f(aT)$.

For all $a,b\in A$, $f(a)L(f(b))=f(a)f(Tb)=f(aTb)=f(aT)f(b)=R(f(a))f(b)$. Hence, $(L,R)$ is a double centralizer.

We show that if $f(a)=f(a^\prime)$ then $f(Ta)=f(Ta^\prime)$ and $f(aT)=f(a^\prime T)$ or equivalently if $a\in ker(f)$ then $aT,Ta\in ker(f)$.

This is true since $f(Ta)$ is in the orthogonal complement of $B$ in $M(B)$: let $b\in B$ then there exists $a^\prime \in A, b=f(a^\prime)$, therefore $bf(Ta)=f(a^\prime)f(Ta)=f(a^\prime Ta)=f(a^\prime T)f(a)=0$.

So $f(Ta)\in B^\bot=\{0\}$. Thus, $\overline{f}$ is well-defined.