We have such a claim:

If $A$ is an ideal of C*-algebra $B$, then there is a unique morphism $f\colon B\to M(A)$ such that $f$ is identity on $A$, here $M(A)$ is the multiplier algebra of $A$.

Now if $A$ is only a C*-subalgebra of $B$, does it also true? How about a hereditary C*-subalgebra?

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    $\begingroup$ If $A$ is a corner of $B$, then $M(A) = A$ and $B$ doesn't embed general. (Corners are hereditary.) $\endgroup$ – Michael Oct 1 '13 at 12:18
  • $\begingroup$ Embed the Cuntz algebra $\mathcal{O}_3$ into $\mathcal{O}_2$ as a corner (this can be done by, for example, noting $M_2(\mathcal{O}_2)$ is isomorphic to $\mathcal{O}_2$ and using Kirchberg's embedding theorem). Since $\mathcal{O}_2$ does not embed into $\mathcal{O}_3$ (I believe), you have a counterexample. $\endgroup$ – Michael Oct 1 '13 at 13:00

No. Take $B=C[0,1]$, and $A = \mathbb{C}\cdot 1$, where $1$ denotes the constant function on $[0,1]$. Then $A$ is a subalgebra of $B$ and $M(A) = A$. Now, for any $x\in [0,1]$ the evaluation map $$ f\mapsto f(x)\cdot 1 $$ is a map from $B \to M(A)$ that is identity on $A$.

I am not sure about hereditary sub-algebras though.


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