I am preparing for an exam and working through Murphy's $C^*$-algebras and Operator Theory and I came across the following text:
Now suppose that $A$ is an arbitrary $C^*$-algebra containing $B$ as a $C^*$-subalgebra. Replacing $B$ by $\tilde{B}$ and $A$ by $\tilde{A}$ if necessary, we may suppose that $A$ has a unit 1 which lies in $B$.
For context, we are proving that any positive linear functional on $B$ can be extended to a positive linear functional on $A$ (using Hahn-Banach). But I don't understand why we can assume the units of $A$ and $B$ to be the same. For example, when $B$ is generated by a projection $p$, that projection acts as the unit in $B$ but not in $A$. So why can we assume this?
Let me add the details of this WLOG argument. Say $B\subset A$ is an inclusion of $C^*$-algebras. Consider the unitization of $A$, namely $\tilde{A}$, where we adjoin a unit regardless of $A$ already having a unit or not. Consider $B_1:=B+\mathbb{C}1_\tilde{A}$, which is a $C^*$-subalgebra of $\tilde{A}$, contains $B$ and $B_1$ has the same unit as $\tilde{A}$. Actually, it is a cute little exercise to show that $\tilde{B}\cong B_1$.
In the first paragraph of the proof that you are reading (namely theorem 3.3.8), Murphy has already shown that, if $\tau:D\to\mathbb{C}$ is a positive functional on a $C^*$-algebra $D$, then $\tau$ extends to a positive functional on $\tilde{D}$.
The reason why there is indeed no loss of generality: assume that we already know that, if $D\subset C$ is an inclusion of unital $C^*$-algebras with $1_C\in D$, then positive functionals from $D$ extend to positive functionals on $C$. Now let $B\subset A$ be any inclusion and $\tau:B\to\mathbb{C}$ positive. Consider the induced inclusion $B_1\subset\tilde{A}$, which satisfies $1_\tilde{A}\in B_1$. Since $B_1\cong B$, by the first paragraph of the proof, $\tau$ extends to a positive functional on $B_1$. By our assumption, since $B_1\subset\tilde{A}$ with the same unit, this extension further extends to a positive functional on $\tilde{A}$. To finish, just restrict this positive functional on $\tilde{A}$ to a functional on $A$, which will still be positive.