$\ast$-isomorphic between two $C^\ast$-algebras

Question

Let's consider the following unitization of the following $C^\ast$-algebra. Denote it $\bar{\mathscr{A}}=\mathscr{A}\oplus\mathbb{C}$ with $$(a,\lambda)(b,\mu)=(ab+\lambda b+\mu a,\lambda \mu) \text{ and } (a,\lambda)^\ast=(a^\ast,\bar{\lambda})$$ We know that this is a $\ast$-algebra with unit $(0,1)$.

Consider now $\bar{\mathscr{A}'}=\mathscr{A}\oplus\mathbb{C}$ and suppose that this IS unital. Then $$(a,\lambda)(b,\mu)=(ab,\lambda \mu) \text{ and } (a,\lambda)^\ast=(a^\ast,\bar{\lambda})$$ This has unit $(1,1)$.

I just want to know if $\bar{\mathscr{A}}$ is a $\ast$-isomorphic to $\bar{\mathscr{A}}'$ in a unit preserving manner.

Answer 1

Some notation first. Let $A$ be a unital $C^*$-algebra (which is equivalent with the direct sum $A\oplus \mathbb{C}$ being unital). Let $A_I$ be its unitisation and $A\oplus \mathbb{C}$ (so the unit is $(0,1))$ be the direct sum of $C^*$-algebras (so the unit is $(1,1)$). A $*$-isomorphism $\Phi: A \oplus \mathbb{C}\to A_I$ must preserve the units, so we must map $(1,1)\mapsto (0,1)$. Thus, the identity map on the underlying vector spaces will not work. What is the easiest way to map $(1,1)\mapsto (0,1)?$ Well, consider

$$\Phi: A\oplus \mathbb{C} \to A_I: (a, \lambda)\mapsto (a-\lambda 1_A, \lambda).$$ Check that this is a $*$-isomorphism.