How to verify the isomorphism between two C*-algebra

Question

Let $B$ be a C*-subalgebra of a unital C*-algebra $A$, how to verify $C^{*}(B, 1_{A})\cong \tilde{B}$? Here, $C^{*}(B, 1_{A})$ denotes the C*-algebra generated by $B$ and $1_{A}$, meanwhile the $\tilde{B}$ denotes the unitization.

Answer 1

This is not true, because a unital C$^*$-algebra is not isomorphic to its unitization. To see this, consider $B=\mathbb C$. Then $\tilde B=\{(z,w):\ z,w\in\mathbb C\}$ with the operations $$ (z,w)+(x,y)=(z+x,w+y),\ \ \ (z,w)(x,y)=(zx+wx+zy,wy). $$ The unit of $\tilde B$ is $(0,1)$, while the unit of $B$ is $(1,0)$. And $$ (1,0)(0,1)=(1,0), $$ so $(1,0)$ is not a unit for $\tilde B$, which guarantees that $B\subsetneq\tilde B$. And it is clear that $\tilde B$ has dimension 2, so $B$ and $\tilde B$ are not isomorphic.

The above, with $B=A=\mathbb C$, is a counterexample to your assertion.

The assertion is true, though, if $B$ is non-unital. In that case you can check that $C^*(B,1_A)=B\oplus\mathbb C\,1_A$ (direct sum, because $B$ and $1_A$ are linearly independent). In this case the map $$ b+\lambda\,1_A\mapsto (b,\lambda) $$ is well-defined (by the linear independence) and a $*$-homomorphism (easy check). It is clear that it is bijective (again using the linear independence) and so it is a $*$-homomorphism.