unitisation of $C^*\!$-algebra $A$

Question

If $A$ is a unital $C^*\!$-algebra, we can construct a $*$-homomorphism such that the unitisation of $A$, $\tilde{A}$ is $*$-isomorphic to $A\oplus\mathbb{C}$. If $\tilde{A}$ is $*$-isomorphic to $A\oplus\mathbb{C}$, can we deduce that $A$ is unital?

Answer 1

Yes, but it may not be what you are asking for.

When you are talking unitization, you consider $A\oplus \mathbb C$, with the coordinate-wise addition and multiplication by scalars, but with the "twisted" product $(a,t)(b,s)=(ab+tb+sa,ts)$. If that's the product you are using, then $A\oplus\mathbb C$ is unital with unit $(0,1)$.

On the other hand, on its own, one usually considers $A\oplus \mathbb C$ with coordinate-wise product. Consider for instance the nonunital algebra $A=\ell^\infty(\mathbb N)\oplus c_0(\mathbb N)$. Since $\ell^\infty(\mathbb N)\simeq \ell^\infty(\mathbb N)\oplus\mathbb C$, we have $A\simeq A\oplus\mathbb C$, while $A$ is nonunital.

Answer 2

Note that if $\tilde A$ and $A\oplus\mathbb C$ are $*$-isomorphic, then in particular $A\oplus\mathbb C$ is unital. Let $(e,\lambda)$ denote the unit of $A\oplus\mathbb C$. Then doing some basic algebraic manipulations, we can show that $\lambda=1$ and $e$ is a unit for $A$, hence $A$ is unital.