[proofplan] The proof constructs the required map by letting each element of $B$ act on the essential ideal $A$ by left and right multiplication. Since $A$ is a two-sided ideal, these actions preserve $A$, and the $C^*$-norm inequality makes them bounded operators. Associativity in $B$ verifies the double-centralizer identities, after which the algebraic properties of $\Phi$ follow by direct computation. Finally, uniqueness is forced because any unital $*$-homomorphism extending the canonical copy of $A$ must have the same left and right actions on $A$. [/proofplan] [step:Construct left and right multipliers from elements of $B$] Fix $b\in B$. Since $A$ is a two-sided ideal in $B$, for every $a\in A$ one has $ba\in A$ and $ab\in A$. Thus the formulas \begin{align*} L_b(a)=ba \end{align*} and \begin{align*} R_b(a)=ab \end{align*} define complex-linear maps $L_b:A\to A$ and $R_b:A\to A$. For every $a\in A$, the submultiplicativity of the $C^*$-norm on $B$ gives \begin{align*} \|L_b(a)\|_A=\|ba\|_B\le \|b\|_B\|a\|_A \end{align*} and \begin{align*} \|R_b(a)\|_A=\|ab\|_B\le \|a\|_A\|b\|_B. \end{align*} Hence $L_b$ and $R_b$ are bounded linear maps from $A$ to $A$. We now verify the double-centralizer compatibility in the convention of [citetheorem:8575]: a pair $(L,R)$ belongs to $M(A)$ when it consists of bounded linear maps $L,R:A\to A$ satisfying the left and right multiplier identities and the compatibility identity \begin{align*} xL(y)=R(x)y \end{align*} for all $x,y\in A$. For $x,y\in A$, associativity in $B$ gives \begin{align*} L_b(xy)=b(xy)=(bx)y=L_b(x)y, \end{align*} and \begin{align*} R_b(xy)=(xy)b=x(yb)=xR_b(y). \end{align*} The compatibility identity is \begin{align*} xL_b(y)=x(by)=(xb)y=R_b(x)y. \end{align*} Therefore $(L_b,R_b)\in M(A)$. [guided] We want to turn an element $b\in B$ into a multiplier of the ideal $A$. The only available operation is multiplication inside $B$, so we define two maps on $A$: left multiplication by $b$ and right multiplication by $b$. Explicitly, \begin{align*} L_b:A&\to A, \end{align*} is the map $a\mapsto ba$, and \begin{align*} R_b:A&\to A, \end{align*} is the map $a\mapsto ab$. These maps really have codomain $A$, not merely $B$, because $A$ is assumed to be a two-sided ideal in $B$. Next we need these maps to be bounded linear maps. Linearity follows from distributivity and complex scalar compatibility in the algebra $B$. For boundedness, we use the norm inequality in the $C^*$-algebra $B$. For every $a\in A$, \begin{align*} \|L_b(a)\|_A=\|ba\|_B\le \|b\|_B\|a\|_A \end{align*} and \begin{align*} \|R_b(a)\|_A=\|ab\|_B\le \|a\|_A\|b\|_B. \end{align*} The equality of the $A$-norm and the inherited $B$-norm is part of the hypothesis that $A$ is a closed $C^*$-subalgebraic ideal of $B$. It remains to check that $(L_b,R_b)$ is a double centralizer. In the convention used for the multiplier algebra in [citetheorem:8575], the canonical embedding sends $a\in A$ to $(L_a,R_a)$, where $L_a(x)=ax$ and $R_a(x)=xa$. Thus the compatibility condition is \begin{align*} xL(y)=R(x)y \end{align*} for all $x,y\in A$. For our maps, associativity in $B$ gives \begin{align*} L_b(xy)=b(xy)=(bx)y=L_b(x)y. \end{align*} Similarly, \begin{align*} R_b(xy)=(xy)b=x(yb)=xR_b(y). \end{align*} Finally, \begin{align*} xL_b(y)=x(by)=(xb)y=R_b(x)y. \end{align*} These three identities are exactly the double-centralizer identities, so $(L_b,R_b)$ is an element of $M(A)$. [/guided] [/step] [step:Define the homomorphism and verify its algebraic properties] Define a map \begin{align*} \Phi:B&\to M(A) \end{align*} by \begin{align*} \Phi(b)=(L_b,R_b). \end{align*} The preceding step shows that this is well-defined. Let $b,c\in B$ and $\lambda,\mu\in\mathbb C$. For every $a\in A$, \begin{align*} L_{\lambda b+\mu c}(a)=(\lambda b+\mu c)a=\lambda ba+\mu ca=(\lambda L_b+\mu L_c)(a), \end{align*} and \begin{align*} R_{\lambda b+\mu c}(a)=a(\lambda b+\mu c)=\lambda ab+\mu ac=(\lambda R_b+\mu R_c)(a). \end{align*} Hence $\Phi$ is complex-linear. For multiplication, let $b,c\in B$. For every $a\in A$, \begin{align*} L_{bc}(a)=(bc)a=b(ca)=(L_b\circ L_c)(a), \end{align*} and \begin{align*} R_{bc}(a)=a(bc)=(ab)c=(R_c\circ R_b)(a). \end{align*} This is precisely the product rule for double centralizers, so \begin{align*} \Phi(bc)=\Phi(b)\Phi(c). \end{align*} Let $1_B$ denote the unit of $B$. For every $a\in A$, \begin{align*} L_{1_B}(a)=1_Ba=a \end{align*} and \begin{align*} R_{1_B}(a)=a1_B=a. \end{align*} Thus $\Phi(1_B)$ is the unit of $M(A)$. For the involution, let $b\in B$. The adjoint of the double centralizer $(L_b,R_b)$ is the double centralizer whose left and right actions are induced by $b^*$. Indeed, for every $a\in A$, \begin{align*} L_{b^*}(a)=b^*a \end{align*} and \begin{align*} R_{b^*}(a)=ab^*. \end{align*} These are exactly the left and right actions defining $\Phi(b)^*$ in the double-centralizer $C^*$-algebra. Therefore \begin{align*} \Phi(b^*)=\Phi(b)^*. \end{align*} So $\Phi$ is a unital $*$-homomorphism. [/step] [step:Check agreement with the canonical copy of $A$] Let $a\in A$. The canonical embedding of $A$ into $M(A)$ from [citetheorem:8575] sends $a$ to the double centralizer $(L_a,R_a)$, where \begin{align*} L_a(x)=ax \end{align*} and \begin{align*} R_a(x)=xa \end{align*} for every $x\in A$. Since the formula defining $\Phi$ gives exactly this pair when $b=a$, we have \begin{align*} \Phi(a)=(L_a,R_a) \end{align*} for every $a\in A$. [/step] [step:Prove that any unital extension has the same left and right actions] Let \begin{align*} \Psi:B\to M(A) \end{align*} be a unital $*$-homomorphism such that \begin{align*} \Psi(a)=(L_a,R_a) \end{align*} for every $a\in A$. Fix $b\in B$, and write \begin{align*} \Psi(b)=(S_b,T_b), \end{align*} where $S_b:A\to A$ and $T_b:A\to A$ are the left and right components of the double centralizer $\Psi(b)$. Let $a\in A$. Since $\Psi$ is multiplicative and $ba\in A$, we have \begin{align*} \Psi(b)\Psi(a)=\Psi(ba)=(L_{ba},R_{ba}). \end{align*} Comparing the left components of the product and evaluating at the unit-free element $x\in A$ gives \begin{align*} (S_b\circ L_a)(x)=L_{ba}(x) \end{align*} for every $x\in A$. Equivalently, \begin{align*} S_b(ax)=(ba)x \end{align*} for every $a,x\in A$. Similarly, since $ab\in A$, \begin{align*} \Psi(a)\Psi(b)=\Psi(ab)=(L_{ab},R_{ab}). \end{align*} Comparing the right components gives \begin{align*} (T_b\circ R_a)(x)=R_{ab}(x) \end{align*} for every $x\in A$, hence \begin{align*} T_b(xa)=x(ab) \end{align*} for every $a,x\in A$. Because $A$ is an essential ideal in the unital $C^*$-algebra $B$, the annihilator criterion [citetheorem:8574] says that an element of $B$ whose product with every element of $A$ on the relevant side is zero must itself be zero. Applying this criterion to the difference between the two left actions shows that $S_b(a)=ba$ for every $a\in A$. Applying it to the right actions shows that $T_b(a)=ab$ for every $a\in A$. Hence \begin{align*} \Psi(b)=(L_b,R_b)=\Phi(b). \end{align*} Since $b\in B$ was arbitrary, $\Psi=\Phi$. [/step] [step:Conclude existence and uniqueness] The map $\Phi:B\to M(A)$ constructed above is a well-defined unital $*$-homomorphism, satisfies \begin{align*} \Phi(a)=(L_a,R_a) \end{align*} for every $a\in A$, and is given by \begin{align*} \Phi(b)=(L_b,R_b) \end{align*} for every $b\in B$. The preceding step proves that no other unital $*$-homomorphism with this extension property can differ from $\Phi$. Therefore the required homomorphism exists and is unique. [/step]