How to see that $S \mapsto e^{i \theta}, \theta \in [0, 2 \pi)$, is well-defined? Let $S$ be the right-shift operator on $\ell^2(\mathbb{N})$, i.e., $S e_k = e_{k + 1}$ where $\{ e_k \}_{k = 1}^\infty$ is the standard orthonormal basis. In a paper, the authors used the following $*$-homomorphism $\varphi: C^*(S) \rightarrow \mathbb{C}$ (here $C^*(S)$ denotes the unital $C^*$-algebra generated by $S$) defined on the generator via $\varphi(S) = e^{i \theta}$ where $\theta \in [0, 2 \pi)$. How does one rigorously show that this is well-defined?
 A: It is well-known that $C^*(S)\simeq C(\mathbb T)+K(H)$, where $S$ is mapped to the identity function (see, for instance, Theorem V.1.5 in Davidson's C$^*$-Algebras By Example). So you have $*$-homomorphisms 
$$
C^*(S)\to C(\mathbb T)+K(H)\to C(\mathbb T)\to\mathbb C,
$$
where the last arrow is $f\longmapsto f(e^{i\theta})$ for a fixed $\theta$.
A: You can directly define it:
$$
    (\mathcal{S}f)(\theta)=e^{i\theta}f(\theta),\;\;\;f \in L^2[0,2\pi).
$$
That is a well-defined bounded linear operator on $L^2[0,2\pi)$, and $\mathcal{S}e_k=e_{k+1}$ where $e_k(\theta)=e^{ik\theta}$. If $p$ is a polynomial in $z$, then $(p(\mathcal{S})f)(\theta)=p(e^{i\theta})f(\theta)$ is straightforward to check.
Let $P$ be the orthogonal projection of $L^2[0,2\pi)$ onto the closed linear span $\mathcal{M}$ of $\{ e_k(\theta)=e^{ik\theta} \}_{k=1}^{\infty}$. Then $S$ is the restriction of $\mathcal{S}$ to $\mathcal{M}$ and $S^*$ is the compression $P\mathcal{S}^*$ of $\mathcal{S}^*$ to $\mathcal{M}$. It is easy to check that the restriction of $[P\mathcal{S}^*,\mathcal{S}]$ to $\mathcal{M}$ is a rank 1 projection onto the subspace spanned by $e_1$, which is compact.
