Proof of universality of Toeplitz algebra It is well-known that the Toeplitz algebra $\mathcal{T}$ (that I view as concrete subalgebra of $\mathbb{B}(\ell^2(\mathbb{N})$) is the universal algebra generated by an isometry, that is, for any $C^*$-algebra $A$ and an isometry $v \in A$, there exists a unique algebra $\mathcal{T} \rightarrow A$ such that $S \mapsto v$, where $S$ denotes the shift operator in $\mathcal{T}$.
The proof of this fact that I know is done using the Wold decomposition (explained e.g. in Wegge-Olsen, 3.F). He then writes "there are probably easier ways (of which I'm ignorant) to do this". Is anyone here not ignorant of easier ways? In particular, is there a "more $C^*$-algebraic" proof?
Turning things around, we can just define $\mathcal{T}$ as the universal $C^*$-envelope of the $*$-algebra generated freely by $S$ and $S^*$. This then has the universal property by abstract nonsense, but the proof I know that this algebra is isomorphic to the "concrete one" again uses Wold decomposition.
 A: The proof for Cuntz-Toeplitz algebras, i.e. algebras generated by some number of isometries with orthogonal ranges, is pretty $C^{\ast}$-algebraic. Let $s$ be the universal isometry and let $\mathcal{T}_{u}$ be the universal $C^{\ast}$-algebra generated by an isometry. We want to show that the $\ast$-homomorphism $\Phi: \mathcal{T}_{u} \to \mathcal{T}$ defined by $s \mapsto S$ is injective. There are circle actions on both of these actions, given just by multiplication by a scalar. They commute with $\Phi$, so there is an induced map between the fixed-point subalgebras. Moreover, there are faithful conditional expectations onto these fixed-point subalgebras, so if the restriction of $\Phi$ is injective, then $\Phi$ itself is injective as well. The fixed point subalgebras are spanned by the elements $t_n := s^{n} (s^{\ast})^{n}$ ($T_n:= S^{n} (S^{\ast})^{n}$). The span of first $m$ $t_n$'s is actually a finite dimensional subalgebra and just by checking the dimension you may prove that the restriction of $\Phi$ to such a subalgebra is injective, hence isometric. It follows easily that $\Phi$ restricted to the fixed-point subalgebra is isometric.
I don't think this is really an easier way, but it definitely works in more general situations. 
