Uniqueness unitization of a non unital $C^*$-algebra

Question

I am trying to show that the unitization of a non-unital AF $C^*$-algebra, $A$ , is again an AF $C^*$-algebra. In order to do so, I tried to claim the following:

Let $A, B$ be $C^*$ algebras and suppose that $A$ is an ideal in $B$ and $B/A$ is isomorphic to $\Bbb{C}$.
In these conditions, is it true that $B$ and $A^+$ (the unitization of $A$) are isomorphic as $C^*$ algebras?

Edit: I think I should require $B$ to be with multiplicative unit.

Answer 1

When $A$ is not unital, we have that $A^+=A\oplus\mathbb C$ with the product $$(a_1,\lambda_1)(a_2,\lambda_2)=(a_1a_2+\lambda_2a_1+\lambda_1a_2,\lambda_1\lambda_2).$$ Consider the map $\phi:A^+\to B$ given by $$ \phi(a,\lambda)=a+\lambda I. $$ This is clearly a $*$-homomorphism. Since $A$ is non-unital, this map is one-to-one. So it remains to check it is onto. Given $b\in B$, we have $b+A=\lambda(b)+A$ for some $\lambda(b)\in\mathbb C$, so there exists $a\in A$ such that $b=\lambda I+a=\phi(a,\lambda(b))$.

Answer 2

The unitization is a left adjoint functor (to the inclusion functor).

As such the unitization commutes with arbitrary colimits.

AF-algebras are colimits of finite dimensional C*-algebras.

The unitization of finite dimensional C*-algebras are finite dimensional.

So the unitization of AF-algebras are AF-algebras.