the induced trivial $*$ homomorphism

Let $A=c_{0}\oplus \mathbb{K}$,$I=c_{0}$ is the closed ideal of $A$,there is an induced $*$ homomorphism $\phi:A/I\rightarrow M(I)/I$,where $M(I)$ is the multiplier algebra of $I$.$\phi(a+I)=(L_{a},R_{a})+I$.I have no idea about showing $\phi=0$.

• What are $L_a$ and $R_a$? What is $\Bbb K$? What is $L$? This question looks tantalizing, but without the definitions everything, I can't really understand it. – Robert Lewis Aug 13 '18 at 17:42
• @RobertLewis: $L_a$ and $R_a$ are the left and right multipliers by the element $a$; $\mathbb K$ is the C$^*$-algebra of compact operators on a separable Hilbert space. This very very standard terminology, particularly taking into account the tags used for the question. – Martin Argerami Aug 13 '18 at 18:31
• @MartinArgerami : Thanks for the clarification! – Robert Lewis Aug 13 '18 at 18:34
• @MartinArgerami,excuse me Pro Argerami,I am still confused that $\phi=0$ is true. – math112358 Sep 14 '18 at 12:13

More generally, let $A,I,B$ be C*-algebras and $A = I \oplus B$. Then $I$ is an ideal in $A$ and the inclusion $I \to M(I)$ can be uniquely extended to a $*$-hom. $A \to M(I)$. However, such an extension is given by the projection $\pi_I : I \oplus B\to I$. So the map $$A \to M(I) /I$$ is already $0$ and therefore also the map $A/I \to M(I)/ I$.
• I don't understand why the extension is given by the projection $\pi_{I}$ – math112358 Aug 20 '18 at 9:35
• The key point is that there is a unique $*$-homorphism $\pi:A\to M (I)$ with $\pi|_I=\operatorname {id}_I$. If you had two such $\pi_1,\pi_2$, then for any $x\in I$ and $a\in A$,$$\pi_1 (a)x=\pi_1 (a)\pi_1 (x)=\pi_1 (ax)=\pi_2 (ax)=\pi_2 (a)\pi_2 (x)=\pi_2 (a)x.$$So $\pi_1 (a)=\pi_2 (a)$. As $\pi_I$ in the answer is such a map, it is "the" map; so, $\pi_I$ is $a\longmapsto (L_a,R_a)$. – Martin Argerami Sep 14 '18 at 13:11