I'm reading Murphys book "C$^*$-algebras and operator theory" and I have a question about a proof in chapter 3.
The statement is (Theorem 3.1.8): Let I be a closed ideal in a C$^*$-algebra A. Then there is a unique $*$-homomorphism $\varphi:A\to M(I)$ extending the inclusion $i:I\to M(I)$. Moreover, $\varphi$ is injective if $I$ is essential in $A$.
Here is $M(I)$ the multiplier algebra of $I$ (which contains all double centralizers of $I$ and $\varphi:A\to M(I)$ is defined by $a\mapsto (L_a,R_a)$, where $L_a(x)=ax$ and $R_a(x)=xa$ for all $x\in A$.
My first question is about the uniqueness of $\varphi$: It starts with "Suppose that $\psi:A\to M(I)$ is another such extension of $i$. For all $a\in A$, $b\in I$ then $$\varphi(a)b=\varphi(ab)=ab=\psi(ab)=\psi(a)b".$$
I dont know, how to read $\varphi(a)b$ because $\varphi(a)=(L_a,R_a)\in L(A) \oplus L(A)$ ($L(A)$=bounded linear operator an $A$). And $b$ is only in $I$. And why is it equal to $ab$?
My second question is how to use that $I$ is essential in $A$ if you prove the injectivity (later..)
Regards.
The whole point of the multiplier algebra is that you should be able to multiply. So $\varphi(a)b$ has to make sense. If you look at page 39 in Murphy's book, you see that the definition of $M(I)$ includes a canonical identification of $I$ as an ideal on $M(I)$.
As Phoenix mentioned, $\varphi$ extends the inclusion $I\to M(I)$. So $$\varphi(a)b=\varphi(a)\varphi(b)=\varphi(ab)=ab,$$ since $ab\in I$. Explicitly, $$\varphi(a)b=(L_a,R_a)(L_b,R_b)=(L_aL_b,R_aR_b)=(L_{ab},R_{ab})=\varphi(ab).$$
When $I$ is essential, if $a\in\ker\varphi$, then $\varphi(a)=0$ so $aI=0$; by definition of essential (above 3.1.8), $a=0$. So $\varphi$ is injective.