I get stuck on Murphy's book $C^*$ Algebras and Operator Theory, Page 95.
First, for a $C^*$ algebra $A$, he defines a closed left ideal $N_\tau = \{ a\in A: \tau(a^*a) = 0 \}$ then, he defines an inner product on the quotient algebra $A/N_\tau$ by $(a + N_\tau, b + N_\tau) = \tau(b^*a)$. Then the Hilbert completion of $A/N_\tau$ is denoted as $H_\tau$. For $a \in A$ he defines the operator $\varphi(a)$ on $A/N_\tau$ by $\varphi(a)(b + N_\tau) = ab + N_\tau$. Then it's bounded and can be extended to an bounded operator on $H_\tau$. Then the map $$ \varphi_\tau: A \to B(H_\tau), a\mapsto \varphi_\tau(a) $$ is a *-homomorphism.
My question is that how to show this map is a *-homomorphism, more concretely, how to show $$ \varphi(a^*) = \varphi(a)^* $$
I try to show for any $c + N_\tau$: $$ a^*c + N_\tau = (ac+ N_\tau)^* = c^*a^* + N_\tau $$ But I don't know how to start.
You have $$ \begin{align} \langle \varphi(a)^*\hat c,\hat c\rangle &=\langle \hat c,\varphi(a)\hat c\rangle =\langle \hat c,\widehat{ac}\rangle\\[0.3cm]&=\tau((ac)^*c)=\tau(c^*(a^*c))\\[0.3cm]&=\langle\widehat a^*c,\hat c\rangle =\langle \varphi(a^*)\hat c,\hat c\rangle. \end{align} $$ As this works for any $c\in A$ and $\{\hat c:\ c\in A\}$ is dense, we get that $\varphi(a)^*=\varphi(a^*)$.