In a $*$-algebra $A$, a map $u : A → A$ is an isometry on $A$ if $u^*u = $, and a co-isometry if $uu^* = $. It is unitary if it is both.
I have to prove that a certain map is unitary, which (in the solutions manual) is done by showing that it is isometric ($\lVert u(x) \rVert^2 = \lVert x \rVert^2 $). I still don't see how that suffices.
The map in question is $u(a + N_\tau) := \alpha(a)+ N_\tau$ for some automorphism $\alpha$ of $A$ and $N_\tau := \{a ∈ A \mid \tau(a^*a) = 0 \}$ for some positive linear functional $\tau$. The objects satisfy $\tau(\alpha(a)) = \tau(a)$ for all $a ∈ A$.
Again, the claim is that since $\lVert u(a+N_\tau) \rVert² = \lVert a+N_\tau\rVert²$, $u$ is unitary, while I can only see that it is an isometry, not also a co-isometry..
(For reference, I am doing Exercise 3.2(c) from Murphy's book $C^*$-Algebras and Operator Theory.)
EDIT
In principle, we don’t know what $u^*$ does, but it will have to, for some $b ∈ A$, map it to $b + N_\tau$. If $u$ is to be unitary (and thus, in particular, a co-isometry), we ought to have $$a+N_\tau = uu^*(a+N_\tau)= u(b+N_\tau) = \alpha(b)+N_\tau,$$ which is true if, but not only if, $b = \alpha^{-1}(a)$. Is this in fact true by necessity? And if so, could we not have applied an analogous argument to $u^*u$? Not having to bother computing the norms?
I think the exercise wants you to show that automorphisms $\alpha$ that leave a state $\tau$ invariant induce unitaries on the GNS space $H_\tau$. The unitary $u$ is defined as $u\pi_\tau(a)\Omega_\tau =\pi_\tau(\alpha(a))\Omega_\tau$. This is isometric because of the invariance. From the definition it is clear that $u$ is invertible with the inverse given by the unitary induced by $\alpha^{-1}$ and the inverse $u^{-1}$ is also isometric.
To prove unitarity, you could now simply show that $u^*= u^{-1}$: \begin{align} (u^{-1}\pi_\tau(a)\Omega_\tau,\pi_\tau(b)\Omega_\tau) &=(\pi_\tau(\alpha^{-1}(a))\Omega_\tau,\pi_\tau(b)\Omega_\tau)\\ &=\tau(b^*\alpha^{-1}(a))\\ &=\tau(\alpha(b)^*a)\\ &= (\pi_\tau(a)\Omega_\tau,u\pi_\tau(b)\Omega_\tau) =(u^*\pi_\tau(a)\Omega_\tau,\pi_\tau(b)\Omega_\tau) \end{align} for all $a,b\in A$. This suffices because $\text{span}{\pi_\tau(A)\Omega_\tau}$ is dense in $H_\tau$.
More elegantly, it follows from $u$ and $u^{-1}$ being isometric that you get unitarity (but you do definetly need the isometricity of both $u$ and its inverse!): $$ (ux,y) = (u^{-1}ux,u^{-1}y) = (x,u^{-1}y) \quad \forall{x,y\in H_\tau}. $$ This argument uses the polarization identity which implies that isometries with respect to the norm are also isometries w.r.t. the inner product.